text:
How to calculate the 15 geometrical isomers of an octahedral complex with all ligands different?
Suppose we have a complex of type $\ce{[Mabcdef]}$ then we know it’ll show geometrical isomerism. According to my textbook (Cengage Inorganic Chemistry II, Page 7.37) there are 15 geometrical isomers possible.
I tried calculating it but ended up with different answers every time.
I think we have to use the knowledge of permutations and combinations for this calculation.

Permutations are irrelevant. Just continue counting until you get the same answer three times in a row, and then call it a day. Sure, there is an easy way via Polya enumeration theorem, but it is a lot harder.

Related: Finding out the number of geometrical isomers in coordination compounds and How many different possible structures (isomers) are there for the complex, [Mn(OH2)3(NH3)3] 3+?

Assign "a" to one of the six equivalent positions.
Note that there are now 5 choices (b,c,d,e,f) for the opposing position.
You are left with four equivalent equatorial positions.  Once a ligand is assigned to one of these four, there are 3 ligands remaining to select for the opposing equatorial position.
Finally, there are 2 ways to assign the last two positions. The two possibilities are mirror images of each other.

So $5\times3\times2= 30$
30 isomers (15 pairs of enantiomers).

Great answer, but I need a clarification. In step 2, if I chose group b anti to group a, I will get four equivalent equatorial positions. However, if I choose group b cis to group a, I'm unable to see how that gives me four equivalent equatorial positions, as you've said. Could you please clarify? Thanks!

@GaurangTandon In step 2, you are choosing which element to put across from group a, so if you are following this procedure you don't have the option of putting anything in the cis positions yet.

Every geometrical isomer would have 3 unique pairs of ligands that would be opposite (or trans) to eachother.
To divide 6 different objects into 3 pairs: $$\frac{6!}{2!^3*3!}$$
In case you don't know this method of division into groups, the answer provided by Dave would be easier.
But if you do want to get into the math part, then the formula can be worked out as:

Permutations of 6 different objects. $$=6!$$
But, two objects belong to the same group, which makes them "similar". So, divide by 2!. Thrice, because 3 groups.
  $$=\frac{6!}{2!^3}$$
The order of the three pairs is not significant to geometrical isomerism. So further divide by 3!.
  $$=\frac{6!}{2!^3*3!}$$

Theoretically, there are 15 possible geometrical isomers for 'Mabcdef' type complex. Possible geometrical isomers (in terms of combinations three ligand pairs) are as follows :

ab cd ef
ab ce df
ab cf de
ac bd ef
ac be df
ac bf de
ad bc ef
ad be cf
ad bf ce
ae bc df
ae bd cf
ae bf cd
af bc de
af bd ce
af be cd


text:
Is there any difference between table salt and common salt?
Iv'e been reading about $\ce{NaCl}$ and this doubt crossed my mind. Are they anything other than just household terms?

in general it’s frowned upon to accept answers before 24hr is up. That way people around the world get to answer.

There are fewer tables in common salt.

Lol.What was that supposed to mean?

They are both the same. If you look at this wiki page, it uses common salt and table salt as the same name. It's mainly just household terms, like people calling hydrochloric acid muriatic acid etc.

The main difference between sea salt and table salt are in their texture, taste and processing.
Sea salt is produced through evaporation of ocean water or water from saltwater lakes. Depending on the water source, this leaves behind certain traces of minerals and elements who add flavor and color to sea salt.
Table salt is mined from underground salt deposits and usually contains an additive to prevent clumping.

So does iodinated salt come under table salt?

Most table salt has added iodine, an essential nutrient.

There is no difference in taste though. Texture, certainly.

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text:
Can Hempcrete be an effective blocker of gamma radiation?
My basic question is if Hempcrete walls or blocks could potentially be effective against gamma radiation. Now I want to say no due to gamma rays being destructive against carbon material in general and well hempcrete is largely carbon based.

All depends on how much radiati0n incident, how much can be allowed through and how thick the barrier. Gamma radiation alone should do little or no damage, even if it creates some free radicals in the organic matter.

For blocking of gamma radiation, materials are mostly irrelevant. It is just the sheer atomic number that matters: the higher, the better.

It will depend on the energy of the gamma photons, if it was something very low energy such as the 30 keV photons from americium-241 then a thick layer of carbon would be usful. But for things like Cs-137, Co-60 and Zr-95 a few inches of carbon or plastic will be close to worthless.
Keep in mind that even with a low density material like water or air it is possible to attenuate even high energy gamma photons like those from Co-60 and Na-24. By using miles of separation from the radiation source you can screen out much of the gamma radiation and also take advantage of the inverse square law.
I have been inside a large scale used nuclear fuel store and I have also stood above a TRIGA reactor and I was able to look down into the ponds, I had about 10 meters of water between me and the fuel. I have seen large amounts of used fuel inside a hot cell. There I had about 3 meters of lead glass between me and the fuel. Both ten meters of water and three meters of glass do give you some funny optical effects. It can distort the shapes of things. I suspect that I would be able to see more clearly without the shielding but I would only be able to spend a very short time in the room before I would have to leave.
There is one use of a plastic sheeting, this is as a radioactivity / nuclear version of an antispall liner.
When you subject high density substances such as lead to high energy gamma rays the surface of the lead emits secondary radiations such as photoelectrons, compton electrons and lower energy photons.
The best thing you can do if you have a wall of lead is to cover it with a sheet of steel. The steel will stop a lot of the secondary radiation from the lead. Then you put up a sheet of aluminium on top of of the steel to block the secondary radiation from the iron. Then if you are very keen you add a layer of plastic to finsih the job off.
Another option would be to use concrete made with either barium added to it or lots of steel shot added to the concrete. This gives good shielding against both gamma and neutrons.
If you think that you will have exceptionally high energy gamma rays, then you need to worry about neutron production as a result of the photonuclear reactions. These only become something to worry about normally above 2 MeV photon energy. For this I would suggest a thick lead shield and then plastic to slow down the neutrons, even if you do not stop the neutrons by slowing them you reduce the amount of harm that they can do. So a thick layer of the carbon based plastic on the human side of the shielding would be helpful under these conditions even if we ignore the secondary radiation from gamma photons at about 1 MeV (like Co-60).
Another use of the plastic sheeting is if you have large amounts of very high energy beta emitters it is better to adsorb the beta particles in a material with a low atomic number as this reduces the amount of X-rays that are generated as the electrons slow down. So for vast amounts of Sr-90 and P-32 / P-33 I would suggest a thick layer of plastic followed by a layer of lead to catch any of the X-rays generated in the shielding. If you wanted to do a perfect job you would use the graduated Z method on the outside of the shielding pot for super high amounts of Sr-90 and other high energy beta emitters.
So while the plastic is close to no use on its own for stopping gamma rays it can be used to optimise the shielding.

Well, that's a lot of talking. You should add a TL:DR, and relate to the specific case of OP.

What is a TL:DR ?

Short for "too long: didn't read" - short clear statement, which your answer lacks. BTW if you don't know something, google it.

Well the problem is with this area that there are few times when an easy answer can be given as gamma photons come with a very wide range of energies. Some as low as 10s of keV some as high as 10 MeV or more. It is like me expecting that an organic chemist can generalise what all organic compounds do.

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Would ionic compounds dissolve in methanol?
Question:

Determine whether liquid hexane ($\ce{C6H14}$) or liquid methanol ($\ce{CH3OH}$) is the more appropriate solvent for the substances grease ($\ce{C20H42}$) and potassium iodide ($\ce{KI}$).

It seems logical that $\Delta H$ to dissociate $\ce{C6H14}$ would be small, since it's non-polar, that $\Delta H$ to dissociate $\ce{C20H42}$ would also be small, since it's non-polar, which would be conducive to a solution forming between those two.
The question that I'm stuck on is if the $\Delta H$ of formation of a solution between $\ce{K+}$, $\ce{I-}$, and the mildly polar $\ce{CH3OH}$ would be greater than the amount of energy it would take to dissociate potassium iodide. How can I estimate how much energy will be released by the attraction of the $\ce{I-}$ to the slightly positive hydrogen on the hydroxide group is greater than energy required to dissociate $\ce{KI}$?
Am I missing something?

You don't need to do compare the said enthalpies. Maybe the process of dissolution of $\ce{KI}$ in $\ce{CH3OH}$ is energetically favorable, maybe it isn't (the latter seems more likely to me); that's irrelevant. Thanks to entropy, everything dissolves in everything, if only in minuscule amounts.
You just need to compare solubilities of $\ce{KI}$ in methanol and hexane, which basically boils down to the comparison of solvation enthalpy in methanol vs. that in hexane, and here the difference in their polarity provides a clear and obvious answer.


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Why are powder XRD axes sometimes plotted in reverse?
Just plotting some powder XRD data and I wanted to know why the axes in some of the literature that I'm looking at is plotted in reverse. Is there a reason it starts at 60 Degrees (2 theta) and then goes down toward zero?

In the days of old it was common to plot these data against $d$ rather that $\theta$, which would kinda explain the reverse direction.

Best reason I can think of is that you're being given data in momentum space rather than position space.
You can convert from "position-time" space, which is what we're comfortable with as humans, and describe the system in "momentum-energy" space, which is simpler from a physics point of view. Position is the reciprocal of momentum and energy is the reciprocal of time. If you convert the diffraction equations to momentum ($k$), which is proportional to $\frac{1}{sin(\theta)}$, then the diffraction equation $\lambda = 2dsin(\theta)$ is transposed to $k' = k + K$ where $K$ describes the peak in momentum space, $k$ describes the momentum of the incoming photon, and $k'$ describes the diffracted photon. This equation is linear and much easier to manipulate than the "real-space" form of $\lambda = 2dsin(\theta)$.
The equation in "momentum" space (k-space) is difficult to understand but is nice and linear from a physics standpoint.
https://en.wikipedia.org/wiki/Crystal_momentum


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Is it possible to form an 8 carbon ring using a Diels-Alder reaction?
I need to convert the starting material to the product on the right but I was wondering if it was possible to use a Diels-Alder reaction and use $\ce{KMnO4}$ to cleave the ring.

A better way would be the tetramerization of ethyne to cyclooctatetraene, then hydrogenation to cyclooctene followed by ozonolysis.

what is tetramerization?

The combination of 4 molecules (monomers, here ethyne) to a single molecule (here cyclooctatetraene).

You can obviously use Diels-Alder Reaction for the synthesis of the desired product but not so directly. But Diels-Alder will give an optimal yield when the ring formed is Six-Membered. After the formation of that Six-Membered product, you have to increase the no. of carbon atoms in the carbon-chain by some way, and the yield of your product will be also better if you no. of steps as less as possible.
I will not suggest you to directly add 1,3-butadiene to acetylene because there will be a chance of aromatisation, and then you cannot do anything, as to break the aromatic rings are highly unfeasible. Instead I can suggest you the following steps:

First you partially reduce the acetylene to ethene, and then perform Diels-Alder reaction with butadiene. You will get cyclohexene.
In the second step, you perform ozonolysis, and then to the resulting compound, perform  Wittig reaction with excess reagents. You will get octa-1,7-diene.
Lastly, you can perform hydroboration, oxidation to get less substituted diol, and then oxidise that to acid by any strong oxidising agent like $\ce{KMnO4}$ or $\ce{K2Cr2O7}$ in slightly acidic medium.

Thus, you will get your desired product. The path of reaction is shown below:

is it possible to use h2 pd/c instead of the witting reaction to reduce the oxygen? My professor never taught that reaction so I don't know the mechanism.

