text:
How do I show that a transition is magnetic dipole allowed with group theory/symmetry?
I am taking an introductory course in symmetry and bonding and have so far learnt about SALCs and various point groups etc... one of my problems asks me to show a $\mathrm{^{2}T_{2g}\rightarrow\ ^{2}E_g}$ is allowed in a $\mathrm d^1$ metal. I am not sure how my knowledge of symmetry can help me answer this, all my searching flags up difficult quantum mechanics/group theory I don't understand! I think a relevant example I could use is $\ce{[Ti(H2O)6]^{3+}}$ as this is an octahedral, $\mathrm d^1$ system.

For a state-to-state transition coupled by an operator to be allowed, the direct product of the irreducible representations (irreps) of all three components must contain the completely symmetric irrep of the point group you're working in. That is, we are determining whether the following integral is forced to be zero by symmetry:
$$
\int \Psi_{i}(\tau) \times \hat{\mathbf{A}}(\tau) \times \Psi_{f}(\tau) \, \mathrm{d}\tau \overset{?}{=} 0
$$
where $\Psi_{i}$ and $\Psi_{f}$ are the initial and final states, $\hat{A}$ is the operator (electric dipole, magnetic dipole, one-electron spin-orbit, there are many more), and $\tau$ is the integration coordinate, usually over electron position. Why bother with this if the integral can be evaluated exactly?

It's no more complicated than this pencil-and-paper calculation to gain qualitative understanding about whether or not a system with some symmetry will have a certain kind of spectroscopic response.
Why bother calculating the integral if it's going to be zero anyway? It's just extra work, though it usually isn't very expensive.

For $O_{\mathrm{h}}$, the completely symmetric irrep is $\mathrm{A_{1g}}$. It's always listed as the first irrep in a character table.
$$
\small\begin{array}{c|cccccccccc|cc}\hline
O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & \begin{aligned}3C_2 \\ \scriptsize=C_4^2\end{aligned} & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & & x^2+y^2+z^2 \\
\mathrm{A_{2g}} & 1  & 1  & -1 & -1 & 1  & 1  & -1 & 1  & 1  & -1 & & \\
\mathrm{E_g}    & 2  & -1 & 0  & 0  & 2  & 2  & 0  & -1 & 2  & 0  & & \begin{aligned}(2z^2-x^2-y^2,\\ x^2-y^2)\,\,\,\,\,\, \end{aligned} \\
\mathrm{T_{1g}} & 3  & 0  & -1 & 1  & -1 & 3  & 1  & 0  & -1 & -1 & (R_x,R_y,R_z) & \\
\mathrm{T_{2g}} & 3  & 0  & 1  & -1 & -1 & 3  & -1 & 0  & -1 & 1  & & (xy,xz,yz) \\
\mathrm{A_{1u}} & 1  & 1  & 1  & 1  & 1  & -1 & -1 & -1 & -1 & -1  & & \\
\mathrm{A_{2u}} & 1  & 1  & -1 & -1 & 1  & -1 & 1  & -1 & -1 & 1  & & \\
\mathrm{E_u}    & 2  & -1 & 0  & 0  & 2  & -2 & 0  & 1  & -2 & 0  & & \\
\mathrm{T_{1u}} & 3  & 0  & -1 & 1  & -1 & -3 & -1 & 0  & 1  & 1  & (x,y,z) & \\
\mathrm{T_{2u}} & 3  & 0  & 1  & -1 & -1 & -3 & 1  & 0  & 1  & -1 & & \\ \hline
\end{array}
$$
The magnetic dipole operator is the angular momentum operator with some constants in front:
$$
\hat{\mathbf{m}} = \gamma_{\mathrm{e}} \hat{\mathbf{l}} = -\frac{e}{2m_{\mathrm{e}}} \hat{\mathbf{l}} \approx \left(\hat{l}_{x}, \hat{l}_{y}, \hat{l}_{z}\right)
$$
The angular momentum operator components have rotational symmetry, so $\hat{l}_{y}$ would use the irrep corresponding to $R_{y}$. Constant prefactors don't affect the symmetry of operators, so all three components of the magnetic dipole operator have $\mathrm{T_{1g}}$ symmetry.
Now we must show that
$$
\left< \Psi_{i} | \hat{\mathbf{m}} | \Psi_{f} \right> \equiv \int \Psi_{i}(\tau) \times \hat{\mathbf{m}}(\tau) \times \Psi_{f}(\tau) \, \mathrm{d}\tau \rightarrow \mathrm{T_{2g}} \times \mathrm{T_{1g}} \times \mathrm{E_{g}} \overset{!}{\ni} \mathrm{A_{1g}},
$$
and to do this we need a product table.
$$\begin{array}{c|ccccc} \hline
 & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\ \hline
\mathrm{A_1} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\
\mathrm{A_2} & & \mathrm{A_1} & \mathrm{E} & \mathrm{T_2} & \mathrm{T_1} \\
\mathrm{E} & & & \mathrm{A_1 + [A_2] + E} & \mathrm{T_1 + T_2} & \mathrm{T_1 + T_2} \\
\mathrm{T_1} & & & & \mathrm{A_1 + E + [T_1] + T_2} & \mathrm{A_2 + E + T_1 + T_2} \\
\mathrm{T_2} & & & & & \mathrm{A_1 + E + [T_1] + T_2} \\ \hline
\end{array}$$
Product tables are symmetric, so only the lower or upper triangle needs to be shown, and inversion symmetry is left out here as long as you consider the rule
$$
\begin{align}
\mathrm{g} \times \mathrm{g} &= \mathrm{g} \\
\mathrm{g} \times \mathrm{u} &= \mathrm{u} \\
\mathrm{u} \times \mathrm{u} &= \mathrm{g},
\end{align}
$$
just like multiplying even and odd integers. Some product tables show all of the redundant information.
Working from left to right, find the product of the symmetry of the initial state with the symmetry of the operator:
$$
\mathrm{T_{2g}} \times \mathrm{T_{1g}} = \mathrm{A_{2g}} + \mathrm{E_g} + \mathrm{T_{1g}} + \mathrm{T_{2g}}.
$$
The product of degenerate irreps with each other leads to multiple irreps, hence the sum. You might think that the next step is
$$
\left(\mathrm{A_{2g}} + \mathrm{E_g} + \mathrm{T_{1g}} + \mathrm{T_{2g}}\right) \times \mathrm{E_g} = \dots,
$$
however we take advantage of two nice rules:

For the transition to be allowed, the final product only needs to contain the completely symmetric irrep; irreps other than $\mathrm{A_{1g}}$ are allowed to be present as long as $\mathrm{A_{1g}}$ is there.
The product of any irrep with itself always gives the completely symmetric irrep; see the diagonal elements in any product table.

From the last equation, we see that $\mathrm{E_g}$ is in the sum, so the other terms can safely be ignored because $\mathrm{E_g} \times \mathrm{E_g} = \mathrm{A_{1g}}$. Additionally, because all three components of the magnetic dipole operator transform as the same irrep, this electronic transition will be allowed if you apply a magnetic field along any of the three Cartesian axes in the molecular frame.

Thank you for your detailed explanation...what does the line "⟨Ψi|m̂ |Ψf⟩→A1g∈!T2g×T1g×Eg" mean? Does it interpret as a magnetic dipole transition from Ψi to Ψf  in A1g symmetry? What about the last part?

It's shorthand for how that integral leads to this product of three irreps, and $\mathrm{A_{1g}}$ must be contained in that product. If there's a better way of representing that, please tell me.

and why are we taking a triple product? are these the "3 components" mentioned earlier and match the integrand?

Ah, are you familiar with Dirac braket notation? I'll expand on in the question if not. It's a simplified way of writing integrals. But yes, it corresponds to the three things you multiply together before integrating (the two wavefunctions and the operator).

i am not but i will research this as it is definitely worth it-seeing it a lot!

@pentavalentcarbon I posted a follow up question https://chemistry.stackexchange.com/q/166089/125277


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Balanced Chemical Equations of the preparation of sodium peroxoborate
I know the chemical equation for the reaction of borax with sodium hydroxide and hydrogen peroxide
$$\ce{Na_2[B_4O_5(OH)_4]\cdot6H_2O + 2NaOH + 4H_2O_2 ->2Na_2[B_2(O_2)_2(OH)_4]\cdot6H_2O + 3H_2O} $$
Now when we were doing the experiment to prepare sodium peroxoborate we added, the $\ce{NaOH}$ first and then $\ce{H_2O_2}$. How do I write the chemical equations for the borax with just the sodium hydroxide and then that product with the hydrogen peroxide. Essentially how do I break the balanced chemical equation that I have into two different steps.

Strongly related: https://chemistry.stackexchange.com/questions/44911/preparation-of-sodium-peroxoborate-equations/44913

It doesn't answer my question at all. I already have that equation posted by Aditya Dev

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Flow of current in graphite
Why current flows through graphite if we join the poles of a battery to it horizontally and why not vertically?

Can you define what those directions mean? You're assuming an orientation that is not explicitly defined for us...

Turn your piece of graphite sideways, and it will conduct vertically all right.

@Zhe I asked the reason. In one orientation it conducts and in the other it does not, why?

@IvanNeretin Why in one orientation it conducts electricity and in the other it does not?

Net electron flux is possible in one direction and not in the other. Graphene has a highly delocalized $\pi$ system roughly within the plane of the sheet, and one might imagine that electron mobility within the conjugated system is quite high.

I'd guess that repulsion between the delocalised electrons and electrons in the molecular orbitals (of each layer of carbon) would keep them between the two layers, as opposed to being able to travel vertically through them?

There is no "between". The conduction electrons are located in those very orbitals.

Wait - so the delocalised electrons in graphite are part of one giant molecular orbital?

Yes, indeed they are.

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Why does water split into ions?
What is the property of a water molecule that allows it to split into $\ce{OH-}$ and $\ce{H+}$ ions?
And subsequently, why - when these ions are attracted towards each other in a neutralisation reaction, for example - can these ions then become a single covalently bonded molecule?

The splitting described doesn't actually happen in bulk water. There is just the autoionization reaction:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
The net transfer of a proton from one water to another creates two ions. This is somewhat disfavored. Hence, the autoionization constant is quite small:
$$\mathrm{K}_{w} = \ce{[H3O+][OH-]} = 10^{-14}$$
Neutralization is a somewhat related process, whereby one species is able to create an excess of hydronium, and another is able to create an excess of hydroxide. This creates an imbalance in the autoionization equilibrium, and it rapidly shifts left to consume excesses of both ions.

The chemical reaction is absolutely correct as shown, but it belies the situation. The bare ions don't just float around in the water. Rather each is solvated. In other words The $\ce{H^+}$ attracts oxygen atoms of adjacent water molecules to form multiple "shells" of water molecules oriented around the  $\ce{H^+}$ ion. Likewise the $\ce{OH^-}$ ion attracts H atoms of water molecules to form oriented layers.

