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Derivation of the Orgel diagram for octahedral d2 complexes
The Orgel diagram for a typical octahedral $\mathrm{d^2}$ complex is shown below:

(source: Kettle, Physical Inorganic Chemistry, p 146)
I understand that the right-hand half of the graph shows how the energies of electronic states of a $\mathrm{d^2}$ complex vary with the ligand field splitting $\Delta$.
However, neither the Wikipedia page nor the LibreTexts (formerly UC Davis ChemWiki) page are helpful in describing how this diagram is constructed. In particular, I have a few questions:

What are the "weak-field" and "strong-field" limits?
How are the symmetry labels for the electronic states constructed?
How are their relative energies determined?
Why are some of the lines straight, but others curved?
Why is the diagram also used for $\mathrm{d^3}$, $\mathrm{d^7}$, and $\mathrm{d^8}$ complexes (both octahedral and tetrahedral)?

1. Weak-field and strong-field limits
I will adopt the description used in Figgis and Hitchman's Ligand Field Theory and Its Applications (p 5), because I cannot really phrase it better:

It is useful to recognise the relative importance of the terms in the Hamiltonian for different systems:
First- and second-transition series: $\mathbf{H}_\mathrm{LF} \approx \mathbf{H}_\mathrm{ER} > \mathbf{H}_\mathrm{LS}$
$\mathbf{H}_\mathrm{LF}$ (ligand field) takes into account the presence of the surrounding ligand molecules [...]. $\mathbf{H}_\mathrm{ER}$ (electron repulsion) takes into account the repulsions within the d or f electron set [...]. $\mathbf{H}_\mathrm{LS}$ (L-S coupling, also known as Russell-Saunders coupling) takes into account the formal coupling between the spin and the [orbital] angular momenta of the d or f electrons.

Spin-orbit coupling (which scales as $Z^4$, $Z$ being the atomic number) plays a nonzero, but relatively small, role in the properties of the 3d or 4d transition metals (compared to the 5d, 4f, or 5f metals). Therefore, it is not reflected in the Orgel diagrams above.
The weak-field limit simply refers to the case where the ligand field is weak, i.e. $\mathbf{H}_\mathrm{LF} < \mathbf{H}_\mathrm{ER}$. In this regime, the spherical symmetry of the free ion (e.g. $\ce{V^3+}$) is almost unperturbed by the six ligands (assuming octahedral symmetry). Therefore, the electronic states are similar to those of the free ion.
The strong-field limit simply refers to the opposite case, i.e. $\mathbf{H}_\mathrm{LF} > \mathbf{H}_\mathrm{ER}$. In this regime, the electron-electron repulsions are relatively small compared to the ligand-field splitting $\Delta$. As such, the electronic states are entirely determined by the electronic configuration $(\mathrm{t_{2g}})^m(\mathrm{e_g})^n$ (for a $\mathrm{d^2}$ complex ion, clearly $m + n = 2$).
As $\Delta$ increases (i.e. going from left to right), we move from the free ion limit where there is no ligand field and $\Delta = 0$, through the weak-field regime where $\Delta$ is small, into the strong-field regime where $\Delta$ is large.

2. The electronic states: their symmetries and their energies
2.1. Free-ion terms
For a $\mathrm{d^2}$ configuration, one can either enumerate microstates (a nasty, tedious method that I do not recommend), or one can use group theory (wait for the MathJax to load, then scroll to direct product tables: $R_3$). The wavefunction of a d-orbital, in the spherical symmetry of the free ion, transforms as $D \equiv \Gamma^{(2)}$. The electronic wavefunction of two d-electrons transforms as the direct product of $D$ with itself:
$$\begin{align}
\Gamma^{(2)} \otimes \Gamma^{(2)} &= \Gamma^{(4)} \oplus \Gamma^{(2)} \oplus \Gamma^{(0)} + [\Gamma^{(3)} \oplus \Gamma^{(1)}] \\
\mathrm{D \otimes D} &= \mathrm{G \oplus D \oplus S \oplus [F \oplus P]}
\end{align}$$
Square brackets indicate spatial wavefunctions that are antisymmetric with respect to interchange of the two electrons; to satisfy the Pauli exclusion principle, these have to be paired with symmetric spin wavefunctions. Adding up the spins of two electrons ($s_1 = s_2 = 1/2$), we can either get $S = 1$ ("triplet") or $S = 0$ ("singlet"). The triplet spin wavefunctions are symmetric, so these go with the $\mathrm{F}$ and $\mathrm{P}$ terms. Likewise, the singlet spin wavefunction is antisymmetric, and goes with the $\mathrm{G}$, $\mathrm{D}$, and $\mathrm{S}$ terms. This yields the term symbols $\mathrm{^1G}$, $\mathrm{^1D}$, $\mathrm{^1S}$, $\mathrm{^3F}$, and $\mathrm{^3P}$ for a $\mathrm{d^2}$ configuration, where the superscript indicates the spin multiplicity $M = 2S + 1$.
The singlet states are higher in energy due to exchange energy (Hund's first rule). As such, they get ignored in our treatment here. This is not always wise, and the more sophisticated Tanabe-Sugano diagrams do plot electronic states of different multiplicity, but as far as we are concerned, we can neglect them here.
Between the two triplet states, the state with the larger value of $L$ (total angular momentum quantum number) is the ground state (Hund's second rule). Therefore, the $\mathrm{^3F}$ state with $L = 3$ is the ground state, and the $\mathrm{^3P}$ state with $L = 1$ is the excited state.
The difference in energy between these two states is solely attributable to electron-electron repulsions.[1] The two states are separated by an energy difference $15B$, where $B$ is a Racah parameter that acts as a measure of electron-electron repulsions. I believe there is a quantum mechanical expression for it, but in practice the value of $B$ is empirically obtained (via Tanabe-Sugano diagrams). It is easier to treat it as an empirical parameter that is experimentally obtained, similar to how one would treat the ligand-field splitting $\Delta$.
The free-ion terms and the difference in energy $15B$ are shown in the middle of the diagram, where $\Delta = 0$.
2.2. Strong-field limit terms: no electron repulsions
In an octahedral $\mathrm{d^2}$ complex, there are three ways of arranging the two d electrons: $\mathrm{(t_{2g})^2}$; $\mathrm{(t_{2g})^1(e_g)^1}$; and $\mathrm{(e_g)^2}$. These are the three electronic states under consideration and are what you see at the right-hand side of the diagram. Hopefully the energy ordering is straightforward; the energy gap between each state is equal to $\Delta$ since it requires promotion of one electron from $\mathrm{t_{2g}}$ to $
\mathrm{e_g}$.
2.3. Weak-field limit terms
As can be seen from the Orgel diagram, there are four states: two $\mathrm{T_{1g}}$ states, one $\mathrm{T_{2g}}$ state, and one $\mathrm{A_{2g}}$ state. (The subscript g, as well as the spin multiplicity, are omitted in the diagram. This is to allow it to be generalised to tetrahedral complexes, where the g/u labels are not applicable, as well as complexes with other d-electron counts, where the spin multiplicity will be different but the symmetries the same.) The question we will have to look at is how these states are obtained, and how they correlate with the two extreme limits described earlier.
Going from the free ion limit, we need to find out what happens to $\mathrm{F}$ and $\mathrm{P}$ terms when the symmetry of the compound is reduced from spherical to octahedral. Luckily, there are group theory tables that do just that, called tables of descent in symmetry. These tell us that:
$$\begin{align}
\mathrm{F} &\to \mathrm{A_2 \oplus T_1 \oplus T_2} \\
\mathrm{P} &\to \mathrm{T_1}
\end{align}$$
and you can see this correlation in the lines spreading out from the centre. Conventionally, the $\mathrm{T_1}$ state arising from the $\mathrm{P}$ state is labelled as $\mathrm{T_1(P)}$ and likewise the other $\mathrm{T_1}$ state is labelled $\mathrm{T_1(F)}$. This isn't shown on this particular diagram, but may or may not appear in other texts (and it is very helpful because it eliminates any ambiguity about which $\mathrm{T_1}$ state you are talking about!).
Now, we need to figure out which final configurations to join them to. This is done by taking more direct products (yes, group theory tables again). For the $\mathrm{(t_{2g})^2}$ configuration we have
$$\mathrm{T_2 \otimes T_2 = A_1 \oplus E \oplus [T_1] \oplus T_2}$$
however since we are only interested in the triplet states, we are only going to take the $\mathrm{T_1}$ term, which has an antisymmetric spatial component to go with the symmetric triplet spin component. Likewise
$$\begin{align}
\mathrm{T_2 \otimes E} &= \mathrm{T_1 \oplus T_2} \\
\mathrm{E \otimes E} &= \mathrm{A_1 \oplus [A_2] \oplus E}
\end{align}$$
So, the $\mathrm{(t_{2g})^1(e_g)^1}$ configuration gives rise to $\mathrm{T_1}$ and $\mathrm{T_2}$ terms,[2] whereas the $\mathrm{(e_g)^2}$ configuration gives rise to an $\mathrm{A_2}$ term. The only remaining question is, which of the $\mathrm{T_1}$ states correlates with which of the strong-field configurations? The answer is given by the non-crossing rule: two states of the same symmetry cannot cross each other.[3] Therefore, we can confidently join the $\mathrm{T_1(F)}$ state to the $\mathrm{(t_{2g})}$ configuration, knowing that the alternative would lead to a forbidden crossing.
Incidentally, this is also why the two $\mathrm{T_1}$ electronic states have curved lines: quantum mechanical mixing introduces an extra perturbation to the energy. The upper energy level is always pushed upwards, and the lower energy level pushed downwards, which leads to the curvature shown in the diagram. The other two states do not have such effects and are therefore depicted as straight lines.

