How is ATP generated by cellular respiration?

One note at the beginning: please do not be scared! You can scroll right down to the bottom of this answer to read the conclusion which in itself should sufficiently answer the question. The bits in between are merely elaboration and need not be known in full detail.

The reaction one typically receives in high school biology classes to explain cellular respiration is already a simplification on many levels. It is presented as $(0)$, but in reality that is not one single reaction. Whether or not $\ce{ATP}$ has already been put in is irrelevant.

$$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}\tag{0}$$

In fact, this reaction is a series of steps, each performed by a single, specific enzyme. The first reaction, that basically immediately happens when glucose enters an animal cell is phosphorylation as shown in $(1)$.

$$\ce{\underset{{glucose}}{C6H12O6} + \underset{{ATP}}{C10H12N5O13P3^4-} ->[enzyme] \underset{{glucose-6-phosphate}}{C6H11O9P^2-} + \underset{{ADP}}{C10H12N5O10P2^3-} + H+}\tag{1}$$

I have written $\ce{ATP}$ and will continue to do so for all products in the form they are in at physiological $\mathrm{pH}$

Glucose-6-phosphate is isomerised to fructose-6-phosphate, which cannot be distinguished by sum formulae alone, so I’m leaving this step out. Another enzyme comes along which phosphorylates fructose-6-phosphate again to give fructose-1,6-bisphosphate; see $(2)$. So far, all we have done is use up $\ce{ATP}$ at no benefit for the cell.

$$\begin{multline}\ce{\underset{{fructose-6-phosphate}}{C6H11O9P^2-} + \underset{{ATP}}{C10H12N5O13P3^4- ->[enzyme] }} \\ \ce{\underset{{fructose-1,6-bisphosphate}}{C6H10O12P2^4-} + \underset{{ADP}}{C10H12N5O10P2^3-} + H+}\end{multline}\tag{2}$$

In a retro-aldol reaction, fructose-1,6-bisphosphate is split into glycerinaldehyde-3-phosphate (commonly abbreviated GAP) and dihydroxyacetonephosphate (commonly abbreviated DHAP). These once again are structural isomers that bear the same sum formula and they are interconverted. Hence I am pretending in equation $(3)$ that only GAP is a product. This assumption is close enough to valid in sum formulae.

$$\ce{\underset{{fructose-1,6-bisphosphate}}{C6H10O12P2^4-} ->[enzyme] 2 \underset{{GAP}}{C3H6O5P^2-}}\tag{3}$$

We have now reached the phase of glycolysis (the consumption of glucose) called the pay-off phase. Previously, as you may have noticed, all we did was consume ATP. It is now almost time to get some back. But first, we need to prepare GAP for its prominent role in reaction $(4)$:

$$\begin{multline}\ce{2 \underset{{GAP}}{C3H5O6P^2-} + 2 HPO4^2- + 2 \underset{NAD+}{C21H26N7O14P2-}} \\ \ce{->[enzyme] 2 \underset{{1,3-bisphosphoglycerate}}{C3H4O10P2^4-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{4}$$

It may seem confusing that I labelled an anion as $\ce{NAD+}$ (its common name). This is because there are two phosphate diesters that can each be protonated once. Fully protonated, $\ce{NAD+}$ is indeed a cation — only not at physiological $\mathrm{pH}$.

In the previous step, we synthesised a phosphoric carboxylic mixed acid anhydride. These are very energetic and consequently its hydrolysis is now used in reaction $(5)$ to synthesise $\ce{ATP}$. We are breaking even.

$$\begin{multline}\ce{2 \underset{{1,3-bisphosphoglycerate}}{C3H4O10P2^4-} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} ->[enzyme]} \\ \ce{2 \underset{{3-phosphoglycerate}}{C3H4O7P^3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{5}$$

After this step, we again have an isomerisation that cannot be shown in sum formulae. The phosphate group is moved from position 3 to position 2 giving 2-phosphoglycerate. This can then be dehydrated as shown in $(6)$ to give phosphoenolpyruvate: a very important cellular intermediate. We have also generated our first two water molecules.

$$\ce{2 \underset{{2-phosphoglycerate}}{C3H4O7P^3-} ->[enzyme] 2 \underset{{phosphoenolpyruvate}}{C3H2O6P^3-} + 2 H2O}\tag{6}$$

Finally, a second set of two $\ce{ATP}$ molecules is formed by dephosphorylating phosphoenolpyruvate to give free pyruvate as shown in reaction $(7)$.

$$\begin{multline}\ce{2 \underset{{phosphoenolpyruvate}}{C3H2O6P^3-} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 H+} \\ \ce{->[enzyme] 2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{7}$$

We can draw an intermediate line here, since the process typically known as glycolysis has reached its close. We can sum it up in reaction $\text{(1–7)}$. We still have not generated any $\ce{CO2}$ and only $\ce{2 H2O}$. We also already have $\ce{2 ATP}$ per molecule of glucose and 2 mysterious molecules called $\ce{NADH}$.

$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 HPO4- + 2 \underset{NAD+}{C21H26N7O14P2-} ->[steps]} \\ \ce{2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+ + 2 \underset{{ATP}}{C10H12N5O13P3^4-} + 2 H2O}\end{multline}\tag{1–7}$$

The next step is still very easy to explain. It is called oxidative decarboxylation and adds a cofactor known as coenzyme A of $\ce{CoA}$ covalently to what was pyruvate before the reaction. See equation $(8)$.

$$\begin{multline}\ce{2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{NAD+}{C21H26N7O14P2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} ->[enzyme]} \\ \ce{2 \underset{{acetyl-CoA}}{C23H35N7O17P3S^3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 CO2}\end{multline}\tag{8}$$

Acetyl-CoA now enters into a magic known as the Krebs cycle or citric acid cycle. As the name implies, this is constantly going around. At the beginning, the two carbon atoms of acetyl-CoA are added to oxaloacetate. After the ten cycle steps, we will have lost two molecules of $\ce{CO2}$, effectively reforming the oxaloacetate substrate. And once again some of the steps are merely isomerisations which do not show up in some formulae. Are you ready? Here goes. Reaction $(9)$ has already been described in the text above:

$$\begin{multline}\ce{2 \underset{{acetyl-CoA}}{C23H35N7O17P3S^3-} + 2 \underset{{oxaloacetate}}{C4H2O5^2-} + 2 H2O} \\ \ce{->[enzyme] 2 \underset{{citrate}}{C6H5O7^3-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-}}\end{multline}\tag{9}$$

After isomerisation of citrate to give isocitrate, this is oxidised to give oxalosuccinate as shown in $(10)$.

$$\begin{multline}\ce{2 \underset{{isocitrate}}{C6H5O7^3-} + 2 \underset{NAD+}{C21H26N7O14P2-} ->[enzyme]} \\ \ce{2 \underset{{oxalosuccinate}}{C6H3O7^3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{10}$$

For each oxalosuccinate, one molecule of $\ce{CO2}$ can be removed to give α-ketoglutarate as shown in $(11)$.

$$\ce{2 \underset{{oxalosuccinate}}{C6H3O7^3-} + 2 H+ ->[enzyme] 2 \underset{\alpha{-ketoglutarate}}{C5H4O5^2-} + 2 CO2}\tag{11}$$

α-ketoglutarate is then once again attacked by coenzyme A in the same manner as in step $(9)$. This can happen because both pyruvate and α-ketoglutarate have the same structural feature. Refer to equation $(12)$.

$$\begin{multline}\ce{2 \underset{\alpha{-ketoglutarate}}{C5H4O5^2-} + 2 \underset{NAD+}{C21H26N7O14P2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} ->[enzyme]} \\ \ce{2 \underset{{succinyl-CoA}}{C25H36N7O19P3S^4-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 CO2}\end{multline}\tag{12}$$

We have now lost all $\ce{CO2}$. What we need to do is regenerate the oxaloacetate from the beginning of the Krebs cycle. The first step of this is to generate succinate from succinate-CoA by hydrolysis as given in $(13)$.

$$\begin{multline}\ce{2 \underset{{succinyl-CoA}}{C25H36N7O19P3S^4-} + 2 H2O + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 HPO4^2-  ->[enzyme]} \\ \ce{2 \underset{{succinate}}{C4H4O4^2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{13}$$

Next, we oxidise the saturated succinate molecule to an unsaturated fumarate. This is typically perfomed by yet another cofactor called flavin-adenine-dinucleotide or $\ce{FAD}$; see reaction $(14)$.

$$\ce{2 \underset{{succinate}}{C4H4O4^2-} + 2 \underset{FAD}{C27H31N9O15P2^2-} ->[enzyme] 2 \underset{{fumarate}}{C4H2O4^2-} + 2 \underset{FADH2}{C27H33N9O15P2^2-}}\tag{14}$$

Water is added to the double bond to give malate (reaction $(15)$):

$$\ce{2 \underset{{fumarate}}{C4H2O4^2-} + 2 H2O ->[enzyme] 2 \underset{{malate}}{C4H4O5^2-}}\tag{15}$$

And malate is finally oxidised by using $\ce{NAD+}$ as shown in $(16)$ to give the starting material of the Krebs cycle, oxaloacetate.

$$\begin{multline}\ce{2 \underset{{malate}}{C4H4O5^2-} + 2 \underset{NAD+}{C21H26N7O14P2-} ->[enzyme]} \\ \ce{2 \underset{{oxaloacetate}}{C4H2O5^2-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{16}$$

Technically, we could end it here. Let’s look again at the final products of reactions $(1–16)$. Thankfully, a lot cancels out.

$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 4 \underset{{ADP}}{C10H12N5O10P2^3-} + 4 HPO4^2- + 10 \underset{NAD+}{C21H26N7O14P2-} +} \\ \ce{2 \underset{FAD}{C27H31N9O15P2^2-} ->[steps] 6 CO2 + 6 H+ +} \\ \ce{4 \underset{{ATP}}{C10H12N5O13P3^4-} + 10 \underset{NADH}{C21H27N7O14P2^2-} + 2 \underset{FADH2}{C27H33N9O15P2^2-}}\end{multline}\tag{1–16}$$

We’re still not there yet. We have generated a lot of $\ce{NADH}$ and $\ce{FADH2}$ which somehow needs to be transformed back into $\ce{NAD+}$ and $\ce{FAD}$ for the next iteration to be performed — remember that cellular energy is, as you noted in your question, typically equated to $\ce{ATP}$. And we still haven’t used oxygen at all and are not producing enough water.

The next set of reactions to bring us closer to our goal is the electron transport chain. By a series of reactions that is not easily described (but basically involve electrons being moved in redox processes from one compound to another), oxygen molecules acquire four electrons and four protons each to become water. The electron transport chain that performs this will, at the same time, use the energy liberated by the generation of water to set up a proton potential: protons ($\ce{H+}$) get pumped from one side of an intracellular membrane to another.

This proton potential of course is a driving force since there is a lack of protons on the other side. Protons may travel back through special trans-membrane proteins. These proteins can be imagined like a watermill and start rotating as protons pass through. This rotation allows them to literally slap a phosphate onto an $\ce{ADP}$ molecule to give $\ce{ATP}$.

As a theoretical maximum, the regeneration of $\ce{NAD+}$ from $\ce{NADH}$ gives around 2.5 equivalents of $\ce{ATP}$ while $\ce{FADH2}$ can only be used to generate around 1.5 equivalents of $\ce{ATP}$ upon the reverse reaction to $\ce{FAD}$. These numbers are, however, generalised, since some energy is also lost in transport and other processes.