As @MaxW says, solutions are infinitely more complicated than what we can capture in a single line of text. The goal of this answer is to help the OP understand the nature of the "split."

Well the point of discussing solvation is that it stabilizes the split. Small amounts of water in different solvents have vastly different dissociation constants.

Water does not really split into ions.  It transfers protons, reversibly, from one molecule to another:
$\ce{2 H2O = H3O+ + OH-}$
The ions represent, of course, a less stable state than molecular water, at least under ordinary conditions, so the equilibrium constant is low.  Yet this reaction strongly affects the chemical prpoeryies and even some physical properties of aqueous solutions.
Such a transfer reaction is possible because of the polarity of the hydrogen-oxygen bonds in water.  But if that is so, then might other solvents with similarly polar bonds to hydrogen transfer protons in a similar way?
Of course they do.  Ammonia does it, forming ammonium and amide ions; although it does so only weakly because you have to chill it to make it liquid and the nitrogen-hydrogen bonds are only weakly polar.  Sulfuric acid does it (and undergoes some other solvent reactions such as oligomerization) to such an extent that "100% sulfuric acid" contains a significant amount of things other than $\ce{H2SO4}$.  Even some organic liquids like ethanol and glacial acetic acid undergo intrinsic proton transfer.
But wait, there's more.  Solvents that contain no hydrogen but do contain halogens with polar bonding can transfer halide ions, like bromine trifluoride ($\ce{BrF3}$) forming $\ce{BrF2+}$ and $\ce{BrF4-}$ ions.
Dinitrogen tetroxide, $\ce{N2O4}$, is an exception, as it really does split molecules to form $\ce{NO+}$ and $\ce{NO3-}$.  The relatively weak nitrogen-nitrogen bond allows that, but even here an oxygen atom must also be transferred between the fragments.  Direct cleavage would produce species that are too strongly acidic ($\ce{NO2+}$) and too strongly basic ($\ce{NO2-}$) for the solvent.
Among common solvents with polar bonds, the one that most stands out is not water but sulfur dioxide.  It fails to autoionize to any significant extent, because it would have to form radicals to make singly charged species and the alternative of forming doubly charged ions ($\ce{SO^{2+},  SO3^{2-}}$) is too highly disfavored for just about any solvent (even water).

The "bare" proton (H+) does exist. But not in equilibrium aqueous solution (near STP). Chemists talk about H+ and Physicists do too, but generally they are talking about very different things. The Physicists actually are (usually) talking about H+, and the Chemists are almost always talking about the various species which have been protonated. In the case of water, the two most common cations are H3O+ and H5O2+ which can be thought of as n(H2O)H+ with n a small integer. We chemists call all of these cations "the (aqueous) proton" although we know 'taint so. Its a short-hand. What is going on is that while liquid water has its molecules moving and dancing around, at any instant of time, most of the molecules are in one or more hydrogen bonds.  With the H atom often bonded to one molecule's O atom and h-bonded to a second molecule's O atom. (a single O atom can be associated with more than one single H-bond at any given instant, due to its much larger size). Let me simplify this: any given HOH molecule  h-bonded to an O of another molecule will have the h-bond looking like this O-H....O, but also like this O-H...O and this O-H..O. That is, it isn't correct to think of the three atoms O,H and O as "exactly" covalently bonded and "exactly" h-bonded but on a spectrum between O-H....O (neutral) and O....H+O ionized. (where I've ignored the other bonds in the two molecules). If 1 h-bond in 10,000,000 results in the second case (the ion) then the pH would be 7, and guess what? it is. So, we can say that the proton (although we don't mean a bare proton)  engages in the H2O→OH- + H+ reaction rarely, but when you've got 6E23 molecules in 20grams of water, there are a LOT of "rare" events. The proton is able to "escape" its molecule because that molecule's negative charge is being partially neutralized by other h-bonds --I mean you would expect OH- to be h-bonding in water, wouldn't you?? So, while the h-bonds "explains" why water dissociates, you shouldn't look at these as stable bonds, it's more of a one night hook-up, or perhaps even an orgy of constantly changing hookups.


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Is water a possible fuel for jet engines?
Jet engines can run on almost any fuel, and the operating temperatures of modern jet engines' hottest sections are anywhere between 3000 and 3150 degrees F (1648 and 1732 degrees Celsius). Does that mean that a hydrogen on-demand system could work on modern jets?
Water is pumped and heated first by the exhaust section, then directed towards the hotter sections of the engine (when hot enough to not cause cooling and lower engine efficiency) where it's broken down into hydrogen and oxygen at a heat above 1472 degrees F (800 degrees Celsius), then those gases are pumped into the engine for combustion.
The advantages are that firstly, water is abundant and therefore cheap. Even sea water could be used because at those temperatures it's easy to design a system that would get rid of the impurities that would otherwise corrode critical engine parts.
Secondly, it would save on manufacturing costs given that non-heat critical parts in the exhaust section would not need to be made of sophisticated and expensive materials and alloys given the cooling effect of water.
Thirdly, the costs of the fuel weight would be reduced given that the energy density of hydrogen is twice that of fossil fuels, so less would need to be carried. And most importantly, the environment problem would be solved in aviation given that there would be little or no carbon dioxide emissions.

It won't work. You can't oxidize water exothermically.

How do you heat your engine to 3000 degrees in the first place?

You use stored hydrogen or excess stored hydrogen from previous flight hopefully

Where the energy is going to come from?

What's the combustion product of burning hydrogen with oxygen? Water. So why have big fuel tanks at all? We can just condense the water coming out of the back of the engine, pipe it back to the fuel pump, and burn the same few liters of water forever. Now that's fuel weight savings.

I feel it's not best site to ask - it's mostly technical problem about jets. In principle this could work, but it wouldn't be practical, but that's for jet engineers to tell not chemists.

@hobbs Not "forever" but partial steam exhaust condensing to reuse it, could be possible in theory, and better then tanking all the water in the beginning.

You're asking, why don't we unburn water and then burn it? The simple answer is that the energy you get from burning it has to be put back in to  unburn it.

This would work if you could crack apart the water molecule outside the plane and before the flight, and store the hydrogen and oxygen in tanks as fuel ready to use.

The fluorine reacts with water. But don't try at home.

@Criggie You don't take oxygen to burn jet fuel with you on a plane, why take it to burn hydrogen? - Still, the whole proposal is just a perpetuum mobile, can't understand why people seem to have such a hard time wrapping their heads around it. :)

Dude, we already told you this is impossible over on Aviation SE. Why did you waste everybody's time by posting it again? (cc @Mithoron, who suggested the question should be elsewhere -- it is!)

@Waylander "How do you heat your engine to 3000 degrees in the first place?" By having a "start cart" at the airport. This isn't the problem with this idea...

Is this correct :https://i.stack.imgur.com/6oIFp.png I'd assume in the h2O mix 2 thirds is hydrogen therefore for 3.5 kgs of water you get 2.33kgs of hydrogen

@AlexanderKosubek fair point, but you're missing my point that the cracking can happen sometime before the flight using ground-based energy.
This might work as a way to time-shift solar energy, for example.     But then again, a tank of Hydrogen might not be the safest fuel in the air.

@Criggie Well, okay, you probably were focussing your comment on the implied question in the title, if hydrogen would be a suitable fuel for jet engines? The full text question is about a perpetuum mobile, though. Due to the specific assumptions layed[! :) ] out in the question, especially cracking up the water to make hydrogen in-flight, I strongly believe that OP should just do the math and realize that what he supposes will not, as we all know, work.

I'm voting to close this question as off-topic because it's was already asked and answered on Aviation, where it belongs.

@securitydude5 3.5 kg of water contains 0.4 kg of hydrogen not 2.33kg. Hydrogen atoms weigh a lot less than oxygen atoms and it's 2/3 of the atoms that are hydrogen not 2/3 of the mass. That's basic chemistry.

@Mithoron Don't vote to close it for that reason. The chemical explanation why it can't work is a good explanation of chemistry and thermodynamics. Probably better explained here than on aviation.

Iranians already have created a hydro engine.

Using an efficient catalyst you could get hydrogen from water with a lower amount of electricity needed in hydrolysis process.
After hydrolysis, you'll have oxygen in hand which could be involved in other energy provider processes.

https://www.youtube.com/watch?v=qSHKt9kWRGg

It can't work because of the fundamental thermodynamics
What you are proposing is, basically, the plane carries water; the water is broken down into its components, hydrogen and oxygen; the components are recombined by burning them as fuel. Burning hydrogen and oxygen is a perfectly good way to create a lot of heat. But it doesn't much matter how you break the water apart into hydrogen and oxygen, the thermodynamics of the reaction won't work.
The problem is simple: you need to have a source of energy to split the water apart. In chemistry we know the energy levels of the reactants and the products and we can work out whether energy is released or stored in a reaction. Burning hydrogen and oxygen releases a lot of energy, but by the rule of thermodynamics, breaking water apart to its components requires the input of exactly the same amount of energy. You can't get round this. Worse, in the real world, there are losses at every conversion step so you can't even break even (ain't things unfair!)
In your plane you could, in principle, split water and burn it in the engine for propulsion. But you would need to have some other source of the vast amount of energy required to split the water. That implies both another fuel and another engine. In reality they would vastly outweigh any imagined savings in weight and cost.
Even if you could build some sort of SciFi engine that both splits water and then burns it again you would still be nowhere: the entire output of the burning would be required for the splitting with nothing left over to generate thrust (and in the real world there are losses so your plane would rapidly drop out of the sky).

Now, if you had a concentrated, compact power source (like, say, a miniaturized nuclear reactor) you could use it to heat water into steam and use that to generate thrust.  You could even do it with just air.  The latter option was experimented with back in the '50s, and the Russians even managed to make one fly, although the crews were not re-usable...

Two questions: 1) Why does this barrier apply to water but not to other fuels? Is it that there's no step to breaking down a fossil fuel before it is burned? 2) I know that fission works because splitting atoms releases more energy than is put into it, so I assumed something similar was possibly for water, but now I'm guessing at that level we're looking at a completely different set of equations, right? If I misunderstand fission you can just say "you misunderstand fission", I'm not asking you to give me a nuclear crash course here.

@LordFarquaad 1) Yes it does apply to other fuels. The (improper) analogy would be that water and CO2 are "separated" into fuel and oxygen, then burned in the jet. Except, of course, this "separation" was done on the ground for a very long time and took a lot of sunlight (and other energy). If you tried to synthesize the fuel on the jet and then burn it again, you'd have the same problem.

It doesn't work this way for other fuels because they release energy when they are burned in air (producing water and carbon dioxide, mostly). Fission is a nuclear process not a molecular one and can't be compared to chemical stuff (except that even in nuclear processes, energy is only released when the products have less than the starting materials: you don't release energy by splitting oxygen nuclei in two).