3. Other geometries and configurations
Since a $\mathrm{(t_{2g})^3(e_g)^2}$ configuration is spherically symmetric, the electronic states arising from an octahedral (high-spin) $\mathrm{d^7}$ configuration are exactly the same as those arising from an octahedral $\mathrm{d^2}$ configuration. (The spin multiplicity is different: $S = 3/2$ gives $2S + 1 = 4$.)
Furthermore, the octahedral $\mathrm{d^3}$ configuration can be thought of as having two holes in a spherically symmetric $\mathrm{(t_{2g})^3(e_g)^2}$ arrangement. These holes give rise to exactly the same symmetry states as the $\mathrm{d^2}$ configuration; however, since holes are essentially "negative electrons", the relative energies are reversed. For example, the $\mathrm{A_2}$ state, which arises from placing the two holes in the $\mathrm{e_g}$ orbitals (i.e. a $\mathrm{(t_{2g})^3}$ electronic configuration), is actually the ground state instead of the highest excited state. This is reflected in the left half of the diagram.
(If the above two paragraphs were confusing, they were covered in more detail in another answer of mine: How do I find the ground state term symbol for transition metal complexes?)
As for tetrahedral complexes, note that the orbital splitting in a tetrahedral complex is exactly the same as in an octahedral complex, except that the relative energies of the $\mathrm{e}$ and $\mathrm{t_2}$ sets are reversed (and the g subscripts are gone).
Therefore, if we consider (for example) the $\mathrm{(e)^2(t_2)^1}$ ground state of a tetrahedral $\mathrm{d^3}$ configuration, this can be considered to have two holes in the $\mathrm{t_2}$ orbitals. These two holes give rise to exactly the same states as the octahedral $\mathrm{d^2}$ case, where in the ground state there are two electrons in the $\mathrm{t_{2g}}$ orbitals. The tetrahedral $\mathrm{d^3}$ ground state therefore has a $\mathrm{^4T_1}$ label, exactly analogous to the octahedral $\mathrm{d^2}$ ground state which is a $\mathrm{^3T_{1g}}$ term. Of course the spin multiplicity is different and the g label is gone, but we've already established that the diagram specifically says nothing on these two points for a reason.
Consequently, tetrahedral $\mathrm{d^3}$ and $\mathrm{d^8}$ complexes fit into the right half of the diagram. Using similar arguments, tetrahedral $\mathrm{d^2}$ and $\mathrm{d^7}$ complexes fit into the left half of the diagram.
Notes
[1] Quoting myself from another answer

The classical argument behind Hund's second rule is that electrons that have the same sign of angular momentum "collide" into each other less and as a consequence repel each other less. It is like two people running around a track; if they ran in the same direction, they would bump into each other less than if they ran in opposite directions.

[2] As an aside: why is the $\mathrm{(t_{2g})^1(e_g)^1}$ configuration split into two terms? The answer lies in, unsurprisingly, electron-electron repulsions (which were neglected in the strong-field limit). Consider the electronic configuration $(\mathrm{d}_{xy})^1(\mathrm{d}_{x^2-y^2})^1$, which has electron density concentrated in the $xy$-plane. We might expect this configuration to experience more electron-electron repulsion than, for example, the $(\mathrm{d}_{xy})^1(\mathrm{d}_{z^2})^1$ configuration. While molecular terms do not necessarily have a one-to-one correspondence with orbital microstates, one can see from this discussion how it is plausible for the degeneracy to be lifted when electron repulsions are taken into account.
[3] The Hamiltonian $\hat{H}$ transforms as the totally symmetric irreducible representation of the relevant point group (which makes sense, since the energy of a system has to be constant under reflection, inversion, rotation, etc.). Consequently, the matrix element $\mathbf{H}_{mn} = \langle m | \hat{H} | n \rangle \neq 0$ if $m$ and $n$ have the same symmetry. This directly leads to the non-crossing rule, that energy levels corresponding to wavefunctions with the same symmetry do not cross under any strength of a perturbation (in this case, the ligand field).

will you please explain how we decide the different excites states in tanabe sugano diagram.or how we can draw it for any tradition metal complexes

@orthocresol what's the physical meaning of the lines we have in the correlation diagram we can construct this way? If a given configuration in the strong field limit is, for example, connected with two terms in the limit of strong repulsion, can this configuration be expressed as linear combination of these two terms?

@Chemistry. I'm sorry, I have to admit I don't remember much of this stuff any more. It's been many years and I've not used it since graduating

@orthocresol don't worry! Thanks anyway for your constant support


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In what circumstances and why do bonds break without an external energy source?
Edit:
I've flagged my own question because it's too broad instead I'll split it up into more specific questions.
So I'm trying to work out exactly how bonds are able to break when there is no external energy source such as heat or light.
I am already aware of displacement reactions but I am unsure exactly why this happens. So in general it seems that a more reactive element will replace a less reactive one in a molecule. Why does this happen? Is it because the more reactive element is more electronegative and yanks electrons away from the previous bond? Does this happen in every reaction?
It seems as if it is a similar case with free radicals. I've read that this is to do with "yanking" electrons also. Do they always attack the weakest bond?
What other circumstances are there and why? Ideally I would like a mechanistic answer.

This question is too general. Specific, indivudual Bonds form and break under specific circumstances, when the geometry and energy allows for the rearrangement.

Heat cannot be an "external energy" in the sense you seem to think. A reaction mixture simply has a temperature.

You don't need a mechanistic answer. You need thermodynamics.
You're missing one key insight which is that a well-defined process has a non-zero probability of happening. Universe entropy changes and Gibbs free energy changes only tell you how likely something is to happen, and while the corresponding probabilities from these values may be close to zero, they are not zero. Therefore, bonds have some nonzero probability of cleavage.

Unless you are at zero kelvin, there is always heat energy. Most reactions have an activation energy or energy barrier between reactants and products that has to be overcome if a reactant is to transform into product.
The Boltzmann distribution shows us that at a given temperature T there is always a distribution of energies between molecules. The chance of having a given energy say E decreases exponentially with energy as $\exp(-E/(k_BT)$ where $k_B$ is Boltzmann's constant.
Thus if the activation energy is large then there is only a small chance in any given second that by random motion of molecules, producing by collision in a solvent  a short lived fluctuation in energy, that the barrier can be overcome and a product formed. If the barrier is smaller then its height (in energy) may be more easily surmounted and thus molecules react many more times /second. (You may associate 'weaker' bonds with lower barriers towards a particular product).
If the energy of the product is lower than that of reactants then the reaction back to reactants will have a larger barrier that that which led to products, thus the equilibrium between products and reactants lies on the product side. Vice versa for endothermic reactions.
Some reactions have a very low or no barrier between reactant and product, radical reactions can be of this form as can some electron transfer reactions. In the gas phase these species react on contact and the reaction rate is very large, e.g. $\ce{Cl\cdot + H2 }$. In solution the products are also formed as soon as reactants come into contact with one another, however, in these cases the rate of reaction is limited by how fast the reactants can diffuse together in whatever solvent is used.
In reactions that proceed by many steps each step has one of the forms described.

Just to repeat what's already been explained to you; your statement implying only "heat" and light are sources of (external) energy is really misguided/wrong. Especially in the context of chemistry. To name two other sources of external "energy", collisions and electricity. Energy, as is obvious once you've learned some Freshman chemistry and physics, is an abstract concept, I can't point to something and say "That's energy" - it's something else which contains energy. Since atoms obey quantum mechanics, and since quantum mechanics is about probabilities, a reaction in which an A-B bond with a near-by C may change to (say) an A-C bond, and will if the A-C species is (has) lower energy, even near absolute zero, in the dark, etc. OTOH, in general a reaction is more likely to occur as the difference in energy between reactants and products increases and as the activation energy decreases (activation energy is the difference in energy between the transition state and the reactants - see Wikipedia). But lets take an example which might be what you were thinking about. Let's imagine a molecule with a X-Y bond (while my example here is conceptual, there are plenty of examples of it). If the "internal degrees of freedom" have sufficient energy to break the X-Y bond, then it may come to pass that the bond breaks. So, what do I mean by "internal degrees of freedom"? Well, conservation of energy and momentum hold (almost perfectly), but say there is a bond in the molecule which is vibrating (all bonds vibrate, even at 0 K, although the "energy" of vibration of a bond in its lowest energy state can't be used). If that energy is above the lowest possible for that vibration, it can "couple" to the X-Y electronic energy and increase its vibration, possibly to the point where the bond breaks. What other "internal degrees of freedom" does a molecule have? Spin, steric strain (conformation), it also has velocity (momentum) and angular momentum, but those are conserved. A molecule (or molecular fragment) may already have enough energy internally to spontaneously decompose. The rate this happens can be fast (boom) or take centuries (diamond converting to graphite). Repeating myself: even a molecule "sitting" on an atomic lattice at 0 K which is near to an atom which tends to substitute can "make the switch", BUT this is typically very slow to happen. Hydrogen is the principle atom which finds this kind of quantum mechanical "tunneling" to be easy. Responding to a different part of your question (is it because...) Well, I suppose you could look at it that way for some reactions. It's better, imho, to look at it as nature's tendency to attain the lowest energy state. So if a "highly reactive" element were to replace a less reactive element BUT the product were even MORE highly reactive, then the reaction wouldn't be likely to occur spontaneously. So it's not about "reactivity" per se, but about lowering the system's energy. Electronegativity is a concept which is well defined for elements in their ground state but what is the electronegativity of Fe(II) compared to Co(III) ? or of Cl(-) compared to OH(-). Chemistry is almost (but not quite) exclusively about electronic (and steric) changes, so all of it (almost) will be "about" electrons and charge (electron density) differences. It may be helpful in some situations to look at it that way. But in others it will not lead to obvious predictions/conclusions. That is, in some situations it will be useful, in others not so much. But that's nothing new. Most of science is like that.

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Electron density in benzene ring
What would be the order of electron density in the benzene ring in the following compounds?
Methyl Benzene
Ethyl Benzene
t-butyl benzene
I would say that t-butyl benzene would have the highest electron density as the alkyl group attached to it is bulkier than methyl and ethyl thus it will exert a greater +I effect pushing the electrons into the ring making it electron rich. Using same analogy ethyl benzene would have a greater electron density in the ring compared to methyl benzene.
However that is not the case. Something counter intuitive seems to be happening here. The order is methyl benzene >ethyl benzene >t-butyl benzene.
I think this phenomenon can somehow be explained using hyper conjugation, but i don't get it how?

The differences are small, but consistent with hyperconjugation arguments.  If we look at the relative rates of electrophilic nitration as a measure of electron density we find benzene=1, toluene=24, tert-butylbenzene=15.7.  See this earlier Q & A.

Well, tetrahydrocannabinol, you said it correct only the correct reason/explanation is hyperconjugation only.
More the number of (alpha) hydrogen atoms on a C=C [isolated as in alkenes or in resonance as in benzene], more the possibility of hyperconjugation.
In benzene, this will shift the electron density from C-H bond of CH3 group towards the ring, making it electron rich.
Contrary to what could have been thought from +I effect of corresponding alkyl groups on the ring, hyperconjugation is more in CH3.
Since hyperconjugation involves complete shifting of electron from valence shell of atoms to the other, it is more "effective" that INDUCTIVE which just pushes electron little more towards more EN (more correctly, towards group with more -I) atom.

First of all: C-C bonds are able to engage in hyperkonjugation, so "NO hyperconjugation possible" is just wrong. And Second: the drawing is terribly wrong, since no bonds are broken in case of hyperconjugation, and for hyperconjugation you need a certain orientation of the orbitals to each other, for a methyl group only one CH bond can really engage in hyperconjugation, the other two aren't aligned very well.

@DSVA I will agree with your ‘first’. Obviously, as ron commented on the question, tert-butylbenzene has the highest electron density so there must be hyperconjugation. Your ‘second’, however, is not correct. The depiction as shown by Che Mistry shows resonance structures connected by mesomery and not reactions. Understand the difference between reaction and mesomery arrows. (However, I’ll admit that the difference could be made clearer in the answer.)

@Jan even if this would be a mesomeric structure (I never seen an actual sigma bond shifted while describing mesomeric structures) it doesn't make sense and is a mesomeric structure that shouldn't be used since it's extremly far away from the actual structure. Hyperconjugation is too weak to justify such a form.