NO! You cannot. The reduction by $H_2/Pd $ will reduce the double bond too and you can't proceed further. In fact this reagent can reduce almost everything to saturated forms , which are not always desired. That 's why different reducing agents are made which can reduce to certain extent according to our choice.

Please also observe that by Wittig reaction, I am increasing the no of Carbons in the chain. You cannot do it by using a reducing agent..


text:
What do you think is a good way to explain electronic configurations?
Understanding electronic configurations is vital, but a huge number of students face difficulties in understanding the why and the how. What approach do you think I should take in explaining electronic configurations to a student who only knows the Bohr model? Thanks.

Then the student needs to learn a lot about orbitals and aufbau principle.

Teach them about orbital theory before you teach them about electronic configurations. Teach them the primitive ideas; that electrons exist in orbitals,  and that orbitals exist in subshells. Electronic configurations are written in terms of subshells so you'd be skipping a fair bit of content if you jumped straight into electronic configurations - maybe this is why students find it difficult?

I seriously think any good general chemistry textbook can explain this, and be supplemented with videos if they learn better that way. See here.

If your motive is to explain how and why $mvr = n\hbar$, then use the de Broglie's wave particle equality. Electrons must form standing waves at each orbit.

As an undergrad student that has recently learnt that the Bohr model is just an approximation at best, and mostly incorrect, I think I am well equipped to answer this.
When I first learnt about the structure of an atom, I was told that "electrons move around the nucleus in fixed paths having fixed energy, called orbits. The angular momentum of the electrons is an integer multiple of $h/2\pi$."
When I was first taught the electronic configurations and the quantum mechanical structure of an atom, my teacher explained it thus (and it was easily understood by most of the class): he made an analogy between a person's location and the location of an electronic.
Just how a person can be located if we know which country they're in, which city they're in, which locality, which room, and finally which area in the room; an electron can be approximately located if we know its principal quantum number, its azimuthal quantum number and its magnetic quantum number. Another quantum number, i.e., spin quantum number tells us which direction it is spinning in.
After this, a brief historical understanding of why the Bohr model was discarded and this model was proposed was given to us. The Heisenberg's uncertainty principle, and the Schrodinger equation were introduced, as the uncertainty principle explained why Bohr's model was incorrect, and the Schrodinger equation explained how to find the probability density of the presence of the electron.
This method helped me understand as it felt like I was transitioning through the same phases that all the people of the scientific community had to when the quantum mechanical model was on its way in, and the Bohr model on its way out. This really felt like a fluid transition, rather than "Bohr was wrong, this is right", which would've caused a lot of confusion (at least for me).
Finally, there isn't a one size fits all answer to this, as every student is different. I am just sharing my experience.


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Is the dimethylglyoximate anion ambidentate or not?
In class, my professor said that dimethylglyoximate is a special ambidemtate ligand that can donate lone pairs via both nitrogen (which it normally does), and oxygen (very rarely).
On googling whether it is ambidentate or not, I didn't find anything. Can anyone cite an example where dmgH is donating lone pairs of electrons via oxygen and not nitrogen?
If this doesn't exist, can someone confirm that dmgH is not ambidentate?

Here is an example, in which the oxygen atoms serve as "bridging" functions http://pubs.rsc.org/en/content/articlelanding/1990/dt/dt9900000469#!divAbstract

Still, I don't know any case of dmgH complexes in which the ligand bounds with O and N to a single metal center (and I wonder if it could even be possible, since I think that a good orbital alignment with both O and N on the same metal center couldn't be possible)

@The_Vinz thanks for that, can you further explain why there won't be a good orbital alignment? I don't know how it works..

The reason why I said that is because (writing the Lewis structure might help) in the O-N-C-C-N-O chain, nitrogen is planar and sp2, and carbon is planar and sp2. Now, if you wanted oxygen to approach the metal center, you should force a rotation around the double CN bond, or hope in a O-M electron interation, but being O so distant, that's highly unlikely to happen, unless I'm missing something.
Also, note that if you rotated the CC single bond to somehow accomodate oxygen, the nitrogen's lone pair would lose its alignment with the metal

I'm not sure if this example qualifies as an example for dimethylglyoximate acting as an ambidentate ligand, because it, for me, is more like bidentate. Nonetheless, I'm presenting this example here only because both nitrogen and oxygen are donating lone pairs within the same molecule. See trinuclear copper complex below (ref: doi.org/10.1107/S160053680301331X):

I was taught that we would call a ligand "ambidentate" when it donates electrons from either one atom or the other atom, but not both at once. For example, either S or N in $\ce{SCN-}$, but not both at once. But, in the example above, both are acting as donors at once. How does that count as ambidentate?

@Gaurang Tandon: That's my point as well.  But I wasn't sure  when it donate lone pairs via nitrogen and oxygen to two different $\ce{Cu}$ atoms.

@GaurangTandon there are many cases (look under Complexes in https://en.m.wikipedia.org/wiki/Dimethylglyoxime) where dimethylglyoximate acts as a donor through 2 nitrogens, and now an example where it donates via one oxygen and one nitrogen. Hence it's ambidentate.

@GaurangTandon SCN- can also be bridging ligand. That's a consequence of ambidentacity.


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Are anions stronger bases than (different) neutral molecules?
I need to compare the basicity of 2 molecules: a negatively charged $\ce{O}$ attached to a t-butyl (t-butoxide), and a neutral $\ce{N}$ with 3 ethyls attached to it (triethylamine).
These are completely different molecules, so by instinct, I would say that the one with the negative charge is probably more basic. But, is there a more formal way of figuring it out (not experimentally)?

I would like to tell you that every negatively charged molecule is more basic than a neutral one, but sadly it's not (for comparison, ammonia and alkylamines are generally more basic than a deprotonated beta-diketone).
I guess that you are at the beginning of your career, but it's highly advisable to learn some key pka values for some compounds/functional groups. You will later learn how to guess if a compound has higher or lower basicity than a similar one, but a mnemonical reference is precious

You need to compare the pKa for the two compounds. This chart has a substantial list of pKa values including those you are looking for http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf

Ok, thanks guys! Yes, seems like a good idea to have some pKa references memorized and guesstimate that of related compounds from there.


text:
Increase in equilibrium concentration of the reactant via decrease in container volume

Consider the hypothetical reaction $\ce{3A(g) -> B(g) + C(g)}$; how to increase the equilibrium concentration of $\ce{A}$, given that the reaction is endothermic?

increasing the volume of the container
decreasing the temperature
decreasing the volume of the container
adding $\ce{B(g)}$ at equilibrium

Answer given - 2, 3, and 4. I have trouble with the option 3.
I know according to the Le Chatelier's principle, equilibrium will shift forward on reducing the container volume, but the solution says it still manages to have an increase in the overall equilibrium molar concentration of $\ce{A(g)}$; and that's what I'm not understanding.

You're right that at first glance, this is indeed self-contradictory. However, on a closer look, here's how you can realize that it isn't!
The catch here is that when the reaction shifts forward - to counter-balance the decrease in volume - the reduction in the moles of $\ce{A}$ is much lesser than the decrease in volume, hence, the overall ratio, $\frac{\text{moles}}{\text{volume}}$ actually increases!
You can demonstrate this via your own reaction. Assume $K_\mathrm{c}=\pu{1M-1}$, and all equilibrium concentrations to be $\pu{1M}$. When volume is halved, the concentrations double, so $Q=\pu{0.5M-1}$. The reaction thus shifts forward as $Q<K$.
Let the new concentrations be $\pu{2M}-3x, \pu{2M}+x, \pu{2M}+x$ for $\ce{A,B,C}$ respectively. Solving for $x$, you'll get $x\approx0.117$. Hence, the new equilibrium concentrations are $\pu{1.64M}, \pu{2.117M}, \pu{2.117M}$. In fact, not only the reactant, but all species show an increase in their concentration!
Moreover, the same result is obtained in the reverse direction. If the container volume was increased instead, all species would show a decrease in their volume. (the overall ratio $\frac{\text{moles}}{\text{volume}}$ decreases, as the moles increase far less proportionately as compared to the volume)
I must state that this isn't a rigorous proof but a layman's explanation for why this is happening, and how to understand it in simple terms. I would leave it up to a more experienced physical chemist, to demonstrate the applicability of the rules I've stated in bold above, via proofs or other methods.


text:
Mechanism for the conversion of an unsaturated alkyl halide to an alcohol
Question:

Predict the mechanism for the following reaction:

Attempt:
I thought it will proceed through SN1 mechanism as $\ce{H_2O}$ is involved (solvolysis). So I proceeded with the formation of a primary carbocation, which then rearranges via 1,2 methyl shift. Then the nucleophilic attack of $\ce{OH^{-}}$ leads to the following product, which is different form the required one.
Can any further steps lead to the required product or am I wrong from the start (most probably)?

For what it's worth, this reaction does not appear in the literature. So, whether it actually forms the product or not is anybody's guess.
However, if I had to suggest a mechanism... I would note first that an SN1 reaction on a primary centre, as you have proposed, is difficult. I think it more likely that the alkene assists in the expulsion of the chloride.

The resulting cyclopropylmethyl carbocation is stabilised and is electrophilic at multiple sites. I have only drawn one resonance form, but there are a few more. Attack of water at the tertiary carbon leads to the desired product after deprotonation.


text:
How expensive are Hartree Fock calculations?
Hartree Fock is a very simple mean field approach for finding the ground state energies of - for example - molecules. But how expensive is it from a computational view? So after calculating all one - and two-particle integrals, which is $O(n^4)$, how expensive is it to do HF then? How large molecules could one in general do with that?

There are lot of graphs about that scaling time, look at the gaussian manual, there is an example for hydrocarbons.

If I'm understanding right, you asking about the complexity after solving for the integrals which while formally is O(N^4) can usually be done O(N^3) or even O(N^2) in practice. Once you have evaluated the integrals, the only other step to the algorithm is to iteratively form the Fock Matrix and diagonalize so solve for the new coefficients. Matrix diagonalization is O(N^3) so with the integrals solved, this is the most expensive part.

@santimirandarp Where exactly is this located? I can't find it.

I wouldn't call the Hartree-Fock methods 'very simple', I would consider it fairly complex both in terms of its derivation, interpretation, computational cost and implementation details. If Hartree-Fock calculations are now commonplace it is also because so many man-hours of study have been devoted to it.
In any case, as you say, its formal cost is $n^4$ ($n=$ number of basis functions): the number of two-electron integrals is $n^4$, and all these integral are necessary to build the Fock matrix, so the overall cost is also $n^4$. However, this is not the end of the story; as a system grows larger and larger in spatial extension many of the required integrals become vanishingly small and can be neglected, and other approximations become possible, at the cost of introducing a small and adjustable error. As long as the error introduced is (much) smaller than the intrinsic error of the Hartree-Fock method, it doesn't really matter. Some keyword often used in this context are 'linear-scaling' and 'density-fitted' methods.
The difficult, computationally expensive part of the Hartree-Fock method is dealing with the electron-exchange part of the Fock matrix, and most studies try to speed up this step. For example, a recent study of this type is:
Christoph Köppl and Hans-Joachim Werner, Parallel and Low-Order Scaling Implementation of Hartree−Fock Exchange Using Local Density Fitting, J. Chem. Theory Comput. 12, 3122−3134 (2016).
In this study the largest calculation they report on is for alpha-cyclodextrin, a polysaccharide with formula $\ce{C36H60O30}$ (126 atoms, 516 electrons); using the aug-cc-pVDZ basis set (2058 basis functions) the authors report that a density-fitted Hartree-Fock calculation took about 4 minutes on a beefy workstation with 20 cores (two Intel Xeon E5-2690 @ 2.8 GHz).