@LordFarquaad 2) Fission is pretty much the same. You have some heavier elements, they fall apart (because they are unstable or because you bombard them with neutrons or both or other) and give off energy. We've found ways how we can trigger said fission using less energy than is released in the process. But that doesn't explain how we got those heavier atoms in the first place. They were (for the most part) created in a star, and used tons of energy. If we tried to "make" them, we'd use more energy than we can get by breaking them apart. Fission is not "free energy", it's concentrated solar.

@Ordous and matt_black, ok, thanks. It sounds like for both those processes then we're taking advantage of energy that's already been introduced into the system, so OP's issue comes from trying to start from scratch?

@LordFarquaad Pretty much! Energy is conserved, so you always have to get it from something or transform something (taking it's energy).

Matt, I'm afraid you're answer is totally invalid: "How do you heat your engine to 3000 degrees in the first place? You use stored hydrogen or excess stored hydrogen from previous flight hopefully" OP isn't talking about any perpetuum mobile, but recycling waste energy from turbine.

@Mithoron But how do you generate excess hydrogen on the previous flight? You need it all to provide enough energy to spit the water (even if you assume a lossless system). The best you in the best possible system is to cycle between water and the oxygen/hydrogen mixture with no energy left over for flight.

@LordFarquaad To avoid confusion on the issue with other fuels: the energy produced from combustion of a fuel has nothing to do with splitting it apart. It comes because the products of the combustion have a lower energy then the fuel and oxidant mixture. Energy release comes from the net difference in bond energy between the products and the reactants: some things split up and release energy; some things combine and release energy.

Geez, you simple have "normal" engine using hydrogen tanked before flight, but also carrying some water, taking off some waste heat and partially braking down to O2/H2 which may be used later.

@Mithoron And how are you going to recover the hot gas to store the fuel?Jet engine thrust is dependent on expelling hot gas at high velocity. Even with out this practical barrier, how do you get more energy out than you put in?

Stealth bombers have exhaust cooled to reduce thermal signature, so I don't see much problem with this. Main obstacle would be probably that it would be very difficult to get actual water dissociation, as probably it would not get to high enough temp.

@Mithoron Cooling the output form stealth engines comes at a big cost in efficiency of the engine. It costs to use the heat so doesn't provide a means to get "free" fuel however you organise the recovery.

Let us continue this discussion in chat.

Iranians already have created a hydro engine. 1. Using an efficient catalyst you could get hydrogen from water with a lower amount of electricity needed in hydrolysis process. 2. After hydrolysis, you'll have oxygen in hand which could be involved in other energy provider processes.https://www.youtube.com/watch?v=qSHKt9kWRGg

The steam engine has already been invented, but you need a source of heat outside the system. It is thermodynamically impossible to use the heat from the steam generated to produce more steam. It gets worse if the heat is converted to something else first, like the heat generated from friction. You are throwing (kinetic) energy and mass out the back of the turbine. Both need to be constantly replaced.

This is possible when the jet engine has multiple stages.
The first low temperature stage (low temperature meaning 2500°C in this case), water is split into hydrogen and oxygen.
In a next stage, the hydrogen is separated from the oxygen (which is discarded) and compressed to a few bar of pressure.
The third stage is kept at a temperature of above 10.000.000°C. In this stage, the hydrogen is exothermally burned into helium, supplying the heat for stages one and three and also propulsion.
This engine also produces large amounts of radioactive radiation, which can be seen as a disadvantage in some situations.
Building a jet engine which can withstand a temperature of 10.000.000°C is left as an exercise for the reader.

Ya beat me to it.   But you did forget Step 3  discover cold fusion.


text:
Miller Indices: How to deal with some weird(-ish) cases?
I've recently been doing a spot of self-learning (Crystallography), and some of the examples provided in the Wikipedia article for Miller Indices have stumped me.
I can't for the world figure out how they were obtained!

Quick introduction to Miller Indices (the basic stuff):
 Including this here, because it was what I was referring to when learning about this stuff

 Miller Indices  (hkl)
The orientation of a surface or a crystal plane may be defined by considering how the plane (or indeed any parallel plane) intersects the main crystallographic axes of the solid. The application of a set of rules leads to the assignment of the Miller Indices , (hkl) ; a set of numbers which quantify the intercepts and thus may be used to uniquely identify the plane or surface.
The following treatment of the procedure used to assign the Miller Indices is a simplified one (it may be best if you simply regard it as a "recipe") and only a cubic crystal system (one having a cubic unit cell with dimensions $a$ x $a$ x $a$ ) will be considered.

The procedure is most easily illustrated using an example so we will first consider the following surface/plane:

Step 1 :  Identify the intercepts on the x- , y- and z- axes.

In this case the intercept on the x-axis is at x = a ( at the point (a,0,0) ), but the surface is parallel to the y- and z-axes - strictly therefore there is no intercept on these two axes but we shall consider the intercept to be at infinity ( ∞ ) for the special case where the plane is parallel to an axis. The intercepts on the x- , y- and z-axes are thus:  $a$ , $∞$ , $∞$ (in that order).

Step 2 :  Specify the intercepts in fractional co-ordinates

Co-ordinates are converted to fractional co-ordinates by dividing by the respective cell-dimension - for example, a point (x,y,z) in a unit cell of dimensions $a$ x $b$ x $c$  has fractional co-ordinates of ( x/a , y/b , z/c ). In the case of a cubic unit cell each co-ordinate will simply be divided by the cubic cell constant , $a$.
This gives:  $a/a$ , $∞/a$, $∞/a$ ;   i.e-    $1$ , $∞$ , $∞$.

Step 3 :  Take the reciprocals of the fractional intercepts

This final manipulation generates the Miller Indices which (by convention) should then be specified without being separated by any commas or other symbols. The Miller Indices are also enclosed within standard brackets (…) when one is specifying a unique surface such as that being considered here.
The reciprocals of 1 and ∞ are 1 and 0 respectively, thus yielding the Miller Indices ($100$)
So the surface/plane illustrated is the ($100$) plane of the cubic crystal.

Further on, it is mentioned:

If any of the intercepts are at negative values on the axes then the negative sign will carry through into the Miller indices; in such cases the negative sign is actually denoted by overstriking (i.e- adding a little bar/dash over) the relevant number.

Now I was going to through the Wiki article on Miller Indices (linked earlier on in this post), and the following examples were provided:

 The corresponding Miller Indices have been provided in the boxes right under each entry. Also, the little "T"s are actually "1"s with a bar on top... kinda small, so it's easy to see it wrong.

Now the first six entries are pretty straightforward (I count left-right), but it's the others that are confusing.

Take for example, the seventh entry:

Okay, the third axis ($a_3$) lies along orange-colored plane, so the plane has no intercept  with the ($a_3$) axis (i.e- intercept here is taken to be $∞$, as the orange plane and the axis in question are parallel). Which would naturally result in the Miller Index $0$ (third digit) which is the reciprocal of $∞$.
Alright, no issues there.
But what's the deal with the intercepts on the the first two axes ( $a_1$ and $a_2$ )?
The "intercept at infinity" logic doesn't work here (since the orange plane isn't parallel to $a_1$ or $a_2$).
According to the (box under the) entry, the orange plane makes an intercept of $1$ (second digit) with the $a_2$ axis. Moreover, it also (apparently) makes an intercept of $-1$ (first digit) with the $a_3$ axis.
This is incredible...I am totally lost.
The eighth and ninth entries, too, are made in the same vein (well, obviously... because the seventh, eighth and ninth entries are essentially the same thing)

My question:
How exactly, are the Miller Indices for the seventh entry (the one with the orange plane), obtained?

I've just asked for the rationale behind the Miller Indices of the seventh entry because I can easily work out the Indices for the eighth and ninth entries after that.

I'm still relatively new to Miller indices and I can't say I have an answer for a general procedure, but the Miller indices define an infinite family of planes. So while the particular plane shown for the $7^\text{th}$ entry doesn't have those intercepts, you can picture shifting that plane up to a location where it doesn't intercept $a_3$, intercepts $a_1$ at $-1$, and intercepts $a_2$ at $1$. The same sort of shifting will give you the correct intercepts for 8 and 9.

a similar question from physics.SE: https://physics.stackexchange.com/questions/62955/i-cant-figure-out-crystal-planes-with-negative-intercepts

This section explains which plane from a set of planes you consider to determine the indices. It has an image that shows planes beyond the unit cell, which resolves the conceptual hurdle of the images shown by the OP.

IFF I remember correctly...
Sometimes you have to "move" the plane, in order to see where the axis crosses.
Take example 7, if you look at the example plane, the one actually shown, it is (0 ∞ ∞) not very useful. The ones making the examples have "moved the planes" (not really, but it is a useful fiction) in order to get the plane into the unit cell so they can show it. If they hadn't moved it, it wouldn't be visible.
The plane (-110) crosses $a_2$ axis at $a_2$ = $1$ and you see, using your spatial imagination, that it then needs to cross $a_1$ at $-1$ as well. It will never cross the $a_3$ axis, so $0$. The 8th and 9th are much harder to visualize but it is basically the same thing over again. The planes won't fit into the unit cell unless you "move" them.
Disclaimer: the (hkl) represents the set of all planes of the same err... "orientation", so it is technically not moving anything at all, it is merely choosing a representation of that said plane. If two planes in a certain crystal are equivalent, they can be represented with curly brackets. The sets (100),(010) and (001) are equivalent in a cubic crystal and can be represented collectively as {100}. Source: Basic Solid State Chemistry, A.R West 2nd ed. (1999)

[OP] How to deal with some weird(-ish) cases?

The cases are not weird(-ish), it is the images that are misleading in some cases, and inconsistent overall. Here is a better depiction that shows more than a unit cell:

Source: https://geo.libretexts.org/Bookshelves/Geology/Mineralogy_%28Perkins_et_al.%29/11%3A_Crystallography/11.10%3A_The_Miller_Indices_of_Planes_within_a_Crystal_Structure
The description for the image says to take the plane (line in this case because they show a 2D crystal) that is adjacent to that crossing the origin. The origin is marked with "O" and the line you have to pick is marked with "x" and "y" in the images above. The isolated unit cell is shown in red on the right.

[OP] Take for example, the seventh entry [...] How exactly, are the Miller Indices for the seventh entry (the one with the orange plane), obtained?

Here is another picture of the seventh entry, showing how planes graze the unit cell at two edges.

The difference to the original image is subtle, but the hand-drawn orange lines show where the two planes would just touch the unit cell. You would have to extend these planes beyond the unit cell to figure out the Miller indices. One crosses $a_1$ at 1 and $a_2$ at -1, the other at -1 and 1, respectively. Neither intersect $a_3$ (which is in plane). The corresponding Miller indices are the equivalent (1, -1, 0) and (-1, 1, 0). Both describe the same set of planes (i.e. same orientation and same distance between adjacent planes).
Trevor's answer has a helpful practical hint: instead of drawing more planes, you can also move the origin by a unit cell vector translation. What is missing from the answer is that you should consider a plane that is adjacent to that crossing the origin. If you consider one further away, you get a Miller index smaller than one, which should not happen, or turn a higher order set of planes such as (2 8 4) into a lower order set such as (1 4 2).
A different approach
Once you know about the reciprocal lattice, you can also do the following. Take a vector perpendicular to the planes, and give it the length of the reciprocal of the lattice spacing. This vector (corresponding to a reciprocal lattice point) will have the Miller indices as its coordinates in reciprocal space. With this definition, it is clear that the distance between planes is important, and the it does not matter which one of the parallel planes you consider for the direction.
With the traditional recipe, considering the distance of the planes comes by choosing the plane that is immediately adjacent to the one going through the origin.