@DSVA Feel free to downvote all answers in which I use these resonance forms to explain hyperconjugation.

@Jan well, I like to learn, any high level textbook or paper doing it this way? It just doesn't make sense to me, I mean the same way I could write H- "ethene" H+ as resonance structure for ethane, but that's just not very useful at all.

@DSVA I’m really not the best person to ask about textbooks, but the professors who taught organic chemistry at the university of Munich (LMU) did so.

Ok, not the case in Vienna and not on a single textbook of mine. Well, I guess I will stay at my point of view until I find any good reasoning for using such structures. Imo it's more useful to explain hyperconjugation in a different way. IUPAC seems to agree since they tend to specify pi-electrons and pi-bonds into their definitions of mesomerism and delocalization. Their definition of resonance would allow for such representation but that's more aimed on wave function which aren't really shown in such structural formulas.

It's simply due to hyperconjugation effect. As hyperconjugation effect is considered stronger than inductive effect therefore before seeing the inductive effect of methyl, ethyl and t butyl, we should check it's hyperconjugation via the number of alpha hydrogens.
In the first case, it's 3,
In the second case, it's 2' and
In the third case, it's 0.
Hence 3 > 2 > 1

Both C-C and C-H bonds can act as hyperconjugative donors. So your reasoning is not justified.

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Why does NF3 have a smaller bond angle than NH3?
I've already read many answers about the reason why $\ce{NF3}$  has a smaller bond angle than $\ce{NH3}$ , but I can't seem to understand them. Here's my understanding of the situation:

$\ce{NH3}$: Here N is more electronegative than H so a large electron cloud is crowded over N. This will push the bond pairs $\ce{N-H}$ away from the central atom. So, the angle of $\ce{H-N-H}$ will decrease.
$\ce{NF3}$: Here F is more electronegative than N, so the lone pair cloud over N is scattered into the $\ce{N-F}$ bonds. Thus the smaller electron cloud over central atom is unable to push the $\ce{N-F}$ bonds away from itself as much as it did previously. So, bond angle should be higher than previous case.

have a look at a previous explanation for related molecules http://chemistry.stackexchange.com/questions/14981/why-does-bond-angle-decrease-in-the-order-h%E2%82%82o-h%E2%82%82s-h%E2%82%82se/57973#57973

Read about Bent's rule

Related answer 1, related question with related answer 2. Not really dupe, though.

The bond angle difference between $\ce{NH3}$ and $\ce{NF3}$ is not easily explained — but that is primarily because ammonia’s bond angles already violate the simple theories that work so well for phosphane, arsane and stibane.
From a nitrogen atom’s point of view, having its lone pair in an s-type orbital and forming bonds with the p orbitals alone would be energetically beneficial. p orbitals can overlap with other atoms much better because they are directional. However, if that was done the resulting ideal $90^\circ$ bond angles would bring the hydrogens far too close together. Therefore, s contribution is mixed into the bonding p orbitals to alleviate the steric stress until an observed ‘equilibrated bond angle’ of $107^\circ$.
The same thing occurs in $\ce{NF3}$. However, fluorine atoms also bond with p orbitals (rather than hydrogen’s s orbitals) and they are larger meaning that the bond lengths are also greater. Therefore, the fluorine atoms are already more spaced out as is and the bond angle needs to be expanded less far from the ideal $90^\circ$ to sufficiently prevent steric repulsion.

This explanation is just your own conjecture right? Although undoubtedly, the N-F bond has a greater length than the N-H bond, the fluorine atom is also a significantly larger atom than the hydrogen atom. In fact, fluorine has approximately twice the covalent radius of hydrogen according to one Internet source I looked at. Thus, your steric repulsion argument may not be so convincing.

@TanYongBoon No, it’s actually what I was taught by Professor Klüfers in the introductory and inorganic chemistry lecture in Munich in 2006/7.

This can be argued on the basis of Bent's rule; concisely stated

Atomic s character concentrates in orbitals directed toward electropositive substituents

What follows below is a crude explanation. Before that, I'll note that we concern ourselves with the hybridisation of the orbitals at the central atom.
Since s orbitals are lower in energy than p orbitals, they are better at stabilising electrons. Which is why molecules like to use them (s orbitals) in bonds where there is "more" electron density to stabilise.
Thus, Bent's rule can just as easily be formulated as follows:  p character concentrates in orbitals directed towards electronegative substituents; (the central atom doesn't have to waste its low energy s orbitals, the orbitals on the electronegative atom, can (possibly) take care of it.
What has all of this got to do with bond angles? Well, more s character leads to larger bond angles. You can think of how s orbitals are spherical in shape, and a large s character would lead to more "spherical" hybrid orbitals, and that would lead to larger bond angles.
Going back to your question, we are supposed to compare $\ce{NF3}$ and $\ce{NH3}$.
Fluorine is more electronegative that hydrogen, and the $\ce{N-F}$ bond would have greater p character than the $\ce{N-H}$ bond, (or $\ce{N-H}$ bond has greater s character, whichever you prefer). Thus $\ce{H-N-H}$ angle is greater in $\ce{NH3}$ than it is in $\ce{NF3}$

P.S I'm sure you can learn more about what I just tried to explain (rather poorly) by reading the Wikipedia article, and/or a textbook and/or using the search function on this site to find related questions/answers.

When you're comparing bond angles between $\mathrm{NH_{3}}$ and $\mathrm{NF_{3}}$, you'd want to take the electronegativities the of hydrogen and fluorine into consideration.
Fluorine hits a 3.98 on the Pauling Scale for electronegativity, while hydrogen does a 2.2 on the same scale.
Hydrogen, with its said electronegativity of 2.2 does a fairly decent job of pulling the N-H electron bond pairs towards it. As you would know, adjacent electron bond pairs repel, and it's this repulsion that gives ammonia an H-N-H bond angle of 107°.
Now fluorine, with an electronegativity of 3.98, does an even better job of pulling the N-F electron bond pairs towards it. Since its got a much greater electronegativity than hydrogen, fluorine's able to pull out the electron pairs even further. The consequence of this is that the N-F bonds can be pushed closer than N-H bonds before they're stopped by the repulsion of their electron bond pairs. This is what gives $\mathrm{NF_{3}}$ a smaller bond angle of 102°.

Both $\ce{NH3}$ and $\ce{NF3}$ molecules formed as the shape of pyramid (bonds pairs-3 lone pairs-1) But the angles between those molecules different to the each other because of the electronegativity of the each atom.
Electronegativity of each atom
Nitrogen-3.07
Hydrogen-2.1
Florin-3.98
Get the molecule $\ce{NH3}$. Actually according to the Folin's law nitrogen's electonegativity is more than the Hydrogen. Therefore the bond electron cloud tries to attract to the direction of the nitrogen atom. There for the whole electron cloud attract to the central atom. Then the each electron pair come closer to the each and because of that electrons tries to repeal.(When electrons come closer to the each other they try to repeal) And then the the angles been increased because of the repultion.
Then get the molecule $\ce{NF3}$. It also making a pyramidal shape. The electronagitivity of Florin is more than the Nitrogen. Therefore the bond pairs attract to the florin atoms, that means they attract to the terminal atoms. Then the distance between electon pairs increase, and not try to repulsion. Because of thar the angles be smaller.
*If the repulsion is very large electrons try to go far. Then the angle increase.
*If the repulsion is small electrons not try to go far. Then the angle is try to be bended.
Therefore the bond angle of $\ce{NH3}$ is more than $\ce{NF3}$.
Now think about the angle of $\ce{NCl3}$ also.


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Gas chromatography analysis - number of peaks?

This is an example in a Princetown Review prep book. I am not understanding the statement "his separation was only partial (because two different peaks are recorded)." This is implying that Compounds A and B did not separate, but um... there were two peaks? Doesn't that imply that it was separated?
Unless there is a solvent peak? But I did a lot of Googling, and didn't find any indication that there should be a solvent peak under normal circumstances.
Why did the example say that the 2 compounds weren't separated despite the fact that there are 2 peaks?

I think it means the distillation was unsuccessful. Hence two peaks in the GC whereas pure compound would have only one

The GC is the instrument used (in this case, although for very small amounts, there is no reason GC couldn't be used to separate materials - assuming collection after detection (via, say, refractive index - for obvious reasons flame ionization, etc. wouldn't be usable)) to determine if his primary separation technique was effective. It wasn't.


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Why do non-polar covalent compounds dissolve in non-polar covalent compounds only?
My textbook says that electrovalent and polar covalent compounds dissolve in electrovalent compounds and polar covalent compounds but not in non-polar covalent compounds. I understand the reason for this. It is because the oppositely charged moeities of the two compounds attract each other and the compounds thus dissociate and ionize into ions thus resulting in dissolution.
Now, it also says that non-polar covalent compounds dissolve in non-polar covalent compounds but not in electrovalent and polar covalent compounds. I do not quite understand the reason behind this.
I would like to know what dissolution is at the molecular level and why non-polar covalent compounds dissolve in non-polar covalent compounds only?

In a way, the two statements are identical if you imagine mixing of two liquids, one polar and one non-polar.  Fundamentally the reason for statement 2 is the reason for statement 1.

Can you edit your question to clarify what you understand the reasoning to be for polar covalent compounds to dissolve in polar covalent liquids? Since the processes in both systems are the same, your confusion might stem from a subtle misunderstanding. We can answer more directly if we knew your line of reasoning.

There is the oft quoted adage in chemistry, which goes something like:

like dissolves like

The reason for this can be understood on the basis of polar/ionic compounds have some degree of charge separation. In polar solvents, the partial charges on on the solvent molecules can interact with the partial charges on the solute molecules. These interactions lower the overall energy of the system, and thus dissolution takes place. This part you get, so it's fine.
First let's address why non-polar solutes don't dissolve in polar solvents. Consider the simple case, of a long chain hydrocarbon (solute) in water (solvent, present in larger excess). Typically, water molecules will form (and break) hydrogen bonds with each other. At any given time, a water molecule would like to maximise the number of hydrogen bonds it can form, simply because it is energetically favourable.
Remember that the water molecule, cannot form hydrogen bonds with the hydrophobic molecule; it can only do so with other water molecules.
So when a hydrophobic is dropped in an aqueous medium, hydrogen bonds between water molecules will be broken to make room for the hydrophobe; however, water molecules do not react with hydrophobe. This is bad energetically. So, water molecules would now try to adopt an arrangement that maximises the hydrogen bonding network; to do so they will arrange themselves in a "cage" like structure surrounding the hydrocarbon droplet. While, this allows the water to make more hydrogen bonds, it leads to overral "ordering" and thus decreases entropy. Again, bad. Thus to minimise this hydrocarbon droplets will clump together, and will be exluded by the solvent--this is why oil and water don't mix. (Read more here)
So what happens when non-polar compounds are tossed into non-polar solvents. True, there aren't any dipole-dipole, ion-dipole, etcetera kinds of interaction that you have with polar molecules, but you still have something called London dispersion, and non-polar entities often interact via these weak interactions.

