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Question on Melting Points of Group 17,18 and 1
So , I have done some searching and found that melting points of groups 17 and 18 increase going down the group whereas group 1 decreases.
However , the explanation seem to all be the same.

Number of shells increase
atomic radii increases
Van der waals force between atoms increases
more heat energy needed to overcome said forces

But for group 1, the explanation is that even when atomic radii increases , the metallic bond between atoms decreases and thus lower melting point.
So , I'm a little confused about this metallic bond , and why doesn't group 1 alkali metals also create Van Der Waals forces?
I may be wrong on some parts so  please do correct me.

Google search metallic bond and you will find that is when the atomic size increase, the metallic bond get weaker and continue the thought process

Metallic bonding is a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons (in the form of an electron cloud of delocalized electrons) and positively charged metal ions
Van der Waals forces include attraction and repulsions between atoms, molecules, and surfaces, as well as other intermolecular forces.
The alkali metals have the largest size in their period. Consequently the distance between nuclei of one atom and the outermost electron of other atom is more and hence metallic bonding is less. But owing to their small size the distance between the nuclei of one halogen atom and the outermost electron of other atom is comparatively less. Hence they have show more metallic bonding.


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Which of the five main branches of chemistry has the most labwork?
Budding chemistry student here.
I am aware of the so called "five branches" of chemistry, namely organic, inorganic, bio, physical and analytical.
Which of these paths lead to work with the most lab time?
I suspect physical chemistry would be fairly academic, considering it's links to physics and maths. Which is also very appealing.

Physical chemistry also involves lab work such as titrations , preparation of various acids, bases, salts etc

Lab work in actual phys chem research is not the same as student lab experiments designed to illustrate concepts in phys chem.

Are we talking about lab work for your thesis, or lab work in an academic position or in industry? Also, are these categories relevant any more in research?

Analytical is not usually a separate branch. Used to be, when analytics were still largely actual lab chemistry with solvents and pipettes and bunsen burners.

The only way to spend your complete graduate studies in front of a computer is doing it in theoretical chemistry, i.e. QM and simulation.

Personally I am in an interdisciplinary field and I have direct or indirect experience with many techniques and type of lab work. I have no doubt that the field requiring most physical presence and work at the bench is organic synthesis. No idea about biochemistry, tough. And though I am categorical, everything need work. There are experiment in which you wait hours, but need to be there at 22 pm, for instance.

All the five so-called branches of chemistry require heavy use of the brain. Modern chemistry is very interdisciplinary. At the research level, the distinction among these branches becomes blurry. Everyone is doing some part of each branch. Traditional inorganic and organic chemistry both require a lot of wet work and synthesis and characterization.
If you ask which branch of chemistry has the highest level of automation, and most of the work is done by instruments-it is analytical chemistry. Many modern physical chemists or theoretical chemists rely on in silico techniques (=computers, modelling, simulations etc). You may not be doing a single real experiment (if your not lucky). Computer is your virtual lab. As a hardcore theoretical chemist, finding a job afterwards in the field, may be difficult. My acquaintance switched to computer programming after a PhD, which was a wise decision after remaining jobless for a year. He learned programming languages in the mean time. In any case, all fields are good. It all depends on your passion.

Here's my take:
There are a LOT more branches and sub-branches of Chemistry than you mentioned. Under inorganic chemistry alone, we have the chemistry of the main groups, as well as transition metal chemistry (and organometallic chemistry) as well as f-block chemistry. Owing to the interdisciplinary nature of modern Chemistry (some call MOT/VBT "The theory underlying all fields of chemistry", and rightly so) its impossible to say "I'm an inorganic chemist and I never deal with organic chemistry" or something along those lines.
There's also, as mentioned, the rise of computational methods like the Density Functional method (DFT) and "in silico" (literally "in silicon", i.e. using a computer) methods are also quite popular, and there are research papers with titles like "DFT analysis of [title]". If you want to go into the field of quantum/computational chemistry, you probably don't need much lab work (which is something I personally despise).
Hope this helped.

source: Stack Exchange

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Removing sodium borate for the test proving thiocyanate ions
I've just learned that when carrying out the identification test for thiocyanate ions borax can induce a false positive and therefore needs to be removed. I couldn't find the answer on google so I'm asking a question here.
I would think you first carry out the test to see if borax is present; but if it is, how would you remove it?

There are numerous methods of identifying thiocyanate anions both quantitatively and qualitatively. Which one are you asking about? The formation of iron(III) complex?

@andselisk yes. Normally there should be a complex with iron(III), thiocyanate ions and water but with Borax ions present the iron(III) makes a complex with borax. How would one then remove the borax ions?

Add more context to your question. Which test are you talking about, since this is not a generic interference. Check these tests, it may not need borate removal. https://books.google.com/books?id=-Z5NP98K7cIC&pg=PA35&lpg=PA35&dq=interference+of+borate+in+thiocyanate+detection&source=bl&ots=y0tfHcw2uY&sig=ACfU3U3zsScOIfWG7VntO-omBdw-FUAVcw&hl=en&sa=X&ved=2ahUKEwjcn--W1oTiAhUBGKwKHXcwAsIQ6AEwBHoECAcQAQ#v=onepage&q=interference%20of%20borate%20in%20thiocyanate%20detection&f=false

Lehrbuch der analytischen und präparativen anorganischen Chemie by Jander et al. [1] suggests to precipitate borate with $\ce{Ba^2+}$ in neutral solution beforehand as barium(II) thiocyanate is one of the few barium(II) salts actually soluble in water, or use an excess of $\ce{Fe^3+}$ to make sure all the interfering anions are precipitated.
From [1, p. 358]:

6 . Nachweis als $\ce{Fe(SCN)3}$
[…]
Störungen: Um den störenden Einfluss von $\ce{F-}$, $\ce{PO4^3-}$, $\ce{AsO4^3-}$, $\ce{H3BO3}$, $\ce{C4H4O6^2-}$, $\ce{C2O4^2-}$ usw., die mit $\ce{Fe3+}$-Ionen Komplexe bilden, auszuschalten, wird $\ce{Fe^3+}$ im Überschuss zugegeben.

From [1, p. 418–419] (emphasis mine):

9. Nachweis von Fe(III) als $\ce{Fe(SCN)3}$
$$\ce{Fe^3+ + 3 SCN- <=> Fe(SCN)3}$$
[…]
Störungen: […] Ferner beeinträchtigen
$\ce{Hg^2+}$-Ionen durch Bildung von wenig dissoziiertem $\ce{Hg(SCN)2}$, $\ce{F-}$ durch $\ce{[FeF6]^3-}$-Komplexbildung, die Anionen organischer Säuren ebenfalls durch Komplexbildung und auch $\ce{PO4^3-}$, $\ce{AsO4^3-}$, Borat-Ionen sowie ein größerer Mineralsäureüberschuss die Reaktion. Es ist
daher ratsam, vor der Prüfung […] die störenden Anionen in neutraler Lösung mit $\ce{Ba^2+}$ abzutrennen.

References

Jander, G.; Blasius, E.; Strähle, J.; Schweda, E.; Rossi, R. Lehrbuch der analytischen und präparativen anorganischen Chemie, 16. Auflage.; Hirzel: Stuttgart, 2006. ISBN 978-3-7776-1388-8. (in German)

Seems like a lovely book. Is it for undergraduate level?

@M.Farooq Pretty much. Bare-bone analytical/inorganic chemistry, that's why I like it:)

I am just finishing an educational article on the reliability of German translations which are translated by free software. Your reference just reinforced my conviction that as scientists we should use and learn at least one more language. Passing a reading test in Russian or French or German was compulsory, decades ago, for a PhD in chemistry.

When you ask how to remove borax, I assume it is Sodium tetraborate decahydrate (Wikipedia). Thus, I agree with @andselisk of using $\ce{Ba^2+}$ solution to precipitate it. Even though it is seemingly an excellent answer, the reference given and the text body are in German, and I didn't understand it much. :-) Thus, I want to give some clues to make it sense.
According to Ref.1, tetraborate-ions ($\ce{B4O7^2-}$) form a white precipitate of barium metaborate $\left(\ce{Ba(BO2)2}\right)$ with $\ce{Ba^2+}$ solution ($\ce{BaCl2}$).
However, if it is a neutral solution, only following reaction occurs:
$$\ce{B4O7^2- (aq) + Ba^2+ (aq) + 3H2O (l) -> Ba(BO2)2 (s) + 3 H3BO3 (aq)}$$
Meaning, removal of borax is only partially completed. Complete precipitation of $\ce{B4O7^2-}$ ion would be reached only in the strong alkaline medium:
$$\ce{2 H3BO3 (aq) + Ba^2+ (aq) + 2 OH- (aq) -> Ba(BO2)2 (s) + 4 H2O (l)}$$
Keep in mind that this white precipitate is dissolved in inorganic acids:
$$\ce{Ba(BO2)2 (s) + 2 H3O+ (aq) -> Ba^2+ (aq) + 2 H3BO3 (aq)}$$
Thus, white precipitate of $\ce{Ba(BO2)2}$ should be filtered off before adding $\ce{Fe^3+}$ solution.
Alternatively, I believe testing of thiocyanate ions can be done in presence of borax by doing two back to back reactions: (i) Check the solution with the blood-red color formation reaction with $\ce{Fe^3+}$ solution as usual; and then (ii) Check again the same solution with the blue color formation reaction $\left(\ce{(NH4)2[Co(SCN)4] (aq)}\right)$ with $\ce{Co^2+}$ solution:
$$\ce{Co^2+ (aq) + 4 SCN- (aq) + 2 NH4+ (aq) <=> (NH4)2[Co(SCN)4] (aq)}$$
Reference:

O. A. Ievtifieieva, V. V. Bolotov, T. A. Kostina, O. M. Svechnikova, T. I. Yuschenko, N. I. Kaminska, A. E. Kosareva, L. V. Slobodyanyuk, O. P. Yashchuk, In Analytical chemistry (Qualitative analysis) Part I: The manual for students of higher schools; O. A. Ievtifieieva, Ed.; Publishing house the CLL: Kharkiv, Ukraine, 2014 (168 pp.).

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


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What would happen if I combined this polymer and this metal (assuming I can)
So let me explain, I recently found out about a material called sorbothane (which is a polymer). Now, sorbothane is special because not only does it highly efficiently absorb shock, but also highly absorbs vibration.
However, when it does this, it temporarily loses its shape, so I was wondering, what would happen if I combined it with a strong yet light metal like titanium? (Note: I am 17 and don’t know a whole lot about chemistry.)

If we could combine stuff like that so easily while retaining individual properties, we'd have Vibranium and Adamantium already.

