Why does phosphorus give a free electron in silicon doping?

First, don't forget that the phosphorus is sitting on a diamond cubic silicon lattice site, which pretty much defines the symmetry of the potential around it - the P cannot change that, but must respond to it. So, you should not think in terms of atomic orbitals.

Second, the fifth electron is not initially a free electron - it occupies a state that happens to be in the band gap.  Now, that state is really close to the conduction band, only 45 meV below it at 300 K, so at room temperature there is a really good chance that it will jump into and roam about the conduction band.  Nor is that the only state in the gap for P - there is another state 72 meV above the valence band (i.e. it is slightly easier to ionize a 4-fold coordinated P than a Si).

Why doesn't the P only bond with 3 Si? Well, that would leave one unhappy Si atom, and one not much happier P atom.  Again, this is occurring in a crystal, not an isolated molecule. The P will occupy the lowest free energy state. Note that there are other impurities that preferentially occupy various interstitial sites. It is just that P can be coerced into pretending to be pretty much a Si atom - peer pressure, perhaps...


Hydrolysis of transition metals' halides?

$\ce{MCl_n(aq)} + n\ \ce{H2O(l)} \rightarrow \ce{M(OH)n(s)} + n\ \ce{HCl(aq)}$ does tend to go to the right. "Basically insoluble" is not quantitative enough to generalize. If the hydroxide is very insoluble, the acid produced by hydrolysis may not keep it in solution. But if the hydroxide is even only slightly soluble, it may not precipitate out. The acidity of the solution is a subtle indicator that the reaction has shifted a little bit to the right. The reaction isn't slower, just not always so extreme.

If you put $\ce{SiCl4}$ in water, it gives $\ce{HCl}$ + silica gel because $\ce{SiO2}$ (or "$\ce{Si(OH)4}$") is so insoluble.

If you put $\ce{MgCl2}$ in water, no precipitate appears because $\ce{Mg(OH)2}$ is slightly soluble (~2 mg/L), and the $\ce{HCl}$ produced by the hydrolysis reaction is acidic enough so that if a significant amount of $\ce{Mg(OH)2}$ were produced, it would just redissolve. $\ce{Mg(OH)2}$ gives a pH of about 10 in water, so at pH's lower than 10, $\ce{Mg(OH)2}$ doesn't precipitate out. BTW, $\ce{MgCl2}$ is slightly acidic because of the hydrolysis reaction.

$\ce{AlCl3}$ is complicated because of ionic complexes, but in water it hydrolyzes to give a very acidic solution and some aluminum hydroxide complexes or gels which are of varying solubility. $\ce{AlCl3}$ has been used as a water treatment because it produces flocculent precipitates that carry down impurities.

$\ce{NiCl2}$ does not precipitate in water because $\ce{Ni(OH)2}$ is even more soluble than $\ce{Mg(OH)2}$.

$\ce{FeCl3}$ is quite acidic in water. You might expect the commercial 40% solution to be a cloudy product because of the production of acid, but it is a clear, deep red color. Oh, but when it is diluted and the pH rises because the acid is diluted, it precipitates in flocs like $\ce{AlCl3}$. It is used to clarify water by precipitating dirt and dust.


Confusion in the name of this organic compound

From here:

"Number the carbon atoms of the double bond as “1” and “2” in the direction that
gives the substituent encountered first the smaller number."

The methyl group right on the double bond is deemed first, so the carbon at that point on the ring gets 1 and the other double-bonded carbon gets 2, forcing 6 for the ethyl substituted carbon.


If for reversible process, ΔS= 0 for universe (system+surroundings), why would there be non-zero change in gibbs energy?

The equation in Atkins' book refers to the system only, and not to the universe, and, moreover, in this equation, w' specifically refers to non-PV work, like electrical.


Yes, that is a perfectly acceptable formula for entropy change even outside of thermal equilibrium.

And, as you say, it's related to the irreversibility of heat transfer. The point about ideal thermal reservoirs always exchanging heat reversibly is not necessary to understand this, so let's park it outside for a moment.

Let's imagine a body in an environment. We'll call the temperature of the body $T_b$ and the temperature of the environment $T_e$. If they exchange heat, their respective entropies, $S_b$ and $S_e$, change according to the formula you mentioned:

$dS_b=\cfrac{\delta q}{T_b}$

$dS_e=\cfrac{- \delta q}{T_e}$

Note the sign (the heat flowing into the body is leaving the environment, or vice versa).

Also note, and this is critical, that the magnitude of $\delta q$ is equal in both equations (one receives exactly the same amount of heat as the other loses), but, if $T$ is different, the entropy change won't be the same. The change in entropy (positive or negative) will depend on the temperature of the body or the environment.

Let's have a look at how the total entropy of the system ($S=S_b+S_e$) changes.

If the body and the environment are at thermal equilibrium, in other words, their temperature is the same ($T_b=T_e=T$),

$dS=dS_b+dS_e=\cfrac{\delta q}{T}-\cfrac{\delta q}{T}=0$

In other words, transferring heat between bodies at thermal equilibrium does not change the total entropy of the system.

However, if they are at different temperatures,

$dS=\cfrac{\delta q}{T_1}-\cfrac{\delta q}{T_1}=\delta q \left( \cfrac{1}{T_1}-\cfrac{1}{T_2}\right)$

Now, the sign of the entropy change depends on the relative values of $T_1$ and $T_2$:

If $T_1>T_2 \rightarrow 1/T_1 < 1/T_2 \rightarrow 1/T_1 - 1/T_2 < 0 \rightarrow dS < 0$

If $T_1<T_2 \rightarrow 1/T_1 > 1/T_2 \rightarrow 1/T_1 - 1/T_2 > 0 \rightarrow dS > 0$

Note that I chose the signs of $\delta q$ assuming that the flux of heat went from the environment to the body. So, this means that if the body is hotter than the environment ($T_1 > T_2$), this would imply a reduction in the entropy of the whole system ($dS < 0$) - which the second principle explicitly forbids. And, to the contrary, if the body is colder than the environment, that flux will be spontaneous ($dS > 0$).

So, outside of thermal equilibrium, you still can use that expression to calculate entropy changes - as long as the heat flux goes in the right direction, from hot to cold. It will also apply to heat transfers in equilibrium (and therefore reversible) - although in that case you'll be having some other energy transfer process of equivalent magnitude siphoning that energy off - otherwise the heat would increase the temperature of part of the system, breaking thermal equilibrium (that won't happen).


Chemist Neil Bartlett in 1962 discovered that although Xenon was a noble gas, it is able to form compounds with other substances even though it is chemically "inert."
Neil Bartlett, at the time in 1961, produced an unidentified red solid and discovered that the red solid was a reaction between gaseous fluoride, platinum hexafluoride ($\ce{PtF6}$) and oxygen that was oxidized. Finally, determining the chemical structure, they found that the red solid was $\ce{O^{2+}PtF6}$.
Oxygen is a powerful oxidizing agent, being consumed in many combustion reactions. However, Bartlett believed that the platinum hexafluoride ($\ce{PtF6}$) was a stronger oxidizing agent than oxygen, and oxygen is instead reduced. Bartlett then believed if the oxidation of oxygen was possible, he could apply this xenon.

Because my co-workers at that time (March 23, 1962) were still not sufficiently experienced to help me with the glassblowing and the preparation and purification of $\ce{PtF6}$ [platinum hexafluoride] necessary for the experiment, I was not ready to carry it out until about 7 p.m. on that Friday. When I broke the seal between the red PtF6 gas and the colorless xenon gas, there was an immediate interaction, causing an orange-yellow solid to precipitate. At once, I tried to find someone with whom to share the exciting finding, but it appeared that everyone had left for dinner!

The discovered new compound was xenon hexafluoroplatinate ($\ce{XePtF6}$). Bartlett's important contribution to chemistry is a series of noble gas compounds with specific and useful properties.
In regards to your question on how xenon can be so diverse, with over "100 noble gas compounds known today..."

Noble gas compounds have already made an impact on our daily lives. $\ce{XeF2}$ has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents. The reactivity of radon means that it can be chemically scrubbed from the air in uranium mines and other mines. Excimer lasers use compounds of argon, krypton or xenon to produce precise beams of ultraviolet light (when electrically stimulated) that are used to perform eye surgery for vision repair.

Material derived from a highly interesting article from the American Chemical Society.
http://www.acs.org/content/acs/en/education/whatischemistry/landmarks/bartlettnoblegases.html
Looking at the literature:

Neuroprotection

For the use of Xeon during mechanical ventilation, an article examining the neuroprotective effects of xenon,
Neuroprotection against traumatic brain injury by xenon, but not argon, is mediated by inhibition at the N-methyl-D-aspartate receptor glycine site.

Anti-Tumor Agent

Noble gas compounds have already made an impact on our daily lives. $\ce{XeF2}$ has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents.
The American Cancer states that

Fluorouracil belongs to the class of chemotherapy drugs known as anti-metabolites. It interferes with cells making DNA and RNA, which stops the growth of cancer cells.

Anesthesia

I'm still looking for literature to add here. Some of these cover very complex physiological processes, and this is not my area of expertise. I will try to find more pertinent articles from peer-reviewed journals with high enough impact and relevance. It will be up to the poster of this question to explore in-depth as much as they need by starting at some of the literature I provide.


Why two ( or more) gases have proportional volumes at constant pressure and temp.? (Avogadro's Law)

So, there are many questions, and I'm not sure if I understand you 100 %:

We have Avogadros Law, which applies for an ideal gas at constant pressure and temperature:
$$\frac{V}{n} = k \quad\Rightarrow\quad \frac{V_1}{n_1}=\frac{V_2}{n_2}$$
With $k$ as a constant.

And we have the ideal gas law:
$$pV = nRT \quad\Rightarrow\quad \frac{p_1V_1}{n_1T_1} = \frac{p_2V_2}{n_2T_2}$$

And this is probably what your first question is about. There are problems when you have a state $1$ and a state $2$, which differ.

For example: You have a container where everything ($p_1,V_1,T_1,n_1$) is given. Now you double the temperature - and the task is to calculate the pressure. Since the volume and amount of substance is fixed ($V_1 = V_2$, $n_1 = n_2$), you can just rearrange everything and calculate the pressure $p_2$.

In the ideal gas-model, gases do not seperate. I may be misunderstanding you, but if you put 1 mol $\ce{N2}$ and 2 mol oxygen in a container, you will have different partial pressures, but no demixing.

A relevant question was posted on physics.SE, that may help you if you have to dive deeper into this topic.


Is there anything that doesn't mix with water OR oil?

In short, yes, many highly fluorinated liquids are miscible with neither water nor organic solvents.
If you look at my answer to this question, you can see that I made a mixture of water, hexanes, and a fluorinated solvent called HT-110.
Likewise, the HT-110 will not dissolve most compounds, apart from fluorinated ones like Teflon AF.
Many compounds will have appreciable solubility in more than one layer and if they are in contact, there can be transfer between them.