There is a important point here. "Dissolves" is an ambiguous term. So as chemists we'd normally think that a mixture of benzene and water would form two exclusive phases, one benzene and the other water. The absolute truth of course is that in such a system a tiny amount of water would dissolve into the benzene, and a tiny amount of benzene would dissolve into the water.


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Why is ammonium a weak acid if ammonia is a weak base?
$\ce{NH3}$ is a weak base so I would have expected $\ce{NH4+}$ to be a strong acid. I can't find a good explanation anywhere and am very confused. Since only a small proportion of $\ce{NH3}$ molecules turn into $\ce{NH4+}$ molecules, I would have expected a large amount of $\ce{NH4+}$ molecules to become $\ce{NH3}$ molecules.

I'd rather avoid using the vague terms like "weak", "strong", "big", "small" without any numerical characteristic. They don't really mean anything. There are layers upon layers of different kinds of "small". See, you are very small compared to an elephant, and elephant is a good deal smaller than the Empire State Building, which in turn is so small it is barely visible from space. Ditto for "weak".

@IvanNeretin Strong and weak in terms of acids and bases is defined as $\mathrm{p}K_x > 0$ (weak) or $\mathrm{p}K_x < 0$ (strong) — at least that’s the definition that followed me through school and uni.

First, let’s get the definition of weak and strong acids or bases out of the way. The way I learnt it (and the way everybody seems to be using it) is:

$\displaystyle \mathrm{p}K_\mathrm{a} < 0$ for a strong acid
$\displaystyle \mathrm{p}K_\mathrm{b} < 0$ for a strong base
$\displaystyle \mathrm{p}K_\mathrm{a} > 0$ for a weak acid
$\displaystyle \mathrm{p}K_\mathrm{b} > 0$ for a weak base

Thus strong acid and weak base are not arbitrary labels but clear definitions based on an arbitrary measurable physical value — which becomes a lot less arbitrary if you remember that this conincides with acids stronger than $\ce{H3O+}$ or acids weaker than $\ce{H3O+}$.

Your point of confusion seems to be a statement that is commonly taught and unquestionably physically correct, which, however, students have a knack of misusing:

The conjugate base of a strong acid is a weak base.

Maybe we should write that in a more mathematical way:

If an acid is strong, its conjugate base is a weak base.

Or in mathematical symbolism:

$$\mathrm{p}K_\mathrm{a} (\ce{HA}) < 0 \Longrightarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) > 0\tag{1}$$

Note that I used a one-sided arrow. These two expressions are not equivalent. One is the consequence of another. This is in line with another statement that we can write pseudomathematically:

If it is raining heavily the street will be wet.
$$n(\text{raindrops}) \gg 0 \Longrightarrow \text{state}(\text{street}) = \text{wet}\tag{2}$$

I think we immediately all agree that this is true. And we should also all agree that the reverse is not necessarily true: if I empty a bucket of water on the street, then te street will be wet but it is not raining. Thus:

$$\text{state}(\text{street}) = \text{wet} \rlap{\hspace{0.7em}/}\Longrightarrow n(\text{raindrops}) \gg 0\tag{2'}$$

This should serve to show that sometimes, consequences are only true in one direction. Spoiler: this is also the case for the strength of conjugate acids and bases.
Why is the clause above on strength and weakness only true in one direction? Well, remember the way how $\mathrm{p}K_\mathrm{a}$ values are defined:
$$\begin{align}\ce{HA + H2O &<=> H3O+ + A-} && K_\mathrm{a} (\ce{HA}) = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}\tag{3}\\[0.6em]
\ce{A- + H2O &<=> HA + OH-} && K_\mathrm{b} (\ce{A-}) = \frac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}\tag{4}\end{align}$$
Mathematically and physically, we can add equations $(3)$ and $(4)$ together giving us $(5)$:
$$\begin{align}\ce{HA + H2O + A- + H2O &<=> A- + H3O+ + HA + OH-}&& K = K_\mathrm{a}\times K_\mathrm{b}\tag{5.1}\\[0.6em]
\ce{2 H2O &<=> H3O+ + OH-}&&K = K_\mathrm{w}\tag{5.2}\end{align}$$
We see that everything connected to the acid $\ce{HA}$ cancels out in equation $(5)$ (see $(\text{5.2})$) and thus that the equilibrium constant of that reaction is the autodissociation constant of water $K_\mathrm{w}$. From that, equations $(6)$ and $(7)$ show us how to arrive at a well-known and important formula:
$$\begin{align}K_\mathrm{w} &= K_\mathrm{a} \times K_\mathrm{b}\tag{6}\\[0.6em]
10^{-14} &= K_\mathrm{a} \times K_\mathrm{b}\\[0.6em]
14 &= \mathrm{p}K_\mathrm{a} (\ce{HA}) + \mathrm{p}K_\mathrm{b} (\ce{A-})\tag{7}\end{align}$$
Now let us assume the acid in question is strong, e.g. $\mathrm{p}K_\mathrm{a} (\ce{HA}) = -1$. Then, by definition the conjugate base must be (very) weak: $$\mathrm{p}K_\mathrm{b}(\ce{A-}) = 14- \mathrm{p}K_\mathrm{a}(\ce{HA}) = 14-(-1) = 15\tag{8}$$
Hence, our forward direction of statement $(1)$ holds true. However, the same is not true if we add an arbitrary weak acid to the equation; say $\mathrm{p}K_\mathrm{a} (\ce{HB}) = 5$. Then we get:
$$\mathrm{p}K_\mathrm{b} (\ce{B-}) = 14-\mathrm{p}K_\mathrm{a}(\ce{HB}) = 14-5 = 9\tag{9}$$
A base with a $\mathrm{p}K_\mathrm{b} = 9$ is a weak base. Thus, the conjugate base of the weak acid $\ce{HB}$ is a weak base.
We realise that we can generate a weak base in two ways: by plugging a strong acid into equation $(7)$ or by plugging a certain weak base. Since the sum of $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b}$ must equal $14$, it is easy to see that both cannot be strong. However, it is very possible that both the base and the acid are weak.
Thus, the reverse statement of $(1)$ is not true.
$$\mathrm{p}K_\mathrm{a}(\ce{HA}) < 0 \rlap{\hspace{1em}/}\Longleftarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) > 0\tag{1'}$$

Regarding the opposite direction it might be worth noting that it works with the negations. $$\mathrm{p}K_\mathrm{a} (\ce{HA}) \geq 0 \Longleftarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) \leq 0\tag{1b}$$  This is known as Modus Tollens.

@JoseAntonioDuraOlmos Yes, the reverse of the negation is always true afaik. So the conjugate acid of a strong base is a weak acid.

First, let's understand the perspective of weak acid and weak base. This is in relation to pure water — like in most general chemistry courses.
Pure Water has a $\mathrm{p}K_\mathrm{a}$ of 14.
$\ce{NH3/NH4+}$ has a $\mathrm{p}K_\mathrm{a}$ of 9.25.
We know:
$$\ce{NH4+ + H2O <--> NH3 + H3O+}$$
Thus, we solve for the $K_\mathrm{a}$ (which depends on the concentration of each of your chemicals):
$$K_\mathrm{a}  = \frac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]}$$
Similarly, we can solve for $K_\mathrm{b}$:
We know the ionization equation for a base is:
$$\ce{B + H2O <--> HB+ + OH-}$$
Which means:
$$\ce{NH3 + H2O <--> NH4+ + OH-}$$
So in order to solve for the $K_\mathrm{b}$, plug the concentration of your chemicals in:
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$
tl;dr: As the math shows, you can think of these as a "weak but not meaningless" acid/base. Don't let your assumptions throw you off — a weak acid generates a weak base, and vice versa. See the Henderson-Hasselbalch Equation.
$\ce{NH3}$ is not in the same class of weak bases as say, $\ce{Cl-}$. The acid to base curve isn't extreme like $\ce{HCl + H2O -> Cl- + H3O+}$.

One warning: if you have both acid and base in sufficient quantities, make sure to use the Henderson-Hasselbalch Equation, in addition to the above.

For example: pH = -log10(kA of NH4+) + log10([NH3]/[NH4+])

Why is there a need to use H-H under "different conditions"? This is not true because the H-H equation is simply a different restatement of the definition of the Ka (take log10 on both sides). As such it is entirely equivalent to the Ka equation you gave above and likewise the scope of use is exactly the same.

I only meant that the requester needs to be careful if he has other solutions in the mixture (e.g. this solution being a buffer for a strong acid or base). Apologies for the confusion.

In case a less abstract answer will help:
A base, like $\ce{NH_3}$, is a base because it has a significant chance of picking up protons in water.  You can almost think of it as a competition between the $\ce{NH_3}$'s and the $\ce{H_2O}$'s to pick up the free protons.
It is a weak base because it is not a certainty that all the $\ce{NH_3}$'s in a given sample will pick up a proton and hold onto it.  In any given sample at any point in time, a certain portion of the $\ce{NH_4^+}$'s ($\ce{NH_3}$'s that won the proton) will give their protons up and a certain number of $\ce{NH_3}$'s will pick up new protons. Eventually, the system reaches a steady state (quantifiable with $K_b$).
Our definition of an "acid" is just something that donates protons, as the $\ce{NH_4^+}$ does above.
It's not that they're really different things: the $\ce{NH_3}$'s a base when its picking up protons and its an acid when it lets them go.


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Why there is a temperature change in adiabatic process?
I am learning the topic thermodynamics and I came to know that in an adiabatic process the heat of the system remains constant. So is it a hard and fast rule that temperature will  change in an adiabatic process?

Just a quick pointer: although no HEAT is exchanged in an adiabatic expansion (or contraction), that does not mean that TEMPERATURE stays the same. For example, suppose that a gas expands by 1 liter against a pressure of 1 kPa. The gas has done 1 joule of WORK on the surroundings. Because we required that no heat be supplied back to the gas, the internal energy of the gas has just decreased. Since internal energy is related to temperature (in fact, it is directly proportional in ideal gases), the temperature must also decrease. Loosely, temperature decreased to provide the energy for work.

Remember adiabatic process implies process in which no heat exchange takes place between the system and the surrounding i.e. neither heat is supplied nor heat is loosed.
Temperature of gases changes only when internal energy of the gases changes.
Net heat of the system constant  means dq = 0
dq= du+dw, to maintain dq = 0, you can have many possible set of values of du and dw. Like whatever amount work you do, say dw, to maintain addition of du and dw always as zero, du = -dw. So any amount of work you do , du becomes negative of that work you perform to maintain net heat of the system as constant (i.e. dq as zero).
So even while maintaining addition of du and dw as zero, you can have infinite possible values of du and dw ( i.e. do any aribitary work and du becomes the negative of the work you performed and hence addition becomes always zero)
Temperature of the gas is related to du by du = nCvΔT. Hence for all those infinite possible values of du, you will have infinite possible temperatures, still maintaining constant net heat of the system .i.e dq as zero.
So in an adiabatic process, if you performed work, then du has to change to a value which is negative of the amount of work you performed (otherwise you won't be able to maintain their addition to be zero) , and as du has to change, hence temperature also changes.
So every time you perform work in an adiabatic container, temperature  has to change.