For the seventh entry/diagram:  The way I understand it, is you can't have the plane cutting the origin at zero.  Therefore you must use a different origin.  Assume the origin is now on the front RHS corner.  Now the plane cuts at a1=-1, a2=1, a3=infinity.  Then take the reciprocal and you get (-1 1 0)

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


text:
Nomenclature for an alkane substituted with a branched alkyl group
Consider, for example:

From what I've learned in chemistry class, this molecule would be unofficially named 3-isopropylpentane.
What would the official IUPAC name for this molecule be? And what's the general way to name these types of molecules?
EDIT: For this specific molecule, the naming can be simplified by looking at the main carbon chain differently.
For a more general case, consider heptane instead of pentane, in this case there's one chain that you're stuck with.

3-ethyl-2-methylpentane, also it would be propan-2-ylpentane and learn the IUPAC rules that's how to name all molecules.

@AS_1000 Wouldn't the name "3-ethyl-2-methylpentane"  suggest that there's 2 branches on the pentane chain, an ethyl on C3 and a methyl on C2 ? That's a different molecule. Also the name " propan-2-ylpentane" doesn't specify which of the pentane carbons is the propyl attached to..

@Chady are you sure it's a different molecule? Draw it out.

@Chady AS_1000 is correct. Once you have attached the isopropyl group, there is more than one possible chain of length 5, and one of those has a methyl at carbon 2 and an ethyl at carbon 3.

@AS_1000, orthocresol, Tyberius,  I see it now, it didn't occur to me to look at the chain differently.

But this only works in this specific case.. What if instead of pentane there was heptane ? There's only one chain to work with, how would the naming be done then ?

Related: How should I name this branched alkyl substituent?


text:
Is there a possible distortion of XeF6 from Oh point group to reduced symmetry?
Below the molecular orbital diagram for $\ce{XeF6}$ (from What is the molecular structure of xenon hexafluoride?)

I am trying to figure out if a distortion from $O_h$ can take place and to which (lower symmetry) point group it would be. I know that it will involve the HOMO $a_g$ interacting with the $t_u$ orbital (bonding). Using a descent of symmetry table for $O_h$, I have found that the $a_g$ must remain completely symmetric. The part I am confused about is, looking at the descent of symmetry table, I am looking for an $a_g$ component of $t_u$ in a point group. I am using the following table:

$t_{1u}$ does not $a_{1u}$ in the table. Literature actually suggests $C_{3v}$ as the resulting distortion, but how would I know this when it isn't in the table? and I was wondering where I am going wrong in my approach?
Also, I am curious where the subgroups fit into the table, as $C_{3v}$ is not there but what does this mean?

Now, there are a few things that must be stated very clearly. Firstly, it pays to be very careful when you are talking about point groups. It's careless (and confusing to others) to write $\mathrm{T_{2u}}$ when you mean $\mathrm{T_{1u}}$, or to write $\mathrm{A_{g}}$ when you mean $\mathrm{A_{1g}}$, so please pay attention to this next time, regardless of whether you are writing online or submitting a piece of work. Good communication in science has to be precise and clear!
Secondly, let's use this reduced descent of symmetry table (I have removed some columns and some rows, which aren't relevant now)
$$\begin{array}{ccc} \hline
O_\mathrm h & T_\mathrm d & D_\mathrm{3d} \\
\hline
\mathrm{A_{1g}} & \mathrm{A_1} & \mathrm{A_{1g}} \\
\mathrm{T_{1u}} & \mathrm{T_2} & \mathrm{A_{2u} + E_u} \\ \hline
\end{array}$$
This tells you that if you decrease the symmetry of the system from $O_\mathrm h$ to $T_\mathrm d$, then anything that previously transformed as $\mathrm{A_{1g}}$ will now transform as $\mathrm{A_1}$. Likewise, anything that previously transformed as $\mathrm{T_{1u}}$ will now transform as $\mathrm{T_2}$.
If you are trying to look for potential mixing of orbitals in the lowered symmetry, then you have to look what the irreps in the lowered symmetry are. From what you have written, I sense that you're trying to "look for an $\mathrm{A_{1g}}$ component", which isn't what you should be looking for, because something that transformed as $\mathrm{A_{1g}}$ in $O_\mathrm h$ symmetry may not necessarily transform as "$\mathrm{A_{1g}}$" in the lower symmetry. Point in case: $T_\mathrm d$.
If you are considering the viability of a distortion to $T_\mathrm d$, then you should be looking for an $\mathrm{A_1}$ component in the second row, not $\mathrm{A_{1g}}$. Of course, $\mathrm{T_2}$ does not contain $\mathrm{A_1}$, so the distortion to $T_\mathrm{d}$ isn't favourable.
Now, in $C_\mathrm{3v}$ the correlations are $\mathrm{A_{1g}} \to \mathrm{A_1}$ and $\mathrm{T_{1u}} \to \mathrm{A_1} + \mathrm{E}$. If we add this column to the reduced table above
$$\begin{array}{cccc} \hline
O_\mathrm h & T_\mathrm d & D_\mathrm{3d} & C_\mathrm{3v} \\
\hline
\mathrm{A_{1g}} & \mathrm{A_1} & \mathrm{A_{1g}} & \color{red}{\mathrm{A_1}} \\
\mathrm{T_{1u}} & \mathrm{T_2} & \mathrm{A_{2u} + E_u} & \color{blue}{\mathrm{A_1 + E}} \\ \hline
\end{array}$$
then it is clear that because the $\color{blue}{\text{blue}}$ $\mathrm{A_1 + E}$ shares one irrep in common with the $\color{red}{\text{red}}$ $\mathrm{A_1}$ component, the HOMO–LUMO mixing is allowed in $C_\mathrm{3v}$ symmetry, and this provides a qualitative argument for why distortion to $C_\mathrm{3v}$ symmetry is favoured.
Again, to drive home the point, you shouldn't be looking for whether the $\color{blue}{\text{blue}}$ bit contains $\mathbf{A_{1g}}$. There is no $\mathrm{A_{1g}}$ irrep in $C_\mathrm{3v}$, so of course you won't find it! You should be asking whether the $\color{blue}{\text{blue}}$ bit contains the $\color{red}{\text{red}}$ bit.

How to get those correlations, is quite a fair question. Well, I got it from this diagram in Orbital Interactions in Chemistry, 2nd ed. by Albright et al. The correlation diagram 14.4 should be self-explanatory, hopefully (you can ignore the pictures of the MOs, 14.5 and 14.6).

But if you want to obtain it without flipping through every book in the library, this is how you need to do it. Because high-symmetry point groups like $O_\mathrm h$ contain tens and tens of subgroups, books don't usually list every single possible correlation to lower symmetry, since there are far too many possibilities. The table you were given has a bunch of omitted correlations, listed under "other subgroups" (the implication being, "other subgroups that I couldn't be bothered to put in"). You'll notice that $C_\mathrm{3v}$ is one of those "other subgroups" that weren't explicitly given.
The strategy therefore is to carry out the correlation in two steps. First you go from $O_\mathrm h$ to $D_\mathrm{3d}$, and then you go from $D_\mathrm{3d}$ to $C_\mathrm{3v}$. The tables you are using don't provide that, but a more complete list of tables is provided in the Appendixes of Molecular Vibrations by Wilson et al. The $D_\mathrm{3d}$ to $C_\mathrm{3v}$ correlation is given as (again I removed the irrelevant rows)
$$\begin{array}{cc} \hline
D_\mathrm{3d} & C_\mathrm{3v} \\ \hline
\mathrm{A_{1g}} & \mathrm{A_1} \\
\mathrm{A_{2u}} & \mathrm{A_1} \\
\mathrm{E_u} & \mathrm{E} \\ \hline
\end{array}$$
so now the correlation from $O_\mathrm h$ to $C_\mathrm{3v}$ is easy: $\mathrm{A_{1g}}$ in $O_\mathrm h$ goes to $\mathrm{A_{1g}}$ in $D_\mathrm{3d}$, which in turn goes to $\mathrm{A_1}$ in $C_\mathrm{3v}$. Likewise, $\mathrm{T_{1u}}$ in $O_\mathrm h$ goes to $\mathrm{A_{2u} + E_u}$ in $D_\mathrm{3d}$, which in turn goes to $\mathrm{A_1 + E}$ in $C_\mathrm{3v}$.


text:
Binary diagram and two phase region
I know that to find the composition in a two phase region of a binary diagram we have to draw the isotherm and to look at the intersection with phase boundaries.

But why? Is there a demonstration?

I will assume that the y-axis in your graph is temperature, and that the pressure of the system is fixed.
Consider the physical system: a mixture of two species in two phases is at equilibrium. Gibbs' phase rule tells us that we have two degrees of freedom, which can be chosen from the set $\{P, T, X_B^{liq}, X_B^\text{solid}\}$. By fixing $P$ (first degree of freedom) and letting $T$ (second degree of freedom) vary over a prescribed range, we produce the functions $X_B^{liq} = X_B^{liq}(T)$ and $X_B^\text{solid} = X_B^\text{solid}(T)$, and this is exactly what is represented by a phase diagram.

I understand this but for $\ X_{b} $ fixed and T fixed (the green dot), why $\ X_{b}^{liquid} $ and $\ X_{b}^{solid} $ gives us the composition in this precise case? I don't know if I'm clear

Actually, here you probably have $P$ and $T$ fixed, and $X_B$ just tells you whether or not you're at equilibrium. $X_B$ is not one of the degrees of freedom available because it doesn't tell you anything about phase compositions. For $P$ and $T$ fixed, I have argued that there is only one possible value of $X_B^\text{liq}$ and $X_B^\text{solid}$, and this can be graphically found by looking at the intersection of the curves with the line $T = T^$, where $T^$ is your known temperature.

The last statement is a simple property of functions: if we have a function $f(x)$, then $f(5)$ is the intersection of $f(x)$ with the line $x=5$.

A critical thing to keep in mind is that the red dot gives the composition of the liquid phase, and the blue dot gives the composition of the solid phase, but neither curve gives you the relative amounts of each phase, i.e. how much liquid there is compared to solid.  To get that, you need the lever rule, which is really just a restatement of the formula for weighted averages.


text:
Excited helium gas from the colision of alpha particles with beta particles?
As alpha particles are made out of 2 protons and 2 neutrons, they are the same as the helium atom with a +2 charge, but with a certain speed. The beta particles are made out of 1 electron at a certain speed. So, theoretically, "mixing" alpha particles with beta particles would create an excited helium atom, which then, would decay to a normal helium atom releasing the kinetic energy through gamma radiation, or simply light. Is this correct? If not, why?