Polarity plays a large role in whether two liquids are miscible.
Basically, for two liquids to mix, it must be energetically favourable to disrupt the intermolecular forces of the individual liquids in favour of the new intermolecular forces between the components of the mixture.
For your example of water and oil, water on its own has significant hydrogen bonding, but since the oil can't participate in hydrogen bonding, there would be a lot less hydrogen bonding in a homogeneous mixture of oil and water—the oil molecules essentially block water molecules from accessing each other.
Because the mixture would greatly reduce the intermolecular forces holding the liquids together, it is more energetically favourable to remain separate.
In contrast, a mixture of water and ethanol is energetically favourable because ethanol can hydrogen bond and is similarly polar, so the disrupted water-water intermolecular interactions are replaced by similar water-ethanol interactions.

In a little more detail, a mixture is favourable when its Gibbs energy of mixing is negative: $$\Delta G_{mix} = \Delta H_{mix}-T\Delta S_{mix}$$

In an ideal mixture of two components A and B, A-B interactions are the same strength as A-A and B-B, so the enthalpy of mixing, ∆H, is 0.
If A and B have similar properties, A-B interactions tend to be similar and ∆H will be close to 0.
If A-B interactions are weaker (e.g. water and oil), ∆H will be positive.
∆S is almost always positive for common mixtures because each component has a larger volume to spread into than if the components were separate. (the exception is when A-B interactions are much stronger than A-A and B-B interactions which prevents A and B from moving randomly through the volume)
This means that mixtures can still form if the enthalpy of mixing is positive, if the entropic contribution can overcome it.
This explains why some mixtures are possible at elevated temperatures but not room temperature.


Electrochemistry - Redox & Electroplating

Here the iron electrode is made by dipping an iron plate or wire in a $\ce{FeCl2}$ or $\ce{FeSO4}$ solution. It is an anode, where $\ce{Fe}$ is oxidized according to $$\ce{Fe -> Fe^{2+} + 2 e-}$$ The chlorine electrode is a cathode made by dipping a platinum electrode in a solution where some chlorine gas $\ce{Cl2}$ is arriving from under the platinum plate, so that a few bubbles remain adsorbed on its surface. This $\ce{Cl2}$ gets reduced to $\ce{Cl-}$ by absorbing the electrons coming from the iron anode according to the equation : $$\ce{Cl2 + 2 e- -> 2 Cl-}$$ The global effect of these reactions is equivalent to the following equation that describes the general change of composition in the cell : $$\ce{Fe + Cl2 -> Fe^{2+} + 2 Cl-}$$


How to tell if a molecule will participate in hydrogen bonding with water?

I figured out the answer. If there is a lone pair of electrons belonging to nitrogen, oxygen, or fluorine, then it is possible to bond to water. Therefore, it is can accept a hydrogen. If there is a hydrogen bonded to one of the three elements listed above, then it able to donate.


Isocyanides undergo addition reactions, but cyanides do not. Why?

Isocyanide can undergo addition reaction due to the fact that it is capable to attach both the electrophile and nucleophile in the same carbon atom while it is not achieved by cyanide. The negative charge on carbon atom accepts the electrophile to form an intermediate species in which the positive charge on nitrogen atom gets neutralized by the addition of the nucleophile to the carbon atoms and the nucleophile gets attached in the same carbon atom. Cyanide is incapable of undergoing such mechanism because of less nucleophilicity of nitrogen atom of cyanide and thus don't undergo addition reaction. The mechanism is as follows:


Are phase transitions reversible when performed at supercooled temperatures?

The entropy change for a system is the same whether or not the process that takes the system from its initial to final states is performed reversibly. That's because the entropy is a state function. The difference between a reversible and irreversible process, all else being equal with the terminal states of the system, lies in what happens to the surroundings.

Therefore, to find the change in entropy of the system you start by describing a reversible path for which you can compute the entropy change. Such a path might consist of the following steps:

(1) Warm supercooled water from the initial temperature ($\pu{-20^\circ C}$) to the normal freezing point ($\pu{0^\circ C}$)

(2) Freeze the water to ice at the normal freezing point ($\pu{0^\circ C}$)

(3) Cool the ice from the normal freezing point to the final temperature ($\pu{-20^\circ C}$)

The path involves reversible heating and cooling of supercooled water  and ice, respectively, and the intermediate step of freezing water to ice at the normal freezing point of water ($\pu{0^\circ C}$).

You can use the heat capacities to compute the entropy changes during heating and cooling of the system, respectively, from the initial and to the final states. For one mole of water:

$$\Delta S_{m,T_i\rightarrow T_f}=C_{p,m}\ln\left(\frac{T_f}{T_i}\right)$$

making sure to use the appropriate heat capacity (assumed constant over the temperature range) for water or ice.
For the first step, you warm up the supercooled water, and compute the entropy change by assuming that the supercooled state is stable. For the final cooling step, you compute the entropy change of the cold ice.

For the intermediate freezing step

$$\Delta_fS_m=\frac{\Delta_f H_m}{T_f}$$

The molar change in entropy for the system undergoing the irreversible freezing process is then just equal to the sum of the above changes for the three steps.

As for the computation of the entropy change of the surroundings: you compute the enthalpy change for the alternate path; then, since the enthalpy change for the actual irreversible process is the same as that for the reversible path, and the process is isothermal and isobaric (the intial and end states have same T and p), you can equate the enthalpy change with the heat, and can compute the change in entropy of the surroundings by dividing the heat by the temperature, making sure to use the correct value ($\pu{-20^\circ C}$).


Products for the hydrogenation of the cycloalkene in (3R,4R,5S)-4,5-diamino-3-(pentan-3-yloxy)cyclohex-1-ene-1-carboxylic acid

Hydrogenation of the alkene introduces a stereo centre, thus the products are:


1) Why does a molecule become achiral if it has a plane of symmetry
  even though it has stereocenters?

If a plane of symmetry bisects a molecule such that one half of the bisected molecule is the mirror image of the other half of the bisected molecule, then the entire molecule must be superimposable on its mirror image, it must be achiral.  Said differently, if mirror symmetry exists within a molecule, then the mirror image of the entire molecule will be superimposable upon the original drawing of the entire molecule.

Here's an example of a molecule ($\ce{A}$) with 2 chiral groups ($\ce{G_{R}}$ and ($\ce{G_{S}}$).  There is a plane of symmetry that bisects $\ce{A}$, it is perpendicular to the plane of the screen and contains the central carbon atom and the 2 methyl groups.  Now, when we draw the mirror image of $\ce{A}$ (let's call it ($\ce{\bar{A}}$), we find that it is superimposable upon the original drawing of $\ce{A}$ by a simple 180-degree rotation.
Even though $\ce{A}$ contains 2 chiral substituents, the molecule as a whole is achiral.


Bond length increase when replacing nitrogen with phosphorus for NH3 and NF3

This answer of mine basically leads the way to answering your question but it is focussed on bond angles, not lengths. So let me rewrite it to focus on lengths.

All four compounds would ideally want a set of $90^\circ$ angles which nitrogen cannot achieve due to steric stress; instead, the angles are expanded to $102^\circ$ ($\ce{NF3}$) and $107^\circ$ ($\ce{NH3}$). This seems couner-intuitive, because the smaller atoms require a larger bond angle. But looking at it closely, we realise that the $\ce{N-H}$ bond is simply shorter so the hydrogens are closer together hence more strained and hence further apart. Hydrogen being a very small atom the bond lengths cannot be extended well.

The fluorine atom on the other hand is much larger, allowing not only for a greater standard bond length but also for greater bond length variations. It can extend the bonds when compared to $\ce{NH3}$ to allow for a smaller bonding angle and in turn allow for a larger s-contribution to nitrogen’s lone pair. Lengthening (that is, weakening) the bonds slightly allows for a greater stabilisation on the central atom and thus a more stable overall molecule.

Another important issue is the fact that a smaller bond length in $\ce{NF3}$ would bring the fluorine atoms closer together causing electrostatic repulsion of their lone pairs (although that is what is commonly meant with steric stress). Hydrogen, having no lone pairs, does not have this issue.

In phosphorus’ case, both bond angles are almost perfectly $90^\circ$, so we can expect an almost unstrained molecule. The smaller hydrogen still calls for an equally short bond but the larger fluorine can now come closer making the bonds stronger, because the angles are still near-perfect and phosphorus’ lone pair has a very high s-contribution. In fact, the $\ce{P-F}$ bonds are rather strong; much stronger than their $\ce{N-F}$ counterparts.

Rationalising the entire thing we come up with:

The difference between $\ce{NH3}$ and $\ce{PH3}$ ($40~\mathrm{pm}$) is approximately equal to the larger phosphorus atom’s size. (The difference between the empirical atomic radii is $35~\mathrm{pm}$.)
In $\ce{NF3}$’s case, the bonds are lengthened by some $15$ or $20~\mathrm{pm}$ to accomodate for steric stress both. This weakens the bonds but stabilises the atoms, especially nitrogen. Also, steric stress between the fluorine atoms is reduced.
$\ce{PF3}$ has the bond lengths one would expect.


Can I successfully use a humidifier to distribute caffeine to wake someone up without the usual morning drowsiness?

It looks like at least one person/organization has thought of this and patented a delivery mechanism ("Delivery of caffeine through an inhalation route").

That said, you might be able to couple the system outline in the patent documentation with an alarm clock of sorts to trigger the release at a specified time. I'd certainly buy one if the price was right.

Edit: The inhalation part has been investigated. As far as I can tell, the business about administering caffeine to a sleeping person and then having them awaken into a fully alert (or pseudocaffeinated) state has not.


How do I determine the temperature at which a dissolved substance evaporates?

First, it is a good idea to clarify the meaning of the word vaporize from here:

  (Physics / General Physics) to change or cause to change into vapour or into the gaseous state
  to evaporate or disappear or cause to evaporate or disappear, esp suddenly
  (Physics / General Physics) to destroy or be destroyed by being turned into a gas as a result of extreme heat (for example, generated
  by a nuclear explosion)

Chemists should be careful when using this word because it can be used to describe evaporation, sublimation, and boiling.  Note the third definition, which suggests that the vaporization of matter typically involves a large amount of energy.

Since the mixture you refer to in your question is in a liquid state, I am going to focus on liquid boiling.  (Note, we can say that boiling is the rapid vaporization of a liquid.)  The boiling point of a liquid is the point at which the vapor pressure of the liquid is equal to the surrounding pressure.  The boiling point of a liquid is affected by intermolecular forces such as hydrogen bonding, dispersion and dipole-dipole interactions.

When a solute is dissolved in a pure liquid (your example of caffeine dissolved in water), these intermolecular forces are altered.  The end result is an elevation of the boiling point.  This change can be predicted quite accurately using concepts typically covered in the first year of a college-level general chemistry class.

So, we know now that a mixture of caffeine and water will have a boiling point higher than pure water; however what happens as the solution continues to boil?  I ignored an important point in the previous paragraph; boiling point elevation due to addition of a solute will only occur if the solute is nonvolatile. Caffeine is such a solute.  Therefore, as you boi your mixture, the water will be turned into a gas, but the caffeine will be left behind.  The volume of solvent is decreasing; therefore the solution concentration is increasing.  Once the caffeine solubility limit is reached, it will start to crystallize.

When you are left with a mound of caffeine you can continue to heat it.  As you heat it above about $230\,^{\circ}\mathrm{C}$ it will start to sublime, or in more general terms, vaporize.

For those interested, the experimental procedure for the extraction and purification of caffeine from tea leaves is described at an undergraduate level here.