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Why can Wurtz reaction be not used for preparation of alkanes with odd number of carbon atoms?
My book says that Wurtz reaction can not be used for preparation of alkanes with odd number of carbon atoms. Why is it so?

https://en.wikipedia.org/wiki/Wurtz_reaction It can be but usually you'd get mixture of compounds.

It's always good to think of what would probably go on in the reaction vessel. Here's a mechanism of the Wurtz reaction:
First, one electron from the metal is transferred to the halogen to produce a metal halide and an alkyl radical.
$\ce{R–X + M → R. + M+X−}$
The alkyl radical then accepts an electron from another metal atom to form an alkyl anion and the metal becomes cationic. This intermediate has been isolated in several cases.
$\ce{R. + M → R^-M+}$
The last step is a $\ce{S_N2}$ reaction; the nucleophilic carbon of the alkyl anion displaces the halide, forming a new carbon-carbon covalent bond
$\ce{R−M+ + R–X → R–R + M+X−}$
Simple. Now, let's take a look at an example.
If we start with a single alkyl halide, in this case a methyl halide,
$\ce{2 CH3-X + Na ->[][-2NaX] CH3-CH3 }$
Now, let's consider two dissimilar alkyl halides, ethyl and methyl halide:
3 possible reactions take place, giving a statistical distribution of products.
$\ce{2CH3-X + Na ->[][-2NaX] X-CH3}$
$\ce{2CH3-CH2-X + Na ->[][-2NaX] CH3-CH2-CH2-CH3}$
$\ce{CH3-X + X-CH2-CH3 + Na ->[][-2NaX] CH3-CH2-CH3}$
Sure, you've produced an alkane with an odd number of carbon atoms, but at the same time you have a mixture of alkanes (this is not really a selective reaction), that you will have a hard time separating.


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On the unit of standard reduction potential and its independence of elements involved
Why is the standard reduction potential in units of volts (Joules/coulomb)? What does this tendency of a material to reduce has to do with the charge in coulombs?
Also, why this number doesn't depend on which elements the reducing material is interacting with? (which material gets oxidized)

I think that it is joule which is then  converted to electron volts . Thats it

A an electric potential (Volt) is equivalent to the unit of J/C. This is because if you move a charge through a potential you gain energy. Volt is kind of a mysterious unit and by the time people reach it the sub-units of J and C are really well understood so it often helps to make sense out of potential. In the same way you can understand speed by thinking of it as distance/time.
The standard reduction potential can be thought of as the potential produced if the reaction happens (if positive) or required to make it happen (if negative). It is determined usually by measuring the total potential (of both reactions) with one half reaction being the one you are interested in and the other which has a zero potential so the standard reduction you are looking for is then just the net potential.
When you have two of those reactions which are both non-zero, the net potential does depend on both.
There is an excellent article on it at CLT :
Standard Reduction Potential


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What is the difference between an ion-induced dipole attraction and just an ion-dipole attraction?
By what I've read, an ion-induced dipole attraction is a weak attraction that results when the approach of an ion creates a dipole in a non-polar atom or in a non-polar molecule by disturbing the arrangement of electrons in the non-polar species. So this is a dispersion/van-der-Waal force, I believe?
On the other hand, an ion-dipole attraction is a weak attraction that results when the approach of an ion interacts with the dipole in a polar atom or in a polar molecule. And so this wouldn't be a van-der-Waal force would it?

If what I said above happens to be true, does the same apply with dipole-induced dipole forces and just dipole-dipole forces?
In this scenario, a dipole-induced dipole attraction would be a weak attraction that results when the approach of an dipole (polar covalent bond) creates a dipole in a non-polar atom or in a non-polar molecule by disturbing the arrangement of electrons in the non-polar species. So this is a dispersion/van-der-Waal force too, right?
And a dipole-dipole attraction would be a weak attraction that results when the approach of an dipole that already exists  in a polar bond attracts with another dipole in another polar atom or in polar molecule.

It is useful to note the definition of a van der Waal force: weak electric forces attracting neutral molecules to each other in gas thus causing a gas to deviate from an ideal gas.
So, there are three types of these forces

Dipole-dipole force
Dipole-induced dipole force
London dispersion force

It becomes clear that forces involving ions do not fit van de Waals forces as the molecules involved must be neutral.
An ion-induced dipole force is not a dispersion force, either (as dispersion forces are a subset of van der Waals), and with ion-dipole forces, ion-induced dipole forces have a category of it's own
A dipole-induced dipole force is a van der Waal force but not a dispersion force

So what exactly is a London Dispersion force? I've been reading that it is sort of like a "Induced Dipole-Induced Dipole" interaction, where if neutral atom "x" is induced by a permanent dipole, then neutral atom "y" which next to "x" also gets induced by "x's" temporary dipole. Then neutral atom "z" gets induced by "y's" temporary dipole. Kind of like a chain reaction? If not, then what is it?

An asymmetric distribution of electrons by chance can cause a temporary dipole, causing a slight positive and slight negative end, and the London Dispersion Forces involves the attraction of such atoms with dipoles. So it is an induced dipole-induced dipole, but not by the mechanism you described


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How to evenly ionize in mass spectrometry
I know that in mass spectrometry they charge different molecules and then run them through a magnetic field and then see where they land.
Depending on this, they calculate their masses relative to each other. To do this however, they have to ionize the different particles evenly. If one molecule is charged +2 and the other +1, then the results would be thrown off.
How do they do they make sure that the charges of all the ions are the same?

You can't. The ion's path is bent according to its mass to charge ratio.

I know that in mass spectrometry they charge different molecules and then run them through a magnetic field and then see where they land.

That is indeed how some mass spectrometers work, but not most of them.  Most mass spectrometers in the world do not use magnets of any type.  Instead they rely on electric fields to move the ions (i.e. charged molecules) and to see where they land.

Depending on this, they calculate their masses relative to each other. To do this however, they have to ionize the different particles evenly. If one molecule is charged +2 and the other +1, then the results would be thrown off.

Multiple charge states definitely happen in mass spectrometry.  Mass spectrometrists like to say that name "mass spectrometry" is a misnomer, since nearly every "mass" spectrometer in the world is really a mass-to-charge ratio spectrometer.  That is, MS doesn't measure the mass of molecules, it measure the mass-to-charge ratio (or $m/z$) of an ion.
The problem of multiple charge states is especially common in the mass spectrometry of proteins, because proteins are very large molecules with many sites that can become charged.  For example, the image below shows the "mass" spectrum (i.e. $m/z$ spectrum) of horse myoglobin, which is a compound with the formula of $\ce{C769 H1212 N210 O218 S2}$.  The spectrum was obtained from this web page which further describes the multiple charge state phenomenon.
The different peaks correspond to different charge states.  The uncharged molecule has a mass of 16951 Daltons.  The peak at an $m/z$ 737.99 corresponds to a +23 charge state, and the peak at 1542.02 corresponds to a charge state of +11.  The peaks in between correspond to charge states from +22 to +12.

How do they do they make sure that the charges of all the ions are the same?

There is no foolproof way to do this chemically or physically.  Some modes of ionization, such as electron ionization or APCI tend to favor the formation of singly-charged ions relative to electrospray ionization, but these modes of ionization are sadly ineffective on large molecules such as proteins.  And there are many exceptions to this tendency.
The mass spectral data can be manipulated computationally through a process called deconvolution in an attempt to computationally estimate the true mass spectrum at a given point.  A 1992 paper describes one computational method for spectral deconvolution.  This and/or very similar algorithms are often provided by mass spectrometer vendors in their software for browsing and analyzing MS data.

Thanks so much! I was puzzled by this question for quite sometime and even my teacher couldn't answer. Thanks! It really makes sense now.


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Density of cyclic hydrocarbons
Why is the density of cyclic hydrocarbons more than that of their acyclic counterparts?
For ex:
Cyclopentane: 751 kg/m³
Pentane: 626 kg/m³

Cyclohexane:779 kg/m³
Hexane:655 kg/m³

In fact, as you can see above, cyclopentane is even denser than hexane.
My reasoning so far is:
Attractions in hydrocarbons depend only on London forces, which increases with surface area.
An alkane will have more surface area for bonding as compared to a cycloalkane (The inside of the ring cant be used), so it will have more forces, and hence will be packed closer (be denser).
But this contradicts the facts.

Very relevant discussion in this question.

A simple explanation might be that the ratio of carbon to hydrogen is higher in the cyclic hydrocarbons, and carbon is a heavier element than hydrogen.


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Chemical Kinetics - change to system conditions
This is a question from a very old AP Chem exam (2006) prep book which I am amusing myself with in my spare time which is kind of humbling. There is something I am missing with this question, one of several similar questions in the kinetics section.

Given the following reaction (which doesn't establish an equilibrium) :
$\ce{A (g) + B (g) -> C (g)}$
which of the following would increase the concentration of $\ce{A}$?

addition of $\ce{A}$ at constant $T$ and $V$
addition of $\ce{A}$ at constant $T$ and $P$
addition of $\ce{A}$ at constant $V$ and $P$

Now my first thoughts are to reason as follows :
If you add $\ce{A}$ and the volume is constant obviously the concentration of $\ce{A}$ increases. If you add $\ce{A}$ and keep $T$ and $P$ constant the $V$ will just expand and the concentration will be the same. But if I ignore the reaction this is a gas law / concentration question so it is out of place. But it isn't an equilibrium so none of that reasoning applies either.
Now if the reaction happens so slow it takes a year for $1\,\%$ of $\ce{A}$ to react I could just consider the reaction not even happening? If the reaction happens so fast that it is essentially over in 1 second then it seems like any change to A would be undone as fast as you could increase it. But it seems like I am missing something basic as there is nothing given regarding the rate.
There are several questions almost identical to this in the same section which just have other than $1:1$ coefficients in the equation.

It's not really kinetics, but concentration question, it seems.

As you said, your statement "If you add A and the volume is constant obviously the concentration of A increases" indeed does not apply.  In fact, the number of mols of A never changes.  Add a mol of A, and it reacts irreversibly with B and is gone forever as part of C.
This means that the total volume must decrease for the concentration to increase, which means that "addition of A at constant T and P" must be correct, as it is the only answer that allows for a decrease in volume and therefore an increase in the concentration of A.
An assumption made here that is not explicitly stated in the question, is that we never run out of B.


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Most stable conformational isomer of 3-methoxycyclohexan-1-ol

Among the following, the most stable isomer is?

I am aware of the fact that equatorial substituents are more stable than axial substituents but couldn't proceed to apply it here. However the answer key gives the answer as d) in which both substituents are in axial position.