You can combine materials in two ways: chemically and mechanically.
On average, if you want to the desired physical properties to add up, you want to go mechanical way, not the chemical one.
Once the reaction takes place, the there is absolutely no reason to assume the product preserves properties of both precursors. It might sometimes, but more often than not it's a completely different beast.
Speaking of retaining physical properties in terms of logic and addition operation, in chemistry both $\text{true} + \text{true} = \text{false}$ and $2 + 3 = -42$ are possible.
Mixing water and sodium won't result in liquid metal.
What you probably want to look into instead is mechanical combination, or, more precisely, composite materials such as fiberglass or carbon plastics alternative where titanium would act as a 2D or 3D supporting mesh for sorbothane, though not reacting with it chemically.
Such poly-phase materials are already used for monolithic composite armor plates.

Mixing water and sodium is a lot of FUN$^{(TM)}$ though.

granted, sometimes mechanically combining materials can create different and new properties.

"Mixing water and sodium won't result in liquid metal." Actually, it will. Albeit only briefly.


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How to detect nM levels of Copper(I) Oxide in blood?
I'm trying to design an experiment to detect nM quantities of Copper(I) Oxide in fish blood. Are there any specific tests for Copper(I) Oxide that can be used to detect trace quantities(~nM)? Any non-destructive tests such as using optical methods would be preferable.For purposes of designing the experiment, let's assume the particles are nano-sized and dispersed in blood.

Add some context to your analytical problem. What is the sample? Cu(I) is not stable in water with respected to oxidation. Cu(I) oxide is a solid which does not dissolve in water as well. Do you have a solid dispersed in water? Are you specifically interested in only one species Cu(I)?

I posted the original question thinking I could simplify the problem by calling it a water dispersion, but I have added what I am trying to do in more detail.

Nano-sized is good, because you can first filter out any blood cells, platelets, etc. Greatly simplyfies your problem. Why Cu(I) nanoparticles, specifically?

@Karl Some farmers use co-culture, which means they culture fish in the water clogged in paddy fields. Cuprous oxide is a cheap and effective fungicide they often use with paddy. For my project I am trying to assess how much of this enters the fish's blood stream. I guess it is tiny, but I wanna validate that.

The main analytical problem in your case of detecting Cu(I) oxide in blood is that you do not know the fate of the particles once they enter blood stream. As I said earlier, Cu(I) is thermodynamically not stable in any aqueous system. It oxidizes to Cu(II), which is water soluble. The non-destructive technique to detect species is X-ray photoelectron spectroscopy, but its detection limit is very poor. It may take a day to collect weak signals from such a low concentration of Cu(I) oxide. The sample has to be bone-dry as well! This is a species specific analysis, i.e. one can distinguish oxidation states of metals by deconvolution.
A practical suggestion is that you do a control test, i.e. measure copper levels before treatment in blood. As a result, all copper originates from your copper (I) oxide dispersion. You would not know the specific species of copper but you will get the total content. For that one can use common techniques such as graphite furnace atomic absorption analysis (only a few microliter of sample is required).

I am having trouble understanding Cu(I) Oxide's thermodynamically instability in aqueous system: don't tests like Benedict's reaction produce copper(I) oxide that precipitate out unoxidized?

Yes, but if you let it sit for sometime, it will be oxidized to Cu(II).

There will be problem with AAS, that supposed cuprous oxide would be far below the physiological copper content. The question "why cuprous oxide ?" remains.

@Poutnik, Karl had the same question, so I provided more context to my problem in the comments


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Can benzene undergo Diels–Alder reactions?
I was reading about the Diels–Alder reaction and wondered that since benzene also consists of two conjugated double bonds, then should it also undergo Diels–Alder reactions? If not, then why not?

The properties of benzene are very different to the hypothetical 1,3,5 cyclohexatriene that you seek to compare it to.

With that said, some benzene lookalikes are amazing Diels-Alder substrates, e.g. triazines and tetrazines. Actually, I've covered this topic before (specifically focusing on benzenes): https://chemistry.stackexchange.com/a/77049/16683 The question isn't a duplicate, though.


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Are the triply-bonded carbons in pyridyne (and benzyne) sp-hybridised?
I was given this molecule (3,4-pyridyne) in a test:

I had to find out if it had any atoms which had a hybridization state other than sp2. The solution said that it doesn't have any. But aren't the triply-bonded carbon atoms sp-hybridized?
Edit:
Benzyne also has a similar structure, with a distorted triple bond -

I also wanted to know what the IUPAC name of that compound would be. But I'm new to the site so I don't know if I should ask this as a new question, altogether. I don't see how that would be a good contribution to the site since it's only a picture.

This molecule does not exist, other than possibly as a short-lived intermediate. If not for this fact, your reasoning would be OK.

As for the name, 3,4-didehydropyridine would describe the structure well.

@IvanNeretin While equally thinking "if existing at all, than briefly only", May and Moody describe it as an easily obtained intermediate from shelf stable solid 3-(3,3-dimethyltriazen-l-yl)pyridine-4-carboxylic acid for moderately working Diels-Alder reactions with dienes like 2,5-dimethylfurane. (Tetrahedron Letters, 26, 1985, 2123-2124 "A convenient precursor to 3,4-didehydropyridine").

Related: Why is the benzyne triple bond distorted?; Structure of Benzyne, although they don't directly address the issue of hybridisation head-on.

Thank you for the question, by the way; I learnt something, and apparently the conventional wisdom that the carbons are sp2 hybridised is not correct, which actually surprised me - I was expecting the calculations to output something intermediate between the idealised sp2 and sp hybridisation models.

TL;DR It is sp-hybridised, but maybe not in the way we normally think of alkynes as being.

There are a couple of things to say before going into the actual answer.
Firstly, hybridisation is derived from geometry, not from the bonding pattern (in other words, how many σ- and π-bonds there are). In most cases, the bonding pattern dictates the geometry: for example, most saturated carbons are tetrahedral, and a tetrahedral geometry usually signals sp3 hybridisation, so saturated carbons are usually described as being sp3-hybridised. So, effectively, the bonding pattern does dictate hybridisation. This is all fine, but only up to a certain point, because it can break down in exceptional cases such as cyclopropane. Because of the geometrical constraints of a three-membered ring, there is not quite enough room for the carbon to be perfectly tetrahedral, and so it is not correct to label the carbons as being sp3-hybridised in the usual sense.
Secondly, hybridisation is usually more nuanced than "sp", "sp2" or "sp3". It is individual orbitals that are hybridised, not atoms; so for example, we can talk about an sp orbital, which is a linear combination (a sum) of one part s-orbital and one part p-orbital. When we say that "a carbon atom is sp-hybridised", what we really mean is that the bonding of that carbon atom can be described using four valence orbitals: two sp-hybridised orbitals, and two unhybridised p-orbital. Likewise, an "sp2 carbon atom" uses three sp2-hybridised orbitals and one unhybridised p-orbital; and an "sp3 carbon atom" has four sp3 orbitals.
Again, in the vast majority of cases, these three categories are sufficient to accurately classify all carbon atoms in a molecule. However, as you may guess, this breaks down in exceptional cases: cyclopropane is the most well-known example. Here, each carbon has two sp5(ish) orbitals and two sp2(ish) orbitals. It uses the sp5 orbitals to form σ-bonds to the carbons next to it, and the sp2 orbitals to form σ-bonds to the two hydrogens on it. "How is it possible to have five p orbitals?" sp5 does not literally mean one s orbital and five p orbitals; it just means the ratio of s-orbital and p-orbital contribution to this particular hybrid orbital is 1:5.

With that out of the way, here is how benzyne works (PBE0/cc-pVTZ gas phase optimisation + NBO analysis). We're going to look at the orbitals that the triply bonded carbon (C1) uses for bonding to its neighbours (C2 and C6, see structure below):†

One sp orbital (50.02% s, 49.75% p) for forming a σ-bond to C2;
One p orbital (0.39% s, 99.38% p) for forming the π-bond to C2 in the plane of the molecule;
One p orbital (0.00% s, 99.84% p) for forming the π-bond to C2 perpendicular to the plane of the molecule;
One sp orbital (49.64% s, 50.17% p) for forming a σ-bond to C6.

On their own, the numbers don't tell you which direction the p-orbitals are pointing in, but you can use visualisation software to figure this out. From left to right, these are the (σ- or π-, not the individual hybrid) orbitals, in the same order as listed above.

Based on the above, the orbitals can be crudely illustrated as such:

This is, in fact, pretty much what an organic chemist would describe as an "sp-hybridised carbon atom". However, it is not sp-hybridised in the same sense as an ordinary alkyne. In an ordinary alkyne, the σ-bond between the two triply bonded carbons comes from overlap of sp-orbitals. This is also the case here, but there is a distinction. In an ordinary alkyne, the sp-orbitals point directly towards each other. Here, they are not quite aligned in the same way; they instead form what is sometimes called a "bent bond". The same can be said of the C1–C2 π-bond.
This is fundamentally because of the geometry of the ring. Since we have already allocated two p-orbitals to two other bonds, the two sp-orbitals must be derived from the same p-orbital. As a consequence, the two sp-orbitals must point in diametrically opposite directions, 180° away from each other. If C1 were to use an sp-orbital pointing directly towards C2, then the remaining sp-orbital would point rather far away from C6, significantly weakening the corresponding bond with C6. The orbital geometry that you see is a compromise.
The triply bonded carbons in pyridyne, for all intents and purposes, can be described in the same way as those in benzyne. The effect of nitrogen on the orbitals above is tiny; qualitatively they look the same (the colours are flipped for the first and last ones relative to the benzyne orbitals earlier, but this is of no physical significance).

† The numbers don't add up to exactly 100% because the tiny d- and f-orbital contributions are neglected.

No.
Hybridization derives from geometry, not bond type.  For instance, the central carbon in allene ($\ce{CH2=C=CH2}$) has $sp$ hybridization like the carbons in acetylene, despite the absence of triple bonds in the allene, because the linear bonding geometry brings this hybridization about.
In pyridyne the formally triple-bonded carbons do not have the linear bonding geometry corresponding to $sp$  hybridization, rather the geometry corresponds to $sp^2$ hybridization like pyridine or benzene.  What seems to be one of the "pi" bonds in the pyridyne structure as drawn on paper, is in fact a "bent bond" formed by the side-by-side overlap of two subsidized orbitals, not a pi bond formed from two $p$ orbitals.

On the same note, this was another compound on the test (which didn't have all sp$^2$ hybridized carbon, interestingly)
https://chemistry.stackexchange.com/questions/114896/what-is-the-hybridization-state-of-carbons-in-10annulene.


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SN1, SN2 and enantiomeric excess

When (R)-2-bromobutane is heated with water, the $\mathrm{S_N1}$ substitution proceeds twice as fast as $\mathrm{S_N2}$. Calculate the appropriate enantiomeric excess and the specific rotation of the product of the mixture. The specific rotation of (R)-2-butanol is +13.5.