Therefore, our theoretical maximum yield of $\ce{ATP}$ is 25 from $\ce{NADH}$, a further 3 from $\ce{FADH2}$ and a further 4 that are directly produced in the reaction sequences. Each of these $\ce{ATP}$s is generated by the reaction of a phosphate anion onto an $\ce{ADP}$ molecule, technically involving the loss of water or an equivalent.

As you may see, writing up a complete reaction for the entire cellular respiration process is very difficult. We can, however, arrive approximately there by a few basic principles:

there are up to $\ce{32 ATP}$ generated over the course of the digestion of one molecule of glucose
each $\ce{ATP}$ is generally synthesised by some form of $\ce{HPO4- + ADP}$ — therefore, the phosphorus part is balanced.
the exact process is not necessarily important; a lot cancels out and the end result counts.

This in total may lead us to question why exactly the reaction taught at schools is often the one presented in equation $(0)$. Well, we can see that exactly six molecules of carbon dioxide are formed in the overall process per molecule of glucose. This makes sense, as it means each glucose carbon atom effectively leaves the body as $\ce{CO2}$, the carbon compound with the lowest remaining energy. The number of oxygen and water molecules cannot, unfortunately, be accurately established for the single process. However, schools want to teach the concept of a balanced equation. And if all carbon leaves the body as $\ce{CO2}$, then all hydrogen must eventually leave it as $\ce{H2O}$. To accomplish that, a certain set of oxygen molecules is needed — exactly 6.

The reaction $(0)$ thus never actually happens in our body. That is a good thing. You can see what it looks like by heating suger to very high temperatures and then adding a match — actually don’t. It’s the same combustion as any other combustion. Part of it can be adequately summarised of many partial reactions, part of it is just there so the summary fulfills the required stoichiometry. And even the $\ce{32 ATP}$ are only an approximation based on a lot of averaging.

What we can definitely say, however, is that each of the $\ce{32 ATP}$s is produced by the reaction of phosphate and $\ce{ADP}$. And therefore, at least consistent with your high-school level, the reaction should better be written as:

$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 6 O2 + 32 HPO4- + 32 \underset{ADP}{C10H12N5O10P2^3−}} \\ \ce{-> 6 CO2 + 6 H2O + 32 \underset{ATP}{C10H12N5O13P3^4−}}\end{multline}\tag{full}$$


Would ABS plastic degrade in a dishwasher?

According to several application guides/compatibility charts for ABS plastic (like this one), ABS is fairly resistant to the conditions present in a dishwasher (mild to strong alkaline, salts, no organic solvents, mild temperatures).

Please note that while the ABS structural backbone may be attacked by nucleophilic agents at the nitrile carbon, and that the nitrile group may be hydrolised in acidic or basic environments, the conditions required are much harsher that a common dishwasher.

For example (p.8), glacial acetic acid was found to cause significant swelling, while 25% w/w sodium hydroxide and 25% w/v HCl barely caused changes.

So, while the concern for release of toxic additives present in the formulation of the original ABS filament may or not be significant, from the chemical point of view it is almost sure it won't degrade noticeably.

Also, anecdotally, I used to wash old LEGO pieces in a dishwasher (kids are dirty), and not degradation was observed at all.


The first thing to note is that nucleophilicity is a kinetic parameter that cannot be addressed easily with thermodynamic parameters such as electronegativity. Instead, you need to inspect all potential nucleophiles by their kinetic merits.

One easy thing that we can use, even though it is thermodynamic, is the effective charge. The centre of negative charge of the acetate molecule is somewhere in the middle between the two oxygens and that of tosylate is somewhere in the middle between three equivalent oxygens. Also, in both systems the oxygens are bound to very highly oxidised atoms ($\ce{C^{III}}$ and $\ce{S^{IV}}$, respectively), which greatly reduces their electron density and thus their desire to attack nucleophilicly. Finally, in acetate’s case the nucleophilic attack would reduce the effectiveness of the carboxylate resonance as one of the two mesomeric structures becomes disfavoured (charge separation) which adds another grain of salt against nucleophilic attacks.

Tosylate is less nucleophilic than acetate because there are three oxygens than two but they only bear a single negative charge between them.

Now onto cyanide and methanolate. For cyanide, first note that it is both $\ce{C}$-nucleophilic and $\ce{N}$-nucleophilic. Since the HOMO, the highest occupied molecular orbital, has a much larger contribution on the carbon side, the $\ce{C}$-attack is strongly favoured over $\ce{N}$-attack and we can ignore the latter.

We can again use the bond-partners argument here. The oxygen of methanolate is bonded to a $\ce{CH3}$ group which, in general, is electron donating, thus increasing its electron density. The cyanide carbon, however, is bonded to the more electronegative nitrogen which slightly reduces its electron density. Thus, we can infer that methanolate should be a better nucleophile than cyanide. (Note, however, that both are very good nucleophiles, especially when compared to the other two.)


For $\ce{CO_3^2-}$ you have the following possible resonance structures:

The structure bellow, is the one which represents better the real ionic molecule, in which all the bonds are equal. You can delocalize the negative charge through all the ion.

For $\ce{H2CO3}$ the structure is:

As you can see, in this case, bonds are fixed, and there is no possible resonance delocalization, so you have two single bonds (less strong) C-O and one double bond C=O (stronger). For $\ce{CO_3^2-}$ you have that all bond lenths the same showing an intermediate situation between single and double bonds.


A red gold sol is a colloidal suspension of gold nanoparticles with average size of particles less than 100 nm. The suspension is quite stable — the nanoparticles don't aggregate because of the existence of an electrical double layer on the surface of particles which causes electrostatic repulsion between the particles.

If you add a sufficient amount of an electrolyte (e.g., NaCl) to the colloid, the ions will disrupt the electric double layer which will cause aggregation (coagulation) of gold particles. A stabilizing agent (e.g., citric acid) is a compound that is able to adsorb on the surface of gold nanoparticles and protect them from aggregation (it acts like a shield), even in a high electrolyte concentration.


What is the physical basis for Hund's first rule?

Disclaimer

My following answer is the "traditional" explanation of Hund's first rule, which is based on a smaller value of $V_\mathrm{ee}$ (electron-electron repulsions) in the triplet state arising from Fermi holes. According to Levine's Quantum Chemistry 7th ed.:

  This traditional explanation turns out to be wrong in most cases. It is true that the probability that the two electrons are very close together is smaller for the helium $^3S$ $1s2s$ term than for the $^1S$ $1s2s$ term. However, calculations with accurate wave functions show that the probability that the two electrons are very far apart is also less for the $^3S$ term. The net result is that the average distance between the two electrons is slightly less for the $^3S$ term than for the $^1S$ term, and the interelectronic repulsion is slightly greater for the $^3S$ term.

  The calculations show that the $^3S$ term lies below the $^1S$ term because of a substantially greater electron nucleus-attraction in the $^3S$ term as compared with the $^1S$ term. Similar results are found for terms of the atoms beryllium and carbon.

Wildcat has an answer which briefly mentions this. Aside from that Levine gives several references:

Katriel, J.; Pauncz, R. Theoretical Interpretation of Hund's Rule. Adv. Quantum Chem. 1977, 10, 143–185. DOI: 10.1016/S0065-3276(08)60580-8.
Shim, I.; Dahl, J. P. Theor. Chim. Acta 1978, 48 (2), 165. The DOI doesn't resolve properly, but here's the link to the publisher's website.
Boyd, R. J. A quantum mechanical explanation for Hund's multiplicity rule. Nature 1984, 310, 480–481. DOI: 10.1038/310480a0.
Oyamada, T.; Hongo, K.; Kawazoe, Y.; Yasuhara, H. Unified interpretation of Hund's first and second rules for 2p and 3p atoms. J. Chem. Phys. 2010, 133, 164113. DOI: 10.1063/1.3488099.

With that said, here is the old-fashioned argument for why parallel spins are favoured over paired spins (all else being equal).

Introduction

The origin of Hund's first rule lies in "exchange energy", which is a way of saying that electrons with like spin repel each other less than electrons with unlike spin.

To see why, let's consider an excited state of the helium atom: $\mathrm{1s^1 2s^1}$. (The ground state configuration $\mathrm{1s^2}$ is of no use to us because those electrons have to be paired with unlike spin.) In the excited state, the electrons can either have parallel spin (the triplet case) or paired spin (the singlet case).

The total electronic wavefunction comprises a spatial part, which describes which orbitals the electrons are in (i.e. it specifies $n,l,m_l$), and a spin part, which describes the, well, spin ($m_s$). According to the Pauli exclusion principle, the total wavefunction must be antisymmetric with respect to interchange of the electron labels: $\Psi(1,2) = -\Psi(2,1)$.

Spin wavefunctions

The spin wavefunctions are simultaneous eigenfunctions of the $\hat{S}^2$ and $\hat{S}_z$ operators. The total spin quantum number, $S$, can be found from the eigenvalue of the operator $\hat{S}^2$:

$$\hat{S}^2\psi = \hbar^2 S(S+1)\psi$$

For more detailed discussion, refer to any QM text on the coupling of angular momenta.

There are three possible triplet spin wavefunctions, which are degenerate.

$$\psi_{\mathrm{spin,triplet}}(1,2) = \begin{cases}
\alpha(1)\alpha(2) \\
\beta(1)\beta(2) \\
\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) + \alpha(2)\beta(1)]
\end{cases}$$

It turns out that $\hat{S}^2\psi_\mathrm{triplet} = 2\hbar^2\psi_{\mathrm{triplet}}$, which means that these have the quantum number $S = 1$. This corresponds to a multiplicity of $M = 2S + 1 = 3$, hence "triplet". These wavefunctions represent the case where the electrons have parallel spin. (Loosely speaking, each electron has a spin of $1/2$, and if these are aligned parallel the total spin is $1/2 + 1/2 = 1$.)

The singlet state is

$$\psi_{\mathrm{spin,singlet}}(1,2) = \frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \alpha(2)\beta(1)]$$

and as you might guess, $\hat{S}^2\psi_\mathrm{singlet} = 0$, meaning that $S = 0$ and $M = 2S+1 = 1$. This represents the case where the electrons are paired - one can think of it as the two $1/2$ electrons spins “cancelling each other out”.

The important thing to notice is that the triplet spin wavefunctions are symmetric with respect to permutation of the labels, whereas the singlet spin wavefunction is antisymmetric. For example:

$$\begin{align}
\psi_\mathrm{singlet}(2,1) &= \frac{1}{\sqrt{2}}[\alpha(2)\beta(1) - \alpha(1)\beta(2)] \\
&= -\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \alpha(2)\beta(1)] \\
&= -\psi_\mathrm{singlet}(1,2)
\end{align}$$

Spatial wavefunctions

The only way of constructing these are the linear combinations below.

$$\begin{align}
\psi_\mathrm{space,symm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(1)\mathrm{2s}(2) + \mathrm{1s}(2)\mathrm{2s}(1)] \\
\psi_\mathrm{space,antisymm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1)]
\end{align}$$

Remember that the atomic orbitals above are functions of the coordinates of the electron: $\mathrm{1s}(1) = \mathrm{1s}(\vec{r}_1) = \mathrm{1s}(r_1,\theta_1, \phi_1)$. Now, the antisymmetric spatial wavefunction has an interesting property. Let's explore what happens when the electrons get very close together, i.e. $\vec{r}_1 = \vec{r}_2 = \vec{r}$.

$$\begin{align}
\psi_\mathrm{space,antisymm} &= \frac{1}{\sqrt{2}}[\mathrm{1s}(\vec{r})\mathrm{2s}(\vec{r}) - \mathrm{1s}(\vec{r})\mathrm{2s}(\vec{r})] \\
&= 0
\end{align}$$

If the wavefunction is 0 when the two electrons are on top of each other, that means that there is no probability of this happening. The average distance between the two electrons $\left\langle\left|\vec{r}_2 - \vec{r}_1\right|\right\rangle$ will therefore be larger, and on average the electrons will repel each other less, since they are on average further away from each other. This is known as a Fermi hole.