Note that there is no space between the substituent prefixes (3-methoxy) and the parent compound (cyclohexan-1-ol). Furthermore, (b) and (c) are diastereomers, not different conformers, of (a) and (d). Anyway, the answer in three words is "intramolecular hydrogen bonding".

@orthocresol There would be a lot of steric hinderence too if both substituents are on the same side.

Not steric hindrance, but steric repulsions. The word "hindrance" means that something is being blocked, such as an incoming nucleophile in an SN2 reaction. Nothing is attacking anything here, and consequently nothing is being blocked here. So, the word "hindrance" doesn't apply. Anyway, the question really boils down to whether the destabilising steric repulsions are bigger, or the stabilising intramolecular H-bond is bigger. The answer is you probably need a computer to work that out, but in this case the H-bond is more important.

On top of that (b) and (c) are the same isomer/conformer...?

@orthocresol yes b) and c) are the same conformers.it must have been misprinted in the book.Sorry for that.

@orthocresol Maybe, the originally intended question was about all combinations: (1R,3S)-, (1S,3S)-, (1S,3R)-, and (1R,3R)-3-methoxycyclohexan-1-ol. However, now there are two (1S,3S) isomers, and the (1R,3R) isomer is missing.

The questions in the title and in the text body are not strictly identical. Is the question about the most stable of the shown stereoisomers of 3-methoxycyclohexan-1-ol with any possible cyclohexane conformation, or is the question about the most stable of the shown isomers with precisely the shown cyclohexane conformation?

I would say that the answer to the question depends strongly on the solvent used.
In case anybody still doesn’t see it: in the orientation (d), the compound can form an intramolecular hydrogen bond from the hydroxy group to the methoxy group. This is especially favourable in solvents that cannot participate in hydrogen bonding, e.g. dichloromethane.
Dissolving the same molecule in methanol, however, could change the entire story. The intramolecular hydrogen bond is favoured in the absence of other hydrogen bond donors or acceptors, but in a hydrogen bonding solvent there is absolutely no shortage of donors and acceptors and it can be assumed that all sites that can participate in hydrogen bonding in any way are saturated with hydrogen bonds. At this point, the steric interaction probably becomes more important and I would assume that the molecule preferentially assumes a diequatorial configuration.

Sometimes, the intuitive rule of thumb that favours conformations with $\ce{OH}$ substituents in equatorial positions miserably fails.

For 1, a cyclization product of geranyl acetate, one might assume that the favoured conformation in solution and in the crystal is 1eq. However, this is not the case! Both NMR spectroscopy and X-ray crystallography indicate that 1ax is the actual conformation.
In the case of 1, the reason most likely isn't a (stabilizing) intramolecular hydrogen bond that would form a seven-membered ring, but the sterical interaction of the methyl groups. The unfavourable interaction of the 1,3-diaxial methyl groups in 1eq is avoided in the conformer with the axial $\ce{OH}$ group.

So why is this the case? Has it something to do with a 1,3-diaxial hydrogen bonding of the hydroxyl and ester carbonyl group?


text:
Oxidation of styrene with permanganate
My teacher told me that when we oxidise styrene with $\ce{KMnO4}$, it will stop at benzaldehyde. But I think that it will continue to go on till benzoic acid. Am I right?

In the heterogenous permanganate oxidation of styrene, that is when the reaction is performed in dichloromethane with potassium permanganate on a solid support (such as aluminium oxide), the reaction stops at the stage of the aldehyde.
The method was published in Synthesis 2001, 11, 1645-1648 and the authors report a yield of 90% for the oxidation of styrene to benzaldehyde.


text:
What will be the final product when the above bridged compound is heated with concentrated sulfuric acid?

What will be the final product when the above compound ((1S,2S,4R,7S)-7-methylbicyclo[2.2.1]heptane-2,7-diol) is heated with concentrated sulfuric acid?
I know that the hydroxyl group in the bridgehead position cannot be protonated and removed to form a carbocation. The other hydroxyl group may be protonated and removed to form a carbocation I suppose. But after that what rearrangements will take place ? What will be the final product(s)?

There isn't any bridgehead hydroxyl in your compound.

Please have a look at 2-norbornyl cation and/or search for non-classical cation on this site.

@orthocresol I meant the OH on the top of the compound (in digram). That cannot be removed by protonation as a positive charge will be left behind on a non-planar center.

Why can't that carbon be planar? It's not a bridgehead, it's a bridge.

I fail to see why elimination from the tertiary alcohol should not be possible. The resulting cation could either add a nucleophile, which would result in the epimerization of that centre, or undergo elimination. The latter would yield an exomethylene group.

@orthocresol Okay, suppose I form a positive charge on the bridge, after removing OH, then what happens ? Can you tell me the steps after that ?

@KlausWarzecha I read the Wikipedia page you linked. Now my doubt is : Which OH will be protonated and removed first ?

@orthocresol Also, how do we know which OH will be removed first ? Will the tertiary OH be removed first as the tertiary carbocation is more stable ? And what next?

The carbocation resulting from the removal of hydroxyl group will not assume a planar structure,however pyramidal structure is possible but in my view the steric repulsions will outweigh the effect of the stabilising effect of a tertiary carbonation and hence the secondary hydroxyl group will be eliminated easily.

@aniline What can't it assume a planar structure ? Which steric effects are you talking about ?

Oh the hydroxyl group is in the bridge not on the bridgehead.ignore my previous comment.

@aniline Do you know how to get the final product? The steps?

Hmm... I wonder if forming the epoxide here is viable...

Protonation of both $\ce{OH}$ groups and subsequent formation of the corresponding cations is possible.

The tertiary cation will yield an exomethylene group
The "secondary" non-classical cation leads to a $\ce{C=C}$ double bond in the ring. With other words, a norbornene is formed.

The combined eliminations will give 7-methylene bicyclo[2.2.1]hept-2-ene as the product.
In J. Am. Chem. Soc., 1973, 95, 7801-7813, the same product was obtained by elimination from the tosylate of 7-hydroxy-7-methylnorborn-2-ene.


text:
At the freezing point, why is the vapor pressure of the liquid phase equal to the vapor pressure of the solid phase?
At the freezing point of a substance, the solid phase is in dynamic equilibrium with the liquid phase. My textbook defines the freezing point of a substance as: "the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase". Why is this so? How does the vapor phase even come into the picture during freezing? Shouldn't it be all about the solid and the liquid?
My book also states that when we add a non volatile solute to a liquid solvent, "the solution will freeze at the temperature when its vapor pressure is equal to the vapor pressure of the pure solid solvent". This doesn't make sense to me as well. Assuming the first statement is true, shouldn't the second statement read, "...is equal to the vapor pressure of the solid solution"? Are the two statements somehow related to each other?And why are they true?

Vapor per se does not really come into the picture, except maybe in some specific setup. It is that we mention vapor pressure as a measure of activity of the substance.

I find the first statement to be a bit of a stretch. It is a certain feature of any substance that the vapour pressure of solid and liquid are equal at the freezing point. The inverse statement is only true because in equillibrium, water does not exist below 0 °C, and neither does ice above. Might even turn out false if you look at a supercooled liquid.

And the second statement is overstretched so far it turns wrong. The actual freezing is a process that is strongly kinetically hindered, to state that it will happen at a certain temperature only from thermodynamic reasoning is nonsense. Ah, and the point is that you will not get a solid solution but nearly pure frozen solvent.

The vapour pressure comes in because via Raoult's law the activities (or free energy /mole) are equal in the crystal and liquid at the freezing point. In a solution on cooling the solvent separates out as a crystal while the solute remains in solution being insoluble in the crystallised solvent. When at a temperature T the solution is in equilibrium with crystal T is the freezing point and the chemical potentials of the two phases (crystal/solution) are equal.

If the vapor pressure of the substance in its liquid phase would be different from the vapor pressure of its solid phase then there would be a transfer of mass from the phase with the higher vapor pressure to the phase with the lower vapor pressure.
This would involve releasing or consuming energy (heat of fusion) even in a closed system in thermal equilibrium, which is obviously a contradiction.


text:
Enantioexcess for a non-chiral molecule?
I am trying to understand this quote in a section on "Biotransformations of Organic Molecules" from a handbook I am reading:

[...] Euglena gracilis could convert acetophenone to 2-phenylethanol; however, its enantioexcess is very poor (10%) (Takahashi, 1994).
Ref.: Takahashi, H., 1994. Biotransformation of terpenoids and aromatic compounds by some microorganisms. PhD thesis. Tokushima Bunri University, pp. 1–115.

What does "enantioexcess" mean in this context? Neither of those molecules are chiral, right?  I cannot spot any asymmetric C atoms. Nor any higher order symmetries.
I looked up Wikipedia entries on both molecules and neither mention any d / l isomers.
What am I missing?

Scifinder search for "Euglena gracilis acetophenone" doesn't bring up anything except for one Chinese patent, CN 103757061, which suggests that the product is 1-phenylethanol.

It should have been 1-phenylethanol, I guess. Look at the molecules again. Suppose you reduce the carbonyl group; where would be your hydroxyl?

I didn't read the patent in detail because technical documents in different languages are really difficult to read. However I did spot that the reduction product is (S)-1-phenylethanol. (These authors also got >99% e.e., so...) Anyway, I would assume it is a typo and the authors meant 1-phenylethanol, because that is the product of direct reduction of acetophenone and it is chiral.

@orthocresol That makes a lot of sense. I think you are right and the Handbook or the original author got his ones & twos mixed up. I was just wondering if talking about an enantioexcess in the context of 2-phenylethylalcohol made any sense. I think not.

@curious_cat You think right; ee in non-chiral molecules wouldn't make any sense.

The Ph.D. thesis that is cited is probably near-impossible to obtain and may very well be in Japanese, which certainly doesn't help.
2-Phenylethanol is not a chiral compound, so as Ivan pointed out, the concept of enantiomeric excess (ee) makes no sense. The fact that the direct reduction product of acetophenone (e.g. via hydride transfer) is 1-phenylethanol strongly suggests that this is nothing more than a typo. However, I thought that it might be fun to go to great lengths to verify the applicability of Occam's razor to this situation.

A SciFinder search for "acetophenone Euglena gracilis" did throw up one Chinese patent, CN 103757061. There is a translation on Google Patents. The patent describes the reduction of various prochiral ketones with various microphytes (also known as microalgae) to form chiral alcohols, with near-perfect enantioselectivity.
I managed to figure out the technical terms in the patent. E. gracilis in Chinese appears to be 纤细裸藻 (xian1 xi4 luo3 zao3). Acetophenone is 苯乙酮 (ben3 yi3 tong2). The patent does not, in fact, describe the reduction of acetophenone by E. gracilis, and the text in Table 2 is so small that I cannot quite make out what compounds they tested (they seem to be aliphatic). Perhaps they simply did not include this combination in the patent if the enantioselectivity they obtained was very low.
However, there are experiments with different species that lead to the reduction of acetophenone to "S-苯乙醇" (ben3 yi3 chun2). This translates to (S)-phenylethanol, and leaves open the possibility of it being either 2-phenylethanol or 1-phenylethanol. Since only the 1-phenyl isomer is chiral, I assume the reduction product is (S)-1-phenylethanol.
This is supported by Figure 1 in the patent. I have scribbled some loose translations. In hindsight I think "Photosystem" should be "Light-harvesting complex". The text translates to "Light-harvesting system" which is a bit of a mix and match of the two, so I wasn't sure what to go for.