I din't understand how to interpret '$\mathrm{S_N1}$ proceeds twice as fast as $\mathrm{S_N2}$' mathematically.

Basically, if 100% conversion occurs, you have $\frac{2}{3}$ of molecules converted by SN1 and other $\frac{1}{3}$ of molecules converted by SN2. That means, you have $\frac{2}{3}$ of $(S)$-isomers and other $\frac{1}{3}$ of $(R)$-isomers.

In this case, the absolute configuration remains same (as R) for $\mathrm{S_N1}$ and changes (into S) for $\mathrm{S_N2}.$ This is because the priority order of non substituted groups (wrt $\ce{C2})$ remains the same before and after substitution of $\ce{Br}$ by $\ce{OH}.$
Let the initial mole fraction of (R)-2-bromobutane be 1. After the reaction, let the mole fractions of R- and S-2-bromobutane be $x$ and $1-x$, respectively.
According to question condition, as $\mathrm{S_N1}$ proceeds twice as fast as $\mathrm{S_N2},$ we get
$$x = 2(1 - x) \quad\implies\quad x = 2/3$$
So we have excess of R-enantiomer, and the specific rotation is
$$\frac{2}{3}\cdot (+13.5) + \frac{1}{3}\cdot (-13.5) = +4.5$$


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Stereochemical outcomes in opening of vinyl epoxides

Explain the outcomes of the following reactions. In each case the major product is shown.

Firstly, I don't understand why the chloride always attacks on the right hand side of the epoxide. Also, regarding the stereochemistry for (ii), I assume the chlorine can open the epoxide as it is antiperiplanar to the left hand side of the epoxide?

This is acid catalyzed ring opening. The intermediate is stabilized by adjacent double bond.

(1) The product in (ii) isn't consistent with the formation of a chloronium ion. (2) Antiperiplanarity doesn't matter here; the substituent is acyclic so in both (ii) and (iii) it can rotate into a conformation where the chlorine lone pair can attack the epoxide, should that really be the mechanism.

Thank you for your comments. I have reevaluated my mechanism for the formation of ii) and the chlorine attacking the LHS of the epoxide couldn't result in the product. I am still not quite understanding the mechanism of the chloride attack however, especially in ii) as it seems to be attacking the top face of the epoxide face as opposed to the bottom face as it normally does

I don't know the answer either. One possibility is that the chloro substituent does attack the epoxide first, but at the allylic (right) carbon, not the left carbon. Then you have a typical SN2 to open that ring. It does give the correct product, although I'm not fully convinced; I've yet to see an example of a 4-membered ring being formed in this sort of reaction. Why it doesn't also happen in (iii) is beyond me at the moment, although I would hazard a guess at there being some problem with achieving the necessary conformation.

Ugh, I can't believe I'm actually correct: https://pubs.acs.org/doi/10.1021/acs.orglett.5b00558 Where is this question taken from? The author(s) obviously took some liberties in changing the substituents on both ends.

While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along the red bond of 1, places the two carbon chains anti to one another. Intermediate 3 undergoes straightforward SN2 displacement at the allylic site.
In the case of epoxide 5 (ii), the two alkyl chains maintain the anti positions in conformation 6 but now the carbon-bound chlorine acts as a neighboring group in an intramolecular SN2 displacement to form intermediate 7. This species suffers a second SN2 displacement to afford 8 with retention of stereochemistry at the reacting site. Compound (i) follows the same pathway as 1 (iii).    The role of water is to cleave the silylether.

From 6 to 7 neighbouring group effect  , which is operative in 3 membred cyclic state.Is 4 membred cyclic state stable ?

No, that is why it is reactive. 4-Membered rings do form.


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What (simple) household chemicals react with oxygen?
I'm interested in making a small "preservation chamber" that I can basically use to store objects without the presence of oxygen. I have a resealable stainless steel canister and the plan is pretty much just to put whatever object I want to be preserved inside, together with a chemical/object/thing that absorbs the oxygen from the air inside the can, protecting the other item. I don't have access to anything fancy, so I need something is/can be made from household materials that will react with/contain oxygen.
Here's the catch: It CANNOT be a candle. The container has a plexiglass cover that will melt if it's exposed to the heat of a flame. (I tried it before; it worked, but the plastic almost cracked. I don't dare to do it again.)

Use a candle but arrange it so the flame doesn't touch the cover? Use a different cover?

@immibis, I can't use a different cover. The cover is attached and has the seal. The problem wasn't the flame itself touching the cover, but the heat above it. I don't want to try anything with a hot flame because if it doesn't work out I very well might ruin the cover.

@Rafael, To offer a practical application to andselisk's answer, you may use simple hand-warmers such as this ones: https://www.onlyhot.it/hand-warmer.html which includes iron salts and do not heat above a few tens of degrees (°C): https://en.wikipedia.org/wiki/Hand_warmer#Air_activated_(iron), they can be found in any sport store (given you are in a country where winter sports is a thing) or ordered cheaply from Amazon and co.

You don't have to reinvent the wheel.
Food industry already uses quite simple oxygen scavengers for decades, and one of the most affordable formulations include iron powder mixed with sodium chloride.
Typically, an oxygen scavenger is packaged in small packets (sometimes labeled as "oxygen absorber" or "antioxidant") that are added with the product inside the pact and sealed.
The absorber itself consists of iron powder and an inert absorbent (sodium chloride, activated carbon, food-grade clay or a composite such as nylon 6/nanoclay).
When placed inside a package with a product intended for storage, it begins to actively absorb oxygen and completely removes it within 12–96 hours.
The drawback of using Fe is that it promotes rapid lipid oxidation, if it's leaked and gets in contact with the product.
However, since you are already using stainless steel container, going with iron-based quencher seems like an optimal choice to me.
More historical background and interesting tech details can be found in Wiley's Encyclopedia of Packaging Technology [1, p. 845]:

[…] The iron-based oxygen scavengers are the most prevalent and they are examined
  first.
The iron technology is based on rapid oxidation of iron powders to prevent headspace oxygen from reacting with foodstuff components or to reduce oxygen permeation rates through container walls. Reduced iron powders (zero-valent iron $\ce{Fe^0}$) are highly reactive and possess a higher capacity to react with oxygen than partially oxidized ferrous iron ($\ce{Fe^2+}$) compounds. Large exposed surface area of iron powders per unit weight, which increases with reduction in mean particle size and especially when using highly porous "sponge" iron morphologies, promotes simultaneous oxidation at multiple reactive sites, resulting in faster rates of oxygen scavenging. Presence of electrolytes and an aqueous interface to facilitate electron transfer during oxidation is known to accelerate oxidation reactions of transition metals. This is exempliﬁed by rapid rusting of ferrous metals and steels exposed to road salt and salted seawater, in contrast with slow rates of corrosion in the dry climates. Ferrous and ferric iron oxides and hydroxides produced by oxidative reactions are basic. As a result, acidic environments promote oxygen scavenging by shifting equilibrium to iron oxides. At low pH values, oxidation of reduced iron proceeds in humid environments even in the absence of oxygen, by deoxygenizing water molecules and resulting in hydrogen gas formation.
Since low pH values and aqueous interfaces require the presence of water, water vapor diffusion and variation in relative humidity of the environment can be used to control activation of oxygen scavenging capacity of iron powders. The speciﬁc formulations are designed to trigger the activation at desired RH levels, control the reaction rate, achieve the full oxidative potential of the iron powder, maintain compatibility with various products, and ensure food safety. The reaction formula for complete oxidation of reduced iron to eventually form ferric oxide trihydrate complex (commonly known as rust) is
$$\ce{4 Fe^0 + 3 O2 + 6 H2O → 4 Fe(OH)3 → 2 (Fe2O3* 3 H2O)}$$
Many transition metals are recognized as being sufficiently oxidative to become candidates for an oxygen-scavenger formulation; however, iron does offer unique advantages that have driven the industry to general use of this medium. Iron has a relatively high afﬁnity to combine with oxygen on a per unit weight basis compared to most alternatives. Complete oxidation of 1 g of reduced iron removes 300 cm³ of oxygen at STP conditions. Elemental candidates that exceed it have drawbacks such as the odor problems with the use of sulfur and the propensity of aluminum to form an almost impermeable oxidized skin layer that limits further oxidation. Iron rust, on the other hand, ﬂakes off as it forms and it is highly permeable to further duffusion of water and oxygen. Of great importance is the food safety of iron powders, which are recognized as nutrients important for healthy blood and commonly used as food enrichments in milled flour, breads, breakfast cereals, and baby foods. The same cannot be said about copper, zinc, aluminum, and many other transition metals. The abundance and relatively low cost of producing elemental iron are also important, especially in comparison to choices such as palladium and platinum metals and catalysts based on them. Then there is the ready ability to manipulate reactive capability of iron powder formulations to adapt to a wide variety of applications for both the rate of oxygen scavenging and the activation by the moisture in a package headspace. One of the few drawbacks is that the oxidation of iron is a temperature-dependent reaction and normally slows dramatically as the temperature approaches freezing.
Electrolytes and acidifying agents in a scavenging formulation are commonly supplied by metallic salts […]. The practical considerations of food safety have dictated that common table salt ($\ce{NaCl}$) is often the best choice for the electrolyte, while various metal and organic salts of strong acids are often used to increase the acidity of the formulation. […] When the packaged product does not have a required water activity to fully activate an iron-based scavenger and provide the desired reaction rate, premoisturized silicas, zeolites, and activated carbons are used as self-contained water carriers in various proprietary formulations to provide a reliable activation.

References

The Wiley Encyclopedia of Packaging Technology, 3rd ed.; Yam, K. L., Ed.; John Wiley & Sons: Hoboken, N.J, 2009. ISBN 978-0-470-08704-6.

Very interesting answer! I'm thinking of using steel wool as the iron. I'm just a little confused about the water. Assuming I'll put the iron and salt in a cup, should just add a little water to dissolve everything? Or maybe not because it will then take oxygen from the water, not the air?

@Rafael AFAIK there is no need to use large amounts of water; I'd add it drop-wise if you think the air itself isn't humid enough.


text:
What happens if you mix two insoluble liquids but with the same density?
Would it make two separate phases? or would it make something like and emulsion?

I'd guess an emulsion. However the "same" density has to be taken with a grain of salt. The "same" to how many significant figures? I'm guess that with a very very small difference eventually gravity and environmental vibrations would cause the mixture to separate.

Have you ever seen a lava lamp? The wax in the middle of the pathway has pretty much the same density as the liquid, and it might be a large blob or fine emulsion depending how you treated the system mechanically beforehand.

BTW, an emulsion is a special case of 2 separate phases. Stability of emulsion is given by properties of phase border layers. It is well illustrated by shaking of amyalkokol/isoamylalkohol with water. The former creates emulsion due linearity of molecules, forming oriented surface layer.. The latter quickly separates.