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text:
Biodiesel synthesis
I want to do the synthesis of biodiesel with the cooking oil (it contains colza and sunflower oils) in ethanol and KOH. After 1 hour of reaction, I let the solution to rest for 3 days.
At this stage, I should obtain different phases of glycerol and ester before the washing step, but there is no decantation I have only one phase of solution.
Can you help me to separate the biodiesel?

My guess would be that 1hr is not enough. Are you following a known procedure?

Thank you for your message, yes i followed a known procedure, indeed, it is mentioned that 1h is enough.

Heating it further will not damage the product so there is nothing to lose there.

I have made biodiesel from sunflower oil, I used methanol instead of ethanol and I used sodium hydroxide in methanol as my cataylst. If you use a little methanol I have found that after heating the mixture under reflux and allowing it to cool. You get a layer of glycerol / methanol which is not miscible with the FAME (Fatty acid methyl esters) which has a lower density than the glycerol / methanol / sodium hydroxide layer.
If you use ethanol then it will be harder to separate the lipohilic layer of fatty acid esters from the alcohol. I know that methanol and hexane are not fully miscible while ethanol and hexane are fully miscible.
Assuming that you have access to a proper chemistry lab with a fume hood, I would suggest you change to methanol instead of ethanol for making biodiesel from cooking oil. Also if you use fresh cooking oil instead of used then you will have less free fatty acid (carboxylic acid groups) in the oil that you start with.
The literature suggests that for used cooking oil that acid cataylsts can be a better than alkaline cataylsts,

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text:
Stability comparison of benzylic carbocation and benzene
Which is more stable - benzylic carbocation or benzene - and why?
Benzylic carbocation may have more resonance structure, but these structures are destroying its aromaticity as well. So, how do we compare its stability with benzene?

Aromaticity lowers the energy of a benzene molecule by a huge amount, which is much larger than any resonance stabilisation possible in the carbocation. So basically, benzene is the stabler one. See here: https://www.quora.com/Why-is-the-phenyl-cation-unstable?

@AbhigyanC you linked to phenyl carbocation, but OP is talking about benzylic carbocation ;)

Comparing a stable molecule to a carbonation (although stabilised) is not the way to think. You might compare other carbonations the to the benzylic one, in these terms.

@GaurangTandon nevertheless it's obvious that benzene is more stable #aromaticity

First refer to https://en.m.wikipedia.org/wiki/Chemical_stability to clean up the situation.
There is not much in comparing the stability of individual different molecules which are not among the reactants or the product of a specific reaction or in contact to a specific and same environment.
The question arises from a common confusion for we are used to compare the "stability" of different limiting forms (resonance structures ) of a hybrid of resonance.
Or, differently, but still analyzing entities with the same constitutional formula, the stability of various conformers.
An analysis in the terms meant by the question could be done to compare the stability of different carbocations, for instance, and see if there could be mesomeric effects able to delocalise the charge. In such a sense the presence of the phenyl ring has a stabilising effect as compared to, e.g., a linear alkyl chain.
This state, but anyway answer the question, the number of resonance structure does not even count as for all those of the benzyl cation are to be supposed much less stable  than those of benzene, the carbonation is obviously charged so this will be the main difference anyway.
You won't ask what is more stable between methane and ethyl carbocation. The fact that in benzyl cation there is a phenyl ring can only tell you that its formation under a specific mechanism will be favoured as compared to the case of ethane.


text:
Graph of compressibility factor vs pressure when real gas is assigned Z=1
Problem:

According to the real gas equation, $Z = 1$ for an ideal gas and $Z$ is variable for a real gas. Suppose, in order to easy our calculations, we fixed $Z=1$ for a real gas and for ideal gas, $Z$ will become variable. $Z$ vs $P$ for an ideal gas will be similar to:

Attempt:
I imagined the answer by thinking the normal graph of $Z$ vs $P$ for a real gas and transposed each point on the locus of the real gas curve on $Z=1$ line. I then imagined where the corresponding point of the ideal gas curve would land.
This gave option (a) as the correct answer, but the book says the answer is (b). I believe then the van der Waals equation must be manipulated to get the answer, but can someone tell me how? Or is the answer given in the book wrong?

For a real gas, $(P+a/V^2)(V-b) = RT$
Now, since we are considering $Z=1$ for real gas, all the real gas equations will be reversed for the ideal gas. Meaning, we had to subtract $b$ for volume correction of real gas, for correction of ideal gas we need to add $b$. Same, with pressure.
The equation then comes out to be: $(P-a/V^2)(V+b) = RT$
Now, for the graph, we can calculate the approximate values of $Z$ (relative to unity) for very low pressure, low pressure and high pressure.
Case 1: Very low pressure: $P \downarrow  \downarrow \implies V  \uparrow  \uparrow $
$(P-a/V^2) \approx P$ and $(V-b)\approx V \implies Z \approx 1$
Case 2: Low pressure $P \downarrow \implies V \uparrow$
$(V+b)\approx V$. On solving, we get $Z= 1+a/(RTV)\implies Z>1$.
Case 3: Very high pressure $P \uparrow  \uparrow \implies V^2\downarrow  \downarrow$
$(P - a/V^2) \approx P$. On solving, we get $Z= 1-Pb/(RT)\implies Z<1$.

The graph would thus be (A), because it is the only one satisfying all these conditions.

I fail to see how the inversion of the sign of the a- and b-terms can be physically justified. To me, this seems actually to be a kind of coordinate transformation. But then, how would it be compatible with the usual definition of pressure an volume?

I thought that if Z=1 is assigned to real gas, we need to see ideal gas relative to real gas. If we see real gas relative to ideal gas, we need to add 'b' to compensate our volume. But, if we assign, real gas Z=1, that's other way round. We need to subtract 'b' from the ideal gas volume to make it equal w.r.t real gas.

The compressibility factor $Z$ of a real gas is defined by the equation
$$p ~ v = Z ~ R ~ T,\tag{1}$$
where $p$ is the pressure, $v$ the molar volume, $R$ the gas constant, and $T$ the thermodynamic temperature. It describes the deviation of a real gas from ideal gas behavior
$$p ~ v = R ~ T.\tag{2}$$
For low pressure real gases behave like an ideal gas, thus $Z = 1$.
For medium pressures the attractive forces between the molecules of the real gas make the volume and thus $p ~ v$ smaller than for an ideal gas. Therefore $Z < 1$.
For high pressures the size the of molecules of the real gas comes into play, $v$ converges to a constant, and lets $Z \propto p$ go to infinity.

Suppose, in order to easy our calculations, we fixed $Z = 1$ for a real gas and for ideal gas, $Z$ will become variable.

If we want to measure the compressibility factor $Z$ in terms of the compressibility factor $Z_R$ of a certain real gas, the resulting scaled compressibility factor $Z_s$ is
$$Z_s = \frac{Z}{Z_R}. \tag{3}$$
In this units the compressibility factor of the real gas becomes $1$ and the compressibility factor of an ideal gas becomes
$$Z_s = \frac{1}{Z_R}, \tag{4}$$
i.e. $Z_s$ is in reciprocal proportion to the compressibility factor of the real gas. This means for an ideal gas we have $Z_s = 1$ for low pressures, $Z_s > 1$ for medium pressures, and $Z_s \to 0$ for high pressures.
With this interpretation answer a is correct.


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Do metal ions of 4d and 5d series always form low spin complex?
I've recently come across a source where it stated that all 4d and 5d metals form low spin complexes irrespective of the strength of ligand.
I'm not sure of the authenticity of the source and couldn't figure out the reason for this phenomenon, thereby enquiring about it.
I feel that the splitting energy is higher than pairing energy in this case, but I am not able to justify it properly.

Please add the source

I've been taught the same fact, and am interested in an answer to this question. Sadly, I wasn't able to verify the fact from both JD Lee and VK Jaiswaal inorganic :(

Related: Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral? (read the answer)

PdF2 is an exception - it is d8, octahedral (rutile structure), and paramagnetic according to Housecroft and Sharpe (Second edition, section 22.11)

I went through my coordination chemistry sources but cannot find an answer to this question (only a statement that it is typically the case presented without counterexample).

While the vast majority of the compounds and complexes of the heavier transition metals are low spin (where that concept has any meaning) $\ce{PdF2}$ is one exception, which is unusual not only this way but also having $\ce{Pd^2+}$ in an octahedral environment (see e.g. Wikipedia). Thus your source is not quite correct.
As for the reason why 2nd and 3rd row transition metals are more likely to form low spin complexes than the lighter elements, the reason is given in the answer linked above in the comments. In fact, while the question may be different, the answer is almost a duplicate.

You can simply remember that CFSE of 4d and 5d series is far more than that of 3d series. Therefore, However strong the ligand, The pairing energy will always be lesser than the CFSE. So mostly, all the complexes of these two series are inner orbital complexes.
And yes the reason for high CFSE is diffused state of 4d and 5d orbitals.
Order of CFSE: 5d>4d>3d.


text:
Bohr energy formula for a single-electron system

A single electron atom has nuclear charge $+Ze$ where $Z$ is atomic number and e is electronic charge requires $\pu{16.52ev}$ to excite the electron from the second bohr orbit to third bohr orbit. Find the atomic number of the element?
A) 1               B) 2              C) 3              D) 4

I wish to ask if we can use energy formula $E= -(13.6)\frac{Z^2}{n^2}$ for this problem.

Using the formula you stated,
Energy difference between the orbits can be evaluated as
$$∆E=-13.6\ \mathrm{eV}\left(\frac{z^2}{3^2}\right)-\left(-13.6\ \mathrm{eV}\left(\frac{z^2}{2^2}\right)\right)$$
Equate this expression to the energy you have given, i.e. $16.52\ \mathrm{eV}$
Solve this for $z$ which comes out to be $2.96$ that may be approximated to $3$.

The charge must have been $+(Z-1)e$, since the species must be monoelectronic to apply Bohr's model.
Coming to your question, the formula you have mentioned in your question is used for finding the energy of an electron in a particular shell of a monoelectronic species. What you are supposed to do is use it to subtract the energies of the given shells and equate it to the energy (16.52 eV) given in the question. This should give the answer as $Z = 3$.
PS: You could have found the answer by little self study.


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Quantum Mechanical Model of An Atom, Shells, Subshells
I've just been introduced to the quantum Mechanical Model of an Atom, so please bear with me on this one.
Are Subshells and shell physical things, like do they really exist, or is it a concept only to understand the energy that an electron has?
I'm kind of confused since until now, we've been drawing the K, L , M , N shells with a nucleus in the middle and 4 shells outside with the respective no. Of electrons in them.
And they're related to distance from the nucleus so I was thinking they must be physical.
But if you start visualising each Subshells in each shell having it's own orbitals, that doesn't really make sense.
And what about orbitals? I'm guessing they've  got to be physical, right?
- 10th grader
edit- My point is can you literally visualise the different things as it's regularly down at middle school level.

A shell isn't a physical thing and neither is an orbital; they're concepts to visualise where an electron would exist in an atom's orbit etc. A shell is better described as an energy level, as the electrons' energies increase with higher energy levels. All a shell does is conceptualise an electron's energy and position in orbit. An orbital is just a region where we think an electron would exist; orbitals exist in their own subshells, and subshells exist in energy shells. But as mentioned, they're just concepts. Electrons in orbit do exist though...