What is the difference between a polar molecule and polar bond?

A polar bond is one where the charge distribution between the two atoms in the bond is unequal. A polar molecule is one where the charge distribution around the molecule is not symmetric. It results from having polar bonds and also a molecular structure where the bond polarities do not cancel. For example, water has polar bonds and its bent geometry means that there is a non-zero dipole moment. On the other hand, carbon dioxide has two polar carbon–oxygen bonds, but they are such that the polarities cancel out, so there is zero dipole moment.


Can you color a flame and what would the effects be?

Unfortunately not, I'm afraid.

I think you might have gotten cause and effect mixed up here. Blue flames are hotter than yellow flames not because of their color, but because hotter flames generate blue light and cooler ones generate yellow/red light.

To take an example of this kind of problem, I know that very red tomatoes tend to be tastier than very pale tomatoes. Does this mean that we could make tastier tomatoes by injecting food coloring into them? Well, probably not, because it's not the color that causes the flavor. Rather, tomatoes that generate more color tend to also generate more things like sugar and flavors, and so flavorful tomatoes tend to have more color.

With flames, it's even more black-and-white because while its possible to have a pale tomato that happens to taste good, the color of the flame is set by the universe: unless you add colors yourself, it is impossible to have a blue flame that is the same color as a yellow one. The color of the flame is determined by something called "blackbody radiation," and if you're curious about why flames are colored the way they are, it's definitely something to take a look at.


A molecular orbital diagram is a schematic representation of how we interpret bonding in certain species. It is as much an accurate representation for a specific bonding situation as a Lewis structure for the the molecular geometry.
When we construct a molecular orbital diagram, usually we resort to shapes of hydrogen like orbitals. We most of the times do not even use mathematical descriptions, accurate energy levels, without changing the descriptive value much. Oftentimes we completely omit core orbitals and don't draw nodal planes/spheres for orbitals with $n=2$.
Such diagrams, however, provide qualitatively good descriptions of what otherwise is almost impossible to fathom. They even provide a solid starting point for quantum chemical calculations.
As such, the choice of atomic orbitals does not matter at all; a picture will never contain enough detail to work out the subtle differences of these choices.

By the use of computational chemistry methodologies we can (1.) produce a molecular wavefunction of a diatomic molecule as a linear combination of basis functions (assume they are centred on each nuclei). This would already satisfy the definition above and generate a suitable MO diagram.

Most basis sets indeed contain basis functions which are centred at the respective nucleus and they resemble atomic orbitals (they don't have to). Through a linear combination of such functions we can obtain suitable molecular orbitals within the respective approximations. The wave function is then obtained as a set of orthogonal molecular orbitals. This in principle is the essence of molecular orbital calculations, which is in turn essentially the same as calculations on the Hartree-Fock level of theory. We know that this level of theory doesn't contain nearly enough precision to accurately describe the most simple compounds because of the neglecting most of the correlation energy. Yet, it is impossible to represent the result of such a calculation as a diagram.

We could express molecular orbitals as a linear combination of optimised atomic orbitals of each atom, which would imply an extra calculation for each element in a molecule. A change of basis for the MOs, from the basis set to the atomic orbitals, does the rest.

Yes you can do that, but apart from adding significant computational cost, this would not really increase the accuracy of the calculation. First and foremost you need to account for a larger portion of the correlation energy, for example with density functional approximations, post-HF methods like perturbations theory MPn, coupled cluster, or configuration interaction. Instead of optimising the basis set for each atom in the molecule, it is much more effective adding more functions to the basis set until you reach the complete basis set limit. There are several types of basis set libraries in use now, which have been designed to effectively handle such calculations. I think it was John Pople who laid out the advantages of this system. And while his basis sets have significant shortcomings, they are still used today.
In any case, what you will not get is a better or more descriptive MO diagram.

The natural bonding orbitals (NBOs) method allows one to produce natural atomic orbitals as well, which could also be used as initial orbitals centred on the atoms.

Indeed natural bond orbital theory uses natural atomic orbitals (and other types), however, these are representations of the atomic orbital within the molecular environment.
The theory itself is a localisation technique of canonical orbitals that are already expressed in the chosen basis set. As a result we get a bonding picture which closely resembles a Lewis structure. While they are different from the actual basis set they still have the same functions and would still lead to the same solution.
It is best to treat MO diagrams as what they are: qualitative tools to describe bonding.


Using enthalpies of formation and combustion to find formation enthalpy of Acetylene (Hess' Law)

Like many past chemistry.se questions, the answer to this one hinges on the presence of the $^\circ$.

$\Delta H^\circ_\text{c}$ is the $\Delta H$ value for a "standard" reaction where acetylene at 1 atm of pressure is combusted isothermally at 298 K with 1 atm of oxygen gas, forming 1 atm of $\ce{CO2}$ and liquid water (which is the most thermodynamically stable form of water at 1 atm and 298 K).

I think it's safe to say that no combusion of acetylene, ever in the history of the world, has occured under those conditions.  But those "standard" conditions are the only conditions where the  $\Delta H^{\circ}_f$ is guaranteed to be -227 kJ/mol.

By the way, in practical and engineering literatures, $\Delta H_c^{\circ}$ of a fuel is commonly called the higher heating value of the fuel.  It corresponds to $\Delta H_{rxn}^{\circ}$ of a reaction like

$\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(l)}$

The lower heating value, conversely, corresponds to $\Delta H_{rxn}^{\circ}$ of reactions like

$\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(g)}$


What can I use as a sacrificial electrode to stop the oxidation of aluminium?

I think that you are interested in the corrosion protection of aluminum. You have basically two choices depending on your experiment setup.

You can use an impressed current protection or a sacrificial anode method. The sacrificial anode method uses a more reactive metal (less noble) in contact with the metal requiring protecting. The less noble metal reacts first within corrosive environment.

The impressed current protection uses electric current to charge the metal requiring protecting with excess electrons. It changes the electrode potential in the negative direction until the metal is not reacting.

If you want to use the sacrificial anode method look in the table where you are looking for metal that is positioned higher than your aluminium (is more electronegative - so non of the mentioned metals).

More useful information can be found here CORROSION PROTECTION OF METALS
 or here metals.about.


How to rationalise the planar structre of I2Cl6?

The bonding situation in the compound $\ce{(ICl3)2}$ is by far more complex than what is depicted in this book. The molecule itself has very high symmetry, i.e. $D_\mathrm{2h}$, that needs to be satisfied in the molecular orbital/ valence bond description.[1,2] That is why one cannot distinguish between the bridging chlorine bonds.

Analysing this bonding situation in the simple concept of Lewis-structures is not trustworthy and will lead to wrong conclusions. I have performed a short gas-phase optimisation at the DF-BP86/def2-SVP level of theory and analysed the resulting wave function with the Natural Bond Orbital theory. Covalent treatment of any of the bonds seems to be wrong. In this case I assumed all single bonds. Due to symmetry reasons, we only need to be concerned with two bonds. The outer $\ce{I-Cl}$ bond is polarised 93% towards the chlorine, but only occupied by 1.25 electrons. The remaining 0.75 electrons can be found in anti-bonding lone pair orbitals. This is not even close to expected Lewis structure behaviour. The bridging $\ce{I-\mu{}Cl}$ bond is still 81% polarised towards the chlorine, and has an occupation of 1.92 electrons. This only tells us, that this bond has a higher covalent character.

The analysis with the Quantum Theory of Atoms In Molecules (QTAIM) gives us more tools to analyse the electron density. In here the picture is quite obvious. All of the bonds in the molecule are of highly ionic character. The electron density at the bond critical point of the outer $\ce{I-Cl}$ is with $0.08~\mathrm{e\,a.u.^{-3}}$ very low and the Laplacian is positive at this point. The electron density at the bridging bond is even lower with $0.04~\mathrm{e\,a.u.^{-3}}$, again with a positive Laplacian.

(Laplacian distribution, solid blue lines indicate charge depletion $∇^2ρ<0$, dashed blue lines indicate charge accumulation $∇^2ρ>0$, red spheres are bond critical points, pink spheres are ring critical points, black lines are bond paths, dark red lines are zero flux surfaces.)

This finding is fairly consistent with the observation, that the melt of $\ce{(ICl3)2}$ conducts electricity, which according to Wikipedia is a hint for partial ionic dissociation.
$$\ce{(ICl3)2 <=> ICl2+ + ICl4- }$$

However, the planarity of this compound in solid state can be attributed to the possibility of having a somewhat quite extensive pi-bonding system as part of the small covalent bonding. For final conclusions about this, the results are not sufficient.

What can be said, judging from the molecular orbitals of $\ce{(ICl3)2}$, is that apart from the occupied core d-orbitals of iodine, no other d-orbitals are involved in any bonding - as it is expected. Shown here are the occupied non-core orbitals of $\ce{(ICl3)2}$. It is noteworthy, that the HOMO is fully antibonding.[3]

Finally a short warning:
While the assumption made by VSEPR may lead to the approximately correct conclusion for the interhalogenide compounds $\ce{XF3}$, where $\ce{X$=$Cl, Br, I}$, that the molecules are T-shaped, it is doubtfully the correct conclusion for the much less electronegative chlorine. This can be mainly explained by Bent's rule. Even in the monomeric compound the bonding situation is somewhat more complicated than you would expect.

Notes and References

The point group $D_\mathrm{2h}$ specifies that there is a main symmetry axis $C_2$ which passes through the centre of the ring. Additionally there are also two $C_2$ axis perpendicular to this and a horizontal mirror plane $\sigma_\mathrm{h}$. This means the molecule is planar. There are also non-planar molecules that can be described by this point group.)
As Max pointed out (Thank you), the structure is taken from the solid state. K. H. Boswijk and E. H. Wiebenga,  Acta Cryst. 1954, 7, 417-423. It mentions that it was purified by sublimation. This is no absolute proof of a gaseous dimer, but at least very suggestive. For our purposes we assume that the gas phase calculations are sufficient.
Orbitals obtained at the DF-BP86/def2-SVP level of theory with Gaussian09 rev. D. The energy increases row-wise from left to right and from top to bottom. The last shown structure is the molecule itself.


Does the largest known cycloalkane contain 288 carbon atoms?

According to Makromol. Chem., Rapid Commun. 1985, 6 (3), 203–208[1], Kwang Sup Lee and Gerhard Wegner, attempting the synthesis and study of $n>100, \ce{C_nH_{2n}}$ cyclo-alkanes and linear alkanes, were successfully able to synthesis a linear alkane of $\ce{C384H780}$ and cycloalkanes with 288 carbons ($\ce{C288H578}$)

Linear alkanes up to $\ce{C384H770}$, (2k; M = 5378) and cycloalkanes up to $\ce{C288H576}$(3g; M = 4040) were prepared by this strategy.

A brief method of preparation is as follows:

The strategy of the synthesis of long chain linear and cyclic alkanes was developed starting from the work of Schill et al. who had shown that the oligomers obtained by oxidative coupling of 1 following Eq. (1) can be separated by adsorption chromatography.
\begin{align}
\ce{\underset{\textbf{1}}{HC#C-(CH2)20-C#CH} ->[Cu^2+] \\ \underset{\textbf{2}}{HC#C-(CH2)20-C#C-\left[C#C-(CH2)20-C#C\right]_m-H}} \\
\end{align}

A similar strategy was used to synthesize the cycloalkanes as well.