In fact, one can also show that the symmetric wavefunction has a larger probability amplitude when $\vec{r}_1 = \vec{r}_2$. This is known as a Fermi heap.

The total wavefunction

As I mentioned earlier, the Pauli exclusion principle dictates that the total wavefunction be antisymmetric. This means that the triplet spin wavefunctions, which are symmetric, must be paired with the antisymmetric spatial wavefunction. Likewise, the singlet spin wavefunction, which is antisymmetric, must be paired with the symmetric spatial wavefunction.

$$\begin{array}{c|c|c}
\text{Spin} & \text{Spatial} & \text{Total}\\
\hline
\text{Triplet (S)} & \text{AS} & \text{S } \times \text{ AS} = \text{AS} \\
\text{Singlet (AS)} & \text{S} & \text{AS } \times \text{ S} = \text{AS}
\end{array}$$

So, there is a decrease in electron-electron repulsion when the electrons are parallel - because this means they have a triplet spin wavefunction, and hence an antisymmetric spatial wavefunction, which has a Fermi hole.

Exchange integrals

There is another, perhaps more mathematical, way of looking at it, and that is to calculate the expectation value of the electronic energy for both wavefunctions. For the helium atom, under the assumption of an infinitely heavy nucleus (essentially an atomic version of the Born-Oppenheimer approximation), the total Hamiltonian (in atomic units) is

$$\hat{H} = -\frac{1}{2}(\nabla_1^2+\nabla_2^2) - \frac{2}{r_1} - \frac{2}{r_2} + \frac{1}{r_{12}}$$

where $r_i$ is the distance of electron $i$ from the nucleus, and $r_{12}$ is the distance between the two electrons. The expectation value of the energy is simply $\left\langle\Psi\middle|\hat{H}\middle|\Psi\right\rangle$ (assuming our wavefunction is already normalised). It turns out that the only important term in our present discussion is the electron-electron repulsion term $1/r_{12}$.

Let's look at the triplet spin wavefunction (of course paired with the antisymmetric spatial wavefunction) first. The operator doesn't depend on spin degrees of freedom, so we can separate those, and the inner product of the spin wavefunction with itself is just unity.

$$\begin{align}
\left\langle\Psi_\mathrm{triplet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{triplet}\right\rangle
&= \left\langle\psi_\mathrm{spin,triplet}\middle|\psi_\mathrm{spin,triplet}\right\rangle\left\langle\psi_\mathrm{space,antisymm}\middle|\frac{1}{r_{12}}\middle|\psi_\mathrm{space,antisymm}\right\rangle \\
&= \frac{1}{2} \left\langle \mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1) \middle| \frac{1}{r_{12}} \middle| \mathrm{1s}(1)\mathrm{2s}(2) - \mathrm{1s}(2)\mathrm{2s}(1)  \right\rangle \\
&=
\frac{1}{2} \Biggr\{ \left\langle \mathrm{1s}(1)\mathrm{2s}(2) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(1)\mathrm{2s}(2) \right\rangle  \\
&\quad\quad\quad - \left\langle \mathrm{1s}(2)\mathrm{2s}(1) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(1)\mathrm{2s}(2) \right\rangle \\
&\quad\quad\quad - \left\langle \mathrm{1s}(1)\mathrm{2s}(2) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(2)\mathrm{2s}(1) \right\rangle \\
&\quad\quad\quad + \left\langle \mathrm{1s}(2)\mathrm{2s}(1) \middle|\frac{1}{r_{12}}\middle| \mathrm{1s}(2)\mathrm{2s}(1) \right\rangle\Biggr\}
\end{align}$$

Ugly, but that's what it is.

We can simplify it however by noting that the first and the fourth brackets are the same (since the variables in the integral are just dummy variables, interchanging the electron labels does not affect the value of the integral). This integral is traditionally called a Coulomb integral, since it simply reflects the average Coulombic repulsion between two independent (uncorrelated) electrons, one in the 1s orbital and the second in the 2s orbital. It is given the symbol $J_\mathrm{1s2s}$ (the subscripts indicating the orbitals that the electrons are in). Because this is an expectation value for a potential energy arising from repulsion, it is necessarily a positive quantity.

The second and third terms are actually also the same, again because the variables are just dummy variables and you can interchange the labels. This integral, however, does not have a physical interpretation; it is a purely quantum mechanical effect. It is called the exchange integral, and is given the symbol $K_\mathrm{1s2s}$. It's a bit harder to show, but the exchange integral is also a positive quantity (there is a question on it here). However, its magnitude is not as large as $J_\mathrm{1s2s}$.

Putting it all together (leaving out subscripts for clarity) we have

$$\left\langle\Psi_\mathrm{triplet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{triplet}\right\rangle = \frac{1}{2}(J-K-K+J) = J-K$$

and if you were to perform the same calculation on the singlet spin wavefunction (not very difficult, just replace the minus signs with a plus!) you would get

$$\left\langle\Psi_\mathrm{singlet}\middle|\frac{1}{r_{12}}\middle|\Psi_\mathrm{singlet}\right\rangle = \frac{1}{2}(J+K+K+J) = J+K$$

Since $J > K > 0$, the expectation value of the energy for the singlet state is higher.

But I have more than two electrons. What now?

Well, you just have to do a ton more ugly maths and perhaps a few hundred more integrals. But the idea is the same. For every pair of electrons with parallel spins, there is a corresponding exchange integral that reduces the expectation value of the energy.

Therefore, in a $\mathrm{3d^5}$ manganese atom (we ignore the closed shells) there are

$$\frac{5!}{2!(5-2)!} = 10$$

different pairs of d-d electrons with parallel spins. Therefore if you evaluate the expectation value of the energy, there will be a $-10K_\mathrm{3d3d}$ term, which is stabilising. This is precisely the exchange energy effect that Geoff describes in his answer.


Safe way to extract hydrogen from water through electrolysis

Some sort of advice, since you are concerned with safety and demonstrating the experiment to children. You should generally not perform an experiment in a badly ventilated space. Please be aware, that the generated hydrogen gas is highly flammable and with the addition of the right amount of oxygen, it can make very loud booms: How explosive is hydrogen gas?

Here are two questions, that are very closely related and already contain the answer to your question:

Can I perform electrolysis without electrolyte? (with distilled water)
How can I produce hydrogen gas efficiently and cheaply?

In general any kind of salt as an electrolyte will do.

In low concentrations about 1-2 mol/L they are usually quite harmless. I personally would prefer diluted sulfuric acid. Any addition of group 2 salts may cause passivation of the electrodes by trapping carbon dioxide from the air.

Diluted sodium hydroxide or potassium hydroxide are also safe to handle.

Probably the safest is sodium sulfate, but it might not be the best performer.

Please wear safety goggles at all times during the experiment, best case: The children wear them, too.


What is the difference between infrared action spectroscopy and conventional IR spectroscopy?

After 30 minutes of Googling, I'm thinking that we both (EDIT: nope, it was just me) misparsed the name.

When reading the name "infrared action spectroscopy," my first instinct was to mentally split it up like

$$\textrm{infrared action} \qquad\qquad \textrm{spectroscopy}$$

that is, spectroscopy of the "infrared action" flavor.

However, after running into more dead ends than a drunk rat in a maze with no exits, I realized that the correct way to split it might be

$$\textrm{infrared} \qquad\qquad \textrm{action spectroscopy}$$

or "action spectroscopy" that happens to use infrared light.

Lo and behold, the first hit on google was a wikipedia page on action spectroscopy. That link seems to have more of a biological tint to it (as do most results), but there are several physical science papers out there as well. (See Edit)

It seems that action spectroscopy in general refers to hitting a system with light and watching some kinetic property of the system (rate of photosynthesis, conversion rate of isomers, etc.) In the case of the paper you cited, they report the relative intensity, so my guess would be that they did something like a scan over frequencies and watched how the ions fragmented at each frequency, which is supported by a cursory glance over the text surrounding Equation (1). Unfortunately, I don't have time to read the paper in full, so let me know if I'm totally wrong.

The moral of the story is that English is hard.

=============

EDIT: It turns out I'm the dolt and everyone else processed this correctly. As Brian mentions in the comments, the action spectroscopy page on Wikipedia is pretty awful, so I'll try to expand on action spectroscopy below, given what little I know about the subject.

In your usual spectroscopic setup, you hit the sample with some form of EM radiation, then look at what's reflected or absorbed to learn something about the system (e.g. UV-vis, IR, NMR, rotational)

It seems that in action spectroscopy, your goal is to use the light to cause some sort of a change in the system, and measure that instead. In the example given on Wikipedia, you measure the rate of photosynthesis as you change wavelengths--this tells you at which wavelengths photosynthesis is most effective.

In the paper I linked above, they seem to mediate the conversion of cis-butene to trans-butene using light. Presumably, you could start off with low-energy photons, then gradually ramp the energy up until you see the isomerization happen. The wavelength at which the transformations first start happening would tell you the activation energy barrier for the cis-trans isomerization. (Note that I can't actually read that paper because of the paywall, so I may be wrong about this.)

In the paper Brian linked, they seemed to use IR to study the fragmentation of an ion sample. They mention "loss paths" as observed through an MS, so it seems like they hit a sample with IR light, then figure out what it breaks up into.

As Brian said in a comment, it seems that action spectroscopy is where light is used "to alter the system under study, not to just probe it passively." I'd like to add that, in general, it seems that the quantity of interest is not the amount of light that comes back out, but something else that happens in the system as a result of the exposure to light (e.g. isomerization, ion fragmentation, glucose production).


Aren't hydrogens on less substituted groups more electrophilic?

Water can attack both carbocation and hydrogen atom. Sometimes it attacks C, sometimes it attacks H.

Whenever it attacks C alcohol forms so the reaction goes back to reactants. When it attacks H the product forms. We are interested in product, so we show only this path.


What is the difference between ΔG and ΔG "with an o on top"?

A simple explanation would be that the "o on top" denotes standard state and the one without the "o on top" denotes conditions that are not standard state. However, that may not allow you to "get it" so let's look at the equations.