@KianStevens, that does help, thanks.  So I shouldn't go on about trying to visualise them right? I also read at a couple of places that the electrons randomly go on about around the nucleus, like a honeybee around a comb, so I was thinking that maybe they don't have orbits. That isn't it?

You can visualise them because that's the whole point of the Bohr model. However, shells and orbitals etc. aren't physical things; you wouldn't see concentric circles in an atom's orbit if you were to look at an atom. The concept of electrons falling into the nucleus is a matter of particle physics that doesn't refer to the Bohr model. It's all to do with nuclear interactions I believe.

@KianStevens, I would hesitate to say that you can visualise them. What you can do is to draw representations of them that are entirely unphysical (note that the Bohr model is unphysical). Whether that counts as visualising them, or not... I lean towards no.

@orthocresol Meh. They're not the best way to visualise shells, but it gives an idea of what a shell is I suppose.


text:
Does the way a reaction with gas moles shifts depend on Q or number of moles on either side of the reaction?
Given a container with the reaction $\ce{C2H4{(g)} + HCl{(g)} -> C2H5Cl{(g)}}$, if we inject $\ce{C2H5Cl{(g)}}$ to the container, what will happen?
I thought of this in two different ways and got two different answers:

If we inject $\ce{C2H5Cl{(g)}}$, $Q>K$ so reaction will shift to the left, leading to the formation of more $\ce{C2H4{(g)} and HCl{(g)}} $.
Injecting  $\ce{C2H5Cl{(g)}}$ also increases the pressure I suppose, and if you increase pressure, the reaction will shift to the side with the least moles of gas. In this case, it will shift right, forming more $\ce{C2H5Cl{(g)}}$

Which answer is correct?

Injecting chloroethane will increase the pressure of it. Once the pressure of the product is greater than the pressures of the reactants, the equilibrium will shift to the left, and so more reactant will react to oppose the change and restore Kp. Your first statement seems correct. I'm not sure though; the reaction isn't in equilibrium, so there isn't an equilibrium to be shifted.


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Why is there an orbital angular momentum if the electron isn't properly revolving around the nucleus?
I'm a tenth grader who's just been introduced to this subject of atomic structure, so please help.
I recently read that an electron doesn't have circular motion around a nucleus, its motion is kind of bizzare, buzzing and basically all over the place. (As long as it's around the nucleus) But we use the azimuthal quantum number to talk about an electrons orbital angular momentum.
Now, to be sure I didn't know what that meant, so I looked it up. I can't figure out why is there a question of orbital angular momentum if the electron isn't revolving around the nucleus.
Also, is the azimuthal number somehow related to the energy an electron has? I can't figure out if it's the radius that matters or the energy of the electron, so can someone please explain that?

The picture you have in your mind is the angular momentum of a particle ("revolving around the nucleus"), which is easy to visualise. The problem is that the behaviour of an electron cannot be solely described by treating it as a classical particle. Quantum mechanics is a huge paradigm shift: it's not just an incremental theory that refines the old Bohr model of orbiting electrons and reconciles some problems with it. It's an entirely new way of looking at things, essentially from scratch, and that's why it has the reputation it has now.

@orthocresol that makes sense, thanks. But it's kind of tricky since my course in physics and maths is barely close to explain  quantum mechanics. Can you please tell me how should I think of angular momentum of an electron, then?

The electron can have zero or non zero but quantised amounts orbital angular momentum, depending on what state it is in the atom.; s orbital, zero orbital ang. mom., p orbital 1 unit , d two etc. Perhaps the easiest way to deal with this is not to imagine the electron like a ball or fuzzy thing but that it  has properties and is described by equations that are indicative of angular momentum.  The electron also has spin angular momentum  but is not' spinning'. As you realise quantum behaviour is often hard to explain in everyday terms and not just for students :)

Though the concept of angular momentum is used to describe an electron's state, if the electron actually were moving in an orbit, it would radiate energy because it would be an accelerating charge (i.e. centripetally). It's somewhat mind boggling, but an electron can be detected here, or somewhere else, without having traversed the space between the points. Tunnel, or Esaki, diodes provide a different but easily observable example of quantum oddity. https://en.wikipedia.org/wiki/Tunnel_diode

In quantum mechanics, the classical concept of trajectory of a particle is abandoned. The quantum state of a particle i.e. the electron is characterized by a wave function $$\psi(\pmb{r}, t)$$ that contains all information about the particle and that extends over space. The simplest case where angular momentum appears in the context of quantum mechanics is the one of a particle in a “ring”. Now you have a wave function which extends over a circular region of space which implies that your particle has angular momentum. When you apply  the same idea to the Hydrogen atom you will find the orbital angular momentum.
The energy levels of the Hydrogen atom is given by
$E_{n} = E_{0}/n^{2} $,
where $E_{0} = 13.6 eV$ and n = 1, 2, 3, ... is the principal quantum number and describes the size of the orbital. Therefore, we see that $E_{n}$ is independent of the orbital angular momentum. If the atom is in the presence of a magnetic field then things change.

The simplest case where angular momentum appears in the context of quantum mechanics is the one of a particle in a “ring”. A ring or a sphere?

In a sphere you have two variables in contrast to only one in a ring. The math to work out the particle in a ring is not complicated. Aside from the introductory part of quantization of matter a tenth grade student can understand the problem quite well.

Ah, misunderstood your expression about the simplest case. You mean the simplest mathematical case and not the ground state of an electron.

The concept of orbital angular momentum can be thought of this way.  The electron cloud has a motion surrounding nucleus and its motion can be described by the 'Wave Function' which is nothing but a function of space and time( $\mathrm{\Psi(x,y,z,t))}$. So, the electron cloud has positions described by these wave functions, and also it has definite momentum which is defined as(also, can be derived from $\mathrm{\frac{d<x>}{dt}=<p>}$) $$\mathrm{\hat{p}\Psi = i \hbar (\frac{\partial \Psi}{\partial x}\hat{x} + \frac{\partial \Psi}{\partial y}\hat{y} + \frac{\partial \Psi}{\partial z}\hat{z}) = i \hbar \vec{\nabla}\Psi}$$ So, if you have the wave function of the electron you already have momentum for it. Now, according to the definition of Angular Momentum, we have $L = (\vec{r} \times\vec{p} ) $, and thus we also have Angular momentum of the electrons by this definition where the individual components are, $$\mathrm{\hat{L_x} \Psi= i \hbar(y \frac{\partial \Psi}{\partial z} - z \frac{\partial \Psi}{\partial y})}$$$$\mathrm{\hat{L_y} \Psi= i \hbar(z \frac{\partial \Psi}{\partial x} - x \frac{\partial \Psi}{\partial z})}$$ and also $$\mathrm{\hat{L_z} \Psi= i \hbar(x \frac{\partial \Psi}{\partial y} - y \frac{\partial \Psi}{\partial x})}$$ and similarly, there is also $\hat{L^2} $  operator( square of angular momentum) = $\hat{L_x^2}+ \hat{L_y^2} + \hat{L_z^2}$Now what you know about the Orbital Angular Momentum, is nothing but when we try to find the eigen value of $\hat{L^2} $  by solving the equation $\hat{L^2}\Psi = E \Psi $ From solving that equation, we get some definite eigenvalue of the angular momentum, which is reffered  as those Azimuthal Quantum numbers( $l$).Those eigen values are $\hbar^2l(l+1)$, which you might have encountered as the orbital angular momentum of an electron possesing that($l$) particular quantum number. In quantum mechanical term these, are expectation values ( Eigen values are always expectation values) of  Orbital Angular momentum.  So, though the electron is not actually circling around the nucleus, it has positions and momentum and that's why Angular momentum according to definition and also, when we are concerned about systems of electrons, the solution corresponds to some finite states of presence of electrons, which are termed as orbitals, and the angular momentum associated to that becomes orbital Angular Momentum.

I dont know if we can expect a tenth grader to have the math background to fully understand this answer.


text:
Reaction of ethylene glycol with acidified potassium permangnate
One would expect the reaction of ethylene glycol with acidified potassium permangnate to result in oxidation of ethylene glycol to oxalic acid. But a Q&A guide book tells me it is oxidized to formic acid. Which is wrong: the book or my reasoning?

Related: https://chemistry.stackexchange.com/questions/18771/what-factors-might-affect-the-reaction-between-potassium-permanganate-and-ethyle

I think the final oxidation product should be carbon dioxide ($\ce{CO_2)}$ if acidified potassium permanganate is present in excess amount.
Firstly, ethylene glycol will be oxidised to oxalic acid. But oxalic acid can still be oxidised. In fact, it is the concept of titration of oxalates with potassium permanganate ($\ce{KMnO_4}$). Under acidic conditions, oxalic acid ($\ce{H_2C_2O_4}$) will be oxidised to $\ce{CO_2}$ by $\ce{KMnO_4}$ as:
$$\ce{2MnO_4^- + 5C_2O_4^{2-} + 16H^+-> 2Mn^2+ + 10CO_2(aq)  + 8H_2O}$$
Thus the final product should be $\ce{CO_2}$
Also, note that, formic acid is an oxidisable acid because it gives silver mirror in reaction with Tollen's reagent $(\ce{[Ag(NH_3)_2]OH})$, and thus, formic acid can not be the end product, as it is still oxidisable and hence it will definitely react with acidfied $\ce{KMnO_4}$ if both are present in  the reaction medium.

according to this https://en.wikipedia.org/wiki/Alcohol_oxidation#Diol_oxidation oxalic acid is not fromed but formic acid is ?

I Know that. But you should carefully look into the reagents. Periodates and Lead tetra acetates can cleave diols.  Only $\ce{KMnO_4}$  can not cleave bonds between diols under Normal conditions. See, Lemieux  Reagent for this concern.


text:
Resonance structures of mercurinium ion
Oxymercuration of alkenes follows Markovnikov's rule, but why? (I got some answers like the resonance structures of mercurinium ion are sometimes likely to have positive charge on more substituted carbon. But I don't know what those structures be like and how this whole phenomenon is happening.)

You should read mechanism section of oxymercuration of alkenes on Wikipedia. For your convenience, see the 3-step mechanism given in there:

Now, if you ask yourself a question, "why $\ce{H2O}$ is attacking more substituted carbon center in second step?" You'd get the answer to your own question.
See step-wise mechanism, which shows why more substituted carbon has positive charge.

the mechanism posted in the last segment does not seem to have applied the concept of formation of mercurinium complex (an intermediate complex formed in the due course of oxymercuration-demercuration process), does it hold authenticity?

+1 for what OP said: the stepwise mechanism does not proceed through the mercurinium ion intermediate.

No body claims these step-wise mechanisms are existed during chemical reactions. They were presented to explain the results. Sometimes, intermediates were isolated to confirm the mechanism.


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Why is 2-ethylpentane is not a valid IUPAC name, but 3-ethylpentane is?
I'm currently writing all the names of alkanes having molecular formula $\ce{C7H16}$. I wish to ask why 2-ethylpentane is not a valid IUPAC name, but 3-ethylpentane is?

Welcome to Chemistry.SE. We’d like you to take the Tour to familiarize with the site. Also, let us know what have you done so far to solve this specific problem. It’d help us to answer your question.

