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Is regioselectivity affected by steric factors during alkylation of naphthalene?
I read that Friedel Craft's alkylation of naphthalene by alkyl halides having more than 2 carbons occurs at 2-position. On the other hand, with smaller halides, it occurs at the 1-position.
Does it have something to do with steric factors or is there some other reason?

There is a steric interaction with the hydrogen at the 8 position that makes 2-alkylation favoured for larger alkyl halides. See this answer  https://chemistry.stackexchange.com/questions/31711/thermodynamic-vs-kinetic-sulphonation-of-naphthalene

Consider rewording your question (or maybe just your title). Generally, all factors are in play, but maybe only one or two are dominating.

The 1-isomer is the generally favoured product as it is formed via a less energetic intermediate, whereas the 2-isomer is formed via a more energetic one.
The stabilization of the 1-isomer intermediate is due to the higher extent of resonance. Substitution Reactions of Polynuclear Aromatic Hydrocarbons[1] nicely sums up the point.

Substitution usually occurs more readily at the 1 position than at the
2 position because the intermediate for 1-substitution is more stable
than that for 2-substitution. The reason is that the most favorable
resonance structures for either intermediate are those that have one fully aromatic ring. We can see that 1-substitution is more favorable
because the positive charge can be distributed over two positions,
leaving one aromatic ring unchanged. Only one resonance structure is
possible for the 2-substitution intermediate that retains a
benzenoid-bond arrangement for one of the rings.
1-substitution
2-substitution

But as the size of the substituent increases there is one more factor to consider. As the substituent becomes bulkier, it's steric hindrance with the hydrogens of the neighbouring carbons increases.
This is an issue for the 1-isomer as it's steric interaction with the neighbouring carbons' hydrogen is way more than it's 2-isomer counterpart. This causes the 2-isomer to be the preferred product when the substituent is bulkier.
Following is an excerpt from intermediateorgchemistry's article[2] about naphthalene:

There is one other factor that will also effect the position of the
substitution; that is the size of the group being added. In the case
of large groups the ß position is favoured. This is a case of steric
hindrance with the hydrogen's on neighbouring carbon atoms.

Reference:
(1) Roberts, J. D.; Caserio, M. C. Substitution Reactions of Polynuclear Aromatic Hydrocarbons https://chem.libretexts.org/@go/page/22330 (accessed Jun 12, 2021).
(2)   naphthalene http://www.intermediateorgchemistry.co.uk/aromatic1.htm (accessed Jun 12, 2021).

I read that Friedel Craft's alkylation of naphthalene by alkyl halides having more than 2 carbons occurs at 2-position. On the other hand, with smaller halides, it occurs at the 1-position.

The above statement from OP clearly indicate that the substitution selection is due to steric reasons. I'd say the above belief is derived by the most of literature evidence, yet you can find a plathora of references contradicting the idea as well. The $\alpha$- and $\beta$-acetylation or alkylation is mainly depend on the conditions, catalysts, solvents, etc. For instance, acetylation of 2-methoxynaphthalene in carbon disulphide, chloroform, or nitrobenzene, has given various yields of 1-acetyl-2-methoxynaphthalene, 2-acetyl-6-methoxynaphthalene, 1-acetyl-7-methoxy-naphthalene, and small amounts of the corresponding phenols (Ref.1). It was reported that in carbon disulphide, 1-acetyl-2-methoxynaphthalene has been obtained in 44% yeild while it was 43% in nitrobenzene.
On the other hand, Zeolite catalyzed Friedel-Crafts acetylation of 2-methoxynaphthalene by acetic anhydride in acetic acid has given 2-acetyl-6-methoxynaphthalene with high selectivity, which is a key intermediate for an anti-inflammatory agent, Naproxen (Ref.2). The reaction has converted 82% of 2-methoxynaphthalene to 2-acetyl-6-methoxynaphthalene (major, 86%) and 1-acetyl-2-methoxynaphthalene (minor, 14%).
Nonetheless, the Friedel-Crafts acetylation of unsubstituted naphthalene is of particular interest, since it was shown that the $\alpha/\beta$-isomer ratio was concentration-dependent. The results obtained for 1,2-dichloroethane solution pointed to a duality of mechanism, in that $\beta$-acetylation is first-order in acylating reagent $\ce{(AcCl.AlCl3)}$, whereas $\alpha$-acetylation is second-order in the acylating reagent. The paper (Ref.3) describeing the kinetic studies of the Friedel-Crafts acetylation of naphthalene in 1,2-dichloroethane states that:

The Friedel-Crafts acetylation of naphthalene, using acetyl chloride and aluminum chloride in 1,2- dichloroethane solution, has been studied kinetically. The $\alpha/\beta$ isomer ratio changes as a function of concentration of reactants, and time, e.g., from an initial 4-5 to a final 0.7. The results point to a different dependence for the two positions on the concentration of acylating reagent, $\ce{(AcCl.AlCl3)}$, being second-order in this reagent for the $\alpha$-reaction and first-order for the $\beta$-reaction. The rate of the $\alpha$-reaction is impeded by the presence of free acetyl chloride, whilst the $\beta$-reaction is unaffected. The latter reaction exhibits activation parameters in the expected range, $\Delta H^\ddagger = ca. \pu{48 kJ mol-1}$ and $\Delta S^\ddagger = ca. \pu{-99 J K-1 mol-1}$, whereas for the a-reaction $\Delta H^\ddagger = ca. \pu{21 kJ mol-1}$ and $\Delta S^\ddagger = ca. \pu{-160 J K-1 mol-1}$ are both very low. Competitive and non-competitive kinetic hydrogen isotope experiments were carried out using $\ce{[^2H_8]}$ naphthalene. The mechanism for $\beta$-naphthyl acetylation is believed to involve a two-stage process, the second (loss of proton) being rate-limiting. The $\alpha$-acetylation is believed to proceed through a $\sigma$-complex, from which elimination of $\ce{HCl}$ to give products is prevented for steric reasons; the reaction instead proceeds through a second $\sigma$-complex, decomposition of which is usually at least partly rate-limiting.