$\Delta G^{\circ} = -RT\ln K$
$\Delta G         = \Delta G^{\circ}  + RT \ln Q$

The first one allows us to find the Gibbs free energy at the standard state, as was mentioned. However, it would be useful to review what exactly the standard state is:

For the standard state of formation:

  The partial pressure of any gas involved in the reaction is 0.1 MPa.
  The concentrations of all aqueous solutions are 1 M.
  Measurements are generally taken at a temperature of 25 °C (298 K).
  The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions.
  (source)

Or, very simply:

Partial Pressure of gases = 0.1 MPa
Concentration of Aqueous solutions are = 1 M (Molar)
Temperature usually 298 K

So if it's so nice and well, why would we need the one without "o on top"? Well, if we meet a problem or situation without the concentrations or pressures as stated above, we need to use our handy equation to relate the standard state which may have been given to the actual situation, with perhaps 0.1 M or some other concentration. This is powerful because it allows you to view different states,perhaps with more or less reactants or products, and realize how one can affect equilibrium.


Synthesis of 2-Phenyl-2-Butanol from Grignard Reagent

Hint: The carbon atom bearing the OH group in the product has three substituents--phenyl, methyl, and ethyl. Any one of these groups could be introduced as a Grignard reagent, with the appropriate ketone. You have already correctly deduced one of the three possibilities.


Discrepancy in MO diagrams for Period 2 homonuclear diatomics

As molecular orbital diagrams are qualitative ways to understand bonding, there are different ways in which they may be drawn.

In the simplest possible cases, molecular oxygen (O=O), for instance, it is okay to assume that s-orbitals mix only with s-orbitals (of similar energy), and that the p-orbitals only mix with p-orbitals. This works fine for molecular oxygen, and explains many of its properties: double bond between O and O, paramagnetic etc.

In other cases, molecular nitrogen being the common example, this is not sufficient, and we have to account for orbitals mixing with orbitals of other kinds. This is called s-p mixing.

Firstly, you construct a MO diagram where orbitals of the same kind mix (the same as you would have done for oxygen), you then allow these MO's to interact with other MO's of the same symmetry to produce a new set of mixed orbitals. When the MO's interact in this way, they produce new MO's of higher and lower energy, which accounts for the different order of the MO's between diagrams where s-p mixing has been taken into account.

In general, any homonuclear diatomic molecules below oxygen (lithium, nitrogen, boron etc.) need s-p mixing to be considered.

For more information about s-p mixing (and other kinds of orbital mixing) do a search here on stack exchange. There are plenty of other similar questions with answers to suit.


Are there any general chemical rules for choosing initiators for a polymerization reaction?

It's been my experience that initiators are chosen for: 1) initiation temperature, 2) solubility, and 3) cost.

1) Initiation temperature - Most radical initiators are stable at room temperature; they need to be heated to dissociate. For example, the half-life of benzoyl peroxide at room temperature is very long, but it is about 30 minutes at 92 °C and 1 minute at 131 °C. However, some monomers boil or decompose at temperatures lower than those temperatures. AIBN (azobisisobutyronitrile) decomposes above 65 °C, so less heat is needed. Both benzoyl peroxide and AIBN are suitable for bulk polymerization of most monomers, including acrylamides, acrylates, styrenes, etc.

2) Solubility - Benzoyl peroxide and AIBN are soluble in organic solvents, which include most monomers for bulk polymerizations. However, if you want to perform your polymerization in water, for example if you are doing an emulsion polymerization, then you need a water-soluble initiator, like a peroxodisulfate salt.

3) Cost - Every initiator has a price, even if you are only using tiny amounts. AIBN may have a lower activation temperature than benzoyl peroxide, but AIBN is also more expensive. There are other more specialized inhibitors that are even more expensive.


Why does an anion not only bind to a single cation?

In the vapour phase where NaCl can exist (obviously at high temperature) then the molecule is a diatomic, with a covalent bond, albeit with a lot of ionic character.

In the crystal, to which I assume you refer, there is exactly one $\ce{Na^+}$ to each $\ce{Cl^-}$. You can see this if you look at the structure of the crystal, however to achieve this in the cubic crystal each ion has to be surrounded by six of the opposite charge as you correctly state. However, the charges are not 'used up' with one neighbour but shared equally between them, so a whole perfect crystal is electrically neutral. (If you want you can think of the charges as electric fields spreading out radially from each ion, they cancel each other out at some point between ions making a complicated shaped surface of zero potential). The bigger chloride ions form a simple cubic array with the sodium in the gaps. The crystal has this structure, as opposed to another (say face centred cubic), because this is the one than minimises the total energy, by keeping (large) chloride ions far enough away form one another so that their repulsion does not destroy the crystal.

In aqueous solution the water, which is very polar, means that it is more energetically favourable to dissociate into ions and solvate them rather than keep NaCl as a diatomic molecule, i.e. form $\ce{Na^+(aq) + Cl^-(aq)}$ where each ion is surrounded by several water molecules.


I suggest to do the following:
Line the bottom of a pan with aluminum foil. Put the silver piece on top of the aluminum foil. The silver piece and aluminum foil must be in contact with each other.
To about 2L of hot boiling water add about a half cup of baking soda (Be careful!). Then add the mixture to the pan. Make sure to cover the whole silver piece. The tarnish will disappear within several minutes.
The chemistry underlying this treatment is the following:
The silver tarnish is the chemical product of the reaction of silver with sulfur-containing substances in the air, $\ce{Ag2S}$. Silver sulfide is black. In this process, we convert silver sulfide to silver according to the following reaction:
$$\ce{3Ag2S + 2Al->Al2S3 +6Ag}$$ In fact, aluminum has a greater affinity to sulfur than silver.  The reaction between silver sulfide and aluminum takes place when the two are in contact while they are immersed in a baking soda solution. And it becomes faster when the solution is hot.The silver piece should touch the aluminum because a small electric current flows between them during the reaction. The formed aluminum sulfide may adhere to the aluminum foil, or it may form tiny, pale yellow flakes in the solution.

The advantage of this method of cleansing is that we do not remove the tarnish with some silver. We simply transfer silver sulfide to silver.


The term doesn't seem to have an official IUPAC definition however several sources agree that the lower organic acids are those that are soluble in water.

In the simplest case this could just refer to the first 5 carboxylic acids (C1-C5): methanoic (formic) acid, ethanoic (acetic) acid, propanoic acid, butanoic acid and pentanoic acid.

What is less clear is whether small water soluble acids such as lactic acid  and tartaric acid are counted. (The context of the quote in your question suggests it may well be, though it's wording isn't overly clear)

Since the phrase has no official definition I think it's probably safe to use it as you see fit, ensuring you define what you plan on calling a lower acid and use it consistently. You will equally see people refer to lower amines or halides, which in the same way isn't a well defined phrase but just lets people vaguely define a group of compounds.


In their 2013 Annual Report titled Evolving to a Specialty Care Biopharma Company, a well-known pharmaceutical company, Bristol-Myers Squibb has introduced the following statement under the title, Developing Innovative Drug Platforms:

  As R&D evolves its focus, it is also investing in technology platforms that concentrate on new ways to affect disease targets, including antibody drug conjugates, which combine the targeted benefits of biologics with the cancer-killing ability of traditional small-molecule chemotherapies. R&D also will further expand the potential use of millamolecules, which are larger than small molecules but smaller than biologics. These millamolecules may be able to better exploit novel targets and mechanisms by retaining the desirable properties of small molecules with the high degree of selectivity, especially against antigens, that biologics and small molecules have had difficulty targeting.

And then, Bristol-Myers Squibb Company has filled a patent under World Intellectual Property Organization (WIPO) claiming Pet-imaging immunomodulators, abstract of which states that:

  The invention relates to the synthesis and use of $\ce{^{18}F}$-labeled millamolecules for imaging various processes within the body, for detecting the location of molecules associated with disease pathology, and for monitoring disease progression are disclosed.

Thus, to my understanding, millamolecules is the word invented by Bristol-Myers Squibb Company's scientists to describe special set of molecules with different activity other than those exhibited by small molecules and higher polypeptides.

Finally, Medicinal & Pharmaceutical Chemistry glossary & taxonomy describes it as:

  Millamolecules: Mid-range compounds that fall in size between small molecules and biologics. Size is not the only consideration for this class, as millamolecules should also be orally available and able to interrupt protein-protein interactions.

Unsurprisingly, their reference has also directed to Bristol Myers Squibb: Areas of Focus.


Can the number of intermediates for a reaction be more than the total number of transition states?

Let's assume your reactive complex (reactants interacting) are optimised, i.e. its gradient equals zero, and your potential energy surface is continuous.

Then any direction is either uphill or constant (otherwise the optimisation hypothesis would not apply).
There are three possibilities: (a) going from reactants to intermediate gives a transition state which has higher energy than both reactive complex and intermediate; (b) reactants and intermediate lie on a plateau where energy does not change; or (c) one hypothesis is wrong.

So yes, for barrierless processes.


Why is the tetraamminecopper(II) ion square planar?

Well it's not really square planar, it's just a streched octahedron. For copper(II) you end up with a $d^9$-system, a $d^8$ system might be able to do a real square plane but that one electron in a $d^9$ would cost too much energy if it was in a true square plane. The effect is called Jahn-Teller effect. You can simply imagine the $z$-axis for example of the octahedron to be elongated. If you did this till it breaks, you end up with the square base of an octahedron. And this is also what happens in a real square planar complex. All $d$-orbitals with a $z$-component are lowered in energy so they become occupied by electrons and, due to repulsion with the ligand's electrons, ligands will not come from the $z$-axis anymore. On the other hand if you lower the $z$-part in energy due to lever principle you will have to increase the energy of other orbitals as well.

You can see this quite beautifully demonstrated here:

Let's imagine the right case with a square plane and a strong ligand so you start filling orbitals with two electrons before continuing to fill the next higher level or orbitals. Then you could add 8 electrons in total and leave the $d_{x^2 - y^2}$ empty. In total this gives you more energy than in the case of an octahedron. If you look to the left at the octahedron, you can see that for 8 electrons you would have to fill the $e_g$ as well and in total you'd end up with less energy which is why a plane geometry is favored by low spin $d^8$ systems.

Now copper(II) with its $d^9$ is in between that, a plane would be favored of course but it would require a lot of energy to put that extra electron into the $d_{x^2 - y^2}$. So you go in between, a really stretched octahedron, this is the second picture. Here the $d_{x^2 - y^2}$ is not as high as in the other case but the other orbitals you fill are already lower than in a symmetric octahedron so you end up with more energy already. And this is why copper(II) tends to do that. On the other hand this also changes the $d-d$ gap and explains that blue color of copper(II) compounds.

The other two ligands which are on the streched positions can be water, it could be a water and a sulfate, like in $\ce{CuSO4 * 5 H2O}$, but there is still something usually.


Number of chiral centers in penicillin V

Yes, here Nitrogen has to be considered as a stereocenter. Also, you can assign R and S to it. The rule is that the lone pair has the least priority and then you can assign R and S as you do for other molecules.


Calculation of the enthalpy change of reactions

You always need to add the equations since the break of bonds or the formation of bonds leads to an exchange in energy.
Here is a simple case

$$\begin{alignat}{2}
\ce{2 B + 3 H2O &-> B2H6 + 3/2 O2}\quad&&(\Delta H = 762\ \mathrm{kJ})\\
\ce{H2 + 1/2 O2 &-> H2O}\quad&&(\Delta H = −286\ \mathrm{kJ}) \times 3\\
\hline
\ce{3 H2 + $\boldsymbol{\ce{3/2 O2}}$ &-> $\boldsymbol{\ce{3H2O}}$}\quad&&(\Delta H = −858\ \mathrm{kJ})\\
\ce{2 B + $\boldsymbol{\ce{3 H2O}}$ &-> B2H6 + $\boldsymbol{\ce{3/2 O2}}$}\quad&&(\Delta H = 762\ \mathrm{kJ})
\end{alignat}$$

Notice the ones bold get canceled since the product is the reactant in the other equation.