There is no "or" in the first place. An emulsion is a two-phased system, it is just that one phase comes in a form of small droplets. Small as they are, however, they are pretty big from molecular point of view.
Now to the point. If you stir the liquids vigorously, you'll have an emulsion. Short of that, they will attempt to minimize the surface, and the shape with minimal surface is a sphere. Indeed, that's what you will end up with. It's been done before. Look up the Plateau's oil drop experiment: they put oil in a water/ethanol mixture with ratio tuned so as to match the density of oil, and behold, it just floats there in a huge spherical drop.
If both liquids get in contact with the wall, the geometry might become more complicated. The phase separation surface will develop certain contact angle where it touches the wall, and the rest of it will form some kind of minimal surface, which can be surprisingly diverse.
So it goes.

That's a good point. I guess the resulting geometry also depends on the surface tension for contact with the walls of the container, which I assumed is favorable.

Given that both liquids get in contact with the wall... Wait, I'd better add that into the answer.

Well, if you are speaking about an equilibrium mixture of two pure substances, then you would not have an emulsion, since an emulsion of two such substances would not be thermodynamically stable. Formation of an emulsion requires an additional energy input to form the unstable dispersion.
The stable structure (for instance a sphere or two side-by-side solutions) is a question for surface chemistry. Insolubility implies that the surface tension between the fluids is greater than zero (molecules in the pure liquids have a lower free energy than at the liquid-liquid interface). The result is that the liquids will attempt to minimize contact with each other. If the densities are identical, this will result not in an emulsion but in the liquids occupying side-by-side volumes, provided (as Ivan's answer points out) contact with the walls is more favorable than between the fluids, and the tension between each fluid and the container is identical. If the tension of one of the fluids in contact with the wall is lower than that of the  other, then the second liquid will form a volume with a minimum surface enclosed by the first liquid (as Ivan points out).
Also, lava lamps are a fun illustration of the physical chemistry at work. Note that here thermal energy and mechanical mixing lead to fusion and breakage of globs of one liquid suspended in another. The surface tension of the glob-forming liquid in contact with the lamp walls is higher than that of the other liquid, which explains why it forms suspended globs. If the globs had a lower tension than the other liquid when in contact with the container walls, they would tend to partition to the walls (although shear due to fluid motion might keep them suspended). The density of the liquids is kept in flux, transitioning via heating and cooling through a density equivalence point that causes the globs to rise and fall.


text:
Do quaternary sulfur dications exist?
We know that sulfur can form sulfides $\ce{R2S}$, with two substituents bonded to it. The simplest example of this would be hydrogen sulfide.
However, sulfur can also form sulfonium ions $\ce{R3S+}$, where 3 substituents are attached to the sulfur atom and a negatively-charged counteranion is present.
What I am asking is whether there is such a thing as sulfur bonded to 4 substituents, with each bond being a single bond, with 2 counteranions (either $\ce{(R4S^2+)(X^2-)}$ or $\ce{(R4S^2+)(X^-)2}$). Is there such a thing as that or something similar?

https://chemistry.stackexchange.com/questions/112087/why-doesnt-h%e2%82%84o%c2%b2%e2%81%ba-exist

There are even tertiary dications = carbene analogues. https://pubs.acs.org/doi/10.1021/ja9073968

Ogawa et al. [1] were first to report a crystal structure (CSD-YAFNOI) of a compound with quaternary sulfur, bis(2,2′-biphenylylene)sulfurane:

Figure 1. Molecular structure of bis(2,2'-biphenylene)sulfurane (CSD-YAFNOI). Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$.

Compound 1 was synthesized as follows (Scheme 1). Dibenzothiophene 5-oxide (200 mg, 1.0 mmol) in anhydrous tetrahydrofuran (THF, 10 ml) was treated with trimethylsilyl trifluoromethanesulfonate (0.25 ml, 1.3 mmol) under an $\ce{N2}$ atmosphere at −78 °C. After stirring at 0 °C for 30 min, the
  mixture was cooled to −78 °C and was treated with $\pu{1.0 mol dm-3}$ 2,2'-dilithiobiphenyl (1.0 ml, 1.0 mmol) in diethyl ether solution. The whole mixture was stirred at −78 °C for 1 h and at 0 °C for 30 min under an $\ce{N2}$ atmosphere. After evaporation of the solvent, the residue was washed with anhydrous diethyl ether (10 ml) and was extracted with anhydrous benzene (10 ml) under an $\ce{N2}$ atmosphere. The solvent was removed under reduced pressure, and the crude product was recrystallized from anhydrous THF at −20 °C to give 1 as orange rods in 96% yield.

Scheme 1 Reagents: i, trimethylsilyl trifluoromethansulfonate in THF;
  ii, 2,2'-dilithiobiphenyl in diethyl ether-THF

Further work by Sato et al. [2] resulted in a synthesis and crystal structure (CSD-NEDCEE) of bis(2,2′-biphenylylene)sulfuranyl bis(tetrafluoroborate).
Structurally, it's a similar compound with a greater, nearly 90° (in contrast to 60° twist angle in neutral bis(2,2′-biphenylylene)sulfurane), twist angle between 2,2′-biphenylylene ligands, however water molecules and $\ce{[BF4]}$-counterions appear heavily disordered:

Figure 2. Fragment of the molecular structure of bis(2,2′-biphenylylene)sulfuranyl bis(tetrafluoroborate) (CSD-NEDCEE) showing the bis(2,2′-biphenylylene)sulfuranyl cation. Oxygen atoms from water molecules as well as tetrafluoroborate anions are omitted for clarity. Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$.

Recently, we have succeeded in the first isolation and structural determination of bis(2,2′-biphenylylene)sulfurane [10-S-4(C4)] (1) as a stable sulfurane(IV) having only carbon ligands.[…] We considered that this sulfurane would be a suitable precursor to provide the desired dication. Therefore, we tried the reaction of bis(2,2′-biphenylylene)sulfurane (1) with xenon difluoride ($\ce{XeF2}$) in the presence of $\ce{BF3 * OEt2}$ and indeed obtained the bis(2,2′-biphenylylene)sulfurane dication, [8-S4(C4)]²⁺ (2) as an amazingly stable bis(tetrafluoroborate) salt.[…] Here, we communicate the first isolation and structural determination of bis(2,2′-biphenylylene)sulfurane dication (2) having only carbon ligands. […]
The sulfurane 1 was reacted with 1 mol equiv of xenon difluoride in the presence of $\ce{BF3 * OEt2}$ in dry $\ce{CH3CN}$ at −40 °C (Scheme 1). After the solvent was removed at room temperature, the residue was washed with $\ce{CHCl3}$ at room temperature, and bis(2,2′-biphenylylene)sulfurane bis(tetrafluoroborate) (2) was isolated as a stable moisture-insensitive yellow powder in 62% yield.

Scheme 1

Subsequently, hexacoordinated derivatives – bis(2,2′-biphenylylene)dimethyl- and diphenylpersulfuranes – were synthesized and their molecular structures were elucidated [3].
References

Ogawa, S.; Matsunaga, Y.; Sato, S.; Iida, I.; Furukawa, N. First Preparation of a Sulfurane with Four Carbon–Sulfur Bonds: Synthesis and Molecular Structure of Bis(2,2′-Biphenylylene)Sulfurane. J. Chem. Soc., Chem. Commun. 1992, 0 (16), 1141–1142. https://doi.org/10.1039/C39920001141.
Sato, S.; Ameta, H.; Horn, E.; Takahashi, O.; Furukawa, N. First Isolation and Molecular Structure of Bis(2,2′-Biphenylylene)Sulfuranyl Bis(Tetrafluoroborate) [8−S−4(C4)]²⁺. J. Am. Chem. Soc. 1997, 119 (50), 12374–12375. https://doi.org/10.1021/ja971336k.
Sato, S.; Matsunaga, K.; Horn, E.; Furukawa, N.; Nabeshima, T. Isolation and Molecular Structure of the Organo-Persulfuranes [12−S−6(C6)]. J. Am. Chem. Soc. 2006, 128 (21), 6778–6779. https://doi.org/10.1021/ja060497y.


text:
How does energy and speed affect sub-atomic particles (protons, neutrons, electrons) in an electric field

(Source: https://gyazo.com/a4d5beed3272a7ca793dd8e2f7e8be3c)
I am having trouble understanding the problem above. Why does it matter if the particles are held at constant energy or speed? How does energy and speed even affect sub-atomic particles. Why do the different constants produce different diagrams?

Such a question belongs rather to the physics SE site.

It literally came from a chemistry website

Surrey, but it is a topic of physics. I am writing an answer.

..but note that:  We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework.
Don't worry, they're not banned. But, we require a minimal effort. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.
See homework

I'm voting to close this question as off-topic because belongs to physics SE

Because different masses and different charges lead to differences in how the particles interact with electric fields.

The energy and speed  do not affect particles themselves, but their interactions with other diatomic objects.
And, of course, they affect their kinematics in electromagnetic field.
Kinetic energy of a particle is
$$E_\mathrm{k}=\frac 12 \cdot m \cdot v^2$$
The force on a charged particle in an electrostatic field of the strength $E$ is
$$\vec F=q\cdot \vec E$$
The perpendicular acceleration is
$$\vec a=\vec F/m=\frac{q\cdot \vec E}{m}$$
For the length of the path between the electrodes $L$, the time of flight is
$$t=\frac Lv=L \cdot \sqrt{ \left( \frac {m}{2\cdot E_\mathrm{k}}\right)}$$
The perpendicular deviation due the  acceleration in the field is
$$\begin{align}
\vec L_\mathrm{p}&=\frac 12 \cdot a  \cdot t^2 \\
&= \frac 12 \cdot \frac{q\cdot \vec E}{m} \cdot  \left( \frac {L^2\cdot m}{2\cdot E_\mathrm{k}}\right) \\
&= \frac 14 \cdot q \cdot  L^2 \cdot \frac{\vec E}{E_\mathrm{k}}\\
\end{align}$$
One can see the bending with the equal kinetic energy is mass independent.
The scenario with the equal speed is a different story, as the perpendicular acceleration has the reciprocal proportionality to the mass, so the path of the much lighter electron is bent much more.


text:
Which of the following procedures will allow the student to determine the rate constant, for the reaction?

Is the answer A or B because C and D won't give me the rate constant?

C or D won't give you the rate constant.