Due to IUPAC conventions, the main alkane name is based on its longest chain. Here is an image showing different chain lengths for both alkanes. The numbers denote the longest alkane chain:

As you can see, 3-ethylpentane is symmetrical; no matter what chain you count along, the longest chain will always be 5 carbons long (pentane), and the ethyl group will always be on the third carbon (3-ethyl). Together, this is 3-ethylpentane.
However, if you look at '2-ethylpentane', there's a new longest chain which isn't 5 carbons long anymore. The image shows a new chain which is actually 6 carbons long (hexane), and there's now a methyl group on the third carbon (3-methyl). Together, this is 3-methylhexane. Because of this, '2-ethylpentane' isn't a real structure, since drawing it will leave you with 3-methylhexane as shown in the image.

Hint:
Draw the structure of what you described as "2-ethylpentane" first. Then name it according to IUPAC rules. Does IUPAC name is same as "2-ethylpentane"?
Now, draw 3-methylhexane structure. How many ways you can name it? But only name count is its IUPAC name, which is 3-methylhexane.

I'd recommend not posting answers which are solely hints. We haven't reached a real consensus on this, but generally hints do not fit into the SE mould, which is about building a library of knowledge in Q&A format. It is not supposed to be a forum with back-and-forth discussion between the question asker and you, nor is it supposed to be a medium for tutoring. If you feel that a hint is more justified than a full answer, please do it as a comment.

@orthocresol That happens on Math.SE all the time

@Abcd Math.SE has their own way of doing things. They can do whatever they wish, but I would rather we not follow them.


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Unsteady state process for first law of thermodynamics with changing boundries
Firstly , this is not an homework question .I tried to solve this question to understand the concept.

I made the following attempt to solve this question.However I couldn't solve after exit valve is opened.Can you give me a hint or say my mistakes?
From which boundry should I start to solve easily ?

Take as the system the air in the tank.  Applying the open-system version of the first law of thermodynamics to this process, you have:
$$\Delta U=n_fu_f-n_iu_i=-p_f(V_f-V_i)+(n_f-n_i)h_s$$where $n_i$ is the initial number of moles of air in the tank, $n_f$ is the final number of moles of air in the tank, $u_f$ is the final internal energy per mole of air in the tank, $u_i$ is the initial internal energy per mole of air in the tank, $p_f$ is the air pressure from the supply line, $V_f$ is the final volume of air in the tank, $V_i$ is the initial volume of air in the tank, and $h_s=u_i+RT_i$ is the molar enthalpy of the air in the supply line.
The equation I wrote leads to a final answer which matches the correct answer given in the book.

Thank you so much for your clear answer.I think that I  am not good at defining the system and its properties .Therefore ,I have to make more practice . .

I commend you for articulation of your value for practice.  In my judgment, the importance of practice in mastering a new scientific subject can not be overemphasized.


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Term symbols: counting electrons. How?
I am having some difficulty in writing the correct term symbols for diatomic molecules in their ground states.
For instance, the $\ce{S2}$ molecule has a MO configuration in its valence shell $\ce{3s\sigma^2_u 3s\sigma^2\mbox{*}_g 3p_z\pi^2_u 3p_y\pi^2_u 3p_x\pi^2_u 3p_y\pi^1\mbox{*}_g 3p_x\pi^1\mbox{*}_g}$
Now, for its term symbol I would say (erroneously) that I have two unpaired pi electrons, hence $\ce{\Lambda}=2$, so I would obtain a $\ce{\Delta}$ symbol.
As for the spin multiplicity, it would be 3. And the spin-orbit coupling term would be therefore S+2=3. Concerning simmetry, I would say that it's g, since the last occupied shell is g (still, I don't understand why Ar2 should be g if the last shell is u. Maybe because having wto electrons in a u shell I should consider u*u=g? If that is the case, it's correct to say that every molecule with an even number of electrons in the last energy level, and hence every homonuclear diatomic, is g?)
Actually, it looks like the correct term symbol for the $\ce{S2}$ molecule is $\ce{^3\Sigma _g}$, but I don't understand why. Can you help me please?

Because $\mathrm{g \otimes g = u \otimes u = g}$, all closed-shell configurations have a g term symbol (assuming, of course, that the molecule itself is centrosymmetric)

Thank you, a new trick in my toolbox! But, if I may ask, which is the correct mental process to derive the "g" term for an open-shell molecule like S2?

Also, I corrected the question: I forgot the "homonuclear" word here: "If that is the case, it's correct to say that every molecule with an even number of electrons in the last orbital, and hence every homonuclear diatomic, is g?".
But maybe "energy level" instead of orbital fits better my question: I was still referring to S2

Finding the parity of the term is nothing more than multiplying the parity label of each orbital together, so all you need to know is $\mathrm{g \otimes g = u \otimes u = g}$. That wasn't a trick or a shortcut, it was precisely the way to derive it. And yes, all ground state term symbols of homonuclear diatomics are g.

Have a look also at this answer, https://chemistry.stackexchange.com/questions/37038/how-to-find-the-number-of-atomic-microstates-for-a-given-electronic-configuratio/54333#54333

This example is probably described quite thoroughly on the internet for the case of $\ce{O2}$ (which is isoelectronic), so it may be worth searching a bit for alternate explanations.
You cannot have a $^3\Delta$ term symbol, because of the Pauli exclusion principle. The π* orbitals come as a set of two: essentially, one of those two is associated with +1 angular momentum, and the other is associated with −1 angular momentum.
Here, to be specific, "angular momentum" refers to the projection of the angular momentum onto the internuclear axis. The "angular momentum" quantum number of each MO is denoted $\lambda$, so in other words, the π* orbitals come as a set of $\lambda = +1$ and $\lambda = -1$.
The quantum number that measures the "angular momentum" of the term is denoted $\Lambda$, and because $\Lambda$ is a projection quantum number, the value of $\Lambda$ is simply the sum of the individual $\lambda$ values. The Greek letter that represents it in the term symbol depends on the absolute value of $\Lambda$: if $\Lambda = 0$ then it is a $\Sigma$ term, if $|\Lambda| = 1$ then it is a $\Pi$ term, and if $|\Lambda| = 2$ it is a $\Delta$ term.
Now, in the ground state configuration of $\ce{S2}$, we can ignore all the orbitals below the π* orbitals because all those $\lambda$'s sum to zero. The only thing of interest is the π* electrons.
If you want to have a triplet state, i.e. the electrons have the same spin, then you cannot put them into the same π* orbital: one will have to go into the $\lambda = +1$ orbital, and the other will have to go into the $\lambda = -1$ orbital. Hence you get $\Lambda = 0$, which leads to a $^3\Sigma$ term symbol, not a $^3\Delta$.
Finding the parity of the term symbol is nothing more than multiplying all the individual parities of the electrons (i.e. for doubly occupied $\mathrm{g}$ MOs you need to multiply by $\mathrm{g}$ twice). This is useful, because $\mathrm{g \otimes g = u \otimes u = g}$, which allows you to immediately discard all doubly occupied MOs. You only need to look at the two π* electrons, which are both $\mathrm{u}$, leading to an overall $\mathrm{g}$ parity and a $^3\Sigma_\mathrm{g}$ term symbol.
For $\Sigma$ term symbols there is an additional detail in that you need to specify the symmetry with respect to reflection in a plane containing the molecule. In this case it turns out that the molecular wavefunction is antisymmetric with respect to this reflection, so the proper term symbol is $^3\Sigma_\mathrm{g}^-$.


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Calculating the pH upon titrating barium hydroxide and hydrochloric acid
Question:

Calculate the pH produced from mixing $\pu{25.0 mL}$ of $\pu{0.420 M} $ $\ce{Ba(OH)_2}$ with $\pu{125 mL}$ of $\pu{0.120 M}$ $\ce{ HCl}$.

Attempt:
I'm learning about acids and bases right now (and not very used to it yet). First I wrote the chemical equation $$\ce{2HCl + Ba(OH)_2 \rightarrow 2H_2O + BaCl_2}$$
Then I calculated the number of moles of each $\ce{Ba(OH)_2}$ has
$$ \pu{0.025 L} \times \pu{0.420  M} = 0.0105 $$ moles
and for $\ce{HCl}$ $$0.125 \pu{mL } \times \pu{0.120 M}  =0.0150$$ So $\ce{Ba(OH)2}$ is the limiting reactant. However, I'm not sure how to proceed so any help would really be appreciated!

You are on the right track just one thing. $\ce{Ba(OH)2}$ is in excess and $\ce{HCl}$ is the limiting reactant, which means all of the $\ce{HCl}$ is used up. That alone tells us the pH is above 7, as $\ce{Ba(OH)2}$ is alkaline and there is no acid left.
To find the exact pH of the resulting solution, first we need to find how much of $\ce{Ba(OH)2}$ is left. Since all of the $\ce{HCl}$ is used, we can find the number of moles of $\ce{Ba(OH)2}$ needed to react with it. As you've stated, the number of moles of $\ce{HCl}$ is 0.0150. From the equation we can see that 2 moles of $\ce{HCl}$ reacts with 1 mole of $\ce{Ba(OH)2}$. So 0.0150 moles of $\ce{HCl}$ will react with $\frac{0.0150}{2}$ = 0.0075 moles of $\ce{Ba(OH)2}$. But there are 0.0105 moles of $\ce{Ba(OH)2}$ present. So we find the difference to find how many moles there are in the solution when the reaction is finished.
This number is 0.0105 - 0.0075 = 0.003 moles. Now, one thing about acids and bases, is that their ions dissociate in solution. The more ions dissociate, the stronger the acid or base is. Here, there is no acid remaining, but only 0.003 moles of said base present. $\ce{Ba(OH)2}$ dissociates as follows:
$$\ce{Ba(OH)2 <=> 2OH- + Ba^2+}$$
This means that one mole of $\ce{Ba(OH)2}$ dissociates to form 2 moles of $\ce{OH-}$ ions. So 0.003 moles of $\ce{Ba(OH)2}$ dissociates to form 0.003 * 2 =  0.006 moles of $\ce{OH-}$ ions. The resulting solution has a volume of 125 + 25 = 150 mL. The concentration of $\ce{OH-}$ ions is found by $\frac{0.006}{0.150}$ = 0.04 M.
To find the pH, it's easier to first find the pOH. That can be found using the formula: $$\ce{pOH = -log(OH^-)}$$
In your case, the value of $\ce{OH-}$ is 0.04 and the pOH is $$\ce{pOH = -log(0.04)}$$
or pOH = 1.397940. FINALLY, the pH can be found by subtracting the pOH from 14 because $$\ce{pOH + pH = 14}$$
In this case, pH = 14 - 1.397940 = 12.60206.
This is my first ever answer on this site, hope this helps. If I've made any mistakes, sorry and feel free to help out, ever-active community of Chemistry Stack Exchange

In my opinion, it'd be easier to find $[H^{+}]$ by dividing $K_{w}$ by $[OH^{-}]$, and then finding the negative log of that. It'd skip a few steps of working with $pOH$ etc.

It's all one and the same, really. That works too, but I personally like pOH. It' s like yin and yang, pH and pOH; two sides of a coin and whatnot.

One thing though. If $Ba(OH){2}$ is in excess, then why does all of the $HCl$ react? Why doesn't all of the $Ba(OH){2}$ react to leave an excess of $HCl$?

@iha99 Maybe this should be completely obvious, but where you say "You are on the right track. Ba(OH)2 is the limiting reactant, which means all of the HCl is used up. That alone tells us the pH is above 7, as Ba(OH)2 is alkaline and there is no acid left", I'm confused because if Ba(OH)2 is the limiting reactant, shouldn't it be used first and leave acid behind?