$\Delta H_\mathrm{r}$ = sum of the $\Delta H$s so it’s equal to $-96\ \mathrm{kJ}$.

And in the second equation you were talking about one $\ce{H2O}$ molecule however in the first you are talking about three.


Some recent developments shown in article Transversely quasicrystallographic 3D photonic chiral lattices: Polarization-controllable complex photonic band gap structures, where autors claimed experimantal results on produsing 3d chiral photonic quasicrystals.

  Figure. Computer simulation of the light intensity distribution of
  the interference pattern for transversely quasicrystallographic 3D
  chiral photonic lattices respectively with rotational symmetries
  (a)5-fold (d) 8-fold (g) 12-fold. (b), (e) and (h) are their respective
  x-y planes. (c), (f) and (i) are their respective x-z planes

  "So the fabricated class of 3D PQC chiral lattices integrate both these wonder features on asingle platform in view of their possible applications in new generation integrated optical photonic devices with tunable polarization band gaps as well as frequency band gaps"

I also found fascinating that some researchers, use combination of structures like simple lattice and chiral lattice. This presentation Photonic Crystals and Metamaterials
has a lot of pictures, which could be good "food for thoughts".

Hope this was helpful, I think your area of research is very interesting.


Is it likely that increased understanding of quantum physics will change our understanding of chemistry?

Well, you asked for an example, I'll give you one.
"Quantum" is undeniably the most effective keyword our science has seen, it's changed our 'understanding of chemistry and physics' drastically over the past century.
For example, between 1900 to 1925 (years vary from source to source, but this era exists, that's what that is important), in which Bohr's model was the main focus of study (as wikipedia puts it), is known to be the era of "old quantum theory". There were series of theories, which started by Max Plank, and partially came to results with the final modifications of Sommerfeld; and ended in Erwin Schrodinger's famous model. New efforts led to new results and therefore the "old quantum theory" was never consistent.
What was all of this about? Scientists are always eager to find out the differential behavior of matter and energy. Sub-atomic particles (in first level, n, p & e) were discovered in order to explain the phenomenons that Roentgen saw (radioactivity of Uranium and other heavy elements in 'pitchblende'); Faraday experimented (the $\ce{SnCl2}$ solution experiment); Greek Leucippus and Democritos tried to explain; and so on. After every discovery, since the 'rules' that were discovered usually were up to a point in explaining what's around us, we change them and make new discoveries. And that was the reason of what Einstein, Bell and people like those did. They corrected the quantum for it to be what we know today.
So you see, the science was always about understanding; so why wouldn't a big discovery like quantization that led to quantum mechanics be able to change our direction of thought, either we are chemists, physicists or simply humans?
Links you might find helpful: About the old quantum theory in wikipedia; About the old quantum theory 2; Bell's theorem(which linked classical quantum to quantum mechanics) and finally quantum mechanics according to wikipedia.


The thing you have to understand is that the enthalpy of dissolution is actually the sum of lattice enthalpy and hydration enthalpy. So being a thermodynamic concept, we will have to turn to Gibbs free energy i.e. ΔG= ΔH-TΔS for completely understanding the solubility trends. For the process to be spontaneous, ΔG(dissolution)<0 . Now, entropy will generally increase while dissolving, so we can safely assume ΔS>0 for now. So obviously, the factor of -TΔS will be negative for T>0, so this contributes to the solvation of the said compound.(From now on, ΔH will be referring to enthalpy of dissolution)

Now, it all comes down to the ΔH part : it's magnitude and sign. If ΔH is positive and initially greater than -TΔS, then ΔG will be positive, and as temperature is increased, the -TΔS factor will ultimately overcome the positive ΔH and then, ΔG will become negative. And indeed, this is why you will find that $\ce{PbCl2}$ is insoluble in cold but quite soluble in hot water. Similarly, you can consider the implications of variations in the magnitude and sign of ΔH on solubility (in a polar solvent) yourself.

Now, coming to your specific problem, you see,  you first of all have to consider the configurations of $\ce{Ag+}$ and $\ce{Ca^2+}$

$\ce{ Ag^+ -  [Kr] 4d10}$ and
$\ce{ Ca^2+ - [Ar]}$

So you see, the calcium ion has 8 electrons in it's outermost shell ( 2 in 3s and 6 in 3p) while the silver ion has 18 electrons in it's outermost shell( 4s2 4p6 4d10). Now, due to the ineffective shielding by the d-electrons, effective nuclear charge on the valence shell and hence it's polarising power increases tremendously. Here, the calcium ion possesses an "inert gas configuration" while the silver cation has a kind of  "pseudo noble gas configuration"  (as it is not exactly a noble gas configuration, but still the cation is quite stable). On moving further down the periodic table,you get cations like $\ce{Tl^+, Bi^3+}$ etc. which have neither 8 nor 18 valence electrons (hence they can be called "non-inert gas configurations" ) and owing to the even poorer shielding by f and d electrons, there polarising power is even higher than the previous two configurations. So, the general order of polarising power of cations can be : Non-inert> Pseudo-inert> Inert gas configuration .( Note: All this can also be derived quantitavely by actually calculating effective nuclear charge on valence electrons by Slater's rule)

So, between $\ce{Ag^+}$ and $\ce{Ca^2+}$, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as compared to $\ce{Ca^2+}$ . As to your question about when to apply which concept: You are basically required to compare the sum of the hydration enthalpy and lattice enthalpy. If the lattice is very strong(example: LiF due to compatibility of ions and less interatomic distance) then the hydration enthalpy can be ignored, while for something like $\ce{LiNO3}$ the lattice is weak and hydration enthalpy is highly negative, so lattice enthalpy can be ignored for the final ΔH. You can use any concept as per your convenience, be it Fajan's rule, or extent of hydration or lattice strength; they will genrally complement each other provided you apply them correctly

For alkali metal chlorides like KCl and NaCl, the cations have only got a unit positive charge (which is not much to cause a lot of polarization) and more importantly, these cations belong to the inert gas configuration type, so here lattice energy will be the dominant factor


Why does whipped cream use nitrous oxide instead of nitrogen gas?

There are two ways to efficiently make an aerosol product:

Use a gas that liquifies under the pressure inside the can. For example, butane lighters. Nitrogen is one of the "fixed gases", meaning it's a gas under most conditions (but take a look at the temperatures and pressures needed for liquid nitrogen—it's not going to ever be found in consumer products).

or

Use a gas that is highly soluble in the liquid (carrier) and that will "substantially" vaporize when the higher pressure inside the can is reduced to atmospheric.

The US government restricts the pressures that can be used in aerosol cans (and requires 100% quality control testing—when's the last time you heard of an aerosol can exploding? [although it does happen]). If you cut up a can, you'll notice that it's pretty flimsy. The higher the pressure (and a gas that has been dissolved in a liquid doesn't exert much pressure), the more expensive it will be to build the container (aerosol can).

Thus, the fixed gases are almost never used, except in some medical products. Why? Because they just aren't soluble enough to help move the liquid (or, in other cases, solid) out of the can and also to disperse it into a very fine mist. The customer wants basically one thing when using an aerosol: uniform, consistent spray from first to last drop. Using a very soluble gas helps, and using one with a boiling point near room temperature also helps. But the laws of thermodynamics say that the temperature of the can will drop as you spray out its contents. This may dramatically interfere with the liquid-to-gas phase change, while solubility is less sensitive to temperature. The trick is to make a product (and I've made some) that sprays out consistently and also doesn't leave so much left behind in the can that the customer feels ripped off.


I think you missed the word "compensated".

  The decrease in specific conductivity is more than compensated by increase in its volume.

This simply answers your question. The reduction in molar conductivity due to the decrease in specific conductivity is  overridden by the increase in volume.

It is saying that the latter overrides the former, and not just overrides, but overrides by a lot.


Said by other words, you do not believe that $3 + 2 = 3 + 2$

There are 2 $\ce{Hg}$ columns, sharing the same pressure at the bottom.
Both have the same height, so the pressure difference is the same.
Yet you have doubts the pressures at their tops are the same.

Pressure at A = Pressure at B +pgh

It should be $p_\mathrm{A} = p_\mathrm{B} + \rho gh$ where $\rho$ is the density of mercury.

The vertical gradient of the hydrostatic pressure in liquid of the density $\rho$ at gravitational accelerationg $g$ is:
$$\mathrm{d}p = - g \cdot \rho \cdot \mathrm{d}h$$
As we can consider $\rho$ and $g$ as constants, the difference of the pressures is then :
$$p_C - p_A = - g \cdot \rho \cdot \int_{h_A}^{h_C} {\mathrm{d}h}$$
$$h_C=h_A \implies p_C=p_A$$

If pressure at the same altitude within the connected liquid of constant density was different, there would be the liquid flow until it is equal.
The pressures at the same altitude would be slightly different, if e.g. mercury at both columns had different temperature and density.

Feedback to comments:
In fact, it does not matter if it is mercury or water, the important thing is the density is constant across the liquid. What other than atmospheric pressure would you expect at C ( regardless of the liquid below ) ? Your scenario description is not exactly clear. Now at A, there is pressure by mercury column which is nothing but it's weight.
If the density along the thought path from A down and then up to C is constant, then pressure at A and C are the same, if the liquid is in the rest. Any pressure difference would cause motion of the liquid.

Let assume the liquid is mercury with density $\rho$.
Let assume the atmospheric pressure at C is $p_0$.
Let assign to C the relative height $h=0$.
Then pressure anywhere below C ( with negative $h$ ) is
$$ p = p_0 - \rho \cdot g \cdot h \gt p_0$$
The same is anywhere below A. At A is the same pressure as at C, i.e. $p_0$.
Between A and B, at height $h$, is the pressure:
$$ p = p_0 - \rho \cdot g \cdot h \lt p_0,$$
but this time $h$ is positive.
B is then at the height
$$ h_\mathrm{B}= \frac {p_0}{\rho \cdot g}$$
with $p=0$
( if we neglect very low vapour pressure of mercury.)


Why does water condense on grass and iron grills in cold weather?

While grass IS a poor conductor of heat, it is super thin and will cool off really quickly because of of that alone. The increased surface area also means a greater area is exposed to the night air during the time when the moist air drops to the ground and will thus collect more dew than maybe a stone or a tree trunk.


It looks like you added the alcohol before the glycerin.  Possibly the aluminum sulfate would be more soluble with the glycerin in there, so try adding that before the alcohol.

Wikipedia calls aluminum sulfate only "slightly soluble" in alcohol, so the alcohol addition would be expected to lower the solubility of the salt.  Nothing is reported there about the impact of glycerin.

Upon further review

Nope, we now find this did not work.  Hence as other answers indicate, we need a more careful approach.


Why do the melting points of Group 15 elements increase upto Arsenic but then decrease upto Bismuth?

A lower melting point is an indication of a lesser organised solid structure. So the question is actually:

  What in the structures of arsenic and phosphorus make them less prone to melting?