Pretty much right "theoretically." Pragmatically nuclear process create the same number of positive and negative charges. You can't really "bottle" alpha particles and beta particles. So you can't have two such bottles that you just mix together. You'd have to have beams of charged particles so mixing would be poor. But you're right that a alpha particle is just a helium nucleus with a lot of kinetic energy.

Both beta decay and alpha decay are used in nuclear "batteries". So, yes, helium is generated in alphavoltaic power sources, though it is probably embedded just below the semiconductor surface. See https://en.wikipedia.org/wiki/Atomic_battery#Alphavoltaics

This is very much a non-issue. Alpha particles run at great speed, so do beta particles, and as soon as they cool down, they cease to be called alpha and beta particles. When that happens, they may react all right, but there is more to it.
Everything on Earth (look around you! yes, that too) is made of atoms. Helium has greater ionization energy than any other atom. So an alpha particle, as soon as it cools down enough to react, will snatch electrons from whatever else is nearby and become a neutral helium atom. Beta particles (provided they are cooled down, too) are as good as anything else.
But wait, they both cool down by bumping into matter, and matter contains electrons. You will have your helium atoms, but you'll never be able to tell whether the electrons come from beta particles or from something else. Nor does it really matter, after all.


text:
Do any efforts to construct artificial biochemistries exist and what compounds are they based on?
I'm a researcher interested in the fields of origins of Life, astrobiology and theoretical biology in general but I must admit I think my knowledge of chemistry outside biochemistry is a little bit limited. I'm well aware of the efforts to construct nucleic acid analogues and to construct expanded genetic codes with amino acids other than those found in the proteins in Nature but I wonder how far can we go as to construct a biochemistry which has nothing to do with our own? Is there anyone who had seriously investigated the question what kinds of compounds can be used to build a chemistry capable of providing the grounds for evolution into something we could classify as a lifeform? What classes of chemical compounds other than nucleic acids, proteins and lipids could develop into a real biochemistry? Are there any chemists who had seriously undertaken an effort to make such an investigation about classes of chemical compounds other than those known in biochemistry and if there are such efforts what molecules are they based on?
P.S. This is a question directed toward theoretical chemistry as much as it is a reference request, so I'm asking anyone who knows something about the matter to put their knowledge in the answer section whether it comes from fields related to biochemistry or not. From what I had researched so far there are some efforts to construct a biochemistry mimicking our own but my question isn't about biomimetics. I wonder could there be a biochemistry radically different from our own and how could I "screen" (or is that even possible) for classes of molecular species which could build it? I'm talking about a biochemistry not based on proteins or nucleic acids or even not having any type of "information" (whatever may stand behind such a word) at all. I'm asking this in the Chemistry SE because my interests are to learn could I select classes of molecular species which could do the job even in theory out of what we presently know about the possible diversity of chemical compounds or is our knowledge of all possible chemical compounds that could generate complex structures still too limited to shed any light on the questions of Life? Are we pass the point where we know enough about chemistry to theorize what molecules we could use to "build" a living organism or is there still a long way to go before we could say we know enough about all the possible molecules that can generate complexity? I want to know the answer to this question from a point of view of chemistry, not biology. I had enough experience with biologists and I could say the ones I met are too narrowed by biochemistry as we know it which I think is far from every possible biochemistry. Can you understand my background? Can you make any suggestions which fields of chemistry I should look into to find the answer to my query?

Long story short, we are not there yet.

Is there no effort at all by chemists to tackle the question? I mean, at lease to suggest some compounds can generate complex structures that could turn out to have multiple functional abilities? I can't imagine the only thing able to produce life to be an ensemble of 20 amino acids and 4 nucleotides when there are so many compounds out there. Can chemists, al least suggest classes of compounds able to generate complex structures like amino acids and nucleic acids do?

@YordanYordanov I think the biggest issue is that until we actually encounter lifeforms that either don't use carbon as their main building block or use it very differently then we have seen, it would be very difficult to determine what is feasible due to the vast space of possible compounds that we could generate. I think until we something remotely different than how life is composed here, it is very difficult to picture an alternative to proteins, DNA, etc. This Wikipedia page might link to some articles of interest: https://en.wikipedia.org/wiki/Hypothetical_types_of_biochemistry

https://chemistry.stackexchange.com/questions/70009/how-far-has-self-organizing-chemistry-gone

I just want to know how can I "screen" for compounds that can participate in a possible biochemistry out of the millions we had managed to create up until now. I think this is an issue which has a lot to do with chemistry and my question is right on the spot here. So, what's wrong with it?

@Mithoron, if you read the comment section to the question you refer to there is the remark there that this question is too broad to receive a definite answer. I try to narrow down the range this time and I think that question of mine here is far more "refined" as to the field I'm referring to. I just want to know did anyone tried to construct artificial biochemistry from a chemical perspective and if such an effort exist what compounds it's based on. How is that too broad? I think it's much better than the previous question and the field it had specified!


text:
Why do methane molecules have 15 degrees of freedom?
I've just been learning a little thermal chemistry including the equipartition theorem. As part of this, my textbook discusses how to figure out the degrees of freedom for different chemicals.
The answer to one of the exercises says methane has 15 degrees of freedom: 3 for translation, 3 for rotations and 9 for vibrations. I am confused by two of these.
Firstly the text says that diatomic molecules like oxygen gas only have 2 rotational degrees of freedom, because they can't rotate about their axis of symmetry. However methane has an axis of symmetry, so why does it have 3 rotational degrees?
Secondly I don't understand where all the vibration degrees of freedom come from. The molecule has 4 bonds, so wouldn't each bond have one degree for kinetic energy and one for potential energy, for a total of 8 degrees, not 9?

Sorry for my bad english, but did you learn about wagging and twisting? Because those can account for the "missing" degrees of freedom.

15 is there because you have 5 atoms which can move independently, and we live in a 3D space, and $3\ \times\ 5\ =\ 15$.
3 of these are translational degrees of freedom, which you seem to agree with, so I'll just leave it at that.
3 are for rotations, because the methane molecule is not linear. Axes of symmetry are irrelevant: indeed, you may rotate methane around the axis of symmetry as well as around any other axis. Being or not being diatomic is hardly relevant either, except for the fact that all diatomic molecules are necessarily linear; however, the converse is not true.
Subtracting 3 and 3 from 15, we get 9, and that's how many vibrational degrees of freedom we have, because they can't be anything else. To visualize them is another story, and a great deal more difficult. Counting bonds doesn't help much. The separation of degrees into kinetic and potential is not a thing at all. You don't have to do that anyway. Write $15\ -\ 3\ -\ 3\ =\ 9$ and call it a day.

So, there's 9 vibrational degrees not because we counted them, but because there has to be 15 total and we only have one category left? Sure, it works, but that doesn't sound right...

Yes, that's right, even if doesn't feel so.

Thanks for the response. The book I've been working through is Schroeder. It says: "A diatomic molecule can also vibrate, as if the two atoms were held together by a spring. This vibration counts as two degrees of freedom, one for the vibrational kinetic energy and one for the potential energy". Wouldn't be the first time, but is this an error in the book?

My understanding is that the remaining 9 don't refer to any one atom, but interactions involving pairs (or even larger groups?) of atoms. This page had some helpful visualizations.  Within the constraints of the molecule, the motion of any individual molecule has some affect on the others.

@chepner You are right. It is not even pairs, but mostly larger groups.

@leob That's plain nonsense. There is no such thing as a degree of freedom for kinetic or potential energy; instead, every single vibrational degree of freedom represents a motion in which potential energy changes to kinetic and back. Moreover, a diatomic molecule has but one vibration.

The axis of rotation for diatomics only matters because the symmetry is $C_{\infty}$, i.e., rotation by any amount gives the exact same thing. Methane's axes are $C_2$ and $C_3$.

@Ivan Ok, that statement from the text had me thoroughly confused. So essentially by definition the rotational degrees of freedom are those that remain left after translational and rotational degrees? They don't correspond with different physical motions or vibration modes that can be counted?

They do correspond to certain physical motions, but it is somewhat challenging to tell which are those in particular. We weren't asked to do that. We had a simpler task: to count the vibrations, and that's how we did it.

@IvanNeretin I believe the idea of two degrees of freedom for a vibration is a poor attempt by Schroeder to explain that the average energy of an oscillator is $k_bT$ using equipartition because the mean value of any quadratically varying term in the energy is $\frac{1}{2}k_bT$.

I have to agree about this being nonsense. Either the total number of degrees of freedom or the vibrational number of degrees of freedom have to come from somewhere.

If someone wants to read more about this, the following article essentially explains the same thing in this answer with a bit more context beforehand and visuals for triatomic linear molecules: https://www.tec-science.com/thermodynamics/kinetic-theory-of-gases/equipartition-theorem/#Vibration

@chepner Yes , I believe the diagrams are true. 6 degrees of Freedom for pairs of atom and 3 degrees of freedom(along X,Y,Z) for entire molecule  would sum to 9.


text:
What is the largest non-polymeric heterostructured molecule possible?
I must admit I'm not a chemist but a biologist, so please forgive me if I'm getting some of my terminology wrong. I want to ask a question about the biggest possible complex molecule that can exist in Nature.
I know that certain crystals like quartz and gems like ruby and diamond can grow very large and they could be called molecules. Also, polymers can grow very long with a complex 3D structure.
However my question is about a chemical compound which isn't a polymer but has a complex non-repetitive 3D structure of many covalently bound atoms. If we rely only on synthesis and try to construct the biggest molecule possible how big can we get?
Does the answer vary if we are talking about molecular size, molecular mass, number of atoms, 3D structure (I imagine there can be some very big molecules containing cavities which are lighter and have fewer atoms than other who have more atoms and are heavier but are dense) and so on ?
Or is there a single class or maybe even a single molecular species which combines all these characteristics and can produce the largest molecules no matter what characteristic we use to measure this quantity?
If there is such a class of compounds what it is and how do we synthesize it?

There is no "the largest". Also, not being a polymer isn't well defined. I guess you disregard proteins and DNA as polymers. Well, maitotoxin is pretty huge.

Brief suggestion: polyoxometalates. From abstract: "in fact polyoxometalate clusters are perhaps the largest non-biologically derived molecules structurally characterised".

Some candidates are humic acids, asphaltenes or lignins. I don't think there is any chemical limitation to the size of a molecule. Also as Ivan Neretin noted, it is difficult to define what is and what isn't a polymer. Even moderately large organic molecules are to some point repetitive, for example they may contain multiple phenyl rings, or methylene groups.