The cyclic oligomers were synthesized by the same reaction according to Eq. (2)
under conditions of high dilution of the terminal diacetylene

Finally, the linear and cyclic oligomers were hydrogenated over a Pd/C-catalyst in a conventional manner to obtain the analytically pure alkanes.

As for the assignment question, I was able to link it to a introductory exercise in Organic Chemistry by Janice Gorzynski Smith.[2].
Reference

Kwang Sup Lee, Gerhard Wegner, "Linear and cyclic alkanes ($\ce{C_nH_{2n+2}, C_nH_{2n}}$) with $n > 100$. Synthesis and evidence for chain-folding," Makromol. Chem., Rapid Commun. 1985, 6 (3), 203–208 (https://doi.org/10.1002/marc.1985.030060316)
Janice Gorzynski Smith, Organic Chemistry, Second Edition, McGraw-Hill book Company Inc.;2008, ISBN: 978-0-07-066720-4


Structural isomers of hexaammineiron(III) hexacyanidocobaltate(III)

I have found the answer:

  In the following case:
  $$\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$$
  Both the constituent entities are neutral and hence will exist independently, and thus will not form isomers of the original compound.


The heat of hydrogenation reflects the thermodynamic difference in energy between starting material and product. It is a thermodynamic measure of the difference in stability of the products and reactants.  Reactivity, on the other, hand is related to kinetics, a reflection of the activation energy  involved in the rate determining step.

Cyclopropane and cyclobutane have roughly the same strain energy (SE=27.5 and 26.3 kcal/mol respectively). To a first approximation, the strain energy of a polycyclic system is the sum of the strain energies of the individual rings in the system.

Each of the 3 starting materials contains a total 3 (cyclopropane plus cyclobutane) rings.  Since the SE of cyclopropane ~ cyclobutane, we would therefore expect the SE's of the 3 starting materials to be similar and larger than the SE's of the less strained products. In agreement with this expectation the calculated SE's of the 3 starting materials are 109, 106 and 103 kcal/mol respectively (see fig. 2.2 on p. 37 here), indeed very similar.

The SE's of the final products are 17, 41 and 62 kcal/mol respectively (ref, see Table 3.2 on page 72).

If we calculate the strain released during hydrogenation it is in the same order and very close to the actual heats of hydrogenation reported in your figure.

$\mathrm{109-17=92}$ kcal/mol
$\mathrm{106-41=65}$ kcal/mol
$\mathrm{103-62=41}$ kcal/mol

Clearly, in this series of propellanes the heat of hydrogenation is a measure of the strain released in the process.

Because bicyclo[1.1.1] pentane (the right-most product) has a much higher SE than either of the two other hydrogenated products, the heat of hydrogenation given off when it is formed will consequently be much less.

[1.1.1]propellane is an extremely reactive compound (see here for example).  Comparing the thermodynamic heats of hydrogenation in your series of compounds cannot be used to meaningfully predict the kinetic reactivity of these compounds.


What chemicals would make lime sulfur dip to produce a green colored smoke?

Yes, mixing lime sulfur (calcium polysulfide) with an acid will react via the following reaction:

$$\ce{S^{2-}\:(aq) +2H+\:(aq) ->H2S\:(g)}$$

Which is both extremely toxic and extremely flammable. You should always be extremely cautious and mindful when mixing any chemicals and always be sure to label everything!


A)SiMe3 Increases the inductivity and reduces Aromaticity,
so by free radical (uv termination) its likely to have p-iodideSi(Me)3Bencene.

B)Dimer crystal solution in Cl- ions.  (Ph)2Si(Cl)Mg+---ClSi(Ph)2
Or Solid crystals of (Ph)2Si(Cl)MgCl.


Reduction of α,β-unsaturated nitro compounds

I'm not aware of a $\ce{NaBH4}$-based reduction of nitroalkenes to saturated amines, such as in:

It is however possible to reduce nitroalkenes to saturated hydroxylamines using $\ce{BH3*THF}$ in the presence of catalytic amounts of $\ce{NaBH4}$. (DOI)

If there's a chance to perform the desired transformation in one step via catalytic hydrogenation at normal pressure: go for it!

The handling of these reactions is nothing to be afraid of.

Typically, I performed hydrogenations in a standard 2- or 3-neck flasks and used ordinary balloons (think childrens' birthday) as gas storage. Advice: Stuff a piece of rubber tube into the inlet of the balloon and secure it with teflon tape.

Either attach this to the glas olive of a valve attached to your flask
or take a plastic syringe, cut off the base, stuff the the plastic cylinder into the rubber tube. Fill the balloon and attach a needle. Hydrogen may now be fed to your reaction mixture through a rubber septum cap.

UPDATE

I still doubt that the intended reduction by $\ce{NaBH4}$ is possible.

There is however another established route from $\alpha,\beta$-unsaturated nitroalkenes to amines which is worth to be mentioned:

Condensation products of aromatic aldehydes and nitroalkenes are reduced by iron in hydrochloric (or acetic) acid to the corresponding arylalkanones.

These ketones can be converted to primary amines by reductive amination in methanol (or ethanol) in the presence of (excess) ammonia and, of course, hydrogen over Raney Nickel or Urushibara Nickel.

In the course of this reaction, the imine intermediate is in situ reduced to the desired primary amine.

The resulting amines are valuable precursors in the synthesis of natural products: Their conversion to imines or amides and subsequent Pictet-Spengler or Bischler-Napieralski cyclisations furnish tetrahydro- and 3,4-dihydroisoquinolines.


Gold(III) chloride does not exist as $\ce{AuCl3}$. According to Wikipedia, the name gold trichloride is a simplification of the name, which is referring to the empirical formula, $\ce{AuCl3}$. The X-ray defraction studies revealed that the chemical exists as a chloride-bridged dimer, $\ce{Au2Cl6}$ (Ref.1):

In $\ce{Au2Cl6}$, each gold center is square planar, and the bonding in $\ce{Au2Cl6}$ is considered somewhat covalent (Wikipedia). Yet, your suggestion is somewhat true, since a recent research has shown that $\ce{Au2Cl6}$ is susceptible to decompose to metalic $\ce{Au}$ and $\ce{Cl2}$ gas at high temperatures or expose to $\ce{e-}$-beam irradiation (Ref.2).

Also, another research group (Ref.3) has shown that:

  Gold chloride, supplied as $\ce{HAuCl4}$ solution, is rapidly adsorbed on the silicate surfaces, the Au(III) is reduced to metallic gold, and gold particles grow on the surface. SEM images show agglomerates of gold unevenly distributed on the surface of the silicates, including in some areas forming agglomerates, especially on quartz and feldspar.

References:

E. S. Clark, D. H. Templeton, C. H. MacGillavry, "The crystal structure of gold(III) chloride," Acta Cryst. 1958, 11, 284-288 (https://doi.org/10.1107/S0365110X58000694).
Keita Kobayashi, Ryo Kitaura, Katsuhiro Sasaki, Kotaro Kuroda, Takeshi Saito, Hisanori Shinohara, "In Situ Observation of Gold Chloride Decomposition in a Confined Nanospace by Transmission Electron Microscopy," Materials Transactions 2014, 55(3), 461-465 (https://doi.org/10.2320/matertrans.MD201308).
Sima Mohammadnejad, John L. Provis, Jannie S. J. van Deventer, "Reduction of gold(III) chloride to gold(0) on silicate surfaces," Journal of Colloid and Interface Science 2012, 389(1), 252-259 (DOI: 10.1016/j.jcis.2012.08.053).


Derivation of formula for Hydrolysis constant

Sorry, I didn't get that. first of all, which hydrolysis are you talking about? I will talk about weak acid and strong base equilibrium. You have three concentrations [HA], [H+], [A-]. For instance, if they took 1L of 1 mM of HOAc (acetic acid) and added 0.3 mmoles of NaOH (concentrated, so no dilution) you get [A-]+[HA] = 1mM Then you say HA <-> H+ + A- has Ka. Then

[H+][A-]/[HA] = Ka. Then you say  [HA] = x, so [A-] = concentration_total- x (in this case 1mM -x). Then you get an equation
[H+]*[concentration_total-x]/x=Ka. Then you solve this square equation.

Regarding your question. I think you are mixing up to things: protonation of weak acid (stuff, relevant to analytical chemistry with a lot of drama, pH swings etc) and the behavior of other salts (they just dissociate completely and we forget about them). In stories like "to a solution of NaCl and K-acetat a drop of acid was added. What happen?" you should first forget about NaCl. It has no impact on the question. You should also forget about K+. Now you have "to a solution of Acetate- a drop of acid was added. What happen to pH?" Now you can easily calculate it using a square equation.

Run away from chemistry. Carcinogens, stinky chemicals, no  jobs... It teaches you how to make poisons and drugs, but how useful is that?


By which mechanism would azide react with alcohols or alkyl halides?

After some reading, I found that $\ce{N3-}$ is a good nucleophile. Hence the $\mathrm{S_N2}$ mechanism is followed.

Unfortunately, it is not as simple as that. Whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ is followed, depends on both partners but is more a function of the electrophile than one of the nucleophile. Typically, primary halides will undergo $\mathrm{S_N2}$ substitutions while tertiary ones will undergo $\mathrm{S_N1}$. Secondary ones can do both. However, note that secondary halides that are not 2-halopropanes will undergo Wagner-Meerwein rearrangements if a carbocation is formed.

In an $\mathrm{S_N1}$ reaction, the rate-determining step is the formation of the carbocation, hence weaker nucleophiles can react equally rapidly as stronger ones. In the $\mathrm{S_N2}$ reaction, the nucleophilicity of the nucleophile directly determines the rate of reaction, i.e. both partners are important. Hence these tend to prefer better nucleophiles.

  But if I use $\ce{HN3}$ on an alcohol, the $\ce{H+}$ ions produced will help in removing $\ce{OH-}$ as water forming a carbocation. So either $\mathrm{E1}$ (if the alcohol is tertiary) or $\mathrm{S_N1}$ (if the alcohol is primary or secondary) can take place.

Again, primary alcohols will never react by an $\mathrm{S_N1}$ mechanism and secondary ones are very unlikely to do so. The $\mathrm{p}K_\mathrm{a}$ value of $\ce{HN3}$ is $4.72$ — which means that alcohols will practically not be protonated (you need strong acids for that, e.g. $\ce{HBr}$).

I would assume that $\ce{HN3}$ will not react with alcohols at all since it is too weak to protonate them. There will also be no elimination side reactions neither by $\mathrm{E1}$ nor by $\mathrm{E2}$ — a hydroxide ion will never be a leaving group in itself.


Explanation for bond lengths in trans-hexatriene

If you derive the π-type molecular orbitals of hexatriene, the three lower-energy MOs which are filled would look something like this (image from p 33 of Fleming's Molecular Orbitals and Organic Chemical Reactions, Reference Edition):

I suspect what Clayden is getting at is that in the second MO, there is some antibonding character between C3 and C4, whereas the C1/C2 and C5/C6 interaction is purely bonding.


Should I have a pair of shoes that I wear only in the lab?