Let’s start off by separating nitrogen from the entire group. It is a prime example of a diatomic gas such as hydrogen with a very inert $\ce{N#N}$ triple bond. As throughout the periodic table, multiple bonds with s and p orbitals between atoms of the same element are magnitudes more stable in the second period than anywhere else due to much better p-orbital overlap (smaller size, smaller interatomic distance). Therefore, we can consider phosphorus, arsenic, antimony and bismuth more or less as similar while nitrogen is an odd one out.

Phosphorus has four stable allotropes with vastly different structures. The white phosphorus allotrope, $\ce{P4}$ molecules, basically falls into the same class as dinitrogen and melts at around 40 degrees. Red, purple and black phosphorus build up network-like structures in two dimensions, either in tube-like chains (red) or as ‘wavy graphite’ (black). Only the black modification is clearly defined. Red phosphorus has melting points of around $600~\mathrm{^\circ C}$ while black phosphorus sublimes in vacuum at $400~\mathrm{^\circ C}$ — probably meaning that its melting point at standard pressure is much higher, maybe even higher than arsenic’s sublimation point. I found no references to liquid phosphorus so I have no idea whether the bonds of the network structure need be broken to liquidify or not.

Figure 1: crystal structure of black phosphorus, taken from Wikipedia.

Henceforth, we have established a structure element which we will be sticking to with only moderate modifications. The transition phosphorus — arsenic — antimony — bismuth is more or less the non-metal—metal transition. What explicitly changes is the distance between the sheets as shown in table 1.

$$\textbf{Table 1: }\text{Comparison of bond lengths and inter-sheet contact distances for pnictogens}$$
$$\begin{array}{lrrrr}
\hline
\text{ } & \ce{P_{black}} & \ce{As} & \ce{Sb} & \ce{Bi} \\
\hline
\ce{E-E}\text{-bond/pm} & 223.0 & 251.7 & 290.8 & 307.1 \\
\ce{E\bond{...}E}\text{-contact/pm} & - & 311.9 & 335.4 & 352.8 \\
\text{d}_\text{contact}/\text{d}_\text{bond} & \approx 1.5 & 1.239 & 1.153 & 1.149 \\
\hline
\end{array}$$

So while the phosphorus structure can be described well as consisting of semi-isolated sheets stacked above each other; when we have reached bismuth, these sheets are much closer to an overall three-dimensional network. Therefore, bismuth atoms are in a distorted octahedric environment as other metals would be. Arsenic, where the sheets are still well separated, exhibits almost metallic condictivity perpendicular to the sheets.

Putting this together and taking the extreme points, we can say:

The phosphorus structure is highly organised in that two directions are strictly discernable. Melting along one of these is no problem (perpendicular to the sheets) but melting within a sheet is a huge problem as $\ce{P-P}$ bonds would need to be broken.
The bismuth structure no longer has a true distinction of directions; all three axes are more or less equal. Therefore, it takes similar energies to break bonds in one direction as it does to break them in another direction meaning the overall process is easiert.
Furthermore, liquid phases are generally those that display a high local ordering but a low long-range ordering. The high local ordering can be preserved in bismuth even if the long-range ordering present in the solid state is lost. Conversely, the local ordering is phosphorus is a lot stricter and cannot easily be transferred into a state where long-range ordering is low.
Finally, note that non-metals whose solid phase does not consist of small, well defined molecules generally have higher melting points than metals. (Compare carbon’s sublimation point of c. $3600~\mathrm{^\circ C}$.)


Before to show you that what you said is false (I'm sorry ^^) be sure you understand the constant of a reaction depends on the temperature.

Let the same reaction you want $$\mathrm{A \rightleftharpoons B}$$

With a constant $\mathrm{K^{\circ}}$. Imagine you heat your system a little with $\mathrm{d}T>0$, then by Van't Hoff's law we get:

$$\frac{\mathrm{d}\ln(K^{\circ})}{\mathrm{d}T}=\frac{\Delta_rH^{\circ}}{RT^2}$$
Where $R$ is the perfect gas constant which is positive. Then $RT^2>0$.
 So $\mathrm{d}\ln(K^{\circ})$ as the same sign as $\Delta_rH^{\circ}\cdot \mathrm{d}T$.

So if you have an endothermic reaction $\Delta_rH^{\circ}>0$ because $\mathrm{d}T>0$ then, $$\mathrm{d}\ln(K^{\circ})>0$$

Then $K^{\circ}$ will increase with the temperature. If you take $\mathrm{d}T<0$ for an endothermic reaction then the constant of the reaction will decrease with the temperature.

I let you do the same final reasoning for an exothermic reaction. So the "Le Châtelier principle" is still true.

Note: If you know the chemical affinity, you can do the same proof, just a bit longer!


You should consider the Nernst equation
$$
\Delta G = \Delta G^{\ominus} + RT\ln Q=- RT\ln Keq + RT\ln Q
 $$
At equilibrium deltaG is zero and Q the reaction coefficient is Keq also note
$$
\Delta G = -nFE
 $$
so E is zero at equilibrium. Outside equilibrium Q will move towards K decreasing E.


A very basic way to do this would be to draw a diagram that shows the p-orbitals going above and below the plane.
For Borole, the bonding orbitals would look like as shown below (taken from the wikipedia page on Borole):

Natural Bonding Orbitals of Borole. Structure optimised using ORCA BP86-D3BJ and def2- TZVPP basis set. The calculated occupencies of the obitals going from left to right are $0.13$, $1.9$ and $1.9$ respectively.

[emphasis mine]
The point to be taken here is the sentence in bold, the occupany in the empty p-orbital of boron [left most diagram and first value given] is small compared to the other two π-bond orbitals. This means that there is no complete delocalisation in borole. Only the double bonds are in conjugation.
A reason for this is that the orbital of Boron is an empty p-orbital and so the total number of electrons that can potentially delocalise is $4$ and so not of the form $4n+2$. If anything, the compound would become "anti-aromatic" if conjugation were to happen and so that delocalisation doesn't take place.
Now, on your second question, the lone pair of nitrogen in pyridine doesn't contribute to the aromaticity since the lone pair is out of plane of the ring. So adding placing boron shouldn't affect the aromaticity.


The question requires you to put your answer in kJ $mol^{-1}$. This means kilo joules per mole. So you have calculated the energy in kJ correctly, to get it per mole you need to first calculate the number of moles of ammonium nitrate you used.

Use the equation: $$n=\frac{m}{M_r}$$

n = moles

m = mass in grams

$M_r$ = molar mass

The mole calculation turn out to be 0.05371642723.

Then its just a matter of division, which I calculated to be -25.50 kJ $mol^{-1}$ (2.d.p).

That answer is using your rounded answer for the kJ, so the answer may be off. When i do calculations, i round right at the end to get the most accurate answer.


Hydroxide as leaving group in Aldol Condensation in base

Quoting ORGANIC  CHEMISTRY by Jonathan Clayden  ,Nick Greeves , Stuart Warren ,these are elimination reactions .You cannot normally eliminate water from an alcohol in basic solution as hydroxide is a bad leaving group.It is the carbonyl group that allows elimination here: these are E1cB reactions, with a second enolization allowing the loss of OH−.

the base-catalysed aldol reaction sometimes
gives the aldol and sometimes the elimination product. The choice is partly based
on conditions—the more vigorous the conditions (stronger base, higher temperature,
longer time) the more likely elimination is to occur—and partly on the structure of the
reagents.

The key to what is going on is the carbonyl group.Negative
charges are stabilized by conjugation with carbonyl groups,The proton that is removed in this
elimination reaction is adjacent to the carbonyl group, and is therefore also rather acidic (pKa about 20).The anion that results is stable enough to exist because it can be delocalized on to the carbonyl group.

This next step is ,

The leaving group is not lost from the starting
molecule, but from the conjugate base of the starting molecule, so this sort of elimination,
which starts with a deprotonation, is called E1cB (cB for conjugate base).


When Iron reacts with any other species, how will we know that the reaction will be of Fe(II) or Fe(III) i.e. of ferrous or ferric?

The stability of the various iron oxides depends on the oxygen fugacity (or oxygen partial pressure, which is a good approximation to fugacity in near-atmospheric pressures).
$\ce{Fe^3+}$ is the stable form of iron in atmospheric oxygen partial pressures, this in such conditions $\ce{Fe2O3}$ (hematite) will always be the stable product. The lower oxidation state oxides (such as magnetite $\ce{Fe3O4}$ and wüstite ~$\ce{FeO}$) can only stably form in deep earth environments, or in special controlled atmosphere reaction chambers.
That said, in normal conditions it is even more likely for iron hydroxides to form hydrated compounds, especially when water is a reactant, or is present. In these cases, you get a zoo of various Fe-oxides-hydroxides, some of which are amorphous (not crystalline), and others are metastable. These compounds are generally referred to as "rust".


This could be due to hydride shift.

In the first image, the pi-bond shifts on the carbon from which $\ce{H}$ is eliminated. As the pi-bond leaves the first carbon, it becomes positively charged due to lose of electron, and the $\ce{H}$ gets converted to $\ce{H^-}$(hydride ion) as it accepts electron. Therefore, it is called hydride shift. It also stabilises the molecule as the resulting molecule is conjugated. The same can be applied for second and third images.

This is for the second image. The third one you can do it yourself. All the resulting molecules are stabilised due to conjugation.


Stereochemistry of hydrogenation with Raney nickel

The hydrogenation of alkenes using heterogenous catalysts such as Raney nickel is syn-stereospecific; i.e. both hydrogen atoms are added to the same face of the double bond.

Therefore, the hydrogenation of (2E)-(2,3-2H2)but-2-ene yields
(2R,3R)-(2,3-2H2)butane and
(2S,3S)-(2,3-2H2)butane.


A book like Feynman's Lectures to Physics, which deals with Chemistry and provides insights

About 20 years ago, I have got the German translation of Atoms, Electrons and Change by Peter W. Atkins. This short and simple book was inspired by the lectures on chemistry given by Michael Faraday (such as The Chemical History of a Candle).


How is the tetrachloridocobaltate(II) anion stable if its crystal field stabilization energy comes to be positive?

You didnt put the right values. Higher orbitals have energy value of $+0.4 \Delta_t$ and the lower orbitals have the energy value of $-0.6\Delta_t$.

This complex has a coordination number of 4, so it can have tetrahedral or square planar geometry. Tetrahedral complexes are favored where attainment of regular complexes is important. For the tetrahedral complexes $\mathrm{d^0, d^2, d^5, d^7}$, and $\mathrm{d^{10}}$ the configurations are regular.

In tetrahedral complexes, mostly $\Delta t < P\ (\Delta_t \approx 4/9 \Delta_o)$ and hence they generally form high spin complexes.

The splitting will be into $3\ \mathrm{t_2}$ (not $\mathrm{t_{2g}}$) higher energy orbitals and $2\ \mathrm e$ (not $\mathrm{e_g}$) lower energy orbitals. The effect of the ligand wont matter here because the value of $\Delta_t$ is small. So the filling will be same for strong field and weak field.

$$\begin{align}\text{CFSE} &= +0.4(3) -0.6(4)\\
&= -1.2\end{align}$$


Does chlorine trifluoride react with candlewax?