Having $\ce{R_1 = Me}$ and $\ce{R_2 = Ph}$ would mean acetophenone is reduced to 1-phenylethanol.
In conclusion, I would therefore assume with confidence that the handbook contains a typo and the authors meant 1-phenylethanol.
(On a side note, the Chinese also got >95% ee for all their acetophenone reductions, go figure...)

It’s always good to have a Chinese around who can answer these types of questions <3


text:
Which stabilizes a carbocation more : Back Bonding or Conjugation?
Suppose we take these two compounds:

The first carbocation can rearrange to produce the second one by hydride shift. Which is more stable ? Why ?

The compound to the left seems like a carbonyl in disguise. How did you get there?

@Variax Hydride shift.

@Variax http://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/1,2-Hydride_Shift

What I meant is how did you obtain the first carbocation in the first place?

By heating a diol with concentrated sulfuric acid and then a hydride shift. I don't think it matters how I got the carbocations. I just want to compare the relative stabilities of the two interconvertible structures.

I am inclined to consider the second one as a "bona-fide" carbocation, if that makes sense. The first one resembles a protonated aldehyde rather than a carbocation, so relatively speaking, the first one should be more "stable".

The question deserves a more thorough answer than ‘I think’.


text:
On baker nathan hyperconjugation
Recently I have been reading about the history of the Baker Nathan reaction.

In addition to the large
  rate increase for Me vs H, all the other alkyl groups
  showed decreases with respect to Me in a regular manner
  as would be predicted by reduced hyperconjugative
  ability.

By drawing resonance structures, it is easy to see the $\mathrm{Me < Et < {}^iPr <} {}^t\mathrm{Bu}$ predicted by the inductive effect:

Therefore the more electron donating (i.e. the stronger the I+ effect) the para group is , the more stabilized the transition structure is (black $\delta^+$ on the left diagram become smaller hence less electrostatic repulsion in the $\pi$ system, hence blue $\delta^+$ and the $\delta^-$ can be larger thus facilitating the reaction). Hence $\mathrm{Me < Et <} {}^i\mathrm{Pr <} {}^t\mathrm{Bu}$.

However, for hyperconjugation, the $\sigma$ bonds overlap with the $\pi$ system resulting in electrons to become more delocalised at the C-C bond between the alkyl group and the $\pi$ system. Therefore the more hyperconjugations present, once again the more stablised the transition structure hence a faster reaction.
Since both $\sigma_\mathrm{CC}$ and $\sigma_\mathrm{CH}$ bonds adjacent to the para carbon can potentially engage in hyperconjugation (and naively the $\sigma_\mathrm{CC}$, being larger in lobe size, should expect to overlap better with the $\pi$ system), does that mean the experimentally observed trend that $$\mathrm{Me > Et > {}^iPr >} {}^t\mathrm{Bu}$$
suggest hypderconjugation of $\sigma_\mathrm{CC}$ is much weaker than that of $\sigma_\mathrm{CH}$ hence justifying the quoted message?

If hyperconjugation of $\sigma_\mathrm{CC} < \sigma_\mathrm{CH}$ what is the underlying molecular orbital explanation?


text:
How can I calculate the percentage of dissociated dinitrogen tetroxide?

When $\pu{36.8 g}$ $\ce{N2O4(g)}$ is introduced into a 1.0-litre flask at $\pu{27 °C}$ the following equilibrium reaction occurs: $\ce{N2O4(g) <=> 2NO2(g)}$; $K_p = 0.1642~\mathrm{atm}$.
Find the percentage dissociation.

I tried the following:
Let $x$ be the degree of dissociation and $P$ be the initial pressure of $\ce{N2O4}$  that is $\approx 9\ \mathrm{atm}$.
Then, $$\frac{9Px^2}{1-x} = 0.1642$$
Neglecting $x$ in $(1-x)$, I got $x$ as $\displaystyle \frac{0.4}{3}$ [approx]
But the answer given as $6.44~\%$.

$\ce{N2O4 (g)}$ say initially is $P\ \mathrm{atm}$.
$$PV = nRT$$
$$P(1)= (36.8/92)( 0.082 )(300 K)$$
$$P= 9.84\ \mathrm{atm}$$
Now we have:
$$\ce{N2O4(g) <=>2NO2(g)}$$
Considering initial pressure of N2O4 as p, at equilibrium, assuming 'x' as the degree of dissociation, we will have pressure of N2O4 to be p(1-x) and pressure of NO2 as 2px.
So, $$K_p = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$
$$K_p= \frac{(2px)^2}{p(1-x)}$$
$$K_p= \frac{4px^2}{(1-x)}$$
assuming $1-x$ approximately as $1$,
$$K_p= 4px^2$$
substituting $p$ as $9.84\ \mathrm{atm}$ and $K_p$ as $0.1642$:
$$0.1642= 4(9.84)x^2$$
$$x^2= \frac{0.1642}{4(9.84)}$$
$$x^2= 0.1642/39.36$$
$$x^2= 0.00417$$
$$x= 0.0645$$
degree of dissociation is 0.0645
Hence % of dissociation is $0.0645 \times (100) = 6.45\ \%$ (near to your answer)

Please use units. 2) I don’t get your answer. 3) Please use units.

I have edited my answer to make it more understandable for the viewers.

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text:
Why are neutrino/anti-neutrino released in Beta decay?
In positive beta decay neutrino are released whereas in negative beta decay anti neutrino are released. From where did they came? Are they produced during the collision? What s it's purpose? Is it to conserve the linear momentum or angular momentum or energy distribution of positrons?

Neutrinos come from the elementary nuclear reactions:
$$n^0\to p^++e^-+\overline\nu$$
or
$$p^+\to n^0+e^++\nu$$
Not sure what do you mean by collision, since beta decay normally does not include any collision (unless there is an electron capture, of course; but neutrino emission occurs anyway, capture or no capture). As for the purpose, they have none, much like other things in nature. Well, by simply being there and having some energy and other things, they somehow participate in conservation of these things. In fact, before the neutrinos were first detected, they brought about a great deal of confusion among physicists by causing an apparent violation of energy conservation law (which ultimately led to their discovery).
Some might say that neutrino's "purpose" is to conserve the lepton number, but that's a post factum explanation.


text:
What type of bond forms between an ionic compound solute and a polar solvent
I understand the idea that the solvation enthalpy (ΔH-) must exceed the lattice enthalpy (ΔH+) for a solution to be more energetically stable, but what type of bond is formed between the disassociated ions and the the polar solvent molecules? Surely it must be even stronger than the ionic bonding in the lattice if it releases more energy than the energy needed to break the ionic bonds in the ionic compound.

https://www.chem.purdue.edu/gchelp/liquids/iondip.html see also: Why do salts such as NaCl dissolve?

Firstly, There is no such bond existing inbetween the polar solvent molecule and the solute ions. It is ion-dipole interaction which exists between the polar solvent molecules and the ions of the solute.
I hope you got your answer.
I wanted to add something more which I think you got wrong.
I agree that if hydration enthalpy is greater than lattice enthalpy then solute will dissolve but it is not compulsory for hydration enthalpy to be greater than lattice enthalpy for dissolving. What i mean is: For dissolving its important to break the ionic bonds between the ions of the solute molecule but that energy required to break the  ionic bond is not necessarily from hydration only. That is sometimes hydration enthalpy couldn't meet the energy required to break the ionic bonds (lattice energy). In those cases the extra amount of energy can be from heat of the solvent. In the cases where hydration enthalpy is larger than lattice energy, ΔH is negative. In cases where hydration enthalpy is lesser than lattice energy, in those cases the more energy needed is driven from the heat of the solvent and in this case ΔH becomes positive. Negative ΔH is preferred to be better for dissolving because it helps in making ΔG more negative. But it doesn't mean solute with positive ΔH doesn't dissolve. If ΔS is highly negative, then even though ΔH is positive still it can make ΔG negative and hence dissolving possible. Dissolving of any solute depends on ΔG. For dissolving, ΔH doesn't matter as long as ΔG is negative. Say for example in case of Nacl, net solvation enthalpy is positive, but still it dissolves in water.
Nacl hydration enthalpy is not greater than enthalpy required to break the bonds in between Na+ and Cl-. As hydration enthalpy is not sufficient to release energy which is required for bond breaking incase of Nacl, hence the extra amount of energy required comes from the heat of the water ,which in addition to the hydration enthalpy is able to break the bonds in between Nacl. As energy released is less than net energy absorbed, hence net enthalpy (solvation enthalpy) is positive.The reason behind the dissolving of the salt is ΔS being highly positive which in turn ends up making ΔG negative as ΔG= ΔH −TΔS.
So for a well dissolving ionic solute, hydration enthalpy is always not greater than the energy required to break the ions apart in the solute molecule. An ionic solute with positive ΔH would still dissolve if ΔG is negative.


text:
How to derive if the reaction is endothermic or exothermic based on rate constants at different temperatures?

Listed in the table are forward and reverse rate constants for the reaction $$\ce{2NO (g)  <=> N_2 (g) + O_2 (g)}$$
  \begin{array}{ccc}
\text{Temperature}/\pu{K} & k_f/\pu{M^-1s^-1} & k_r /\pu{M^-1s^-1} \\\hline
1400 & 0.29 & 1.1\times10^{-6}\\
1500 & 1.3  & 1.4\times10^{-5}\\
\end{array}
  Is the reaction endothermic or exothermic?

In my book the solution is given as

$k_r$ increase more than $k_f$, this means that $E_a$ (reverse) is greater than $E_a$ (forward). The reaction is exothermic when $E_a$ (reverse) > $E_a$ (forward).

I can not understand what they mean by the given answer.
Can anybody explain it in more detail?

My hint is use the van't Hoff equation.

@Zhe I am not understanding how to use that

The van't Hoff equation tells you what direction the equilibrium will shift as you change the temperature. It depends on whether the reaction is endothermic or exothermic. For endothermic reactions, as you increase temperature, the equilibrium shifts more towards products. The opposite is true for exothermic reactions.