Standard safety rules (from a pharma and semiconductor processing POV):

To follow industrial hygiene practice absolutely to the letter, you should have dedicated outer layer garments that remain in a quarantined area that would not pose a gross cross contamination risk to other things that cannot be controlled for exposure risk. This means that you should not have been able to take those boots home with you after you wore them in the lab even one time. I do not think any company enforces such a rule, in fact, it is the employer's responsibility to provide PPE to workers and the worker's responsibility to use and maintain PPE (if applicable.) If the risk of cross contamination was assessed and determined to be moderate or higher, you should have been provided shoe covers at your solid state systems lab or dedicated work shoes that had to be kept in a quarantined area.

At the lab bench level, it is my experience that people wear their daily shoes unless you work at a larger site that has a safety shoe/shoe covering policy.


Spontaneous reactions simply mean that the products will be lower in free energy than the reactants ($\Delta G$ is negative). If $\Delta G$ is negative, then it is energetically favourable for the reaction to occur - in other words, there will be a release of energy due to the reaction.

The magnitude of $\Delta G$ (how large it is) does not say anything about the reaction speed. For example, the reaction between gasoline and oxygen in the atmosphere at room temperature will be extremely slow, despite a large, negative $\Delta G$. This is because this reaction has a large activation energy, $E_A$.

Hold on. When two reactants collide, they interact to form an extremely unstable structure called the transition state. The transition state is high in energy and is NOT energetically favourable, and thus the transition state quickly collapses again and becomes either the reactants or the products. You can see this in the following picture.

Depending on the kinetic energy of the reactants, they might not reach the top of the curve, in which case they will fall back to being reactants. But if they collide fast enough and are positioned appropriately with respect to each other, the top of the curve is reached, and the products will form. The energy required to form the transition state is the activation energy.

With gasoline and oxygen, the activation energy is high. So even though a tremendous amount of energy is released during the reaction, it needs a spark to overcome its activation energy. From here on, the heat generated from the reaction supplies the activation energy.

If this reaction was not spontaneous, we might be able to force the reaction along, but the reaction would not continue on its own.

As you know, not all reactions run to completion. This is because Gibbs free energy is dependent on concentrations of reactants and products, so as the products accumulate, and the reactants are used, $\Delta G$ becomes closer to 0, eventually reaching equilibrium, where $\Delta G = 0$. Visualise this scenario on the above picture. If there is no difference in Gibbs free energy, the reactants will still get to the transition state and turn into products. But the energy required for products to reach the transition state and becomes reactants is equivalent, and thus the rate of the forward reaction is equal to the rate of the reverse reaction.

The difference between reactions that will reach an equilibrium and the ones that run to completion is the magnitude of the $\Delta G$ at standard conditions (denoted $\Delta G°$). This is a measure of how energetically favourable the reaction "inherently" is. If $\Delta G°$ is large and negative, the reaction will run to completion. If it is smaller, the reaction will establish an equilibrium at some point.

That's at least generally true for reactions, where all of the species stay in the same phase I think. If you envision a reaction where one of the products is a gas, that escapes the solution of reactants, then the products will be unable to reform into reactants.


Enthalpies of formation of alkali metal halides

The equation for the formation of $\ce{MX(s)}$ from its constituent elements is:

$$\ce{M(s) + 1/2 X2(g) -> MX (s)}$$

In general, $\ce{X2}$ could be liquid ($\ce{X} = \ce{Br}$) or solid ($\ce{X} = \ce{I}$). However, since your question is asking about the fluorides and chlorides, we can just limit it to the gaseous state. The principles that will be described can be applied just as easily to the bromides and iodides without any problem.

For any question that asks about trends in thermodynamics, the best way to go about solving it is always to draw a cycle. This will allow you to identify which terms contribute to the change in the trend, and from there rationalise the observed trend. So, without further ado:

$$\large\require{AMScd}\begin{CD}
\ce{M (s) + 1/2 X2 (g)} @>\color{blue}{\Delta_\mathrm{f}H(\ce{MX})}>> \ce{MX (s)} \\
@V\color{red}{\Delta_\mathrm{atm}H(\ce{M})}VV  @AA\color{red}{-\Delta_\mathrm{L}H(\ce{MX})}A \\
\color{red}{\ce{M (g) + 1/2 X2 (g)}} @. \color{red}{\ce{M+ (g) + X- (g)}} \\
@V\color{red}{I_1(\ce{M})}VV @AA\color{red}{E_\mathrm{a}(\ce{X})}A \\
\color{red}{\ce{M+ (g) + e- + 1/2 X2 (g)}} @>\color{red}{\frac{1}{2}B(\ce{X-X})}>> \color{red}{\ce{M+ (g) + e- + X (g)}}
\end{CD}$$

By Hess's law, the enthalpy change for the blue path must be equal to the enthalpy change for the red path. This follows from the fact that enthalpy, $H$, is a state function. So:

$$\Delta_\mathrm{f}H(\ce{MX}) = \Delta_\mathrm{atm}H(\ce{M}) + I_1(\ce{M}) + \frac{1}{2}B(\ce{X-X}) + E_\mathrm{a}(\ce{X}) - \Delta_\mathrm{L}H(\ce{MX})$$

Note the sign convention I am using for the electron affinity and lattice enthalpy (these seem to be defined differently in different places). The lattice enthalpy used here is always positive (endothermic).

Clearly, the terms $\Delta_\mathrm{atm}H(\ce{M})$ and $I_1(\ce{M})$ only depend on the identity of the metal $\ce{M}$. Therefore, it is helpful to lump them together and consider them as one term, called $\Delta H_\ce{M}$. Similarly, the next two terms only depend on the halide $\ce{X}$ and we can collect them together and call it another term, $\Delta H_\ce{X}$. It is by no means necessary to do this; however, it saves some clutter in the expressions:

$$\Delta H_\mathrm{tot} = \Delta H_\ce{M} + \Delta H_\ce{X} + \Delta H_\ce{MX}$$

with

$$\begin{align}
\Delta H_\mathrm{tot} &= \Delta_\mathrm{f}H(\ce{MX}) \\
\Delta H_\ce{M} &= \Delta_\mathrm{atm}H(\ce{M}) + I_1(\ce{M}) \\
\Delta H_\ce{X} &= \frac{1}{2}B(\ce{X-X}) + E_\mathrm{a}(\ce{X}) \\
\Delta H_\ce{MX} &= -\Delta_\mathrm{L}H(\ce{MX})
\end{align}$$

The fluorides

As shown in your graph, the trend for the metal fluorides is that going from $\ce{LiF}$ to $\ce{CsF}$, $-\Delta_\mathrm{f}H(\ce{MF})$ decreases, i.e. $\Delta_\mathrm{f}H(\ce{MF}) = \Delta H_\mathrm{tot}$ increases (it becomes less negative). Here, the halide ion $\ce{X-}$ is not changing, so $\Delta H_\ce{X}$ is a constant and we only need to consider the other two terms:

Going from $\ce{Li}$ to $\ce{Cs}$, the atomisation enthalpy decreases and the first ionisation energy decreases. That means that $\Delta H_\ce{M}$ decreases.
Going from $\ce{Li}$ to $\ce{Cs}$, the radius of the metal ion increases, and so the lattice enthalpy decreases. This means that $\Delta H_\ce{MX}$ (which is the negative of the lattice enthalpy) increases.

So, there are two opposing factors. The first factor says that $\Delta H_\mathrm{tot}$ should decrease, and the other says that it should increase. Since the observed trend is that $\Delta H_\mathrm{tot}$ increases, this means that the second factor is greater than the first. That is to say, the increase in $\Delta H_\ce{MX}$ outweighs the decrease in $\Delta H_\ce{M}$, leading to an overall increase in $\Delta H_\mathrm{tot}$.

The chlorides

The analysis here - in terms of the individual trends in $\Delta H_\ce{M}$ and $\Delta H_\ce{MX}$ - is exactly the same. However, this time, the overall trend is reversed: $\Delta H_\mathrm{tot}$ decreases. That can only mean that, when $\ce{X} = \ce{Cl}$, the decrease in $\Delta H_\ce{M}$ now outweighs the increase in $\Delta H_\ce{MX}$. Why is that so?

First, we note that $\Delta H_\ce{M}$ is entirely dependent on the identity of the metal ion. It is independent of the identity of the halide. That means that the decrease in $\Delta H_\ce{M}$ when $\ce{X} = \ce{F}$ must be exactly the same when $\ce{X} = \ce{Cl}$ - there is absolutely no reason for it to change.

Logically, since the decrease in $\Delta H_\ce{M}$ is not changing in magnitude, it must mean that the increase in $\Delta H_\ce{MX}$ must have become smaller in the chlorides. Otherwise, there is no way for the decrease in $\Delta H_\ce{M}$ to suddenly outweigh the increase in $\Delta H_\ce{MX}$. And we can explain this very easily using what we know about lattice energies, because that's exactly what $\Delta H_\ce{MX}$ is: a lattice energy (with a minus sign).

The lattice energy of an ionic salt can be estimated using the Kapustinskii equation. For a metal halide, this reduces to the equation;

$$\Delta_\mathrm{L}H \propto \frac{1}{r(\ce{M+}) + r(\ce{X-})}$$

Clearly, going from $\ce{LiX}$ to $\ce{CsX}$, $r(\ce{M+})$ increases and $\Delta_\mathrm{L}H$ decreases. The difference is that when $\ce{X} = \ce{F}$, $r(\ce{X-})$ is smaller than when $\ce{X} = \ce{Cl}$. Mathematically, his means that in the fluorides, the lattice enthalpy is more sensitive to changes in $r(\ce{M+})$. This makes sense - if we went the other way and took the limit of $r(\ce{X-}) \to \infty$, the lattice enthalpy would not depend on $r(\ce{M+})$ at all.

That means that the decrease in $\Delta_\mathrm{L}H$ - which is equivalent to the increase in $\Delta H_\ce{MX}$ - must be larger in the fluorides, and smaller in the chlorides. That is why in the chlorides, the trend in $\Delta H_\ce{M}$ overrules the trend in $\Delta H_\ce{MX}$, and the reversal of the overall trend is observed.

Ending words

How would you know all of this if you were not given the graph? Well, the truth is, without looking up values of thermochemical radii, ionisation enthalpies, etc., there is no real way to know. The fact that the overall trend is reversed in the chlorides means that the variations in $\Delta H_\ce{M}$ and $\Delta H_\ce{MX}$ must be pretty similar in magnitude. However, if you are given data and asked to rationalise it, you can definitely do so by constructing a diagram and relating the variation in the various energetic terms to variation in physical properties that you are familiar with (ionisation energy, ionic radii, and so on).


How are molecular orbitals obtained in quantum chemistry?