Also, the total solution volume won't be 0.150 mol dm^-3. It will be 0.025 mol dm^-3 - you forgot to convert 0.125 mL to L.

@KianStevens No, I didn't. I added up the volume in mL then converted to $dm^3$ or L. And, yeah you're right. I meant to say $\ce{Ba(OH)2}$ is in excess. How silly of me!

But there are more moles of $HCl$ than what there are of $Ba(OH){2}$, so how is $Ba(OH){2}$ in excess?

Although $\ce{HCl}$  has more moles, it reacts in a 2:1 ratio. So 0.0150 moles of $\ce{HCl}$ reacts with only 0.0075 moles of $\ce{Ba(OH)2}$. The number of moles of $\ce{Ba(OH)2}$ is greater than 0.0075 so it is in excess.


text:
What is an exocyclic double bond?
I have been tasked to find the $\lambda_{\text{max}}$ of 3,4-dimethyl-1,2,3,4,4a,5-hexahydronaphthalene.

This was supposed to be fairly easy. I used the base value for homoannular diene ($\pu{253 nm}$) and added the increments for the three ring residues attached to the diene ($\pu{+(3\times 5) = +15 nm}$). This should get me a value of $\lambda_{\text{max}}=\pu{268 nm}$. However, checking the solution this question, they have also added the increment for an exocyclic double bond, and have got the answer to be $\pu{273 nm}$ instead.
As far as I know, an exocyclic double bond means exo+cyclic, where exo means "outside" and cyclic means, well "cyclic", or more simply, a double bond that is outside the ring. I'm looking at this compound, and I dont see any exocyclic double bond at all.
Am I dealing with a key mistake here or is there something that I'm not aware of?

Surely you are aware that the calculation of this sort is but a very rough estimate, yet still I feel the urge to stress that once more. Having said that, let's look at the double bond in the first cycle: isn't it exocyclic with respect to the other cycle?

I like your comments a lot, as they are often very  helpful. In this case I don't think that you answer it properly, as in my eyes you simply repeat the obvious but why is it this way? I don't use this increment method often so its not very obvious for me on which ring to choose and my analytical chemistry courses are quite some time ago.

What's obvious for us may be less than obvious for the others, so the OP and possibly the later readers can benefit from reading it; isn't that what the answers are for? Come to think of it, most of my answers on Chem.SE simply restate the obvious.

The double bond B is exocyclic to ring 2 as it is attached to an atom which is shared between ring 1 and ring 2 while the double bond A is not connected to any ring 2 atoms and is within just one ring, hence making it endocyclic
for further references visit https://pharmaxchange.info/2012/08/ultraviolet-visible-uv-vis-spectroscopy-%E2%80%93-woodward-fieser-rules-to-calculate-wavelength-of-maximum-absorption-lambda-max-of-conjugated-dienes-and-polyenes/


text:
Does negative pH mean it is a stronger acid than one with positive pH?
I've searched this site but I don't think this has been asked yet. Does negative pH indicate a stronger acid than a positive pH (since it's further down the scale)? I'm asking this because in one of the problems from my homework set, they asked "which of the following solutions contains the strongest acid?" and one of the options yielded a negative value.

Negative pH does indicate a strong acid for sure, but it isn't necessarily stronger than that in the other (less acidic) solution. Maybe the latter solution is just more dilute.


text:
is syringaldazine soluble in water?
Would it be better to use water or ethanol as a solvent for syringaldazine? Why? Would water interfere with the oxidation of syringaldazine if laccase enzyme was added?

I refer to a post, which refers to a paper:
https://www.researchgate.net/post/How_do_I_dissolve_Syringaldazine_SGZ_to_measure_laccase_activity
"The assay mixture contained 0.2 mM syringaldazine (dissolved in 60% ethanol) and 40 mM citrate-buffer, pH 5.6. Oxidation of syringaldazine was followed by an absorbance increase at 525 nm (ϵ525 = 65 000 dm3 M−1 cm−1) at a temperature of 25°C"
Ignoring what is stated in the link (which confirms the idea) I doubt that water would interfere, since the molecule is stable in water, and since the enzyme works in a environment in which water is present

The use of ethanol is necessary in order to dissolve the otherwise poorly soluble substrate (21.06 mg/L in water at 25°C)


text:
calculating pH - is my answer correct?
I'am working on this exercice, Is my answer correct? and can you please help me to solve the second part?
exercice:
1/Calculate the ionization coefficient and pH (with two methods) of a 0.1M ammonia solution of pkb = 4.7
2/ to one liter of the previous solution, an equal volume of water is added. Calculate the new ionization coefficient and deduce the pH of this new solution

1/ For the first part my answer is :
$\ce{NH3  +  H2O = NH4+   +  OH^-}\tag{1}$
$Ka = \dfrac{\ce{[NH4+][OH^-]}}{\ce{[NH3+]}}\tag{2}$
The coefficient of ionization $\alpha$
$\alpha = \sqrt{Ka/C°}\tag{3}$
$Pkb = \log{kb}\tag{4}$
$Kb= 10-pkb\tag{5}$
$\alpha =0,014\tag{6}$
Two methods:
First one:
$pH= \frac{1}{2} (pka + 14 + \log{C°})\tag{7} $
$pka + pkb=14\tag{8}$
$pH = 11,15\tag{9}$
second method:
$14 + \log (\alpha C°) = 11,15\tag{10}$
for the second part, i am lost
Thank you

I tried to edit this to make sense of it, but I left the OP's mistakes and sloppy terminology. Also note that OP uses a comma for a decimal point which I left in.

Trying to clear up a terminology problem. Is this problem from a book written in English or did you translate it? I'm assuming that ionization coefficient was mean to mean the fraction of the ammonia that ionizes. The common term for that in English is ionization degree.

The general method to solve problems of this type can be found in the answer here, How to set up equation for buffer reaction?. In your case you need to modify this equation. To follow the text below you need to look at the main result in the linked answer (rather than repeat all this here). The derivation is straightforward although the equation is a bit complicated.
The base constant $K_\mathrm{B}$ is given but as $K_\mathrm{w}=K_\mathrm{A}K_\mathrm{B}$ we can start with  main eqn in linked page and convert to $k_\mathrm{B}$ later on. In this case as ammonia is a base (with concentration $c_b= 0.1$ M) we may suppose that any acid concentration $c_\mathrm{a} \approx 0$ then,
$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} + \mathrm{[H^+]} - \mathrm{[OH^-])} }{(- \mathrm{[H^+]} + \mathrm{[OH^-]} )}  $$
Additionally we can reasonably assume that $\mathrm{[OH^-] \gg [H^+]}$ and so
$$K_\mathrm{A}= \mathrm{[H^+]} \frac{(c_\mathrm{b} - \mathrm{[OH^-])} }{ \mathrm{[OH^-]} }  $$
This can be solved by substituting for $K_\mathrm{A}=K_\mathrm{w}/K_\mathrm{B}$  and for $\ce{[H^+]}$ as $K_\mathrm{w}=\ce{[H^+][OH^-]}$ and solve the quadratic equation.  But as $c_\mathrm{b} \gg \ce{[OH^-]}$ we can simplify by letting $c_\mathrm{b} - \ce{[OH^-]} \to c_\mathrm{b}$ then the equation becomes
$$K_\mathrm{A}= \mathrm{[H^+]} \frac{c_\mathrm{b}  }{ \mathrm{[OH^-]} }  $$
We can now get
$$\frac{K_\mathrm{w}}{K_\mathrm{B}}= \frac{K_\mathrm{w}c_\mathrm{b}} { \mathrm{[OH^-]^2}}   $$
from which
$$\ce{[OH]^- } =\sqrt{K_\mathrm{b}c_\mathrm{b}}$$
and so you can obtain the pH.
When you dilute, the concentration of base $c_\mathrm{b}$ is halved.

The first part of the question ask for the calculation of the pH two different ways and the ionization coefficient for a 0.1 molar ammonia solution. Also given that $pK_\beta = 4.7$.
First calculation of pH using $K_\beta$ of ammonia
$K_\beta = 10^{-pK_\beta} = 2.00\times10^{-5}\tag{1}$
Note: I'm using two extra significant figures for intermediate calculations. The answer should only have one since only one is given $pK_\beta$.
$\ce{NH3 + H2O <=> NH4+   +  OH^-}\tag{2}$
$K_\beta = \dfrac{\ce{[NH4+][OH^-]}}{\ce{[NH3]}}\tag{3}$
Little of the ammonia will ionize to form the ammonium cation, but enough will that the ionization of water itself doesn't need to be considered. So $\ce{[NH4^+] = [OH-] }$ and $\ce{[NH3] = 0.1}$ molar. So
$K_\beta = \dfrac{\ce{[OH^-]^2}}{\ce{[NH3]}}\tag{4}$
$\ce{[OH-]} =\sqrt{\ce{K_\beta[NH3]}} = \sqrt{(2.0\times10^{-5})(0.1)} = 1.414\times10^{-3}\tag{5}$
!!Check!!
We assumed that little of the ammonia would be ionized. There was 0.1 moles originally and $1.414\times10^{-3}$ did ionize. So there is 0.098586 moles of ammonia left. So the assumption is reasonable, especially since the given $pK_\beta$ only had one significant figure.
$\ce{[H^+]} = \dfrac{K_w}{\ce{[OH^-]}}= \dfrac{1.0\times10^{-14}}{1.414\times10^{-3}} = 7.072\times10^{-12}\tag{6}$
$pH = - \log{\ce{[H^+]}} = - \log{7.072\times10^{-12}} = 11.1504 \ce{->[rounding(1)] = 11.2}\tag{7}$
Second calculation of pH using $K_\beta$ of ammonia
For this calculation let $x$ be the molarity of $\ce{NH4^+}$. Again assume that autoionization of water is negligible. Hence $\ce{[OH^-]} = x$ also, and $\ce{[NH3] = 0.1 - x}$. Substituting into equation (3) yields:
$2.0\times10^{-5} = \dfrac{x^2}{0.1 - x}\tag{8}$
rearranging we get the quadratic equation
$x^2 + (2.0\times10^{-5})x- 2.0\times10^{-6} \tag{9}$
for which the real solution is $x = 1.424\times10^{-3} $. Calculating the pH using method given above yields $pH = 11.2$ again.
Calculation of pH using $K_\alpha$ of ammonium
$pK_\alpha = 14.0 - pK_\beta= 14.0 - 4.7 = 9.3\tag{10}$
$K_\alpha = 10^{-pK_\alpha} = 10^{-9.3} = 5.012\times10^{-10}\tag{11}$
$\ce{NH4^+ <=> NH3  +  H^+}\tag{12}$
$K_\alpha = \dfrac{\ce{[NH3][H^+]}}{\ce{[NH4^+]}}\tag{13}$
but $\ce{[H+]} = \dfrac{K_w}{\ce{[OH^-]}}$ so
$K_\alpha = \dfrac{\ce{[NH3]K_w}}{\ce{[NH4^+][OH^-]}}\tag{14}$
Again $\ce{[NH4^+] = [OH-] }$ and $\ce{[NH3] = 0.1}$ molar. So rearranging and solving for $\ce{[OH^-]}$ gives
$\ce{[OH^-]} = \sqrt{\dfrac{\ce{[NH3]K_w}}{K_a}} = \sqrt{\dfrac{(0.1)(1.0\times10^{-14})}{5.012\times10^{-10}}} = 1.413\times10^{-3}\tag{15}$
Solving for pH as above will yield 11.2.