My query is about a molecule or a class of molecules that can generate very complex structures without common underlying pattern. Polymers and crystals can grow very large and diamonds can be considered "one molecule" but I wonder is there or, at least, could there be a very large molecule with unique structure which isn't based on common structural pattern? Can we synthesize molecules which can grow so large they could create very complex structures but aren't repetitive? Is there such class of chemicals?

https://chemistry.stackexchange.com/questions/80718/what-is-the-most-complex-molecule-we-can-manufacture-without-the-help-of-protein

I just want to know what is the largest molecule with "unique", e.g. not a polymer and not something repetitive, structure we can make. This is why I asked the question. What's wrong with that? @Mithoron, I well understand the problems associated with the word complex, this is why I deliberately omitted it. The same is the reason why I think my question is different from the one you cite. I didn't made the mistake of involving a discussion on complexity in the issue and I think the question is clear enough as it is.

Proteins are "polymers" and have unique structure, genetic sequence also is is quite specific thing.

We know of no inherent limitation concerning the size of a molecule. As you pointed out, certain macroscopic entities like diamonds, graphite or charcoal can be considered fully macroscopic molecules to a certain extent (especially diamond). If chemists let their imagination run loose, they can draw whatever size a molecule you wish to create and with the desired complexity.
Whether it can be synthesised is another matter. Even very small molecules, especially natural products, can be very resistant against attempts to synthesise them. The larger a molecule gets, the more mass a rather small sample will have. It will get increasingly tedious but by no means impossible. However, unless you have oil money, Silicon Valley money or a billionaire funding you, you will likely not get the chance to finish what you want.
One of the most well-known large and not inherently repeating (or so Nicolaou says) molecules is maitotoxin whose total synthesis is famously still being attempted by the Nicolaou group.


text:
Are triglycerides macromolecules?
I've previously been taught that triglycerides are macromolecules. However, a current professor of mine gave the definition of macromolecules as "polymers of covalently linked monomers". Since, triglycerides aren't formed from repeating monomers, are they considered macromolecules?

Depending on who you ask you will receive a different concept to define macromolecules. These concepts typically arise from the sub-field of those who made the definition, so a polymer chemist is more inclined to consider repeating monomers of exactly one type since that is what plastics are. At the same time, the branch of macromolecular chemistry usually deals with systems that are much smaller than your average multi-domain protein.
The definition most can agree on (to a certain extent) is that macromolecules must be ‘larger’ than ‘small molecules’ (note the pairs of inverted commas). In this definition, it is perfectly reasonable for lipids to be grouped as macromolecules as they are much larger than most secondary metabolites. However, it is true that they are not made up of subunits linearly connected (whether repeating or not) so one may not consider them biopolymers.


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Why is the ground state electronic configuration of Group 10 (Ni group) elements so anomalous?
While studying about d block elements I came across this table showing outer shell electronic configuration of group 10 elements

$\ce{Ni}$ $ $ $ $ $ $ $ $ $ $ $3d^8$ $4s^2$
$\ce{Pd}$ $ $ $ $ $ $ $ $ $4d^{10}$ $5s^0$
$\ce{Pt}$ $ $ $ $ $ $ $ $ $ $ $5d^9$ $6s^1$

(Source Concise Inorganic Chemistry, JD Lee, adapted by S. Guha Pg 563)
Now I cannot understand the reason behind this. If we consider that $\ce{Pd}$ has changed its config. to attain stability then what's wrong with $\ce{Ni}$ and $\ce{Pt}$? And should the outer shell electronic configuration of the ground state not be same in elements down a group?
Edit
To those who have marked my question as a duplicate of the one mentioned, I humbly request to explain me which answer to the above mentioned question (of which mine is marked as a duplicate of) tells about the anomalous behavior of group 10 elements and why $\ce{Zn}$ assumes $d^{10}$ config and none of the others in the group do the same and again $\ce{Pt}$ assumes a $5d^9 6s^1$ config, while $\ce{Ni}$ does not. Also please do point out  where can I find an answer to why in this particular group there is no similarity in ground state electronic configuration b/w elements of the group. Thank You.

There are relativistic effects in Pt, leading to stabilisation of 6s and destabilisation of 5d

@orthocresol if $\ce{Pd}$ can show $d^{10}$ config due to increased stability why is this not the case with $\ce{Ni}$. Moreover isn't the energy required to pair electrons in d orbitals greater than pairing electrons in s.

related https://chemistry.stackexchange.com/questions/2469/how-can-one-explain-niobium-s-weird-electronic-configuration

possible duplicate of https://chemistry.stackexchange.com/questions/2660/explain-platinums-electronic-configuration and of https://chemistry.stackexchange.com/questions/35487/anomalous-configuration-for-palladium

@Mithoron but why then $\ce{Ni}$ is different, it should also have followed $\ce{Pd}$ or $\ce{Pt} $. Moreover we see that elements belonging to a certain group show similar electronic configuration, but why in this particular group different elements take different paths to attain stability. Don't mind but I have asked about the group as a whole not the single exceptions. Btw the links u provided are helpful .

I think the relevant comparison should be with the earlier transition metal groups where $s^2$ configurations are more predominant.
In multielectron atoms, electrons interact in such a way that for a given shell quantum number $n$, the orbitals with higher total angular momentum $l$ are raised, thus for example $3d$ ($n=3, l=2$) becomes higher than $3p$ ($n=3, l=1$).  This angular momentum effect is most important when the orbitals are well shielded with a low effective nuclear charge, where the electron-electron interactions don't get swamped by electron-nucleus interactions.
In earlier transition groups the valence shells have this low effective nuclear charge and the angular momentum effect is so strong by comparison that the $d$ valence subshell is raised above the $s$ subshell even with one higher $n$ quantum number.  So, for instance, in the fourth period we see $3d$ orbitals filled only after $4s$ in neutral atoms.  Not so much in the ions, however, where ionization leaves the remaining electrons with more effective nuclear charge, hence we often see $d$ instead if $s$ subshell electrons remaining in the early transition ions.
But by the time we get to the later transition metals the effective nuclear charge has increased even in the neutral atoms, the lower-$n$ $d$ subshell drops relative to the higher-$n$ $s$ subshell, and we begin to see a greater preference for the $d$ subshell being filled first.  By the time we get to Group 12, this transition is complete and only the $s$ electrons remain valence electrons except, possibly, in extreme circumstances.

Here Platinum shows d9 s1 configuration due to the relativistic effect of 6s orbital. Which is inert towards oxidation.
But in case of Pd it is d10 configuration due to the full filled highly stable 4d orbital.  But Ni cannot show this
configuration.  When it comes to the case of Ni , it has 3d orbital in its valence shell so it is not so much large and   cannot supply spin pairing energy. So the configuration show d8 s2.That's why this group element show analomous behavior.

Welcome to Chemistry.se! From these statement, I cannot really connect the dots. What do you mean by 'cannot supply spin pairing energy'?


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Is there a simple assay to determine the elemental carbon content of biochar?
A little bit of background. I make charcoal for use as biochar: putting carbon into the ground where it retains water and soluble nutrients. It also provides a good habitat for the friendly soil bacteria and fungi, along with low-tech carbon capture and storage for the masses.
I would like to know how much carbon is going back into the ground, and be able to say with confidence what difference I'm making to my carbon footprint, but the two – at least – unknown factors are the ash content of the char, and the water content from quenching and dust abatement sprays. I can say with confidence that a kilo of dry char will retain 3 kilos of water, which makes it difficult to judge by the look and feel.
I'm looking for a simple assay which will tell me with reasonable confidence how much elemental carbon a sample contains.
FWIW, my understanding of chemistry is rusty: I've got an A-level (the UK university entrance qualification from 1976) but my degree is in Physics, and my background is in engineering. Thanks for any help.

Is your charcoal mostly carbon, water and ash ? Or are there other contaminents you care about ?

Carbon, -with the cellular structure of the wood still intact- , water and ash are the significant components by weight- the hydrocarbons are long gone.

Assuming your charcoal is mostly:

water (evaporates)
carbon (burns)
ashes (does neither)

I would :

weigh a sample of your charcoal to get the mass $\ce m_0$
dry it in an oven at low-ish temperature (50°-80° or more, whatever doesn't burn your charcoal) for a few hours.
weigh it again to get $\ce m_1$
burn the sample in air as thoroughly as possible
weigh what's left to get $\ce m_2$

Now you can calculate :

mass of water (+ other volatiles) = $\ce m_0$ - $\ce m_1$
mass of carbon (+ other non-volatile burnables) = $\ce m_1$ - $\ce m_2$
mass of ashes = $\ce m_2$

Isn't your mass of carbon calculation backwards? In other words, shouldn't you be subtracting the ash content (m2) from m1 (carbon + other non-volatile combustibles + ash)?

Nice and simple idea but m1-m2= c + (o, h..) as well. It doas not tell how much "clean carbon" you have in your char.

Thanks for the reply- the dry and burn assay is along the lines of what I was thinking. The challenge that I can see is supplying the air to burn the carbon away without carrying away any of the fine ash dust.

I don't think oxygen needs to be supplied. Charcoal burns well in air once ignited (that's the point). You could just set it on fire in a crucible some place protected from any wind. If time in a hurry, you could also experiment with crushi NM  the charcoal beforehand.


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How can Hydrogen NMR be used to distinguish between these two isomers?
I am having trouble explaining how and why the spectra of these two compounds would differ:

I am sure it has something to do with how the protons will couple. In the first isomer (left), the protons are anti-periplanar whereas in the second (right), I think they are staggered? So the first may be able to couple stronger than the second?

I think this will be very difficult because both groups are quite small, so chair inversion may be facile. If the fraction of both conformers is significant, the peaks will just be a smear and you likely will find it difficult to measure any coupling constants...

Also these are anomers so be careful with pH...

The difference is through space.

what does this mean?

See the nice answer

In the absence of any other configuration anchoring factors, I would not expect the conformation of the two structures to be reasonably stable at room temperature to be able to distinguish both by NMR. You may well need lower temperatures to freeze chair inversion and create a set of distinct, non-averaged peaks.
This issue aside, it is very easy to distinguish the two once you know you have a stable conformation. The key is the Karplus relationship which allows you to link a vicinal coupling constant to the dihedral angle between the two hydrogens. In general, a $^3J_\ce{HH}$ coupling constant will be very high ($^3J_\ce{HH} > \pu{9Hz}$) if the dihedral angle is either $0^\circ$ or $180^\circ$ — typically, if the hydrogens are anti-configured. On the other hand, the coupling constant becomes minimal for values around $90^\circ$ ($^3J_\ce{HH} < \pu{2Hz}$). For a $60^\circ$ gauche arrangement, a coupling constant of $^3J_\ce{HH} \approx \pu{3 Hz}$ is expected.
So the coupling between $\ce{H{1}}$ and $\ce{H{2}}$ would be around $\pu{9Hz}$ or $\pu{10Hz}$ in the first case but only around $\pu{3Hz}$ in the second — if the configuration is long-living on the NMR timescale.