C will only let you identify the order of reaction of X in this reaction and D will change the rate constant as the temperature change

Tty to develop an straight equation equation from the rate equation

From the graph it is clear that the reaction follows first order kinetics ($\ce{[X] = [X_0] e^{-kt}}$) where $ [X_0]$ is initial concentration. Taking $ln$ on both sides we get
$\ce{ln[X] = ln[X_0] +(-kt)}$
So if you plot a graph between $ln[X]$ and $t$ then you have the magnitude of slope as k ie the equilibrium constant.
So correct answer is A.


text:
Calculating pH of weak base and strong acid: Why is K(a) not required?
Question: 35mL of 1.5M $\ce{HCN}$, a weak acid(K(a)= $6.2*10^{-10}$) is mixed with 25mL of 2.5M $\ce{KOH}$. Calculate the pH of final solution.
The solution given in my textbook is as follows:-

$\ce{KOH}$ completely dissociates to give 0.062 mol $\ce{OH-}$(The $\ce{K+}$ becomes irrelevant)
0.052 mol of $\ce{HCN}$

3.The equation is $\ce{HCN + OH- ⇌ CN- + H2O}$

So 0.062-0.052 mol=0.01 mol of $\ce{OH-}$ is left over
so 0.01 mol/(0.035l+0.025l)=0.17M of $\ce{OH-}$

6.pOH=0.76 so pH=13.24
7.Also, $\ce{CN-}$ is a weak base relative to $\ce{OH-}$ so we consider it negligible in the calculation.

Finally, it says that K(a) was not required and it was just to distract.

However, I think K(a) is actually required because $\ce{HCN + OH- ⇌ CN- + H2O}$ is a reversible reaction and I don't understand how all of the 0.052 mol of $\ce{HCN}$ can react with $\ce{OH-}$(and in turn, 0.01 mol of $\ce{OH-}$ is left) when there is also a backward reaction producing $\ce{OH-}$ and $\ce{HCN}$ back.
Please help.

You can directly calculate, what is HCN/CN- ratio in 0.17M NaOH, And What would be pH of NaCN alone Consider more thorough elaboration of your effort to avoid closing future questions.

Your doubt is legitimate because you are assuming that $K_{(a)}$ is a parameter that you can plug in every reaction. Actually $K_{(a)}$ is calculated using this equation in a solution of the acid with water:
$$
\ce{K_{(a)} = \frac{[H+][CN-]}{[HCN]}}
$$
That is derived from this ideal equilibrium:
$$\ce{HCN ⇌ CN- + H+}$$
Even though it reflects the strength of that specific acid in a pure water solution it's practical usage is limited when the equilibrium is not perturbated by others factors.
Anything that will interfere with this equilibrium can cause problems to your assumptions.
In your specific case, the conditions are different there is an excess of $\ce{OH-}$ in the solution due to the strong base, so every $\ce{H+}$ produced will react with the $\ce{OH-}$. Due to the Le Chatelier principle the equilibrium will respond shifting to the right: dissociating more and more $\ce{HCN}$ from producing more $\ce{H+}$.
In simple words, if one reaction is favourable the solution doesn't care about your $K_{(a)}$ it will go following the most thermodynamic favourable path to the equilibrium.
Possible practical learning aim of this question?
So if you want to have high products amount (i.e. $\ce{CN-}$) and you have cheap reactant (i.e. $\ce{KOH}$) you can use an excess of reactant to shift the equilibrium where you want.
Bonus question: what can you do if you want to have HCN?

Really good answer, your explanation of the equilibrium shifting to the right in a continuous cycle until no more HCN is left clarified my doubt in an instant. Thanks for your help.


text:
What is the hybridization state of the carbon atoms in [10]annulene?
I know that one of the more stable structures of [10]annulene or cyclodecapentaene is planar boat-shaped. But that means the hybridization state of the carbons must have changed to suit the geometry. What does it change to? And how are the hybridized orbitals distributed such that the double bonds remain intact?
Edit: I have encountered this molecule only once and am not familiar with the naming convention. For much the same reason, I don't know the exact structure of the most stable isomer of the compound. So even though I was more interested in the boat-shaped isomer of the compound, I think to do away with the ambiguity of the question, I should mainly ask for the most stable isomer of the molecule, viz., trans,cis,cis,cis,cis-isomer.

A mess.  The molecule wants to be aromatic according to its pi electron conjugation but sterically it can't get there.  Bring on the methylene bridge!

Which isomer and which atom? Also by the time you know about such molecules, you should know that molecules distort as they need, not as "imaginary" hybridisation tells you.

@Mithoron That's true (but I'm not at that level). I saw this molecule on a test that required me to comment if it had any atom which had a hybridization other than sp2. And surprisingly, the answer said that it did.    The question is meant for its most stable isomer. And for any atom which might not be sp2 hybridized.

@Mithoron If the question has any false assumptions, feel free to mention them (or edit the question). Thanks!

I could do pretty much the same analysis as I did for your previous question. However, because there are ten carbons here it will take more time...

You do need to clarify the question, though. It is something that Mithoron has already pointed out: cyclodecapentaene has various possible isomers; the most stable (according to Wikipedia) is the trans,cis,cis,cis,cis-isomer, so presumably this is what you are interested in, but the picture you have chosen is the cis,cis,cis,cis,cis-isomer. The Wikipedia article has quite a nice summary of the isomers.

@orthocresol I couldn't find a picture of the trans,cis,cis,cis,cis-isomer. But I have edited the question, accordingly.

This here is the trans,cis,cis,cis,cis-isomer of [10]annulene or cyclodecapentaene, according to a few textbooks and some websites, this here is the most stable isomer of [10]annulene.
As you can see,this is not planar and contains only one trans structure.Here I don't see any carbon with hybridization other than sp2.


text:
Comparing strength of the C-H, N-H, and O-H bonds

$$
\begin{align}
\ce{CH4(g) + Cl(g) &-> CH3(g) + HCl(g)} &\quad ΔH^\circ &= \pu{-14 kJ mol-1}_\mathrm{rxn}\\
\ce{NH4(g) + Cl(g) &-> NH2(g) + HCl(g)} &\quad ΔH^\circ &= \pu{-36 kJ mol-1}_\mathrm{rxn}\\
\ce{H2O(g) + Cl(g) &-> OH(g) + HCl(g)} &\quad ΔH^\circ &= \pu{+40 kJ mol-1}_\mathrm{rxn}
\end{align}
$$
41. Based on the data above, what can be concluded regarding the strength of the $\ce{C-H}$, $\ce{N-H}$, and $\ce{O-H}$ bonds in the molecules shown?
(A) The $\ce{C-H}$ bond is the strongest.
  (B) The $\ce{N-H}$ bond is the strongest.
  (C) The $\ce{O-H}$ bond is the strongest.
  (D) Nothing can be concluded without knowing the strength of the $\ce{H-Cl}$ bond.

I am not sure how to approach the problem 41. In exothermic reactions, more energy is released when the bonds are formed in the products than is used to break the bonds in the reactants. So would the answer be B because the $ΔH$ is exothermic and is the greater than $ΔH$ of -14 for A?

"ΔH is exothermic" – no, it's the process/reaction that is. Also, you are not only supposed to compare the signs of the enthalpies, but also per how many bonds it is provided.

The answer should be C.
One way would be to write the thermochemical equations, however since we just want to compare, we can do it without going into that.
The formation of the H-Cl bond, (like all bonds) is exothermic. And the breaking of the C-H, N-H and O-H bonds is endothermic (since their formation is exothermic too).
Since the contribution of the H-Cl bond to the enthalpy is the same for all three, we can focus on the C-H, N-H and O-H bonds alone.
Now, the stronger the bond, the more energy it would need, and hence make a positive contribution to the enthalpy change. Hence we can see that since the total enthalpy of the reaction is highest(most positive) for the reaction involving O-H bond, the O-H bond is the strongest. (Infact, it is stronger than the H-Cl bond too as the enthalpy change is positive).
Note that this question also demonstrates the order of electronegativity, which is
O>Cl >~ N> C>H. Since the bond with most electronegativity difference would be the strongest, it would be the O-H bond.


text:
Why is the fragrant molecule linalool volatile when it has hydorgen bonding?
I've done a bit of research on volatile molecules and from what I've read, highly volatile molecules should have weak intermolecular forces (easy to go from liquid to gaseous state)....however it seems like the molecule linalool is an exception to this....I'm wondering if anyone can reconcile this contradiction (?) for me?

L. is not highly volatile. You should have searched for its boiling point (198 °C).

But isn't it the fragrant molecule in many flowers? Many of the papers I read said is was volatile...?

Perhaps you might ask why L. has a strong scent. It is volatile enough to reach our nose.

I was under the impression that molecules contributing to a strong scent are volatile...is this incorrect? Could you say that its pleasant scent is due to the relatively low volatility....so it's not overpowering...or am I oversimplifying?

Pls read comment above

Yes I did, but I'm still a bit confused. Are you saying that linalool is volatile...but only at a relatively low level?

I say that L. Is not highly volatile, at least much less than water so that there is no the contradiction you asked for. For why its concentration in air is enough for us to smell it should depends on its shape and interaction with our olfactory receptors. It is a completely different story. Don't make the equation smelling good (bad) thus highly volatile. All we need might be in a few ppm ranges.

Thank you very much! I'll keep reading :)


text:
Explanation of metallic crystal structures using orbitals and electron configuration
There has to be a direct relation between metallic crystal structure formation and electrons in orbitals. For instance, we know that Fe2+ forms body centered cubic or bcc structure with the coordination number of 8, which means each Fe atom bonds with 8 others after getting rid of the two electrons in the 4S orbital.
The electron configuration of one Fe2+ would look as follows:
1s2  2s22p6  3s23p63d6
The problem is, that I do not see how Fe2+ would react with 8 others, based on the electron configuration and the octet rule, to form bcc structure.

$\ce{Fe^2+}$ might form virtually any crystal structure depending on the anionic part. And no, there is no direct relationship between the electronic configuration of a single atom and the crystal structure it takes a part of (at least, I'm not aware of any).

I second that. There is no relation.

@IvanNeretin maybe I should rephrase that. I mean the relation between the electronic configuration and coordination number, because there is a direct one.

@andselisk what do you mean the anionic part? I'm talking about metallic bond of pure iron. I need the relation between the electronic configuration and coordination number. From that I can guess the crystal structure.

@vahidseo There is one, but it is applicable to coordination compounds and not to pure metals.

@vahidseo Pure iron doesn't consist of $\ce{Fe^2+}$. Crystal structure of a material is a property of a neutral substance, hence there must be an anionic part.

@andselisk I'm not taking about just Fe2+. The two extra electrons are still there. It's a metallic bond.

@IvanNeretin could you please redirect me to a scientific paper or article, which discusses this matter?

There is no matter and nothing to discuss. You can't look at the electronic structure and deduce crystal structure, period. Then again, you may find a good deal of papers on crystal structure prediction, but they deal with the subject at an entirely different complexity level.