I don't think paraffin would last long, since PTFE doesn't seem to last in the presence of CTF.  From IGNITION! (John D. Clark), which you really, really need to read if you are interested in chlorine trifluoride or rocket propellents in general:

  The results were excellent, but the difficulties were infuriating.
  Ignition was beautiful —so smooth that it was like turning on a hose.
  Performance was high —very close to theoretical. And the reaction was
  so fast that you could burn it in a surprisingly small chamber. But.
  If your hardware was dirty, and there was a smear of oil or grease
  somewhere inside a feed line, said feed line would ignite and cleverly
  reduce itself to ashes. Gaskets and O-rings generally had to be of
  metal; no organic material could be restrained from ignition. Teflon
  would stand up under static conditions, but if the CTF flowed over it
  with any speed at all, it would erode away like so much sugar in hot
  water, even if it didn't ignite. So joints had to be welded whenever
  possible, and the welds had to be good. An enclosure of slag in the
  weld could react and touch off a fire without even trying. So the
  welds had to be made, and inspected and polished smooth and
  reinspected, and then all the plumbing had to be cleaned out and
  passivated before you dared put the CTF into the system. First there
  was a water flush, and the lines were blown dry with nitrogen. Then
  came one with ethylene trichloride to catch any traces of oil or
  grease, followed by another nitrogen blow-down. Then gaseous CTF was
  introduced into the system, and left there for some hours to catch
  anything the flushing might have missed, and then the liquid chlorine
  trifluoride could be let into the propellant lines.


Can we equate nucleophilicity and Lewis basicity?

I appreciate your perseverance to think so deeply about this!
But...Yes, you have overlooked the main difference between Basicity & Nucleophilicity ,i.e., Basicity (i.e. Lewis basicity) is a Thermodynamic term whereas Nucleophilicity is a Kinetic term. But here you have compared them under the belt of only thermodynamics.
Basicity is measured on the basis of the stability of the species, whereas Nucleophilicity is mainly aimed at the 'donating ability' of the species.
e.g. In polar protic solvents, the order of nucleophilicity in case of halide ions is:

  $\ce{F-}$ < $\ce{Cl-}$ < $\ce{Br-}$ < $\ce{I-}$

Which contradicts with the basicity order.

Whereas in polar aprotic solvents, the order is:

  $\ce{F-}$ > $\ce{Cl-}$ > $\ce{Br-}$ > $\ce{I-}$

Which follows the basicity order.

I hope this might help you facing your confusion a li'l more logically!


What is the chemical composition of a Rubbermaid Roughneck container and is it feasible for me to make my own molds?

This is a pretty broad question and might be better split into several, however a general comments on polymers, such as poly(ethylene), is that the properties of polymers can be tweaked quite a bit by control over the mean molecular mass and dispersity (probability distribution of molecular mass) of the polymer chains, as well as the addition of all kinds of stabilisers and plasticisers. Plasticisers especially may dramatically alter the flexibility of plastics. The exact composition of the Rubbermaid Roughneck is unlikely to be forthcoming.

Molding of thermoplastics by injection molding or casting is carried out by melting down plastic pellets and squirting the melt into or onto a mold. A mold release agent is often used to help get the cooled plastic out of the mold. The safety of the plastic depends on its specific composition (read the material safety data sheet for the pellets) and how hot it is (don't get molten plastic on yourself, wear protective clothing etc.). I don't know anything about the safety of mold release agents except for the fact that miniature model kits typically still have the agent on the plastic when purchased. This suggests they are fairly safe, though their presence may interfere with paint adhesion which is why pros scrub kits down before painting.

The geometric simplicity of your custom part, however, suggests that you could cut it from a plastic sheet, obviating molding and the potentially expensive equipment that goes with it.


Are there any reactions with no activation energies?

Marcus-Hush theory describes electron/charge transfer rates using conventional transition state theory, e.g.:

$\ce{A+ + B -> A + B+}$

Now one might naively assume that there's no activation barrier in a simple electron transfer, but that's not true. Since the geometry of $\ce{A+}$ and $\ce{A}$ as well as $\ce{B}$ and  $\ce{B+}$ are not identical, there is a reorganization energy of both the molecular species and the solvent or environment. This reorganization energy ($\lambda$) serves as an activation barrier for the charge transfer.

Now the convenient thing about charge transfer reactions is that you can tune the $\Delta G$ of the reactants.

Marcus predicted that the maximum rate would occur when the $\Delta G^0 = -\lambda$ and thus the $\Delta G^\ddagger = 0$. If the reaction becomes more thermodynamically favorable, you enter the inverted region and the activation barrier goes up and the reaction rate goes down.

This is a long-winded answer to say there are some chemical reactions with no activation barrier, in addition to the other cases here. Usually the key phrase describing these in papers is "barrier-less reactions"


Arranging the phenol derivative in decreasing order of acidity

The acidity order should be 1 > 2 > 3 as they have pKa values of 9.48, 9.98, and 10.21 respectively.

+R effect of $\ce{-OCH3}$ is more significant than +R effect of $\ce{-OH}$

In case of 1, -I effect works at -ortho position and also +R effect works at -ortho position.

In case of 2, -I effect works at -ortho position and also +R effect works at -ortho position. Net effect is +R effect as +R effect is dominant than -I effect.

In case of 3, -I effect works at -para position and also +R effect works at -para position. Net effect is +R effect as +R effect is dominant than -I effect. But this +R effect is more dominant than in case 2 because -I effect is more dominant at -ortho position than at -para position but+R effect of equally effective at both -ortho and -para positions.

As a result of which, acidity is reduced to greater extent in the third compound compared to the second compound.


You have only three isomers with the configurations (1R,2R), (1S,2S) and (1R,2S). The one with the configuration (1R,2S) is achiral, as it has a symmetry plan, and it's superimposable on its image by a mirror (i.e. it is identical to its enentiomer (1S,2R)).

This isomer is known as "meso compound".

Meso compounds are a special case of stereoisomers which occur when a molecule has two chiral centers and each chiral center has identical substituants.


Van der Waals equation at high temperature and high molar volume

First, let's look at this from a purely mathematical standpoint. This isn't going to be rigorous at all, but I'm happy to clarify anything if asked.

If we assume we have 1 mole, and we say that $V_\mathrm{m}$ and $T$ become very large, then the question becomes what happens to $P$ (our only remaining variable)? Well as $T$ and $V_\mathrm{m}$ become very large with respect to (w.r.t) our excluded volume parameter, $b$, (also written as $T$, $V_\mathrm{m} >> b$), the -b term in the denominator becomes negligible and we approach the conventional ideal gas law:

$$
P \approx \frac{R T}{V_{\mathrm{m}} - b} \implies P \approx \frac{R T}{V_\mathrm{m}} = \frac{n R T}{V}
$$

If this is not clear, try plugging in some numbers to the equation and calculating what $P$ would be for small $P$ and $T$ (same order of magnitude as $b$) vs. large $P$ and $T$ (100 times larger than $b$). I will let you determine if this leads to greater or less pressure.

Now let's switch gears to a chemistry point of view. What Johannes van der Waals did, is add this little $b$ parameter in the denominator of the ideal gas law. This parameter is a correction to account for the actual volume that each mole of particles can occupy. You can think of it equivalently as the actual volume each particle can occupy, since a mole is just a really big number of particles.

This $b$ correction is taking into account the fact atoms can repel each other at a relatively large distance. If we approximate these atoms to be hard sphere bouncing off of each other, the radii of these hard sphere is what we call the van der Waals radii - the distance at which atoms "bounce" off of each other if they come into contact. So the  parameter $b$ in the denominator is derived purely from the van der Waals radii of the atoms in question. So in the case of large $T$ and $V_\mathrm{m}$, we find that this correction is no longer necessary to make a reasonable approximation, because its presence makes such a small difference.

I recommend looking at the Wikipedia page for more information.


What exactly is formed when K2Cr2O7 is reduced?

The dichromate ($\ce{Cr2O7^2-}$) ions are strong oxidizing agents at low pH. During the redox process, each chromium atom in the dichromate ions (oxidation state = +6) gains three electrons and get its oxidation state reduced to +3. In redox reactions in aqueous acid solution, the aquated $\ce{Cr^3+}$ ion is produced according to following half-reaction (Electrochemical Series):
$$\ce{Cr2O7^2- + 14H3O+ + 6e- <=> 2Cr^3+ + 21H2O} \space \space \space \pu{E^0 = 1.36 V}$$

Assume that the $\ce{Cr^3+}$ ion in aqueous solutions is in its simplest ion form: the hexaaquachromium(III) or $\ce{[Cr(H2O)6]^3+}$. This ion is violet-blue-grey in color. However, when $\ce{Cr^3+}$ ion is formed during a reaction in aquous acid solution, it is often appeared in green color. Thus, we always describe the green color due to $\ce{Cr^3+ (aq)}$, implying it as the hexaaquachromium(III) ion ($\ce{[Cr(H2O)6]^3+}$), but that's actually not the case.

What's really happening in the solution is that one or more of the water molecules in each of $\ce{[Cr(H2O)6]^3+}$ ions get replaced by other negative ions in the solution, typically by $\ce{SO4^2-}$ (if the acid used is $\ce{H2SO4}$) or $\ce{Cl-}$ (if the acid is $\ce{HCl}$). For example, if one of the water molecules is replaced by a $\ce{SO4^2-}$ ion, $\ce{[Cr(H2O)6]^3+}$  becomes $\ce{[Cr(H2O)5SO4]+}$, which is green (notice also the change in the charge on the ion). You can check this by yourself: warm freshly prepared aqueous  chromium(III) sulphate hexahydrate solution, which would change its color of purple-blue or violet ($\ce{[Cr(H2O)6]^3+}$) to green ($\ce{[Cr(H2O)5SO4]+}$) due to the ligand exchange discussed above [Chem Guide].

In $\ce{CrCl3.6H2O}$, on the other hands, the green color of the solid itself or its aqueous solution is due to the $\ce{[Cr(H2O)4Cl2]+}$ ion. Hence, actually more proper way of writing the formula is $\ce{CrCl2(H2O)4Cl.2H2O}$. In contrast to previous case, the green $\ce{CrCl2(H2O)4Cl.2H2O}$ solution would slowly becomes violet in color by standing at room temperature overnight, by slow ligand exchange with solvent (water)[Chem Guide].

Interesting fact: Chromium was discovered by Louis Nicholas Vauquelin in 1797 who named his new element after the Greek word chroma — which means color.

Late Additions:

Although $\ce{Cr^3+}$ ions complexes with most ligands to give green color, it gives vivid line of colors with other ligands (note it names after chroma!). Some interesting examples are illustrated in the picture below:

Also see:

Chromium(III) nitrate: Blue-violet crystals (anhydrous), and purple crystals (nonahydrate), according to Wikipedia (see the picture below).


Why are amines more soluble than ethers in water?

Diethyl Ether can accept hydrogen bonds readily, and you are correct in pointing out it would be better than diethylamine at doing such.  However, diethylamine can also donate a hydrogen bond, giving it two modes of solvation, possibly both simultaneously.

Also, there is some minor acid-base reactions that may occur forming diethylammonium hydroxide, which being an ion, would certainly be more soluble than diethyl ether.


Why does the amount of CO2 produced decrease with more baking soda?

This graph is a very good example of experimental errors which may give rise to a pseudo-phenomenon i.e., an observation which apparently looks like a scientific phenomenon but it is an artifact of the experimental design.

First of all work out the moles of acetic acid (5% acetic acid) and moles of NaHCO3. As you can see vinegar has become a limiting agent. And it should have become a flat curve only if your system was a closed system.