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atomic orbital probability graph for p and s orbitals
In the probability density graphs of s and p and even other orbitals start from nucleus as shown in the picture given below

they all start from nucleus i.e. there is good probability of electrons of the particular orbitals being there.
my doubts:

there is proper energy variation and differences of energies of electrons being in particular orbital, example, energy of 2p will be greater than 2s etc. and as i think energy of electron is due to its position with respect to nucleus. as in the diagram probability regions of 1s 2s and 2p is same {though value of y coordinate is small}
then how can there energies be different?
don't the electron clouds of 1s , 2s , 2p intermix? moreover electron cloud- a 3D SPACE which must be wide enough to get electrons at different distances from nucleus doesn't violate the rule that electron has one constant energy in one orbital.

Energy is associated to the orbital not to a specific r


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How can I figure out which coupling constants match which hydrogen?
I am looking for some guidance assigning the $J$ coupling constants to the corresponding hydrogens. Here is my data:

I can obviously see that shift 3.56 is for the methyl group. I can also see that the 4.63 is for the hydrogen adjacent to the methoxy group as it can only couple ($J=\pu{0.8Hz}$) to one hydrogen next to it. Is the approach then to see that the hydrogen at shift 4.04 also has $J=\pu{0.8Hz}$ so this must be adjacent (going clockwise from C of methoxy, next C along)?

Yes, use the connectivity provided by matching coupling constants. Note that the methylene on the C6 carbon bears a pair of diastereotopic hydrogens, so these will be different. That should also give a hint about which hydrogen is at $\delta$ 3.43.

Coupling constants are mutual. If nuclei $\ce{A}$ and $\ce{B}$ couple and this coupling results in a coupling constant $J_\ce{AB} = \pu{7.4Hz}$ on $\ce{A}$ then $\ce{B}$ will also be split into two signals with a coupling constant $J_\ce{BA} = \pu{7.4Hz}$.
Therefore, you can indeed start at the anomeric hydrogen (which you can identify not only by the coupling pattern but also by its chemical shift) and work your way along the ring, noting whether two hydrogens are in gauche or anti configuration along the way.

I am confused how to assign the diastereotopic hydrogens to their shifts, please could I have advice on this?

@gamma1 You can’t distinguish the two diastereotopic hydrogens on $\ce{C{6}}$ unless you have some sort of external reference. Typically, one would label them H6.1 and H6.2 or something along those lines. The only small hint you have is the coupling constant which can be used together with the Karpluls curve to determine a dihedral angle.

So out of 3.79 and 3.98 shifts, I can assign them to either? thank you

@gamma1 Yes, there’s no way to tell.

As an aside, if you wanted to know the conformation of the molecule, could you use the data to find this do you think?

@gamma1 you can tell that H3, 4 and 5 are anti-configured due to the high coupling constants. Then, H2 must be gauche to H3 also from the coupling constant. H1 could be both axial and equatorial because both positions are gauche from an equatorial proton. And there is no way in NMR to tell absolute configuration.


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Which of the following is more stable?

In this image, does the arrows indicate correct rate of reactions? The upper arrow is longer and lower is shorter.
According to me Chair is more stable than Twist Boat. Does axial and equatorial arrangement of the two -C(CH3) groups play a role?

Check out this paper: http://pubs.acs.org/doi/abs/10.1021/jo051654z. Two reasons are mentioned: 1. Steric repulsion between an axial
methyl group and the axial hydrogens on carbons 3 and
5 of the ring; 2. Hyperconjugation interactions involving the methine hydrogen. I don't have time to post an answer now, so if anyone can do it using this reference as a starting point -- please feel free to do so.

Basically, a tert-butyl substituent is so bulky that it will never occupy an axial position.

Also note that the thin end of a hashed wedge bond should always go away from the ‘stereocentre’ i.e. from the quarternary carbon and not the methyl group here.


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creating an inert gas from air
I have asked this question in Beer, Wine & Spirits however I think the lack of chemists meant I received zero answers (two upvotes which was encouraging).
At the moment I and looking at ways of removing the oxygen from a small vessel (500cm3) of ambient air. My thought is that if I remove oxygen what's left behind is an inert mixture of gasses that can help to protect my wine. I have done some maths and if there is on average 20.95% oxygen in the air and 78.08% nitrogen, does this mean that if I completely remove oxygen then the percentage of nitrogen in the remaining gas mix increases to 98.7% (78.08/(100-20.95)? If not then can someone explain what I'm missing. If I can inject 98% nitrogen into my bottle of wine I'd be a happy chap.
If you need to know how I'm going to achieve this then it will be via an oxygen scavenger(s) placed within a sealed bag of ambient air.

Then what you don't know? What O2 absorbing substance to  use?

@Mithoron I know that gasses can be complicated, wondering if I am calculating these things correctly or if it really is as simple as subtracting O2 from the mix?

Well, water vapor and CO2 also should be absorbed for if it's supposed to be inert. The problem is probably that you won't completely get rid of O2 with any chemical absorbent so it may or mayn't bee good enough.

@Mithoron good point about the water vapour, though some studies have suggested a little co2 may be useful. I'm going to have to order a few scavengers and then find a way of accurately measuring oxygen levels.

I would think it's a lot cheaper to buy a tank of nitrogen than to use a scrubber for O2. Then just flush the container with nitrogen and it should be pretty close to an inert atmosphere for this purpose.

Will that protect wine much longer than if there were a bit of air left? Secondary, anaerobic, fermentation continues for weeks after free O2 is used up. See https://eckraus.com/wine-making-101/ . Just cap it and let the little animalicules gobble up the O2.


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What is the correct name for 2,4-diethyl-4-ethoxyhexane?

A classmate asks you to draw the structure of 2,4-diethyl-4-ethoxyhexane, which he cannot find in a chemical reference manual. Draw the structure of 2,4-diethyl-4-ethoxyhexane.

(Question continued...) The reason your classmate cannot find the compound in a chemical reference manual is that 2,4-diethyl-4-ethoxyhexane is the incorrect name for the compound. What is the correct name?

Here are my steps:

I count the longest carbon chain = 7 carbons = -heptane
I look at all the substituents: 1 ethoxy group, 1 ethyl group and 1 methyl group
I number the carbon chain starting from the end closer to the ethoxy group (alphabetical order)
I add all the substituents and the carbon chain together

In this case, this is the name I come up with: 3-ethoxy-3-ethyl-5-methylheptane, which is incorrect. Can somebody tell me where I went wrong, please?

Your answer is correct, it looks like the answer source you are looking at is wrong. I found this on Chegg and this from a google search.

@Tyberius That would make sense to me, as I really do not understand what I might have done wrong. I guess, my school chemistry software is wrong for once... Thanks!

@FriedaJewel I would always be on the lookout for that with online chemistry homework. Every once in awhile you will run into questions that have the wrong answer or require the correct answer in some strange format.

(This one was easy, but) If you absolutely can't find a solution, you can always search for the sum formula (might be tedious) or use scifinder et al. who let you search for a structure sketch. Also some of the common computer programs to draw chemical structures can generate names, often correctly.

In short, your answer is correct. The steps that you have taken are correct, so it's likely the online homework tool has a key mistake.

This CW answer exists for the OP to accept it, and mark this question as resolved.


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Comparison of Resonance Energies of some Carbocations
We were given the following carbocations and asked to compare their resonance energies.
$\ce{L-CH2+}$, where $\ce{L}$ is:

$\ce{NH2}$
$\ce{OCH3}$
$\ce{Ph}$
$\ce{F}$
$\ce{Cl}$
$\ce{NO2}$

and the order was given as:
$1 > 2 > 3 > 5 > 4 > 6$
I am clear with the other orders but I am unsure about $\ce{F}$ and $\ce{Cl}$. Why does $\ce{Cl-CH2+}$ have a higher Resonance Energy than its $\ce{F-CH2+}$ counterpart even though in resonance, $\ce{Cl}$ forms $\ce{{3p}-{2p}}\ \pi$ bonds compared to $\ce{{2p}-{2p}}$ ones in $\ce{F-CH2+}$, which are less stable due to less efficient overlapping? $\ce{F-CH2+}$ does have a higher electronegativity, but is it enough to overcome the higher stability of the $\ce{{2p}-{2p}}$ bond over the $\ce{{3p}-{2p}}$ one?

It is not like π bond between fluorine and carbon is very stable and that is where your line of thinking is wrong. While it is true that $\ce{{2p}-{2p}}$ π bonds are more stable in general than their $\ce{{2p}-{3p}}$ counterparts, one should not take that alone as a decisive factor.
The question is comparing the electron-donating capabilities of fluorine and chlorine, and that italicised word there alone should ring a bell. Fluorine is much more electronegative than chlorine which means that a carbon bound to fluorine has a much smaller electron density which destabilises cations much more than could be offset by the overall weak π donating capacity of fluorine.
Remember that for electrophilic aromatic substitutions, halide substituents are classified as deactivating — mainly due to their $-I$ effect — and only weakly ortho/para-directing — due to the smaller $+M$ effect. This should serve to show which effect predominates and it only gets more pronounced if instead of a large aromatic system you only consider the atom directly adjacent.

So, is the bond stability factor overlooked when dealing with such examples

@Ayushmaan It’s not overlooked, it’s simply minor.


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Does H+ in the ocean bind with water to form H3O+ or with carbonate ions to form HCO3-?
I've read so many references on the Internet about ocean acidification and basically it can be described by the following equations
This is when $\ce{CO2}$ dissolves into ocean to form $\ce{H2CO3}$: $$\ce{CO2 + H2O <--> H2CO3 <--> H+ + HCO3-}$$
This is when excess $\ce{CO2}$ reacts with pre-existing carbonate ions to form additional bicarbonate: $$\ce{CO2 + H2O + CO3^2- -> 2HCO3-}$$
Based on the two equations, I can tell that $\ce{H+}$ which are positively charged are consuming $\ce{CO3^2-}$ which are negatively charged and good for shells to make unnecessary bicarbonate, so my first thought was to remove $\ce{H+}$ only. But then my chem teacher said that $\ce{H+}$ can also bond with $\ce{H2O}$ to form $\ce{H3O+}$ and that needed to be separated back into water and $\ce{H+}$ and only then can $\ce{H+}$ be removed.
I'm so confused, are both methods the same? And any suggestions on how I can attract hydrogen ions together?

Acidic protons are hard to capture. They will jump wherever they feel like until they have found a reasonably strong base.


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Why we don't talk about unit cells in hcp/fcc structures?
I'm studying basics of Solid State Chemistry from this source.
So there are 14 types of Bravais crystal lattices (Primitive + Centered) in 3D. Every solid in this universe is made by the unit cells of these lattices. OK so far so good.
Later on there was the topic of Close packing and Structure of Metallic Crystals and we are told about hcp /fcc types of structures.
Problems:

Why do we remain silent about the unit cells and crystal system in this context? It seems like a huge disconnect.
Is there any relation between the Hexagonal Unit cell and hexagonal close packing?