Solving for the "ionization coefficient"
I don't remember encountering the term ionization coefficient ($\alpha$) before so I'm assuming it means the fraction of the ammonia which is ionized. Let $C_N$ be the nominal concentration of ammonia. Then the mass balance of ammonia species yields:
$\ce{C_N = [NH3] + [NH4^+]}\tag{16}$
so the desired calculation is:
$\alpha = \dfrac{[NH4^+]}{\ce{C_N}}\tag{17}$
Solving equation (2) for $\ce{[NH3]}$ and substituting into equation (16) yields
$\ce{C_N} = \dfrac{\ce{[NH4^+][OH^-]}}{K_\beta} + \ce{[NH4^+]}\tag{18}$
$\ce{K_\beta C_N = [NH4^+][OH^-] + [NH4]K_\beta}\tag{19}$
$\dfrac{\ce{[NH4^+]}}{C_N} = \dfrac{K_\beta}{\ce{[OH^-]} + K_\beta} = \dfrac{2.000\times10^{-5}}{1.414\times10^{−3} + 2.000\times10^{-5}} = 1.4\%\tag{20}$

For the second part the 0.1M ammonia solution has been diluted to 0.05 molar. Just redo calculations above.

Maxw: According to the charge balance equation:$$\ce{[H+] +[NH4+]=  [OH-]}$$ Why we assuming $\ce{[NH4+]=  [OH-]}$

@AdnanAL-Amleh - $\ce{[NH4^+] \gg [H^+]}$


text:
What is the citric acid concentration, [HCit], in the lemon juice?

The pH of lemon juice is about 2.1.  What is the citric acid concentration, [HCit], in the lemon juice?

I am given that $HCit \leftarrow \rightarrow H^+ + Cit^-$ and that $K_a = 8.4 \times 10^{-4}$.
$K_a = \frac{[H+][Cit]}{[HCit]} = 8.4 \times 10^{-4}$, and $[H+]$ is $10^{-2.1}$. However, without knowing $[Cit]$, how can I get [HCit]? Any help would really be appreciated!

For any mono-protic acid, when the acid dissociates in water, every H+ created will create one CB-.

H-Cit  <--> H+ + Cit-

For each 1 H-Cit. 1 H+ and 1 Cit- is produced upon dissociation.
Therefore, the [H+] = [Cit-]
For a Ka expression, set these two [] equal to each other, and the expression for [H+]*[Cit-] = x^2
Use this to determine the amount of [Cit-] created when you create x amount of [H+].

Nice. Down vote with no comment?

I didn't downvote or upvote but your answer is not complete it's more of a guide of now.

@AvnishKabaj Thanks. For homework type questions, I believe we don't want to just answer, but give information about the general principle at work?

Yes that makes a complete answer.


text:
How to find the ionization constant of an unknown weak acid?

A $\pu{0.45 M}$ solution of a weak acid, $\ce{HX}$, has a $\pu{pH}$ of $4.5$.  What is the ionization constant, $K_a$, of the acid?

$\ce{[HX]}$ is already given as $\pu{0.45 M}$.
$\ce{[H+]}$ is given by $10^{-4.5}$.

Since $\ce{pH}$ is $4.5$, $\ce{pOH} = 14 - 4.5 = 9.5$. Then we can get $\ce{[OH^-]}$ by computing $10^{-9.5}$.
Then computing $K_a$, I get
$$\frac{10^{-4.5} \cdot 10^{-9.5}}{0.45} = 2.22 \times 10^{-14}.$$
But that is apparently not the right answer. What did I do wrong?

$$K_\mathrm a = \frac{[\ce{H+}][\ce{X-}]}{[\ce{HX}]} \neq \frac{[\ce{H+}][\ce{OH-}]}{[\ce{HX}]}$$

Look at the answer given here and modify it:
How to set up equation for buffer reaction?
In your case the concentration of base $c_B \approx 0$ and $[\ce{OH^-}] = K_\mathrm{w}/[\ce{H^+}] \approx 0$, so that the equation reduces to $$K_\mathrm{a} = \frac{[\ce{H^+}]^2}{c_\mathrm{a}-[\ce{H^+}]} \approx \frac{[\ce{H^+}]^2}{c_\mathrm{a}},$$
as $c_\mathrm{a}$ is so large.

Forget about $\ce{[OH^-]}$. This is $K_\mathrm{a}$ we're talking about - the acid dissociation constant. Remember that the equation for $K_\mathrm{a}$ is:
$$K_\mathrm{a} = \frac{[\ce{H^{+}}] [\ce{A^{-}}]} {[\ce{HA}]}$$
As you can see, $\ce{[OH^{-}]}$ doesn't come into the equation.
In terms of data, $\ce{[A^{-}]}$ hasn't been provided, and so we can assume that
$$[\ce{H^+}][\ce{A^-}] \approx [\ce{H^{+}}]^2.$$
Also, the $\ce{[H^{+}]}$ you provided looks more like a $\ce{pH}$ conversion, so $[\ce{H^+}] = 10^{-4.5} = \pu{3.16x10^-5 mol dm^-3}$. Putting all of this together means that:
$$K_\mathrm{a} \approx \frac{[\ce{H^{+}}]^2}{\ce{HA}} = \frac{(3.16\times10^{-5})^2}{0.45} = \pu{2.22x10^-9 mol dm^-3}.$$

It is evident from $\ce{HA <=> H+ + A-}$, that $c(\ce{H+})=c(\ce{A-})$, hence $[\ce{H^+}][\ce{A^-}] = [\ce{H^+}]^2$. Nevertheless, this answer is still not quite correct, as it includes the approximation $c(\ce{HA})-c(\ce{H+}) \approx c(\ce{HA})$, which should be proven to hold here.

But c(HA + A-) = 0.45 M. Since we say cH+ = cA- , then cHA = 0.45 - cA- = 0.45 - cH+ = 0.45 - 3.16 e-5, which is ~0.45.


text:
How to find the compounds that form the solid (chemical formulas)
If we have a solution that contains water and  potassium sulfate  (K2SO4).
And we add to this solution :
Du magnesium chloride (MgCl2) and lead nitrate (II) (Pb(NO3)2)
What would  the compounds that form the solid (chemical formulas)?
Where is my mistake?
2 H2O + K2SO4 --> 2 KOH + H2SO4
MgCl2 +Pb(NO3)2 -->  PbCl2+Mg(NO3)2
I feel something that doesn't make sense

Your first reaction is wrong. You can't just form sulphuric acid from its salt, or more simply sulfate. It will just form bisulfate in some extent, with an equilibrium reaction, not a one way arrow.
PbCl2 formation is expected since it is poorly soluble it will be precipitated.
Look for the most insoluble union in similar cases, it will be precipitated.

In the first reaction with water and potassium sulfate, Postassium Sulfate (K2S04) does not react with water(H2O) but rather dissolves in it. And since it readily dissolves in water, it is considered readily soluble and there is no hydrolysis. It neither releases or absorbs and energy and is considered a dissociation reaction. In the second reaction with the magnesium chloride and lead nitrate, the products will be PbCl2(s) and Mg(NO3)2(aq). Lead chloride is a solid because lead and chloride and form a nonsoluble compound whereas NO3 is very soluble so it makes Mg(NO3) soluble in an aqueous solution.

Sulfate does react with water to form HSO4- to some extent, you should reconsider the first part of your answer. Check the analytical chemistry textbooks for polyprotic acids dissociation constants.


text:
What would be the 19-F decoupled proton NMR of this complex [PtH(PF3)4]-?
The proton is in the equatorial position and the geometry of the complex [PtH(PF3)4]- is trigonal bipyramidal.  The NMR is 19-F-decoupled proton NMR.
My thoughts were that the proton would couple to the platinum centre, but then I wasn't sure about the phosphorous chemical environments.
Are the two axial PF3 groups in a different chemical environment to the two equatorial PF3 groups?
If so would you end up with 6 sets of triplets? Is that a doublet of triplet of triplets?

Pt-195 (spin-1/2) only has 34% natural abundance, so you need to consider that the spectrum will be an overlay of two sub-spectra: one without Pt-195 and one with. You've identified the sub-spectrum with Pt-195 as being a dtt correctly (you can afford to be more confident - your logic is correct!). However, that's only 34% of the total spectrum.

Thank you very much for your help.  So the second sub-spectra would be a triplet of triplets? @orthocresol

Yup, that's right. Lastly, you would want to consider how to arrange the two subspectra together.

You might also want to consider that the 195Pt satellites are broadened significantly due to CSA relaxation. At higher fields (certainly above 400MHz), the 195Pt satellites appear as broad lumps only - you probably only see the H-P couplings in the 195Pt satellites at lower fields.


text:
What is the SNi mechanism of the reaction of moist silver oxide with a halide?
I was told by my teacher that the following reaction proceeds through SNi:

Moist $\ce{Ag2O}$ is basically $\ce{AgOH}$, what role does it play in this reaction's mechanism? I think this is what happens, but I am not sure:

$\ce{AgOH}$ attacks $\ce{Cl-}$ and during this process the $\ce{Ag-O}$ bond breaks and an intimate ion pair is formed between the carbocation and $\ce{Cl-}$.
Since a partial bond still exists between $\ce{Ag}$ and $\ce{O}$, $\ce{O}$ is forced to attack from the same side from where $\ce{Cl-}$ left and the reaction proceeds to completion without Walden inversion.

Is my reasoning correct? Or is there a different mechanism?

As for the reaction mechanism, I would rationalize it as follows: we know that AgX bond is very strong. A transient bond between Ag and Cl readily forms. A "curved arrow" goes from C-Cl to Cl-Ag (charge is transfered from C-Cl to Ag-Cl). At the same time, a curved arrow goes from the Ag-O(H) bond to a new C-O(H) bond, in a sort of internal sn2, being diatomic, fashion with a 4 elements reaction intermediate (C-Cl-Ag-O). The four members cyclic intermediate probably could favor a syn attack, with no stereochemical inversion.

@The_Vinz Thanks, it looks very similar to my proposed one.

This would be reasonable if there were actually reacting AgOH molecules. In such molecule coordination number of Ag is too low.

If you suspect that there is something different going on here, and I believe you are right.
My take, in the presence of lab light, $\ce{Ag2O}$ and $\ce{H2O}$ actually results, in the formation of the hydroxyl radical ($\ce{.OH}$) and the hydrogen atom radical ($\ce{.H}$) via the creation of a powerful photocatalyst couple $\ce{Ag2O-Ag}$ as outlined in this article: Ag2O as a New Visible‐Light Photocatalyst: Self‐Stability and High Photocatalytic Activity.
The photoelectric introduced electron hole $\ce{h+}$ and electron $\ce{e-}$ can act as follows in the present of water:
$\ce{H2O <=> H+ + OH-}$
$\ce{H+ + e- -> .H}$
$\ce{OH- + h+ -> .OH}$
The hydrogen atom radical is likely responsible for the extraction of the chlorine atom from the organic forming a product of $\ce{HCl}$.


text:
Will piperine endure heating for several days of soxhlet extractions?
I'm extracting piperine for use with curcuminoid supplements using a soxhlet apparatus and 91% isopropyl alcohol. My first extraction is right about done but I still need to do two more because of the limited size of my soxhlet.
My question is, instead of distilling off and reusing the distilled IPA could I just use the same solvent that contains the black pepper extract for three subsequent extractions?
I tried googling but could not determine if piperine is stable enough.