So where is the culprit in the experimental design?

As you stated "It was about a 1-1.5 second delay between poring the vinegar and capping it."

By the time, someone puts a syringe, a large volume of gas has already escaped when acetic acid was poured on a heap of sodium bicarbonate. How big is the heap of 3 g NaHCO3!

There is a nice trick to avoid these problems. You put the sodium bicarbonate in an open plastic pouch so that acid does not immediately encounter the salt. You can stopper and then shake to mix (quite well).

So your hypothesis is right for a flat curve, sadly there is an experimental design issue.


How to find the number of atomic microstates for a given electronic configuration?

You really have to calculate out things exactly, and particularly carefully if electrons are in the same orbital, e.g. $p^2$, when the Pauli principle has to be checked. Its a bit tedious but it can be very quick with a bit of practice especially if electrons are in different orbitals.
You start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number s = 1/2 and magnetic quantum number $m_s$ = +-1/2. Orbitals have angular momentum of s=0, p=1, d=2, f=3 etc units.
The total spin angular momentum is the series of values $S=|s1+s2|...|s1-s2| $ where s1 and s2 =+1/2 are the quantum numbers of each electron. The total angular momentum is   $L=|l1+l2|...|l1-l2| $  and total angular momentum $J=|L+S|...|L-S| $
Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by 1. Note that sometimes there is only one value and all values are positive. e.g. J=5/3, 3/2, 1/2 or J could be 0 or 3 etc.
The term symbol has the form $^{2S+1}L_J$.
The super-prefix is the spin multiplicity, for spin angular momentum $S$ this is  $2S+1$ or in general $2X+1$ for angular momentum X.
 In your question $s^1f^1$ the total spin using the series above is  S = 0 and 1. The spin multiplicity is thus 1 and 3, singlets and triplets terms only. The total orbital angular momentum L = 3 since the s orbital has zero angular momentum. Thus the terms so far are
$^1F$ and $^3F$ The number of microstates (multiplicity) in each term is $(2S+1)(2L+1)$. This is 7 for the $^1F$ and 3.7=21 for the $^3F$ making 28 in total. These states are also degenerate in energy. You should calculate that $s^1d^1$ has 20 microstates.

The total angular momentum J series associated with S=1 is $J=|3+1|.... |3-1|$ or 4, 3 and 2 and with S=0 , J=3.  Thus only triplet states have J = 4 and 2 and both singlet and triplets J=3. The terms symbols for the levels are
$^1F_3$, $^3F_4$, $^3F_3$,  $^3F_2$
 The total multiplicity from the J values is is 7+9+7+5 = 28 the same as from terms.

When the electrons are in the same orbital then the only secure way of calculating terms is to list all the individual quantum numbers in a table, with columns $m_{l1},m_{l2},m_{s1},m_{s2}$, S and L and remove those for which the Pauli principle removes a row. (See table in Steinfeld 'Molecules & Radiation' chapter 2 and Engel & Reid, 'Physical Chemistry' chapter 22). In particular you have to look out for when two rows have the same set of quantum numbers but in a different order; for example $m_{s1}$=0, $m_{s2}$=1 in one row and $m_{s1}$=1 and $m_{s2}$=0 in another but the $m_l$ values are the same, and vice versa. Doing this you should find, for example that $p^2$ has 15 terms, as $^1D$, $^3P$ and $^1S$ Hope this rather long answer helps you understand things.


This is a practical matter of making sure that there is a common basis for comparing values from all different types atoms and molecules. If it were not the case there would be a range of different values for the same type of atoms or molecules.

Depending on temperature the ionisation energy would be less due to a molecule's rotational and vibrational energy, and vary with temperature because the number of levels populated depends on the temperature. For some types of atoms there may be low lying electronic levels that could be populated at higher temperatures, say at 1000 K in a high temperature gas. Again the measured ionisation energy would depend on temperature.

Also, if this was not done the $\Delta H$ values would in effect have double counted the effect of temperature, once in the ionisation energy and again in the heat capacity term.


What reactions occur when mixing copper sulfate and sodium hydroxide?

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.


why lithium is less reactive than sodium?

Reactivity decreases down the group for halogens and increases down the group for alkali metals.

As you have said, Lithium is smaller than Sodium, the valence electron on sodium is further away from the positive nucleus and therefore it has a weaker electrostatic attraction and is more easily lost in a chemical reaction, making it more reactive. This is due to the shielding effect of the second energy level (electron shell) which makes the valence electron on Sodium further away from the nucleus and hence the larger atomic radii.


How does the equilibrium shift when concentration of reactant and product are increased simulatenously

Can someone explain this phenomenon in terms of rate of reaction or in terms of molecules?

In terms of rate of reaction, at equilibrium the forward rate is proportional to the concentrations of the reactant (as if this were an elementary reaction). Likewise, at equilibrium the reverse rate is proportional to the concentrations of the products (as this were an elementary reaction). If you change the concentration of a single reactant and a single product by the same factor (i.e. multiply by the same number, not add the same number) and if the stoichiometric factors are the same (i.e. the order of reaction in this special case), and the reactants are sufficiently dilute, the system will still be at equilibrium. In the second change described by the OP, neither is the case (one concentration increases by 50%, the other doubles, and the exponents are different).

In terms of thermodynamics, if the Gibbs free energy of the reactants is changed by the same amount (i.e. add or subtract a number) the system will remain at equilibrium. This will happen if one species on either side is changed by the same factor because the Gibbs free energy changes with the logarithm of the activity of a species. Again, this only works if the stoichiometric coefficient of the two species are the same.

Nomenclature K vs Q

In the equations shown by the OP, different concentrations are substituted into the equilibrium constant expression even when the reaction is not at equilibrium. The correct name for these expresssions is reaction quotient Q. If Q matches K, the system is at equilibrium. In the examples provided, Q stays the same in one case (still in equilibrium) and changes in the other case (no longer in equilibrium). What the equilibrium would be needs not to be calculated. Knowing that Q is unequal to K is sufficient for showing that the system is in fact not in equilibrium.


How many grams of CH3OH can be made from 30.0g of CO if it all reacts?

Moles of CO in equals moles of methanol out (at 100% yield), typically over copper-doped zinc oxide catalyst.  Do the arithmetic.

One gram-mole of a chemical formula is [the magnitude of its atomic mass units] grams.  A nitrogen atom is 14.0067 amu.  A mole of nitrogen gas, $\ce{N2}$, is then 28.0134 grams.


How long must a sample be irradiated before all 59-Co atoms are converted to 60-Co?

At a constant neutron flux, this is a first-order reaction:

$ - \frac{{{\text{d}}\left[ {{}^{59}{\text{Co}}} \right]}}{{{\text{d}}t}} = k \cdot \left[ {{}^{59}{\text{Co}}} \right]$

$\left[ {{}^{59}{\text{Co}}} \right] = {\left[ {{}^{59}{\text{Co}}} \right]_0} \cdot {\operatorname{e} ^{ - k \cdot t}}$

Theoretically, $\left[ {{}^{59}{\text{Co}}} \right]$ will never become zero.


No this is not a job for an IR machine. It all depends on the machine you are using, but provided the monochromators will extend that far all that may be necessary is to change the detector for one that is more red sensitive. If your machine will not do the job, and if its a one-off type of measurement  try to get time on another machine somewhere else. If this is not possible or you need to measure many samples over a long period you may need to experiment by using a lamp , interference filters and a detector etc and make your own machine to measure transmittance.


I will be using the following equation from my previous answer

  $$I = (z \mu \nu c F E)A$$

You are interested in an electrolyte of the type $\ce{A^{2+}}$ $\ce{B^{2-}}$.

So, the "cation" current is

$$I_+ = (z_+ \mu_+ \nu_+ c F E)A$$

and similarly the "anion" current is

$$I_- = (z_- \mu_- \nu_- c F E)A$$

Here, $z_+ = z_- = z$  and $\nu_+ = \nu_- = \nu$. The mobilities, however, are different, so $\mu_+ \neq \mu_-$

So combining the contribution from the cations and anions in the solution, we get a total current

$$\mathrm{I}_{\text{tot}} = (z \nu c F EA)(\mu_+ + \mu_-)$$

After substituting  $\ E = \frac{V}{L}$,  one gets the desired result

$$\mathrm{I}_{\text{tot}} = \left (z \nu c FA \frac{V}{L}\right ) (\mu_+ + \mu_-)$$

In general, if $z_+ \neq z_- $  and $\nu_+ \neq \nu_-$ then simply write down the individual expressions for cationic and anionic contributions to current and add them.


Reactions on alpha tetralone to synthesize naphthalene

Moinak Banerjee: Your concern about the mechanism of the conversion of tetrahydronaphthalene (tetralin) 1 to naphthalene 5 has been addressed in a comment by Waylander. It is important to recognize that the process is a 4-electron oxidation of tetralin 1 and a four electron reduction of the hydrogen that is liberated as 2 equivalents of hydrogen gas. Ideally, the reaction is catalytic in palladium (Pdo). Mechanistically, the reaction likely proceeds by oxidative addition of Pdo into the labile benzylic C-H bond of tetralin 1 to afford Pd(II) species 2, which, in turn, isomerizes to π-complex 3. These steps are shown as reversible because complex 3 is the likely intermediate in the hydrogenation of 1,2-dihydronaphthalene 4 to tetralin 1. Collapse of complex 3 regenerates the catalyst Pdo and an equivalent of hydrogen gas. A repetition of these steps at either the benzylic or allylic C-H bonds of 1,2-dihydronaphthalene 4 leads to naphthalene 5 and a second equivalent of hydrogen gas. These reactions are driven to the right by the loss of hydrogen gas in an open vessel. --continued--

This reaction played a critical role in early attempts to synthesize estrone since it served to reduce the number of asymmetric carbons, thereby facilitating stereochemical analysis. An example of this process is illustrated in the sequence 6 --> 8 from the work of Bachmann and Chemerda1. The double bond in tetracycle 6 is isomerized by Pdo to the tetrasubstituted position in isomer 72. The conversion to the naphthalene 8 follows the same steps as in the case of 1,2-dihydronaphthalene 4.3

1) W. E. Bachmann and J. M. Chemerda, J. Am. Chem. Soc., 1948, 70, 1468.
2) The double bond migration involves a similar C-H insertion and subsequent π-allyl complex. But that is another story.
3) See also: J. Heer and K. Miescher, Helv. Chim. Acta, 1949, 32, 1572.


Do we really have technology with the resolution to distinguish between layers of single atoms of materials?

X-rays do not allow you to look  at the material just like an microscope. You get the spacing information from the diffraction pattern of only crystalline materials and a lot of mathematics. On the other hand, atomic force microscopy may actually be able to tell you where one atomic layer starts and where it ends on a surface. For example, imagine a gold surface, onto which you deposit a layer of molecules (say on local spot by some means). AFM would allow you to create an image (again by indirect means, as it is not an optical technique) of that molecular layer.
See google images generated by AFM and you will see how fancy they are.

To the OP: A good way to studying is to rely on multiple references of the subject. Do not take everything literally. The authors are human beings after all with limited knowledge and of course they are not the experts of $every$ topic covered in the textbook. When I was a student, I used to spend a lot of time trying to decipher what the author was saying in a particular sentence. With maturity you learn that the better way (read: more scholarly way) is to look at the same topic in another book and see what the other author is saying.