In solid state physics, generally one does not care a whole lot about the atomic electronic structure.


text:
half equations for a lead-acid cell
we are learning about galvanic cells. I would like to know the half equations and overall equation that occurs in this lead acid cell. so far my understanding for the oxidisation section is that in a lead acid battery oxidisation occurs in lead which has the equation Pb →Pb2+. The equation for the overall reaction at the anode is Pb + SO42- → PbSO4 + 2e-. is this correct because i am confused about the sulphuric acid as i thought it was H2SO4 but a lot of sites have it as SO4 or HSO4.

See https://chemistry.stackexchange.com/q/89562/9757

Remember that both $\ce{SO4^{2-}}$ and $\ce{HSO4^{-}}$ are from $\ce{H2SO4}$


text:
Why is a LCAO necessary within Hartree Fock?
As I understand it, the electronic Schrödinger equation cannot be solved for polyelectronic systems. To circumvent this problem in the Hartree-Fock method, it is assumed that the polyelectronic wavefunction can be written as a combination of single-electron, molecular orbitals. To satisfy the anti-symmetry requirement, this combination takes the form of a Slater determinant.
What is the value of then going on to express these molecular orbitals as a linear combination of atomic orbitals? Would a Hamiltonian written to describe a single electron in the molecular environment (a molecular orbital) not lead to an analytically solvable Schrödinger equation?

(1) The value is that you know what the atomic orbitals are. (2) n-body problems are not analytically solvable (except in trivial cases).

LCAO is not necessary for HF! There are alternatives like plane waves or representing the orbitals on a spatial grid. You may even make up your own functions as a basis of the orbitals. Atomic orbitals are simply an established choice because it works well.

Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function, I would surmise that it is impossible to solve it analytically. The same goes with the one-electron operators in Hartree–Fock theory, because you have the mean-field potential to contend with.
Therefore, we need to bring in numerical methods. One way of doing that is to express the eigenstates as a linear combination of certain basis functions. If the set of basis functions – the basis set –  is complete, then in principle we can express any function we like as a linear combination of these basis functions. Mathematically, this is expressed by a "completeness relation":
$$\hat{1} = \sum_i |i\rangle\langle i| \quad \Longleftrightarrow \quad |f\rangle = \hat{1}|f\rangle = \sum_i |i\rangle\langle i | f \rangle = \sum_i c_i|i\rangle \text{ where }c_i = \langle i | f \rangle$$
Unfortunately, in the case of an atom or molecule, we need an infinite number of basis functions to have a complete set. This is obviously not possible, so we need a finite number of basis functions, judiciously chosen such that the result obtained using this limited basis is close enough to the result obtained with the infinite basis. The AOs (or approximations to them) are extremely convenient in this regard, and thus LCAO is born.

The hydrogen-like ion, listed as a problem with analytical solutions, is how I was imagining molecular orbitals. Would it be unacceptable to consider the entire “collection” of nuclei throughout the molecule as a singular point charge?

I think you might be asking too much there. :) It is not just the nuclei, but also the average field of all the other electrons. You could simplify it into something that is analytically solvable, but then it loses all physical meaning.


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Dehydration of carboxylic acids with phosphoric acid

To find Z one must find X. Finding Y is of no use to find Z.
I get X as:

After addition of $\ce{H2/Pd-C}$, the unsaturated portion becomes saturated and the carboxylic acid remains unchanged.
However I am unsure what does $\ce{H3PO4}$ do to this (new) carboxylic acid? I was unable to find any information. I did find information regarding dehydration with $\ce{P2O5}$, but unsure what to make use out of it.
After looking at (unofficial/3rd party) solutions, I saw that $\ce{H3PO4}$ protonates the $\ce{-OH}$ part of the carboxylic acid (since it is a much stronger acid than the organic acid?) then $\ce{H2O}$ leaves, rendering a carbo-cation at the carbonyl carbon, followed by EAS at the ring.
However, I do not see how $\ce{H3PO4}$ exactly works with the carboxylic acid. I have merely guessed that the $\ce{-OH}$ portion gets protonated (since, this was the only thing I predicted would happen). I really do not know why this happens and how it happens.
Hints and reference material or examples would be appreciated.
The original question is taken from the JEE (Advanced) [2018] Paper-1.

Friedel-Crafts acylation...

Sure I do know that happens, however for that to happen, a carbonyl cation has to form. I do not know how exactly the carbonyl cation forms here by H3PO4. @orthocresol

Ok, but you already (correctly) described what happens. What else do you really want?

@orrthocresol Please see that I said the sources are third party and unofficial. I would like to know H3PO4 exactly causes dehydration of the carboxylic acid - because I have never seen this before. I will reframe my question to make it more clearer.

phosphoric acid, usually in the form of polyphosphoric acid, is a powerful dehydrating agent. It can be regarded as the anhydride of phosphoric acid capturing OH by formation of an HO-P bond. A brief review is here: http://reag.paperplane.io/00002328.htm   Its major advantage over sulfuric acid for cyclisations is that it is not an oxidising agent.

@Waylander Could you please kindly elaborate on what may have happened had sulfuric acid been used? (As in what type of oxidation might have occured?)

Sulfonation of the benzene ring   https://www.chemguide.co.uk/organicprops/arenes/other.html  also oxidation to phenols and hydroquinones

The $\ce{-OH}$ part will not get protonated, since its lone pair is in resonance with the carbonyl part of the $\ce{-COOH}$ group. Instead, the lone pair of the carbonyl part will give its lone pair to $\ce{H+}$.
Something like this :-

It may look insignificant because the product is same, but this is the correct mechanism.

Welcome to Chemistry.SE! You might want to get acquainted with MathJax/mhchem formatting and drawing chemical structures, it helps make posts look more presentable. Have a great day. -From Review

And what is the reason why H3PO4 does this? Its dehydration properties or its much stronger acidity or both? What about acids such as H2SO4?


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Reaction of HF in water
I was wondering, if we put $\ce{HF}$ into a glass of water the acid will be dissolved according to this reaction:
$$\ce{HF->F- + H+}$$
Now there is a proton travelling across the water.
All $\ce{H2O}$ molecules have equal chance of getting the proton so it travels through the liquid.
Now for the reaction
$$\ce{F-  + H2O ->HF + OH-}$$
the bond strength between O and H is lower than the bond strength of F and H (fluorine is has more  electronegative and has a smaller atomic radius than O).
According to Bronsted-Lowry theory for acids and bases the second reaction happens more often than the first one. But how is this happening? Why does the second  reaction happen at all? Oxygen is more negatively charged than fluorine so the proton would have a very small chance to reach fluorine and then the electrons would be rearranged to form the HF bond.

Off the record: HF usually tries to dissolve the "glass" in the "glass of water". Just nitpicking :D

I'm sorry but after reading several times I find it difficult to process what exactly your query is. Could you please try to elaborate a tad bit clearly? I find several statements in your question slightly off.

I tried editing for clarity, but I'm not sure I succeeded.

Why is the question tagged with "quantum chemistry"?

Yes you did it.

@NilayGhosh Yes it bothered me earlier, but I had made enough jokes for the day.

@DonaldDuck NightWriter succeeded in clarifying formatting, but it's up to you to clarify the content.

F- + H2O -> HF + OH- it is very unlikely to happen , HF -> H+ + F- is more likely to happen but ,  F- is a stronger base than HF is an acid.

So at the end, what is your question? I it "Why does the second reaction exist?"

No why F- is a stronger base than HF is an acid?HF should be stronger than its conjugate base.

Possible duplicate of Why is HF the least acidic hydrogen halide?

The reason why $\ce{HF}$ doesn’t tend to disassociate in aqueous solution can be rationalized as follows:
In aqueous solution, all ions (cation or anion) are immediately solvated by a solvent “shell” that stabilizes the ion by drawing some charge density away from it. This stabilizes larger ions such as $\ce{Cl-}$ or $\ce{K+}$ more than smaller ions like $\ce{Li+}$ and $\ce{F-}$; as a matter of fact the solvation energy for $\ce{HF}$ is less than the energy needed for the $\ce{H-F}$ bond to break (bond enthalpy, if you will).
As such, $\ce{HF}$ doesn’t tend to disassociate in aqueous solution. It should be highlighted that pure, liquid $\ce{HF}$ is a much stronger acid than its aqueous form as the acid disassociates into $\ce{FHF-}$ and $\ce{H2F+}$ instead of $\ce{F-}$ and $\ce{H3O+}$.
EDIT 7 MAY
It appears that what you are asking is “why the H-F bond is weaker than the H-O bond”. If that’s what you’re asking, here’s what I have to say:
We will consider the resonance integral, $\cs{Hij}$, between atomic orbitals i and j. This resonance integral can be interpreted to be the energy of an electron associated with these 2 orbitals.
The one-electron relation (Wolfsberg-Helmholz Equation) tells us that $\cs{Hij = 0.5K(Hii +Hjj) Sij}$, and it so happens that the F orbitals are more contracted and can overlap less well. In addition, since the degree of stabilisation of the bonding MO is inversely proportional to the energy difference between the AOs, we see that there is less stabilisation from the original F 2p Orbital in a MO involving HF compared to a MO involving HO, and the HF bond is weaker.

Quick tip: avoid using all-caps, it gives a general appearance that you are yelling. For emphasis, it is better to use italics or bold..!

Ah, got it. What’s the format for italics/bold?

_italics_ or *italics* and **bold**

???????????????

I dont say it diassosciates .

Just the proton as an subatomic particle has no definite position or momentum if not measured and the propability cloud can be displaced based on potential of points .So it actually swims through the solution.Arrhenius was right ! :)

@DonaldDuck What are you taking about? Is our focus on why the ions are stronger acid/ base or is our focus on the proton "swims" through the solution?

The focus is at the explanation of acids/bases using quantum mechanics.

I’m sorry, but how does QM have got anything to do with this?


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LiOH hydrolysis of methyl 2,2-dimethoxyacetate not giving product?
I am forming 2,2-dimethoxyacetic acid from its methyl ester, and below is the experimental protocol I am following (from Tetrahedron 2016, 72 (3), 420–430):

2,2-Dimethoxy Acetic Acid Formation

Add lithium hydroxide monohydrate ($\pu{89.5 mmol}$, $\pu{3.75 g}$, $\pu{1.2 equiv}$) to a solution of methyl dimethoxyacetate ($\pu{74.6 mmol}$, $\pu{10 g}$, $\pu{1 equiv}$) in dioxane ($\pu{45 mL}$) and water ($\pu{45 mL}$) at $\pu{0 ^{\circ}C}$.
Stir the mixture for $\pu{1 h}$ at $\pu{0 ^{\circ}C}$ and one additional hour at room temperature.
Add sodium hydroxide ($\pu{1M}$ aq soln, $\pu{75 mL}$) to the mixture.
Extract the mixture with diethyl ether ($\pu{375 mL}$).
Acidify the recovered aqueous layer with hydrogen chloride ($\pu{6N}$ aq soln) until a pH of 1-2.
Extract the organics with diethyl ether ($\pu{375 mL}$).
Dry the combined organic layers with magnesium sulfate and filter.
Evaporate the solvent slowly under reduced pressure to obtain 2,2-dimethoxyacetic acid.