I think you are close to the right idea, but let me try to clarify some points.
The full Schrodinger equation for a molecule should depend on the $3N$ nuclear coordinates, as well as the $3n$ electronic coordinates, where $N$ and $n$ are the number of nuclei and electrons respectively. This is challenging both due to the number of coordinates and, in particular, due to the coupling between electronic and nuclear coordinates. However, with the Born-Oppenheimer approximation, we assume they are decoupled and solve for the electronic part of the wavefunction at a fixed nuclear geometry.
The issue here is that we still can't solve the Schrodinger equation for an $n$ electron system with $n>1$. To simplify further, we can make the Hartree-Fock approximation, which (in one way of phrasing it) assumes that the $n$ electron wavefunction can be formed from the antisymmetrized product of $n$ one electron wavefunctions/orbitals. These orbitals can be determined by solving a nonlinear system of equations and then stitched together in the form of a Slater Determinant to make an approximate $n$ electron wavefunction.
While chemists like to frame discussions in terms of orbitals, strictly speaking orbitals are always an approximation (except for hydrogen) and the electronic structure of a molecule is determined by a $n$ electron wavefunction. Even though orbitals aren't quite real, they still can give a pretty good qualitative description of a molecule's behavior.
For a bit more about the meaning of orbitals, you might want to check out this previous question: Is it correct to talk about an empty orbital?


Equivalent of Szabo and Ostlund book for DFT

I can recommend the book of Yang and Parr, although I am not sure it deals  with plane wave bases at all. In principle, a DFT code is not really different from a HF code. In fact, you can use exactly identical codes, just replacing the Coulomb and exchange integrals by numerical integration of some functional. E.g., if you choose $X\alpha$-LDA, than your exchange functional is
$$K = C_\alpha \left(\frac{3}{\pi}\right)^\frac{1}{3} \int \rho(\mathbf{r})^\frac{4}{3} \mathrm{d}\mathbf{r}$$
You can evaluate your density numerically, choose any numerical integration scheme and obtain the exchange energy. That works even, if you still calculate the Coulomb integrals the HF way. However, you wouldn't save much time since you still have the $\mathcal{O}(N^4)$ step in the Coulomb integrals, where $N$ is the number of basis functions. But once you integrate the Coulomb integrals numerically, you have basically a working DFT code.

There is a number of papers about choosing the correct grids and integration schemes for numerical integration. Once, I read one that compared the radial grids quite well, but sadly I don't remember the reference. But the buzz words would be e.g. Becke grid, (Treutler?-)Ahlrichs grid or simply Gauss quadrature for the radial integration (most of the grids are a form of Gauss quadrature, though). For the angular integration there is e.g. the Lebedev grid, but its calculation is a little more involved.

My suggestion, based on teaching experience, is to write first a HF code in a Gaussian basis. Then you pimp it with a numerical integration scheme and something like $X\alpha$. In the next step, you replace your Coulomb integrals. Lastly, you can switch from a Gaussian basis to the basis of plane waves. The last step only changes your matrix elements but not the algorithm (HF/DFT) itself. Since you first implemented a HF code for a Gaussian basis, you can always check your algorithm before going into uncharted waters.


The standard free energy of formation of graphite is $0\ \mathrm{kJ/mol}$, and the standard free energy of formation of diamond is $2.9\ \mathrm{kJ/mol}$.  So, at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{bar}$, graphite is favored thermodynamically over diamond.  Let's see what pressure would be required for $\Delta G$ to be zero at $25\ \mathrm{^\circ C}$.  Since the temperature is constant, the change in $G$ is described by $\mathrm dG=v\,\mathrm dP$, where $v$ is the molar volume.  The density of graphite is $2.26\ \mathrm{g/cm^3}$ and the density of diamond is $3.51\ \mathrm{g/cm^3}$.  So the molar volumes are $5.31\times 10^{-6}\ \mathrm{m^3/mol}$ for graphite and $3.42\times 10^{-6}\ \mathrm{m^3/mol}$ for diamond.  So, at pressure $P$ (in Pa), the Gibbs free energy for graphite and diamond are:
$$G_\text{Graphite}=0+5.31\times 10^{-6}P$$
$$G_\text{Diamond}=2900+3.42\times 10^{-6}P$$
So, $\Delta G$ at pressure $P$ is $$\Delta G_\mathrm{G\rightarrow D}=2900-1.89\times 10^{-6}P$$So the change in free energy is zero at $P=1.53\times 10^{9}\ \mathrm{Pa}=15\,300\ \mathrm{bar}$

Diamonds are formed naturally at even higher pressures and higher temperatures.  But this is the approximate order of magnitude.

So at $25\ \mathrm{^\circ C}$, graphite is highly thermodynamically favored over diamond.  The actual presence of diamonds at $25\ \mathrm{^\circ C}$ must therefore be a matter of kinetics, rather than thermodynamics.


This functional group is an imide. Imides are stable to water and require significantly more effort to hydrolyze than anhydrides (from personal experience). The acyl nitrogen group is a better resonance donor and poorer leaving group than acyl oxygen groups.

From the wikipedia page:

  Being highly polar, imides exhibit good solubility in polar media. The N-H center for imides derived from ammonia is acidic and can participate in hydrogen bonding. Unlike the structurally related acid anhydrides, they resist hydrolysis and some can even be recrystallized from boiling water.

As an example, succinimide, a cyclic imide has a solubility in water of 1 g for every 3 mL - if succinimide reacted with water, such a high solubility would not be possible.

The related N-bromosuccinimide (show below) is also water stable despite that unfriendly looking $\ce{N-Br}$ bond. It is prepared from succinimide in water and an be recrystallized from near-boiling water.


What is the correct nomenclature for Glucuronyl C5-epimerase?

I have not found this particular enzyme in Enzyme Nomenclature. However, following the Recommendations of the Nomenclature Committee of the International Union of Biochemistry and Molecular Biology, the name should be glucuronyl 5-epimerase. See also this sublist for some analogous names of other epimerases.


Which isomers (cis or trans) are more water soluble?

That's not quite correct and can be quite easily to be shown as wrong by looking up data tables and/or MSDS. Fumaric acid (the common name of the trans-isomer) is less soluble, with a solubility in water of 4.9 g/L whereas maleic acid (the cis-isomer) has a solubility of 478.8 g/L.

Maleic anhydride would also react in water to form maleic acid.


How do you recognize a carbohydrate molecule?

The name "carbohydrate", which literally means“carbon hydrate,” arises from their chemical composition, which is roughly $\ce{(C.H2O)_n}$, where $n\ge 3$. The basic units of carbohydrates are monosaccharides.

Monosaccharides or simple sugars are aldehyde or ketone derivatives of straight-chain polyhydroxy alcohols containing at least three carbon atoms. Such substances, for example, D-glucose and D-ribulose, cannot be hydrolyzed to
form simpler saccharides.

Examples of monosaccharides:

Classification

Monosaccharides are classified according to the chemical
nature of their carbonyl group and the number of their C atoms.

If the carbonyl group is an aldehyde, as in glucose, the sugar is an aldose. If the carbonyl group is a ketone, as in ribulose, the sugar is a ketose. The smallest monosaccharides, those with three carbon atoms, are trioses.

Those with four, five, six, seven, etc., C atoms are, respectively, tetroses,
pentoses, hexoses, heptoses, etc. These terms may be combined so that, for example, glucose is an aldohexose, whereas ribulose is a ketopentose.

However, there are similar molecules with different atoms such as nitrogen which are also considered carbohydrates. A class of polysaccharides known as glycosaminoglycans (unbranched polysaccharides of alternating uronic acid and hexosamine residues).Practical examples are Hyaluronic acid(linked disaccharide
units that consist of D-glucuronic acid and N-acetyl-Dglucosamine) and Heparin (sulfated disaccharide units):

Nevertheless, when carbohydrates bond with other molecules to form other newer bonds, they are given different names and are not necessarily carbohydrates but are given new names - a good example are glycoproteins (proteins covalently associated with carbohydrates.)

Reference

Biochemistry Voet and Voet, Grisham


What is the product of the chemical reaction between phenol and ferric chloride?

After doing some looking around, and going back to check what I thought I knew about coordination complexes, I think I understand why you're seeing two equations. I first recommend reading the pages on Chemguide that talk about coordination chemistry, particularly the first half of this page on acidity, and the pages on reactions with hydroxide and ammonia. Both were helpful to me in understanding an investigation I'm doing on copper (II) complexes, but the same ideas apply to 3+ compounds.

The important concepts to keep in mind are that coordination compounds (in situations like this, at least) will either undergo acid-base reactions or ligand-exchange reactions. Acid-base reactions result in neutralization and a precipitate, ligand-exchanges cause a color change. Phenol is a relatively strong acid (stronger than alcohol at least), and so is ferric chloride. I had to look around but this page says at the very bottom that all oxygen-containing compounds act as bases in the presence of Lewis acids (like ferric chloride).

$\ce{3ArOH + FeCl3 -> Fe(OAr)3 + 3HCl}$ is the acid-base reaction, and you get a ferric phenolate salt as a precipitate.

$\ce{6ArOH + FeCl3 -> [Fe(OAr)6]^3- + 3H+ + 3HCl}$ is the ligand-exchange. It's not immediately obvious, but you can rewrite the two reactions a different way and it becomes clear.

$$\ce{3ArOH + [Fe(H2O)6]^3+ -> Fe(H2O)3(OAr)3 + 3HCl}$$

$$\ce{6ArOH + [Fe(H2O)6]^3+ -> [Fe(OAr)6]^3- + 6H2O + 3H+ + 3HCl}$$

Those three $\ce{H+}$ ions are from the excess phenol. The ligand-exchange occurs only when the phenol is in excess and more water molecules can be replaced, making the iron compound ionic and soluble again.

So depending on the context, you'll see the one with three or the one with six. Both are correct, but $\ce{[Fe(OC6H5)6]^3-}$ is the new complex and $\ce{Fe(OAr)3}$ is an intermediate product.

Hope that helps.


Expression for Gibb's free energy change for a cell reaction

According to Nernst's equation,
$$E = E^o - \frac{RT}{nF}\ln Q$$
As you can see, $E$ is independent of $q$, therefore your proposed integral becomes,
$$\Delta G = - \int_{q_i}^{q_f}{E\cdot dq} = -E\int_{q_i}^{q_f}{dq} = -E\Delta q = -nFE$$

Alternatively, you can prove the the relation between Gibbs energy and cell potential as,

Multiply Nernst's equation by $nF$, we get,

$$nFE = nFE^o - RT\ln Q$$

Swapping the sides, we get,

$$-nFE = -nFE^o + RT\ln Q$$

As we know,

$$\Delta G = \Delta G^o + RT\ln Q$$
$$\Rightarrow \Delta G = -nFE$$


Why does the temperature remain constant during freezing?

Creating an ordered solid from a disordered liquid releases energy

At the molecular level this is what is happening when you cool something like water.

At first the cooling reduces the average kinetic energy of the water molecules so they slow down. Basically, temperature is the average kinetic energy so removing energy lowers the temperature.

At some point, though, water starts to form ice where the molecules are not wandering about randomly and get rigidly stuck together in ice crystals. But this process releases energy: the ordered ice crystal has a significantly lower energy than water at the same temperature. Converting liquid water to ice releases that energy.

What this means is that when extracting energy from warm water the result is simply a lower temperature. But one you get to 0°C the extra energy extracted is used to convert the liquid to the solid rather than reducing the temperature. Once all the water has solidified, further removing energy reduces the temperature again.

The key is that the process of freezing itself involves energy changes so instead of reducing the temperature, the energy comes from the conversion of liquid too solid.


What are actually the differences between Sidgwick & Powell theory and Nyholm & Gillespie Theory?