According to http://www.pharmacyjournal.net/archives/2016/vol1/issue7/1-7-21 piperine is indeed subject to thermal decomposition, but the conditions indicated are different than yours (heating without solvent at 80°C for 2 hours).
Maybe, in order to lower the exposition to high temperature, you could dry your IPA first, and use it absolute: that should also improve your extraction yield at the same extraction time of the azeotropic ipa.
A TLC analisys could give you an hint on what is the best procedure, but I would run multiple extractions, followed by vacuum distillations

Hint: in order to dry IPA you don't need a dehydrating agent. You just need table salt and a separation funnel in order to "salt it out".

Thanks for the tip on salting out IPA! That will come in handy. I let the first extraction run for over six hours. The second and third extraction will both be done in 2 hours + about half an hour to distill off about 200 ml of excess IPA to be reused. I will compare the amounts of piperine after I've purified it and hopefully remember to announce my results here as an answer to my own question.

So, I promised to answer this question based on my experiences if I'd remember. And I did. Here we go:
I did three soxhlet extractions, each with 175 grams of black peppercorns.
The first extraction took 6 hours and the cycles were quite slow due to the vapor pathways of the apparatus not being insulated.
The second extraction took 2 hours, again with an uninsulated apparatus.
The third extraction also took 2 hours, but the cycles were much faster because I used aluminum foil for insulating the apparatus.
Each extraction used 400 ml of 99% isopropyl alcohol of which 200 ml was distilled off after extraction. Something to note is that this reclamation for the first run took at least two times as long as it did for the two later ones.
The yields were as follows:
- Run 1: 8,6 grams of piperine
- Run 2: 6,8 grams of piperine
- Run 3: 7,5 grams of piperine
From this I concluded that with sufficient insulation, something like 3-4 hours could potentially provide the optimal efficiency and that the yields of piperine aren't significantly affected by prolonged periods of heating in an alcoholic solution.
I hope this write-up will help or be of interest to someone!


text:
Order of halide nucleophilicity in DMF
The order of nucleophilicity for halide family in DMF is:  $\ce{Cl- > Br- > I-}$. I understand the reason for this must be that DMF, being a polar aprotic solvent, is unable to solvate the halide ions.
I was told by my teacher that this is true only for DMF and DMSO, while in other aprotic polar solvents, the order is as usual: $\ce{F- < Cl- < Br- < I-}$.
I'd like to know what is special about DMF and DMSO (and which is not possessed by other aprotic solvents like acetone) that leads to this order.
I have verified the order in DMF from March's 7th edition (page 427). The book mentions the usual order ($\ce{Cl- < Br- < I-}$) for acetone (another aprotic polar solvent).

Related: Explanation for the trends in nucleophilicity—Orbital interactions or electrostatic attraction, Nucleophilicity/basicity in protic and aprotic solvents, Why do we use ethanol in SN2 reaction?

It depends on the ability of the solvent to solvate anions. If you draw structures of $\ce{DMF}$ with its resonance structure and check the side of the molecule that should interact with anions, you will see, that methyl groups prevent the interaction pretty much. Same for $\ce{DMSO}$. Thus these two solvents are commonly used to enhance the reactivity of nucleophiles. They solvate cations well but anions poorly. Maybe a bit strange, but it works well and it's pretty reliable.


text:
Can CaF2 exist in solution (aq) or only as a solid?
One man says that   Calcium Fluoride solution  can not exist,  because  it is only hydrated  cations (N+) and anions (F-)  surrounded  by  water molecules and that CaF2   can exist only in solid State,  not in a solution.  I  think  he is  wrong  because  if  the water contains Calcium and  Fluoride ions dissolved,  then it is a  CaF2 solution  the same way that  if it contains  Sodium  and Chloride ions it is  a saline solution.    Who is  right?

Welcome to ChemistrySE! Your question is quite unclear. What do you mean by "N+"?

If you add any ionic compound into water, it will dissolve and dissociate into ions... up to a certain extent.
For $\ce{NaCl}$, pretty much all of it dissociates into $\ce{Na+}$ and $\ce{Cl-}$. We say that this compound is soluble.
However, only a little bit of $\ce{CaF2}$ will dissociate into ions. This number is so small that we say this compound is insoluble.
Still, a $\ce{CaF2}$ solution does exist. it is just in a really small concentration. So basically, both of you are wrong; a solution with this compound does exist, but it by no means acts like an $\ce{NaCl}$ solution.
If you want to know how much dissolves, google its $\ce{K_{sp}}$, or its solubility product. The $\ce{K_{sp}}$ of a compound $\ce{X_aY_b}$ is equal to $\ce{[X]^a[Y]^b}$. This way, you can calculate the amount of the compound dissolves and dissociates.

I agree with the previous answer. Almost all group 1 metals are soluble in water and are considered spectator ions. Also compounds containing chloride, bromide, and iodide are almost always soluble. This explains why NaCl is soluble because Na is a group 1 metal and Cl is chloride. In the situation of CaF2, Calcium is a group 2 metal and is most of the time not soluble and fluoride is not part of the soluble halogen group. So, most likely CaF2 cannot exist in an aqueous solution because neither of the elements are soluble so CaF2 will always exist as a solid inside of an aqueous solution.


text:
Reduction half-reaction for the reduction of elemental oxygen

Write the reduction half-reaction for the reduction of elemental oxygen in the presence of water, producing hydroxide ions

This does not make sense to me - a half reaction (from what I understand) is when you have an "element + some number of electrons -> element charged" or "charged element -> element + some number of electrons"
Reaction:
$\ce{HOH + O -> 2OH}$
Oxidation States: They all seem to be neutral to me - but I may be missing something.
So, how is a half reaction being introduced into this and what would it look like? Also, how can i know which element/compound is reduced/oxidized and is the reducing agent/oxidizing agent?
I am really sorry if this seems simple but I am a chem noob and I was out for a month and my teacher is behind schedule so he is giving us worksheets without teaching and i do not understand a lot. Thank you for your time

Why the downvote? I don't get how to do the half reaction from the reaction but I put a lot of effort into making this question and trying to figure the solution.

Sometimes people downvote for inexplicable reasons. Don't worry too much about it all of us have been there.

O2 + 2HO+ 4e- —> 4OH-
That is the correct answer

I think you need to reconsider the equation, I think it is O2 + 2H2O + 4e ---> 4OH-


text:
Can there be two precipitates in a double displacement reaction?
On one of my chemistry homework's, I am doing a double displacement reaction and the question is having me predict the products and make a net ionic equation for copper(|I) sulfate and barium acetate reacting.
$$\ce{CuSO4(aq) + Ba(OH)2(aq) -> Cu(OH)2 + BaSO4}$$
I was able to predict the products of it, but when I was starting to do the net ionic equation, I came across the problem that there were no spectator ions and I am not sure if two precipitates can form out of a double displacement reaction.

Yes, you were right in concluding that there are no spectator ions. How is that a problem? There is no commandment that says "Thou shalt not have two precipitates in one reaction". Indeed, such double precipitations are very much possible, though rare. The most famous example is probably the formation of lithopone.


text:
Why is cis-1,2-dichloroethene more stable than trans-1,2-dichloroethene?

Why is cis-1,2-dichloroethene more stable than trans-1,2-dichloroethene?

Usually, trans compunds are more stable than cis ones, due to less strain and its non-polarity. But, in this case it's quite the opposite. I tried to search the reason behind it, and found that it's supposedly due to the cis effect?

Here's a pre-MOT rationalisation of the cis form being more stable than the trans form for 1,2-dihaloethylenes:
The lone pair of chlorine atoms is involved in resonance with the double bond, as it does so positive charge appears on one chlorine and negative on another. A cis geometry allows for a stabilizing interaction (attraction) between these positive and negative charges, which makes this resonance form a greater contributor to the resonance hybrid than it is for the trans isomer.

MOT-based explanation:
Eyring's work suggests that there is maximum delocalisation possible in trans form.
Quoting, from the same paper:

In a recent paper Binghams proposes that the main effect responsible for the “cis effect” of ethylenes is an effect which tends to destabilize the trans form rather than provide extra stabilization for the cis form. In short, this is an unfavorable conjugation effect which stems from the fact that the conjugation in these cases is between two filled orbitals (halogen lone pair and $\pi$ bond). The interaction gives rise to two new orbitals, one bonding and one antibonding, the latter being more antibonding than the former is bonding. Consequently, the total interaction is energetically unfavorable, in contrast to interaction in hydrocarbon polyenes. This would be equally true for both the cis and trans forms but, following Eyring’s argument, the delocalization is greater in the trans form.
  Consequently this form will be the less stable one.

Reference: Structure of 1,2-difluorocyclopropane and the "cis effect"

In this answer, I will point out an inadequacy of the reasoning based on resonance structures and I will also provided another MO perspective, which I believe is more convincing, on the issue.
A flaw in the resonance argument
As presented by Abcd, the phenomenon can be explained without use of MO theory and purely based on resonance structures. Please refer to his answer for the argument. Based on this argument, we may actually expect the $\ce {C=C}$ double bond to become weakened. Since if the two zwitterionic resonance structures were to be significant, the central double bond should naturally have less double bond character and more single bond character. However, data presented by Skancke & Boggs (1979) show that the $\ce {C-C}$ bond length is in fact shorter in the cis isomer ($\pu{1.311 Å}$) than in the trans isomer ($\pu{1.320 Å}$)$^1$. This suggests that the resonance stabilisation may not be the cause of this phenomena.
Another MO perspective
Fleming (2009) provides an explanation which I feel is more convincing. Essentially, the idea is that extra stabilisation in the cis form is derived from the interaction between the antiperiplanar $\ce {C-H}$ $\sigma$ and the low-lying  $\ce{C-Cl}$ $\sigma^*$ MOs. Jan provides more details on this here when he explains the cause of a similar effect responsible for the gauche conformation being favoured in 2-fluoroethanol. Another phenomenon that Fleming touches on which he uses the same explanation for is the preference for the gauche conformation in 1,2-difluoroethane. It is also explained in the same paragraph on p. 89.
The same explanation can be used for 1,2-dichloroethene. This is because chlorine is also a rather electronegative atom, although less so than fluorine. Thus, we can still expect the $\ce {C-Cl}$ $\sigma^*$ MO to be low enough in energy to interact with the the $\ce {C-H}$ $\sigma$ MO.

References

Skancke, A.; Boggs, J. E. Structure of 1,2-Difluorocyclopropane and the "Cis Effect". J. Am. Chem. Soc. 1979, 101 (15), 4063–4067.
doi:10.1021/ja00509a009
Fleming, I. Molecular Orbitals and Organic Chemical Reactions. John Wiley & Sons, Ltd. United Kingdom, 2009.


text:
Which is more conducting - molten sodium chloride or molten magnesium chloride?
Which is more conducting - molten sodium chloride or molten magnesium chloride?
I'm pretty sure I know the answer to this but my problem is putting into words. Magnesium is more conductive due to its 2+ charge, and the $\ce{NaCl}$ having a neutral charge. Right?
Would I be right in saying that as the compounds are heating up and melting the ions break away from each other and the ratio of ions in $\ce{MgCl2}$ is 2:1 so therefore the solution would be more highly charged than that of the $\ce{NaCl}$?
I know I'm on the right track, just struggling to explain it.

The solution is not charged; it is neutral, since it contains equally many positive and negative charges. Also, you don't have any solution at all. Other than that, you may or may not be right. True, $\ce{Mg^2+}$ has greater charge than $\ce{Na+}$, so what?