I repeated this reaction many times but every time I check the $\mathrm{^{13}C}$-NMR, I only see a single peak at $\delta \: \pu{67.12 ppm}$. I believe this corresponds to C-OMe. By theory (and by literature) I should see 3 peaks.
$\mathrm{^{1}H}$-NMR does show 6:1 ratio of singlets at $\delta \: \pu{3.704 ppm}$ (6H) and $\delta \: \pu{3.458 ppm}$ (1H) but it is not well aligned with literature; By theory, a single hydrogen atom adjacent to $\ce{C=O}$ group should be more deshielded than $\delta \: \pu{3.458 ppm}$ (i.e. be placed at higher ppm). I thought through what I have done and I still can't figure out which step has gone wrong. Any thoughts and ideas? Would appreciate any advice.

What weight of material do you get after concentrating the ether extract? I can't access the paper but I see it comes from a reputable group. Are you required to reproduce this procedure exactly or can you change it?

@ Waylander: OP has followed exactly what experimental section given in the paper. Only change is the volume of ether extracts. Paper recorded 99% yield of colorless liquid as a product: $\mathrm{^1H}$-NMR are reported as $\delta$ 3.44 (s, 6H), 4.85 (s, 1H), and 10.32 (broad s, 1H). $\mathrm{^{13}C}$-NMR are reported as $\delta$ 54.1 ($\ce{2CH3}$), 98.6 ($\ce{CH}$), and 170.7 ($\ce{C_q}$).

I assume you have mixed up your ether extracts and had an NMR made of the cleaved methanol. A single peak in 13C !?!

The step that draws my suspicion is the acidification with 6N HCl to pH 1-2. It would be easy to add a bit too much acid, leave the mixture too long, or let it get a bit too warm and hydrolyse the dimethyl acetal. The resulting glyoxalic acid, if formed, would not easily extract from aqueous.

Ok, the CS doesn't fit, and of course also the 1H spectrum you report.

Points 6+7 sound like they're written to trip up undergrad students (not OP). ;-)  6 should be extract twice (at least), and 7 dry combined extracts from step 6

The answer from @Mathew Mahindaratne in my opinion correctly identifies what the OP has isolated. It remains to suggest how to get the required product. It may be a matter of salting out the aqueous phase, but there are other possibilities. Some observers may feel that given a literature prep from a respected group there is no need to alter it, however the OP has not been able to replicate it. Accordingly I offer this procedure.
Dissolve the methyl dimethoxy acetate in a mixture of 9:1 THF/water at room temperature. Add $\pu{1.1eq}$ of $\ce{LiOH}$ monohydrate. Leave to stir at room temperature overnight. Concentrate under reduced pressure. Dissolve the residue in a minimum amount of water and cool in ice bath. Wash with $\ce{Et2O}$. Acidify using a slight excess of $\pu{2N} \; \ce{HCl}$ ($\mathrm{pH}$ really doesn't need to go below 3.0 as $\mathrm{p}K_\mathrm{a}$ of product is probably around 3.7). Extract $\times 3$ with $\ce{Et2O}$. Dry the combined organic extract, filter and concentrate. This should minimise any change of hydrolysing the acetal.

Should I add 2N HCl in an ice bath? Would adding an acid at r.t. increase the chance of hydrolysis of dimethyl acetal?

You said In only should use a 'minimum amount' of water- is it to reduce the chance of hydrolysis of acetal?

Minimum amount of water as the product appears to have a significant water solubility, you may need to salt it. Keep the aq phase cold as you add the acid as this is exothermic, swirl it too so that you don't get a locally high concentration of acid. This is to minimise the chances of the acetal hydrolysing.

so I add NaCl to the aq. layer to salt my product out?

Correct. This increases the ionic concentration and forces less water soluble species out of solution.

it worked. I appreciate it. I now see peaks at δ 4.86 (s, 1H) and δ 3.46 (s, 6H). I wasn't able to detect the hydrogen in -COOH but I guess that's normal.

and a broad singlet at 2 ppm. I believe it's water.

@chemrese Good to hear it worked. LiOH in THF/H2O (9:1 or 19:1) is my go to sytem for mild hydrolysis of esters.

sorry to ask again, but I was running the rxn again with THF/H2O (19:1) this time and LiOH hydrate is not getting dissolved. Is it okay to add more water in the middle of rxn? it's been about 6 ish hours.

Can add water to adjust to 9:1.

I see a broadened singlet at 2.27 ppm. Do you have any idea what would that be? It integrates to be 11H!

Water?. Redissolve your product in DCM, dry with MgSO4, filter, reconcentrate and rerun the NMR

Diethyl ether will hold a surprising amount of water.

If your reaction has worked and you get the expected product, you should see following NMR data as the sought literature reported:

$\mathrm{^{1}H}$-NMR: $\delta$ 3.44 (s, $\ce{6H}$), 4.85 (s, $\ce{1H}$), and 10.32 (broad s, $\ce{1H}$).
$\mathrm{^{13}C}$-NMR: $\delta$ 54.1 ($\ce{2CH3}$), 98.6 ($\ce{CH}$), and 170.7 ($\ce{C_q}$).

Instead, you got two peaks, $\delta$ 3.704 (s, $\ce{6H}$) and $\delta$ 3.458 (s, $\ce{1H}$), in your $\mathrm{^{1}H}$-NMR. And, you got only one peak at $\delta$ 67.12 in your $\mathrm{^{13}C}$-NMR. All this data telling me is your reaction has not worked as expected. What you have is 1,4-dioxane during the ether extract, $\mathrm{^{1}H}$-NMR of which gives only a singlet at $\delta$ 3.71 when $\ce{CDCl3}$ is the solvent. It also gives one peak at $\delta$ 67.14 in $\mathrm{^{13}C}$-NMR under same solvent (Ref.1).
I believe your compound may still be in the aqueous phase. You may need to saturate the aqueous phase with salt ($\ce{NaCl}$) before extract the product to organic phase. Note that acetic acid is very soluble in water. Therefore, I expect 2,2-dimethoxyacetic acid to be highly soluble in water as well (a $\ce{C4(H2O)4}$ compound like a monosaccharide).
References:

H. E. Gottlieb, V. Kotlyar, A. Nudelman, “NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities,” J. Org. Chem. 1997, 62(21), 7512–7515 (DOI: 10.1021/jo971176v).

And maybe not add the NaOH soln. I don't see why that was done, you're just increasing the volume of the aq phase.

Dioxane, of course! And a spot of water, which probably comes from the CDCl3 bottle.

@Waylander It lowers the dioxane content in the aqueous phase, helping to get a better phase separation for the extraction of methanol and residual ester.

Not working is it


text:
Why does Methyl Carbocation have p orbital
For a carbocation to form, it needs to lose an electron. In the example below the leaving group is Br and the products are a methyl carbocation and a bromide ion.

What I am unsure about is why the carbocation needs an empty p orbital. For a carbon atom with 6 valence electrons, the molecular orbital diagram looks like the one below. So when an electron is lost in heterolysis the $\ce{2p}$ orbital will be empty and I assume that is why the p orbital is empty. However, why do you include the empty p orbital in the shape. Does it serve a purpose?

You might want to read up on hyperconjugation.
Basically, the R groups (if NOT a hydrogen) can donate some electron density into the carbocation’s empty p orbital, stabilising the positive charge and increasing the ion’s stability. This is the reason for Markovnikov’s Rule when it comes to Sn1 mechanisms; the tertiary carbocation is most stable, followed by a secondary one, while a primary/methyl carbocation hardly exists at all.


text:
Rotating dihedral and Angles And Generate File For Each Degree
I need to optimize a force field I generated. For this, I should apply Genetic Algorithm (G.A). To apply G.A I should do this for 220 molecules:
1) Find all dihedrals for each molecule ( I did this already with parmchk)
2) Rotate all dihedrals (10-20 degree step in 90 or 360 degree range) and save a mol2 file at each steps.
3) Change all angles (1-2 degree step in 20 or 30 degree range)and save a mol2 file at each step.
Now, I need a program to do this because 220 molecules are too much. I tried "tleap" but it was unstable. I tried pymol but I couldn't run command directly via "cmd" or "terminal" (if there is a way I don't know). So which program should I use?

Have you tried to automate these with ase (https://wiki.fysik.dtu.dk/ase/index.html)?

I think it would be beneficial, if you were to tell us which operating system you are restricted to (or none of that applied).

If you know Python you can modify this program https://github.com/jensengroup/get_conformations

Martin, I am using vbox so os doesn't make any difference. I am using win10pro + ubuntu bash in windows and kali linux and ubuntu and opensuse in my vbox.

Greg and jan jensen, i will try your advices. Thank you by now.


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Condensation point of molecule compounds?
Which of the following gasses will condense at the lowest pressure (Assuming temperature is held constant)?

$\ce{CBr4}$
$\ce{CH2Br2}$

I believe that compound 2 will condense at the lower pressure because it can form dipole dipole forces with other molecules. I believe that London dispersion forces are weaker than dipole dipole ones so 2 should condensed at a lower pressure.

Hint: At ambient conditions, $\ce{CBr4}$ is a solid and $\ce{CH2Br2}$ is a liquid.

I know that you are right but why isn’t the melting point of CH2Br2 higher than CBr4 since CH2Br2 has stronger intermolecular forces.

A weak man can be weaker than a strong cat. Same thing with dipoles vs London forces. And the dipole moment of C-Br bond, mind you, is pretty weak.

Since dipole moment is rather weak in $\ce{CH2Br2}$, molecular weight difference dominates.


text:
Relation between ΔH and ΔE for the reaction in gaseous phase at constant temperature and pressure

Which of the following statements is correct for the reaction at constant temperature and pressure:
$$\ce{CO(g) + \frac{1}{2}O_2(g) -> CO2(g)}$$

$\Delta H = \Delta E$
$\Delta H > \Delta E$
$\Delta H < \Delta E$
None of these.

Since this occurs at constant temperature, $\Delta E = 0$ and $W < 0$. So, $\Delta H$ turns out to be less than $\Delta E$, but this is not the answer. Why?

Constant T says nothing about delta E or delta H., as the system is not isolated.

Constant temperature does not make the change in energy zero

For a chemically reacting system (or a single component change of phase), the internal energy of the system changes even if the temperature does not.

Wanted to confirm if $\Delta E$ represents internal energy. Sorry, I'm not used to this notation (I've always used $\Delta U$ as internal energy).