Perhaps you should read the original papers: N.V. Sidgwick and H.M. Powell, "Stereochemical types and valency groups", Proceedings of the Royal Society of London Vol 176, No. 965, pp 153-180 (1940), and R.J. Gillespie and R.S. Nyholm, "Inorganic Stereochemistry", Quarterly Reviews, Chemical Society 11 (4) 339-380 (1957).

To quote out of Gillespie's paper:

  Sidgwick and Powell assumed that lone pairs of electrons and bonding pairs were of equal importance and that they arranged themselves symmetrically so as to minimise the repulsions between them. The ligand(or crystal)- field theory is concerned with the influence of the non-bonding electrons upon the arrangement of the ligands which are assumed to be negative ions or dipoles surrounding a central metal atom. Recently the influence of covalent binding has also been considered. These various theories are in fact closely related and an attempt has been made in this Review to integrate them. It appears that the simple theory of electron-pair repulsions receives considerable justification from quantum-mechanics and when refined to take into account differences between the repulsions between lone pairs and bonding pairs of electrons provides a very satisfactory theory of the stereochemistry of inorganic molecules.

That should get you going...


Do Palladium(II) Chloride and Palladium (III) Chloride follow 18-Electron rule or octet rule? How to determine this?

Palladium (II) Chloride follows neither the 18 electron rule nor the octet rule.

Instead, like many Pd(II), Pt(II) and Ni(II) complexes, palldium (II) chloride follows the 16-electron rule.

Pd(II) generally forms square planar complexes, and in palldium (II) chloride each Cl bridges two Pd, with 4 Cl coordinating each Pd in a roughly square planar arrangement.  8 d-electrons plus 2 electrons from each of 4 sigma bonds equals 16 electrons.

I have never heard of "Cl3Pd", which would involve the rare oxidation state Pd(III), but the 16 and 18 electron rules should only be applied to diamagnetic complexes.  Pd(III) complexes are paramagnetic.


Disclaimer: the following is based on using DSC data for organic materials.
Is the melting temperature of Indium (156 C) just at the temperature where your sample exhibits a phase transition / decomposition?  In general, I suggest to use the flat indium pellets and to record their melting for calibration of the DSC, rather than their recrystallisation. This is likely described in the manual of your DSC.  (I does not hurt to have a look at DOI: 10.1007/s10973-015-5186-8, or similar, either.)
Because of its transient character, it is worth to keep an eye on the heating rate / cooling rate while recording DSC data.  Provided enough sample material, sometimes a quick screening with 10 K/min is followed by a second recording at a rate of 4  K/min (or slower).  While the endothermic peaks related to melting may be sharp, the ones for the recrystallisation are often shifted towards lower values because without proper seeds of crystallisation, supercooling may occur, especially organic samples and polymers.
I recommend to consider cyclic DSC, instead of a simple single-heating experiment, as this offers you additional insight about the sample in question. It is applied to characterize pure metals (like DOI: 10.1098/rspa.1996.0113), and may shed a light on the degree of crystallinity and processes of reorganisation of your polymer sample, too.  In the following, a DSC of polylactic acid:

(source)
Sometimes these differences about recording the melting point (pristine sample, after 1st / 2nd heating cycle) are spot on small molecule materials, too.  They may appear even if the sample did not incorporate solvent molecules.


Keto-enol tautomerism in a bicyclo compound

Most of the times, the keto-tautomer is lower in energy than the enol-tautomer of a given compound. In all cases this is also very dependent of the actual conditions.
There are few cases, where the enol-tautomer can be dominant. One of the most common examples is malondialdehyde, and, of course, its derivatives.
I am fairly certain that the original question is trying to trick you into thinking this way; and testing your 3D awareness. Or something like this; I'm not sure.
With these structures it's always a good idea to visualise then. I recommend a molecule kit for hands on approaches, but clay and matches are great, too. Even chestnuts and toothpicks are great tools. If you are technically inclined, there are great programs, such as Avogadro (avogadro.cc), which can do basic modeling.
For the following, I have used gfn2-xtb as an optimizer; that is a semi-empirical level of theory, see xtb-docs.readthedocs.io.

You can clearly see that if you migrate the bridgehead-hydrogen, the resulting lone pair would not be able to form resonance with either of the carbonyl groups. That is known as the empirical observation named Bredt's rule.
On the other hand you are quite correct that you could form a tautomer from the other α-carbon. This carbon can easily distort to have planar coordination and the lone pair can be in resonance with the carbonyl groups. (Provided of course that the framework is localised bond orbitals; and that the model is a valid approach.)

On this level of theory, the enol form is about 64.5 kJ/mol higher in energy than the keto-form. So while there certainly is a theoretical possibility, the likelihood is very small.
(Just for giggles: the bridge tautomer is 246.6 kJ/mol higher in energy.)

Appendix

Keto-Tautomer
17
symmetry c1
C       -0.232194517     -0.455059284     -0.108971891
C        0.435695625      0.521858137      0.853516654
H        0.456616659     -1.075279085     -0.674274662
C       -0.652035075      0.924669592      1.832141805
O        1.572314637      0.893979605      0.826125495
C       -1.888025507      0.221717603      1.265101729
H       -0.393829692      0.553384752      2.825493466
H       -0.740240162      2.008868613      1.896308913
C       -2.240914528      0.890781562     -0.065758217
H       -2.719883618      0.149774125      1.962761267
C       -1.080374859      0.487832943     -0.956781929
H       -3.167688540      0.496107389     -0.485905106
H       -2.336524789      1.975006913     -0.012902862
O       -0.862082222      0.837168772     -2.079995835
C       -1.289181703     -1.103984849      0.787745832
H       -1.992066775     -1.719717994      0.228564710
H       -0.851574934     -1.695108793      1.590840632

Enol-Tautomer
17
symmetry c1
C       -1.252989400     -1.048025267      0.775079043
C       -0.235131476     -0.354975263     -0.135182749
C        0.316010839      0.694441373      0.817231157
C       -0.666608713      1.048338273      1.645393318
C       -1.895763330      0.268945971      1.228776635
C       -2.331880748      0.853259922     -0.128837161
C       -1.153613858      0.494907935     -1.014278443
O        1.568858497      1.186824174      0.741278311
O       -0.952399200      0.822318914     -2.150244908
H        0.493388572     -0.958265049     -0.672260873
H        2.056130434      0.748941273      0.034835696
H       -0.640184178      1.790739752      2.416980529
H       -2.686913718      0.203266237      1.970705846
H       -3.231277420      0.373221210     -0.518677687
H       -2.502755602      1.928070454     -0.108020068
H       -1.934290934     -1.708837720      0.239732785
H       -0.777205764     -1.591075191      1.590309569

Bridgehead-enol-tautomer
17
symmetry c1
C       -1.375174317     -1.103606241      0.843045628
C       -0.248454950     -0.489532946     -0.004463589
C        0.359288830      0.442230533      0.837984097
C       -0.605216725      0.791584850      1.950720322
C       -1.905351826      0.244045543      1.301925006
C       -2.166063721      0.867188742     -0.087057960
C       -0.935024925      0.450040319     -0.933532366
O        1.456979798      1.143062588      0.565484195
O       -0.594744149      0.929934770     -1.975783635
H        1.590334610      1.860425911      1.199389402
H       -0.422954467      0.294599984      2.906280842
H       -0.604773151      1.872841199      2.115837534
H       -2.773451234      0.269475098      1.961354663
H       -3.086355462      0.469282331     -0.513923137
H       -2.229917288      1.956721084     -0.092669863
H       -2.077588583     -1.677749691      0.243281460
H       -0.985680441     -1.728640075      1.644336401


The $\ce{O2NNH^-}$ is formed very rapidly but reacts slowly in the second step. The rate limiting step is therefore loss of $\ce{O2NNH^-}$ in which the back reaction with $\ce{N2O}$ is considered (in the question) to be so slow as to be insignificant when writing the rate expression. Thus the rate involves just $\ce{O2NNH^-}$, however, as the fist reaction is a fast equilibrium substituting for $\ce{O2NNH^-}$ with this equilibrium is valid, provided the $k's$  are marked as being different, which they are not, and in the second reaction $k$ should be $k\times K_{eq_1} $.


Why does neopentane have a higher melting point than n-pentane?

Many symmetrical molecules undergo a phase change before they melt.  This is usually interpreted as the ability to rotate without disrupting the lattice structure.  This has a much greater effect on the entropy of the solid than on its energy.  This reduces the entropy of fusion more than it reduces the enthalpy of fusion, thereby raising the melting point which equals delta-H/delta-S.  This is very dramatic in the comparison of melting points of octane, cyclo-octane, and bicyclo-octane.


You need the vaporization $\Delta H_\text{vap}$ and you need to use this equation: $$Q = m~c~\Delta T$$

Where $Q$ equals the energy to heat $1\ \mathrm{mol}$ of water up to $\mathrm{100~^\circ C}$. $m$ is the mass of water, $c$ is the specific heat capacity and $\Delta T$ is the difference between the initial and final temperature.

In this stage you have the energy that is used to heat the liquid water up to $\mathrm{100~^\circ C}$ but the water is still liquid and you need steam water.

So you need to calculate the energy for phase transition from liquid to steam and the equation for this is: $$\Delta H_\text{vap} \cdot n = Q$$

$Q$ is the necessary energy for the phase transition, $n$ is the amount of substance and $\Delta H_\text{vap}$ is the energy used for phase change of $1\ \mathrm{mol}$ of the substance.

You also need to convert the mass to amount of substance (or use latent steam heat instead) and to calculate the $Q$ in the two stages.

This is the energy used for heat and change the phase.


The properties of a cluster depend on the exact number of constituents, and the geometry of the arrangement.

The properties of a bulk solid otoh are just indifferent to the exact size of the specimen (e.g. crystallite).

(In between you have nanocrystalline substances, where properties change with size, but only gradually.)

I would say that clusters are much like molecular compounds in that they have a distinct structure and absolute composition, however they are held together not by covalent bonds but by polar or dispersive forces (also metallic), and they usually consist of a number of identical particles.


Is it possible to construct a 3D molecular model using x-ray powder diffraction data?

What do you hope to achieve?  It is possible to obtain accurate interatomic distances from powder data (here is an example, the structures of several alkali metal cyclopentadienides), but the method is non-routine, and the data-parameter ratio is low, which means that you cannot refine many parameters and must put much information in through appropiate restraints.  You still need accurate peak shape data.

If you just want the general shape of the molecule to illustrate Wikipedia, you are better off with obgen from the OpenBabel project.  It usually gives good results.


What is the geometrical structure of Pb3O4?

Maybe below 3d structure of Pb3O4 will help?

3d structure of Pb3O4

Also in the below link of a pdf, just before the discussion section the paragraph clearly states that there are both ionic and covalent bonds present in Pb3O4.

REsearch Paper on Pb3O4 structure

Thus we can conclude that it is Tetragonal, tP28 .(as per the link provided by Maxw in the comment to your question)

Also metals can form covalent bonds. Refer to below question and its answers:
Metals forming covalent bonds


Reaction mechanism of a terminal alkene with an anhydride

This is the  Friedel-Crafts acylation reaction of alkene. Catalyzed by Lewis acid. Addition-elimination mechanism.