So @Zhe can we write -$\Delta$H =E$_a$(forward) - E$_a$(reverse)

https://i.stack.imgur.com/GiJ0M.jpg

@orthocresol so the answer of this should be exothermic . Am I correct

@user123733 Seems fine, but computing $E_{a}$ is none trivial. Are you assuming the same prefactor? For what it's worth, there's no reason to calculate $E_{a}$. It's much easier to just look at the rate constants.


text:
Why is Helmholtz free energy often defined as the useful work attainable at constant volume if constant volume implies no expansion?
From Wikipedia:
"In thermodynamic systems where the temperature and volume are held constant, the measure of useful work attainable is the Helmholtz free energy function"
https://en.wikipedia.org/wiki/Work_(thermodynamics)#Free_energy_and_exergy
By definition, constant volume means that no PV work (expansion work) is done by the system. Because of this, is it not confusing to define Helmholtz free energy as a measure of the maximum useful/non-expansion work attainable at constant volume and temperature rather than just the maximum work attainable at constant volume and temperature, since all work done at constant volume is inherently useful/non-PV work? Adding the "useful" makes it seem like the useful work is separate from work done in general for a constant volume process. Am I missing something?

I don't understand. Isn't that the same thing?

No.  You are not missing anything.  There are other kinds of work besides PV work, such as electrical or work in driving a shaft within a constant volume system.

What I mean is that adding the "useful" is confusing because all work done at constant volume is inherently useful because constant volume means no PV work.

I see your issue. But you can't just change the definition of "work" which is not inherently useful in general. Even though it is in this case, we might add the qualifier "useful" to disambiguate and be as clear as possible. But I agree, it's redundant given the context.


text:
There's an absolute zero, is there an "absolute hot"?
This might be more of a physics question, but is there a ceiling on how hot things can get?
What happens at this temperature?

No, there is not. Kinetic energy of particles can be made arbitrarily high.

@Ivan Neretin. Are you sure? Isn't the velocity limited to the velocity of light? So that would be an upper limit, though the energy would be very high indeed. I found some online resource (see https://futurism.com/science-explained-hottest-possible-temperature/) where they say the maximum temperature is $\mathrm{10^{32} K}$, which is indeed very high. Though they don't give a real explanation as to how they obtained that value. I'm guessing: calculate the kinetic energy a particle obtains at velocity of light. That should be a good first bet for the maximum temperature achievable...

@IvanNeretin For all practical means, there isn't a real upper temperature limit though.

@deusexmachina Yes I am quite sure. Kinetic energy of any "real" particle at velocity of light equals infinity. Then again, there might be some "Planck temperature" at which the very fabric of universe will bend, and fundamental laws as we know them will turn into something entirely different. But that's another story.

@IvanNeretin I see. Still there is a limit to temperature then.

@deusexmachina That's not a limit to temperature. That's a limit to our knowledge.

@Jan I understand your point, though I think that knowledge about temperature is key to a good understanding of chemistry and this also includes knowing about the upper and lower temperature limits. In fact, I could expect this question to come up quite often while teaching chemistry, e.g. physical chemistry lectures where we talk about the absolute temperature zero a lot. So I think this question is at least somehow relevant to chemistry...

Vsauce made a video on this topic. https://youtu.be/4fuHzC9aTik

In the actual theories of physics the highest temperature which has a physical meaning is the Planck's temperature.
$$T_\mathrm{P} = \frac{m_\mathrm{P} c^2}{k} = \sqrt{\frac{\hslash c^5}{G k^2}} \approx \pu{1.4e32 K}$$
For the moment no theory predict higher temperature because of the limit of our theories.

There is a Wikipedia article about absolute hot with some references. You must have a look.

Contemporary models of physical cosmology postulate that the highest possible temperature is the Planck temperature, which has the value $\pu{1.416785(71)e32}$ kelvin [...]. Above about $\pu{10^{32} K}$, particle energies become so large that gravitational forces between them would become as strong as other fundamental forces according to current theories. There is no existing scientific theory for the behavior of matter at these energies. A quantum theory of gravity would be required. The models of the origin of the universe based on the Big Bang theory assume that the universe passed through this temperature about $10^{−42}$ seconds after the Big Bang as a result of enormous entropy expansion.

This Wikipedia article got deleted... and it's not very surprising.

It depends on what you mean by ceiling. Are we talking about a practical or theoretical limit?
At a high enough energy, the stress-energy tensor will be large enough that you're going to make a black hole. I'm not sure we understand the astrophysics well enough to know what this will look like in the limits you refer to.
Also, at some temperature, you're going to mess up the separation of forces, and fundamental forces are going to start to merge again, like in the primordial mix present in the brief moments after the big bang. What does this look like? No one knows for sure.

Depends on what you mean by "temperature".
In statistical mechanics, a system of interacting parts is in thermal equilibrium if the probability of finding a given part in a state with energy $E$ is proportional to $e^{aE}$ for some constant $a$ that is the same for all of the parts. Usually, $a$ is negative, and this becomes a Boltzmann distribution if we declare the temperature of the system to be the $T$ such that $a=\frac{-1}{kT}$.
If the system has an upper bound of the energy of possible states, there may be equilibria where $a$ is zero or positive, which formally corresponds to "infinite temperature" or "negative temperature". It turns out that $T=\infty$ is hotter than finite $T$, and negative $T$ is hotter yet -- in the sense that if you let two systems which are internally in equilibria with $T$s of different signs interact, then energy will flow from the negative-$T$ one to the positive-$T$ one.
My understanding is that this can be observed experimentally with systems of spin states whose interaction with their environment is slow enough that they have time to reach a (good approximation of a) Boltzmann equilibrium internally.
In such a system the "hottest possible temperature" is reached when its energy is maximal, in which case all the parts must be in their most energetic state at the same time. For this macrostate, the best value to assign to $T$ would be "$-0\,\rm K$".

From a mathematical point of view, it might perhaps be neater to measure the equilibrium in terms of $a$ rather than $T$. Then the direction of the scale would make more sense: $a=-\infty$ corresponds to $T=+0$, as the system gets hotter we move towards $a=0$ (corresponding to $T=\infty$), and hotter yet sees $a$ go from $0$ towards $+\infty$ while $T$ increases from $-\infty$ to $-0$.
For practical purposes, this is a non-starter, however, because the everyday temperatures an unbounded system can have would correspond to negative $a$. Macroscopic laws would look rather strange in terms of such a scale.

related: https://chemistry.stackexchange.com/q/36885/23561

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text:
Why do we always represent only the hydrogen evolution?
In corrosion of aluminium or magnesium for example, we have the anodic reaction which is the dissolution of the metal but for the cathodic reaction we only represent the hydrogen evolution.
How about the reduction of oxygen gas and water?
Why don't we mention these reaction and we focus only on hydrogen evolution?

There is (almost) no oxygen gas around there, just water. As for the reduction of water, guess what happens and what gas evolves as a result?


text:
Why and how carboxylate ion has greater stability than carboxylic acid?
I found it written in my textbook. I think it's a misprint, I can't think of any logic behind this statement.
I've only learned that ion is less stable specie so carboxylate ion cannot be more stable than carboxylic acid.

It's not a misprint; it's a misuse of language. You can't really compare ions to molecules in terms of "stability". Ion is neither more stable, nor less stable, nor equally stable. It makes no sense.

First of all, it would have been clearer if it's a misprint or not if you had attached a screenshot of the wording in your textbook (you still can).
At any rate, the subject of stability of carboxylic acids and carboxylate anions usually arises when discussing the level of acidity or in other words, the strength of the acid. It is pointed out in most organic chemistry books (where electron delocalization is explained) that electron delocalization or resonance is a concept that contributes to a certain structure's stability. That is to say, when a negative charge can be delocalized over several atoms (as is the case with a carboxylate group), instead of being localized on a single atom (as is the case with an alkoxide group), it is considered more stable energetically. (see here, section B - Resonance).
The relative stability of the carboxylate anion means that a carboxylic acid will favor (again, due to energetic considerations) the release of a proton in a basic environment (this is important, since in an acidic environment, the equilibrium will shift towards the acid, and much less carboxylate will exist in the solution), since the conjugate base (the carboxylate anion) is relatively stable, and therefore can exist in the solution in this form.
 (Taken from Organic Chemistry, by Paula Yurkanis Bruice, any edition really...)
To sum up, Carboxylic acids and their conjugate carboxylate anions are not (and cannot be) compared in terms of stability. Under the right conditions, i.e. in a basic solution, a carboxylate anion will be favored over a carboxylic acid because:
a. There is a base to take a proton (be it a hydroxide ion, an aminic group or any other H+ acceptor).
b. The carboxylic group will release the proton rather easily, because what's left (the carboxylate anion) is relatively stable due to electron delocalization.

A carboxylate is certainly less stable than the corresponding carboxylic acid. the two conjugate forms are related by the equilibrium:
$$\ce{RCOOH <=> RCOO- + H+}$$
I think of this (and equilibria in general) as a tug of war, or a see-saw; the total amount of $\ce{RCOOH + RCOO-}$ is fixed, but one form can become more plentiful at the expense of the other. The molecule will found mostly in whichever of the two forms is more stable.
Experimentally, we observe that carboxylic acids are weak acids with $\ce{pKa}$ around 5. This means at a pH of 5, the population of molecules is 50% in the protonated $\ce{RCOOH}$ form and 50% in the deprotonated $\ce{RCOO^-}$ form. At a neutral pH ' $\ce{RCOOH}$ is about 100 times more plentiful than $\ce{RCOO-}$.
Thus, we might draw the conclusion that the neutral $\ce{RCOOH}$ carboxylic acid is more stable than the anionic $\ce{RCOO^-}$carboxylate (at neutral pH and in a standard temperature range). This is directly related to, and in a way the very cause of, the fact that carboxylic acids do not fully dissociate, i.e. they are weak acids.


text:
Constructing a reactivity series, magnesium more reactive than sodium?
I reacted about $0.25 \,\mathrm{g}$ of magnesium, sodium and lithium in $20 \,\mathrm{ml}$ of $2 \,\mathrm{M} \,\ce{HCl}$ at room temperature. This was all done in a $100 \,\mathrm{mL}$ beaker.
I found the biggest temperature change in the reaction mixture with lithium, followed by the reaction mixture with magnesium and finally sodium.
I measured the temperature change over the course of the reaction with a Vernier temperature probe and accounted for the heat loss by interpolating a fitted line based on the heat loss signature after the reaction has completed.
I calculated the heat of the reaction and divided by the number of moles and get this:

Li reaction: $120 \,\mathrm{kJ/mol}$
Mg reaction: $91 \,\mathrm{kJ/mol}$
Na reaction: $33 \,\mathrm{kJ/mol}$

I did notice the evolution of a lot of gas when reacting $\ce{Na}$ with the $\ce{HCl}$. As expected a flame was produced with the sodium reaction.
My questions are:

Why is it that the reaction that apparently seemed to be the most vigorous produced the least temperature change?

I was initially thinking that the sodium had a lower temperature change as it reacted so vigorously that the hydrogen gas produced had more kinetic energy, and as it leaves the system, the temperature of the system does not change by much. But I figured this explanation to be incomplete as lithium is supposed to be more reactive than sodium, so if this was the case I would have an even lower temperature change for lithium?

Why is it that some reactivity series have lithium at the very top while others just omit it entirely or put it below sodium and potassium? Who is right?

I guess what I am really asking is for ideas on where I went wrong, because this data does not seem to be right.