By definition, $$\Delta H=\Delta E+\Delta(PV)$$For a reacting ideal gas mixture at constant temperature, $$\Delta(PV)=(\Delta n)RT$$where $\Delta n$ is the change in the number of moles between reactants and products.  Therefore, $$\Delta H=\Delta E+(\Delta n)RT$$For the reaction under consideration, $$\Delta n=-\frac{1}{2}$$

$$\Delta H=\Delta E + p \cdot \Delta V$$
$\Delta E$ is energy change of a system at constant volume.
$\Delta H$ as is energy change of a system at constant pressure.
As the molar amount of gases is decreased by the reaction, $p \cdot \Delta V \lt 0$ at constant $T,p$.
Therefore, $\Delta H \lt \Delta E$
Neither that $\Delta E<0$ even at constant $T$, as energy released by the combustion reaction is dissipated in the system surrounding.


text:
Hydroxide as leaving group in Aldol Condensation in base
My professor said that in the base-catalyzed aldol condensation, because there is a lot of hydroxide already in solution, hydroxide on the beta position on a ketone can act as a leaving group. What's the logic/reasoning behind this?

This elimination has a different mechanism than an SN2 reaction. You might ask your professor what s/he meant.

Quoting ORGANIC  CHEMISTRY by Jonathan Clayden  ,Nick Greeves , Stuart Warren ,these are elimination reactions .You cannot normally eliminate water from an alcohol in basic solution as hydroxide is a bad leaving group.It is the carbonyl group that allows elimination here: these are E1cB reactions, with a second enolization allowing the loss of OH−.
the base-catalysed aldol reaction sometimes
gives the aldol and sometimes the elimination product. The choice is partly based
on conditions—the more vigorous the conditions (stronger base, higher temperature,
longer time) the more likely elimination is to occur—and partly on the structure of the
reagents.
The key to what is going on is the carbonyl group.Negative
charges are stabilized by conjugation with carbonyl groups,The proton that is removed in this
elimination reaction is adjacent to the carbonyl group, and is therefore also rather acidic (pKa about 20).The anion that results is stable enough to exist because it can be delocalized on to the carbonyl group.

This next step is ,

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


text:
Is there an easy explanation of MP2 calculation
for my master thesis I calculated some ring strains with MP2 calculation. It was not hard to calculate these things but I'm struggling to understand what the theory is about. Now I'm defending my work and I'm affraid that my professor asks questions about it.
So is there an "easy" way (not really easy but maybe without some hardcore quantum mechanics) to expain the MP2 calculation so an organic chemist can understand it?
Are there any good reasons to use MP2 instead of B3LYP or other DFT calculations?
A friend in theoretical chemistry suggested MP2 but we never talked about why.
Thank you for your answers.

MP2 and perturbation theory should be easy to find in standard quantum chemistry textbooks. For an "easy" explanation you might just consider the initial idea in the beginning of the chapter and the final results at the end. If you have any specific questions about the derivation in between, the right place to ask is here ;)

Related in terms of derivation: How to find the second order perturbation to wave function?


text:
Why does CH2F2 have such a low melting point?
The molecule is highly polar but the melting points are way lower than non polar compounds such CCl4 or other similar non polar compound. Do even dipole dipole  have an affect on boiling point or is the size of the molecule all that matters.

Perhaps a better comparison would be to CF4, which is non-polar and has a far lower b.p and f.p. As for chlorine vs. fluorine, the more massive the halogen attached, the higher the b.p and f.p.

As you go down the family of halogens it isn't the greater mass per se that gives the higher MP and BP but the fact that the higher atomic number the bigger the atom and the more polarizable the halogen is. So none of the $\ce{CX4}$ molecules has a net dipole, but each C-X bond does have a dipole and can polarize a nearby X atom on a different $\ce{CX4}$ molecule, resulting in a net attraction.


text:
Correct Generalization for Adair Equation?

I know how to derive the Adair Equation for n binding sites the long way, writing n+1 mini equations and linking them to relate all sub $K_b$ values to $K_{b-intrinsic}$.
Tried to find a general formula for the adair equation and found different versions online.
If I raise $K_i$ to the ith power, does that make the generalization correct?
Thanks - studying for an exam.


text:
Given Kd find equilibrium concentrations of the reactants

$$\ce{Biotin + Avidin → Complex}$$
Biotin is a vitamin derivative that binds extremely tightly to avidin (a protein found in egg white) with a $K_\mathrm{d}$ of $\pu{1e-15 M}$.
If you make a $\pu{2 mL}$ solution of $\pu{10 μL}$ biotin and $\pu{2 μL}$ avidin, what fraction of biotin molecules will remain free at equilibrium?

Answer Key
The TA found the
$$K_\mathrm{d,new} = \frac{[\text{Biotin}][\text{Avidin}]}{[\text{Complex}]}$$
using the given initial reactant concentration values and making an ICE table to find the equilibrium concentrations and found $K_\mathrm{d,new} = \pu{1e-9}.$
$$K_\mathrm{d} = \frac{(10 - δ)(2 - δ)}{δ} = 10^{-9}$$
Where does the TA get that $K_\mathrm{d} = \pu{1e-9 μL}$? Don't you need to either know the delta term or have the value of $ΔG$ of reaction? Neither of which was given in the problem statement. Also, why did TA find another $K_\mathrm{d,new}$? Doesn't the provided $K_\mathrm{d}$ already describe the dissociation, which gives the concentration of reactants on top?
I understand that $K_\mathrm{d}$ is the equilibrium constant which describes dissociation.
My solution
My solution was to find
$$f_\mathrm{empty} = \frac{[\ce{RL}]}{[\ce{R}] + [\ce{RL}]}$$
I rewrote the $f_\mathrm{empty}$ relation to be in terms of the known values, the provided $K_\mathrm{d}$ and the concentration values.
I found $f_\mathrm{empty}$ to be $0.0009999$ which is also essentially zero, meaning that $f_\mathrm{bound}$ is essentially $1$, which is also what the TA found.
The TA's approach wasn't intuitive for me, so if someone could help clarify it?

Why not to ask TA directly? TAs are here to help you understand stuff, not to invent riddles:)

They didn't respond to my questions so that's why I'm asking the internet. They've been spending the last week grading the last hw, and haven't responded to questions regarding that either.

If you invert the equation then $\displaystyle K=\frac{\delta}{(10-\delta)(2-\delta)}$. If you solve with different $K$ then even if $K$ is small, such as $K=100 ,, \delta=1.998$. So with a huge $K,, \delta=2$ and the amount of reactant present is infinitesimal. Thus it makes no difference $10^9, \to 10^{15}$

It's a question of units: $K_d=10^{15}$ is defined with a M reference state. Presumably it has to do with you defining concentrations in "μL" which leads to a factor of $10^6$ somewhere, but I am not familiar with such use of volume units in place of concentrations, so I am a little confused how you can use the equation as written without further assumptions. Sometimes "units" is used as the unit for enzymes, but in this case I am guessing "L" means liters.


text:
T-shaped and Seesaw geometry hybridization
Why are the filled orbitals s and px (in the T-shaped case) also part of the hybrid orbitals of the central atom in sp3d hybridization?
In other words, why should the hybrid orbital not be a p3d, for the seesaw case, and a p2d for the t-shaped case?*
*Considering the original eletronic distribution to be a s2 p4 and s2 p5 respectively.

This is a complicated issue, and there are many problems with what you are asking.

Hypervalency (expanded octets) don't appear to actually exist. They can be a useful model.  This Wikipedia article does a reasonably good job at looking at the issue: hypervalent molecule.
Basically, the even though we can draw Lewis structures of hypervalent compounds, it doesn't mean that they exist as such.

Sulfur hexafluoride, for example appears to be a covalent-ionic hybrid.
At the same time, Lewis drawings of hypervalent species such as sulfate ion are still commonly used in organic chemistry, biochemistry and biology.
The next issue is hybridization.  As others have mentioned, it has been shown that d-orbitals do not participate in hybridization [E. Magnusen, J. Am. Chem. Soc. 1990, 112 (22), 7940–7951]. As such, the sp3d and sp3d2 hybridizations are incorrect.
One of the problems here is that instructors (and textbooks), especially at an introductory level will have a tendency to over-generalize the rules for Lewis diagrams and for hybridization.
Hybridization needs to be invoked to rationalize the molecular geometries around small atoms (O,N,C, etc) with the geometry of the atomic orbitals in these atoms.
In larger atoms however, we do not always need to invoke hybridization.  For example, the $\ce{H-S-H}$ bond angle in $\ce{H_2S}$ is $92^\circ$.  This corresponds well to simple overlap between the p orbitals of sulfur with the s orbital of hydrogen.  No need to hybridize.

I know this doesn't actually answer your question,  but I hope that it explains why your question is flawed.

I didn't knew about this fact of hybridization, as you said many instructors and textbooks teach us that way, and I was one of this cases.
That said, could you bring some light to the problem of finding the hibryd orbital in a T-shaped molecule, we can say C$\ell$F$_3$.

Although it's not the main point of your answer, your statement that H2S bonding involves only p orbitals is not correct. It is a common misconception. If you look at the MO's of XH2-type molecules, you'll see that there is a large s orbital contribution to the lowest energy bonding MO.

@Andrew, interesting point.  The reason I brought H2S up in particular is because, in spite of  the s-orbital may contribution, from a qualitative point of view, we don't need to invoke hybridization to explain the observed geometry.  At an introductory level, the kind of fine distinction you are pointing out is besides the point, and usually quite beyond the scope of the course and the student's abilities. VSEPR, VB and MO are all models of bonding, and all models have their limits.  My point was that hybridization is usually over-generalized to situations where it doesn't need to be applied.

@liuzp I am not an expert in bonding theories.  However,  $\ce{ClF3}$ is hypervalent when drawn in the traditional way, so the traditional models like Lewis diagrams, VSEPR and VB don't correspond with reality (even if we can draw "valid" Lewis diagrams). Bonding models are just that, models.  They have their limits, and don't work in every situation.

Firstly, hybridisation is absolutely not what you should be using to explain structure and bonding, period.
It has been proven conclusively that the d orbitals do not participate in bonding whatsoever.

There is a multitude of cases, where d orbitals do take place in bonding. Almost all transition metal complexes need d orbitals. You are generalising to the other extend of what is useful. On another note, I'd consider this more like a comment than an actual answer.

Hybridisation should be used even less in transition metal complexes, though.


text:
Nucleophilicity methanol vs water
I have a short question: Which among Methanol and water will be a better nucleophile for a reaction carried out in some polar protic solvent .
Searching over the internet suggests that methoxide ion may be more nucleophilic than hydroxide ion, is that true? If it is, does that automatically imply methanol being more nucleophilic than water?
Also, on what precise factors does it depend?