That's a nice bit of kit to get started. I've annotated the picture a little to aid discussion:

I've labelled the "male" and "female" joints that should be connected with the same letter (e.g. B inserts into B'). The thermometer goes in the top of the blue plastic screw joint.

Joint A serves the purpose of connecting your system to a vacuum source. If you are doing vacuum distillation, you should connect the vacuum tubing at this point. However, if you are just distilling at atmospheric pressure you should leave this open. You should NOT seal this joint, as this would mean there is no pressure relief in the system - when you start to heat your sample, gases expand and they need somewhere to go, otherwise you will at best pop a joint, and at worst cause an explosion.

As you say, the joints on the liebig condenser (G) can be left open since they aren't open to the interior of the apparatus. However, you will get better results if you do hook up the condenser to water using the tubing provided, especially if you are distilling at relatively low temperatures.

The clips are there to secure ground glass joints so they don't fall out. You only really need them on junctions that arent held in place by gravity (e.g. you need one on D/D' since the joint could fall apart under gravity, but not on B/B' since gravity holds the male joint in place).

You shouldn't need to vacuum seal your joints - ground glass quick-fit joints should be manufactured to a reasonably high tolerance so the leakage is minimal. It would only become a problem if you needed very strong vacuuum.

Note: if you are going to use vacuum for your distillation you should carefully inspect every piece of glassware for cracks, imperfections etc. - if the glassware is damaged then when you apply the vacuum it can implode. You should always wear eye protection when doing any chemistry at all.

Hope this helps, please ask questions if you aren't sure of anything - people on here will be happy to help a new recruit to develop their chemistry skills!


Why use potassium iodide instead of copper iodide in Sandmeyer reaction?

Sandmeyer reactions using iodides, water, and thiols do not require the copper(I) ion as a catalyst for the reaction to proceed whereas your other nucleophiles do (though not necessarily stoichiometrically).

As you can see the copper(I) ion is important in reducing the diazonium to nitrogen and forming a radical.  Iodine, oxygen, and sulfur must be able to be oxidized by diazonium and therefore do not require copper.


MOPAC: can we run THERMO immediately after the geometry optimzation?

OK, in between I found it, it is the GEO_DAT keyword, for example, reading in previously converged PM7 geometry of methane:

 AUX LARGE CHARGE=0 SINGLET  GEO_DAT="metan_geomopt_pm7.arc" THERMO(298,298)  PM7


Chair conformations of cyclohexane with two substituents

Here are some structures to help you. You noted that one group was a solid wedge and the other a dashed wedge without mentioning the substitution pattern. I presume that this was a flat, 2-D structure. You also note that in the chair structure, one group is axial, the other equatorial. Only the 1,3-isomer 2 fits these conditions. The conformation 2a is more stable than 2b.
ADDENDUM: The 1,1-disubstituted isomer would also fit the conditions of the question. The questioner did not designate the structure.


The acetyl group itself does not have any significant charge. But the residue that gets acetylated is lysine in this case, which has a positive charge. The acetylation results in a neutral amide, removing the positive charge.

From Wikimedia Commons

I'd add that the effect of histone acetylation is likely more subtle than just the charge effect. This kind of posttranslational modification often affects the protein structure in some way, or is recognized by other proteins interacting with that site.


Can orbitals in a molecule be occupied via excitation like they can in atomic orbitals?

Most of what applies for atomic orbitals in atoms can be transferred to molecular orbitals in molecules. This includes that electrons can be excited to higher lying orbitals, creating excited configurations.

Also, just as for the hydrogen atom, there are in principle infinitely many orbitals available. They are simply truncated since the higher lying ones are less important (and for numeric reasons).


Referring to numbered atom (e.g. C-1 vs C1)

How about using the label plus the number encircled? Would be hard to confuse and misinterpret. I actually use this myself when writing on paper or typesetting on computers.


We know that, at temperature T, $$\Delta G^0(T)=\Delta H^0(T)-T\Delta S^0(T)$$
From Hess's Law, we also know that
$$\Delta H^0(T)=\Delta H^0(298)+???$$
and
$$\Delta S^0(T)=\Delta S^0(298)+???$$

That should be a good enough hint.


Correctness of writing H3C as opposed to CH3 in a carbon chain

Your answer is the same as the book's. The book wrote the first carbon as $\ce{H3C}$ to stress that the chain continues via a carbon to carbon bond, not bonded a carbon to hydrogen to carbon bond.

As for why it's not like this in other examples, technically it is not necessary to write it this way, but it's strange your book lacks that consistency.


Why is the Haber process carried out at such high temperatures?

As others have pointed out, it is purely kinetics, but you may still wonder, why.

For a reaction to actually occur (in both directions) and thus for an equilibrium to be reached, you need to overcome the activation energy. In the case of the Haber-Bosch process, this involves breaking the highly stable $\ce{N#N}$ triple bond. Even with the catalysts used, the energy required to break apart $\ce{N2}$ is still enormous. Therefore, a lower temperature may give a better yield of ammonia theoretically (i.e. based on equilibrium and Le Châtelier considerations) but the reaction speed would be a lot slower.

Even if you consider a batch-wise process of generating ammonia (which, as orthocresol points out, isn’t the case), it is more efficient to run two batches in half the time for two sets of $15~\%$ than to run a single batch for twice the time to get an overall yield of maybe (note: This yield is a ballpark estimate) $25~\%$ — with a batch size where $1~\%$ yield is equivalent to $100~\mathrm{kg}~\ce{NH3}$, two runs at higher temperature give $3~\mathrm{t}\ \ce{NH3}$ and one run at lower temperature $2.5~\mathrm{t}\ \ce{NH3}$ in the same time frame.


Finding identities of diatomics from Spectroscopy and Mass Spec Data

Is this problem conceptual, based on hypothetical data?

  Through [m]ass [s]pec, I found 2 strong spectral lines at $30.0077086\ \mathrm{g\ mol^{-1}}$ and $41.9766928\ \mathrm{g\ mol^{-1}}$.

These mass-to-charge ratios or $m/z$ values are implausibly precise.  The significant figures you show imply a precision of about 30 parts-per-billion (ppb) mass accuracy.  That is well outside the specification of any commercial instrument I am aware of.

Also, if you are right about the first ion being, $\ce{NO}$, the value is inaccurate.  The neutral $\ce{^14N^16O}$ molecule has a mass of 29.99799 Da.  The positively charged ion formed by electron loss from the neutral species would have a mass of 29.9974 Da.  This latter value is off from the value you quote by about 341 parts-per-million.  So if the mass spectrometry data really is precise to within 30 parts-per-billion, and it is accurate, then the molecule can't be $\ce{NO}$.

In fact, with mass spectrometry data that precise (and accurate), we shouldn't need any other data to identify the diatomics, other than the fact that they are diatomics.  We can add the mass of every possible nuclide to every other possible nuclide and see which give the right mass.

Here's how to do that using the R packages ecipex and dplyr.

require(ecipex)
require(dplyr)
nistiso %>%
    mutate(foo = 1) %>%
    left_join(., ., by='foo') %>%
    filter(mass.x > mass.y) %>%
    mutate(diatomic_mass = mass.x + mass.y) %>%
    filter(diatomic_mass < 30.00771, diatomic_mass > 30.00770)

The result is that only the $\ce{^11B^19F}$ diatomic molecule has the requisite mass. Even if you loosen the mass tolerances to allow for e.g. for electron loss or gain, this is still the only possible diatomic with the right mass.

You can repeat this exercise for the other data point, and find that $\ce{^40Ca^2H}$ is the only possible ion consistent with the data.  ($\ce{^40Ar^2H}$ is close in mass but is more than one electron's mass too light.)

I'll leave it to you to decide whether these ions are chemically meaningful or possible for you, and if they are consistent with the rotational spectra.


Confusions about inductive effect and acid strength

The result is correct but the explanation is bad.

The main reason is that when you compare CH3COOH and CH3ClCOOH , the main difference is that the alpha carbon in the former has more electron density than the latter one because of which when the conjugate base gets formed, it is more stabilised than ethanoic acid because of low electron density.

You are correct about this "In my mind, to borrow some charge density you have to be more electronegative than what you're attached to, and the more deficient carbon should have a greater pull".

Edit 1 - The orbitals of CH3COOH can provide more electron density than CH2ClCOOH due to the former having more electron rich orbitals which interferes constructively with the orbitals of other C to satisfy its demands. Think in terms of how Hyper conjugation or resonance works in terms of interference of wave functions.


What will eventually happen to uncovered vegetable (seed) oil exposed to the air?

Some of the lighter constituents could evaporate. There are a lot of minor components in olive oil, and they are largely responsible for the fragrance and flavor. As these evaporate, the fragrance and flavor would change. It is unlikely that they would all evaporate though, as many of these components likely have high boiling points/low vapor pressures.

The oil itself would not evaporate, but it would not solidify either. At most you might notice a thickening of the oil.

Over long periods of time the oil may react with oxygen to form various oxidation products. This process is so slow, however, that it is definitely not a concern for olive oil you might leave out in your kitchen. The oil will start tasting and smelling stale long before significant oxidation occurs.


How can Arrhenius bases also be Brønsted-Lowry bases?

Sure it can.  You're just incorrect about the products when $\ce{NaOH}$ accepts an $\ce{H+}$ in aqueous solution.  The result is $\ce{H2O(l) + Na+(aq)}$ instead of $\ce{HNaOH+(aq)}$.  That is, the $\ce{NaOH}$ doesn't stay together.

Now mind you, most people would point out to you that the $\ce{NaOH}$ doesn't stay together as soon as it's dissolved, meaning $\ce{NaOH(s) -> Na+(aq) + OH-(aq)}$ happens right away, and then as soon as you have some $\ce{H+(aq)}$ around it reacts only with the $\ce{OH-(aq)}$ and the $\ce{Na+(aq)}$ doesn't get involved, probably isn't even anywhere near.  But this doesn't change the main point, which is that if you add $\ce{NaOH}$ to water, you have a solution that can react with $\ce{H+(aq)}$, which means you added a Brønsted-Lowry base.


If hydrogen bonding in water was weaker, what happens to H+ ion concentration?

The article by Martin Chaplin 'Water's Hydrogen Bond Strength' (arXiv:0706.1355 [cond-mat.soft], no evidence of peer review) is discussing how changes to the properties of water would alter life.

If the hydrogen bonding in water was stronger, there would be less ionization (less $\ce{H+}$ and less $\ce{OH-}$).

If the hydrogen bonding in water was weaker, there would be more ionization (more $\ce{H+}$ and more $\ce{OH-}$).


In each case the permanganate oxidises the most electron rich ring.

In the nitro-substituted case, that is the unsubstituted ring, probably first at the position alpha to the nitro group, forming a phenol, then para to that to a quinone but permanganate keeps on chewing until the phthalic acid is formed.

In the aniline case the most electron rich ring is the one with the NH2 attached. Oxidation of anilines by acid permanganate give a polymer called "aniline black" Wiki link. Though I can't find a reaction specific to 1-Naphthylamine other than one that documents the destruction of it by sulfuric/permanaganate without identifying the products here. The site of first oxidation is p to the aniline. This creates a reactive intermediate which is attacked by a second aniline as the first step of polymer formation.