Certainly many books are not silent on unit cells, how to construct them, and the difference between customary unit cells and ones that reflect the full unit cell symmetry (such as Wigner-Seitz unit cells). So I'm quite unclear about just what you are asking.

We do not remain silent about the unit cells in this context. The hexagonal close packed structure is called so because it has hexagonal unit cell, and the other one is called fcc precisely because its unit cell is face-centered cubic.

(source)


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How much energy is needed to convert 20.0 grams of water at 15.0 degrees C to steam at 100.0 degrees C?
How much energy is needed to convert 20.0 grams of water at 15.0 degrees C to steam at 100.0 degrees C?

Show your work.

Enthalpy of vaporization of water: $40.65\pu{\frac {kJ}{mol}}$. This is the amount of energy to convert one mole of water at boiling point to gas.
One calorie = $4.184$ $\pu{J}$ = Energy required to raise the temperature of one gram of water one degree Celsius.
Molar mass of water: $18.01528\pu{\frac{g}{mol}}$. This is the mass of one mole of water.
That being said, $\pu{E_{Total}}$ = $\pu{E_{h}}$ + $\pu{E_{v}}$, where $\pu{E_{h}}$ is the energy used in heating it and $\pu{E_{v}}$ is the energy used to vaporize it. $\pu{E_{h}} = 4.184$ $\pu{\frac{J}{cal}}$ $\times$ $20$ $\pu{g}$ $\times$ $85$ $\pu{°C} \approx 7112$ $\pu{J}$. $\pu{E_{v}}$ = $40.65\pu{\frac {kJ}{mol}}$ $\times$ $20$ $\pu{g}$ $\times$ $1000\pu{\frac{J}{kJ}}$ divided by $18.01528\pu{\frac{g}{mol}} \approx 4513$ (significant figures). $7112 + 45130 = 52242$ $\pu{J}$.

Good answer, but it would be preferable if you didn't give a complete answer. We don't like to show ourselves as a homework completion site.


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Carbon footprint rules of thumb
Could someone perhaps check the rules of thumb that I use for estimating the carbon footprint of hydrocarbon fossil fuels- I'm using schoolboy chemistry theory, and possibly falling into schoolboy traps.
2 litres of LPG weigh 1 kilo
4 litres of petrol/gasoline weigh 3 kilos
6 litres of diesel fuel weigh 5 kilos
If I assume- can I?- that the above fuels are saturated hydrocarbons, propane or longer, with two hydrogen atoms for every carbon atom, then 7 kilos of fuel correspond to 6 kilos of carbon, which in turn correspond to 22 kilos of CO2.
This is a simplistic view, which doesn't take account of the production and distribution impact of road fuels, but if say, I fill up my 1980 Land Rover- whose environmental impact of manufacture ended 37 years ago- with 100 litres of LPG, I can think : 100 litres of fuel, 50 kilos of hydrocarbons, 42 kilos of carbon. I've got to deploy 42+ kilos of biochar before I can drive on with a clear conscience.
Thanks for any help.


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What dynamical features allow for the existence of a liquid-liquid transition?
Recently it has been shown theoretically[1] and then experimentally[2] that a liquid-liquid transition (LLT) takes place in pure supercooled water. That is to say, two distinct liquid phases have a direct transition between each other on the phase diagram. This was a very controversial result as I understand it (and is something people have argued about for a while apparently). It is not the existence of an LLT that is controversial, but rather the existence of an LLT in water.
I am rather confused by this whole idea, however. Ordinarily, when we talk about phase transitions, there are very clear dynamical and energetic reasons why the transition should take place. For instance, water has hydrogen bonds that get weaker and weaker as the temperature increases because the molecules deviate from the minimum energy structure more often which leads to repulsion. Thus, the existence of a phase change from liquid to vapor must exist because at some point the repulsions overcome the attraction. Similar kinds of arguments can be thought of when going from the liquid to solid, and even from solid to solid.
I cannot think of a similar simple competition which makes it clear to me why a liquid-liquid transition should ever happen. I have looked at ref. [3] which makes it clear that the two phases (of this mixture, not water) display different average hydrogen-bonding arrangements, one of which is bimodal. It isn't so surprising that there could be two different hydrogen-bonding arrangements in a phase diagram. What does surprise me is that the path between two hydrogen-bonding arrangements is not smooth.
So, to say it as simply as I can, what competing interactions are responsible for the existence of a LLT in water?
One important point is that all the qualitative features of this transition were described by the TIP4P 2005 water model in ref. [1]. TIP4P has a potential energy function described by Lennard-Jones forces. This confounds me even more because the physics exists in a simple Lennard-Jones fluid with only pairwaise interactions.

References:
[1]: Singh, R. S., Biddle, J. W., Debenedetti, P. G., & Anisimov, M. A. (2016). Two-state thermodynamics and the possibility of a liquid-liquid phase transition in supercooled TIP4P/2005 water. The Journal of chemical physics, 144(14), 144504.
[2]: Zhao, Z., & Angell, C. A. (2016). Apparent First‐Order Liquid–Liquid Transition with Pre‐transition Density Anomaly, in Water‐Rich Ideal Solutions. Angewandte Chemie, 128(7), 2520-2523.
[3]: Bruijn, J. R., van der Loop, T. H., & Woutersen, S. (2016). Changing Hydrogen-Bond Structure during an Aqueous Liquid–Liquid Transition Investigated with Time-Resolved and Two-Dimensional Vibrational Spectroscopy. The journal of physical chemistry letters, 7(5), 795-799.

Phase transitions occur when not only a microscopic feature changes but the ensemble of particles develops a new macroscopic property. Gas to liquid: Surface tension. Liquid to crystalline: Macroscopic order. If a thousand particles fall into that lower state, you cannot lift a single one out of it any more so easily. It's a bit less clear what happens here, but it must be something similar. We have liquid crystals, we have quantum effects like in liquid helium, hydrogen bonds can play a similar game and form low-energy ensembles that segregate from the rest.


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If double and triple bonds are stronger than single bonds then why are unsaturated hydrocarbons less stable than saturated hydrocarbons?
I have learnt that unsaturated hydrocarbons are more reactive and less stable.  However if double and triple bonds are stronger and shorter bonds than single bonds, aren't they supposed to be more stable and inert?

Aren't you forgetting the fact that for every $\pi$-bond you break, you add two $\ce{C-H}$ $\sigma$-bonds?

Double bonds are stronger all right, but their $\pi$-components are not.

Although they are stronger, the presence of more electron density also makes them more attractive to electrophiles

When I was a first year, this concept was hard for me to grasp as well. The idea is that, even though the bond itself (double or triple) is stronger than a single bond as a whole, we must consider the components of a double or triple bond. A double bond consists of one sigma and one pi bond.
Let's take ethene for example. The $\ce{C=C}$ double bond has one sigma and one pi bond. The sigma exists because of two, sp2 hybridized orbitals overlapping head-to-head. This type of covalent bonding is very strong. But since it is only sp2, there is one p orbital leftover that was not hybridized. So, each carbon takes their left over p orbital, and forms a pi bond from the p orbitals overlapping side-to-side (the p orbitals are parallel). The unhybridized p orbitals are perpendicular to the sp2 orbitals.
Pi bonds, because of the way they are constructed, are much weaker than a sigma bond due to side-to-side overlap, versus head-to-head. Having said, that pi bond can be easily cleaved by an incoming group. This properties of the pi bond is what makes an unsaturated hydrocarbon so much more reactive and unstable compared to a saturated hydrocarbon.


text:
Stereo chemistry of Hoffman elimination reaction?
Does Hoffman elimination go through syn or anti elimination?
I referred Solomons and Fryhle,Paula bruice,L G Wade but none mentioned about the stereo chemistry of the reaction.
Does it vary as per the reaction conditions or is it unknown?

My guess is that it's an anti elimination, considering the fact that it's used to make trans-cyclooctene.

I wondered why all the authors I mentioned in the post mentioned about the stereo chemistry of cope elimination but not that of Hoffman elimination.


text:
Unknown alcohol sample. Chromic acid negative but iodoform positive?
I have a unknown sample of an alcohol (ceric nitrate positive) but when I did chromic acid test it was negative since it never turned into blue/green and against all odds iodoform was positive... yellow precipitate and a lot...
I have no clues of how to make sense into this. So I did lucas test as well but noticed later that it didn't help because my sample is not very soluble in water.
Judging by the great amount of precipitate I got from the iodoform test, I'm thinking that maybe my sample is indeed a primary or secondary alcohol but it has something that won't react with the chromic acid. Are there any considerations to account for when doing this test? or some limitations?
Please help!

I would recommend to repeat all the experiments one more time. The reasons for such results may be:

You had dirty test tube for ceric nitrate test, and it was really false positive. You could have a methyl ketone, which gives negative chromic acid test and positive iodoform test.
You added too much chromic acid, and had low amount of your "alcohol". That caused all alcohol to be oxidized, but that blue-green color was too faint, and you didn't notice it because of chromic acid excess.

Moreover, if your "alcohol" is immiscible with water, that means it is one of the higher alcohols. You could also try to identify it (if it's really an alcohol) by smell: those alcohols have pretty distinctive smells. x.x

Reminds me of the time when my Russian co-worker took the flask off the rotavap, sniffed it and proclaimed ‘Smells like product’.

It indeed smelled different than any other alcohols I have smelled. It had a faint mint smell.

You could have a tertiary alcohol that also contains a methyl ketone functionality.  Ceric ammonium nitrate will oxidize tertiary alcohols (because it oxidizes to an alkene rather than a carbonyl), whereas chromic acid cannot oxidize a tertiary alcohol, since that'd result in a "Texas carbon".

What idls a Texas carbon?


text:
A question on spin of an electron
Can an electron have zero spin?
In para hydrogen, since the nuclei have opposite spins, the they add up to 0.
So is it true that one electron can have zero spin?

No, it can't.$,$

Note that your example (para hydrogen) has two electrons, not one.

@VicLineal para/ortho hydrogen is about nuclear spin. Number of electrons does not matter there.

Welp, brain fart on my behalf.
Number of spin particles is still relevant to the OP's misunderstanding: note that your example (para hydrogen) has two particles with spin, not one. [The fact that both electrons and hydrogen nuclei are fermions is what makes both cases comparable.]

No, electrons (and other leptons) are fermions and therefore must have half-integer spin. It is possible to couple the spin of several electrons together to obtain an integer total electron spin. Depending on the total electron spin of the atom or molecule the states are referred to as singlet, doublet, triplet, etc. for $S=0,1/2,1$, etc.
The hydrogen molecule has 2 electrons and the total electron spin can thus be zero (singlet) or 1 (triplet). The ground state of hydrogen is a singlet state, which is the case for most molecules, a notable exception is the oxygen molecule which has a triplet electronic ground state.
It is important to realise that the terms para and ortho have nothing to do with the electronic spin, but are used to distinguish between two different total nuclear spin values, para hydrogen has $I=0$ and ortho hydrogen has $I=1$.