The initial formation of the $\alpha$-isomer is rapid but its curve soon flattens out, whilst that of the $\beta$-isomer climbs more steadily and overtakes the $\alpha$-isomer after $ca. \pu{4 min}$. Formation of the $\alpha$-isomer is virtually complete after $\pu{10 min}$, but the $\beta$-isomer continues to increase for $\pu{1 h}$. The total yield of acetyl isomers was usually 90-95% (for all experiments).
Computating their data on different computer programs, the authors have concluded that the only set of equations which gave satisfactory results, i.e., internal consistency and replicability, involved the equations $(1)$ and $(2)$:
$$\text{Rate}_\alpha = k_\alpha \ce{[naphthalene][AcCl.AlCl3]^2} \tag1$$
$$\text{Rate}_\beta = k_\beta \ce{[naphthalene][AcCl.AlCl3]} \tag2$$
Following shows the isotope effect $(\frac{k_\ce{H}}{k_\ce{D}})$ of acylation of two positions:
$$
\begin{array}{c|ccc}
\hline
\text{Product} & \ce{[AcCl.AlCl3]}, \ \pu{mol L-1} &\ce{[AcCl]_{free}}, \ \pu{mol L-1} & \frac{k_\ce{H}}{k_\ce{D}} \text{ uncorr.} & \frac{k_\ce{H}}{k_\ce{D}} \text{ corr.} \\
\hline
\text{1-acetylnaphthalene} & 0.0041 & 0.0942 & 2.17 & 2.22 \\
 & 0.0337 & \text{trace} & 1.09 & 1.09 \\
 & 0.0168 & 0.0725 & 1.236 & 1.243 \\
& 0.0174 & 0.0751 & 1.70 & 1.61 \\
\hline
\text{2-acetylnaphthalene} & 0.0041 & 0.0942 & 4.11 & 4.35 \\
 & 0.0337 & \text{trace} & 5.07 & 5.39 \\
 & 0.0168 & 0.0725 & 5.04 & 5.40 \\
 & 0.0174 & 0.0751 & 5.01 & 5.35 \\
\hline
\end{array}
$$
The $\frac{k_\ce{H}}{k_\ce{D}}$ clearly shows the two mechanism for each substitution at $\alpha$- and $\beta$-possitions.
References:

R. B. Girdler, P. H. Gore, and J. A. Hoskins, "A re-examination of the Friedel–Crafts acetylation of 2-methoxynaphthalene," J. Chem. Soc. C 1966, 181-185 (DOI: https://doi.org/10.1039/J39660000181).
Tomoyoshi Yamazaki, Makoto Makihara, and Kenichi Komura, "Zeolite catalyzed highly selective synthesis of 2-methoxy-6-acetylnaphthalene by Friedel-Crafts acylation of 2-methoxynaphthalene in acetic acid reaction media," Journal of Molecular Catalysis A: Chemical 2017, 426(Part A), 170-176 (DOI: https://doi.org/10.1016/j.molcata.2016.11.012).
David Dowdy, Peter H. Gore, and David N. Waters, "The Friedel–Crafts acetylation of naphthalene in 1,2-dichloroethane solution. Kinetics and mechanism," J. Chem. Soc., Perkin Trans. 2 1991, (8), 1149-1159 (DOI: https://doi.org/10.1039/P29910001149).


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Ionic compounds with hydride anion
Ionic compounds are formed when an atom $\ce{A}$ donates one of it's electron to another atom $\ce{B}$ to form an ionic compound $\ce{A+B-}$. This is significantly more likely to happen as the electronegativity difference of $\ce{A}$ and  $\ce{B}$ increases.
Hydrogen has a Pauling electronegativity of $2.20$. So, it appears to me that $\ce{H_2}$ should be able to form ionic bonds both as
$$\ce{H+ + A- -> H+A-}$$
$$\ce{B+ + H- -> B+H-}$$
What are compounds that contain $\ce{H-}$ called and what are some examples?

H+ can't form a true ionic bond. H- can and does.

Hydrogen does form ionic bonds. Although not as bare $\ce{H+}$, it can form ionic bonds in the form of $\ce{H-}$  (hydride anion). $\ce{H-}$ forms ionic bonds with alkali metal ions and alkaline earth metal ions (except $\ce{Be^{2+}}$).  These type of ionic compounds formed by hydride ion are called saline hydrides, as they are salt type. These hydrides like $\ce{CaH2}$ (also known as hydrolith) can also be used for production of dihydrogen, as reducing agents, desiccant.

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Could the pH in the equivalence point of strong acid and strong base titration be greater than 7?
I have HCl and NaOH titration and in the equivalence point I get pH=8.6, but the pH must be 7 in this kind of titration, then why this happen? Does it possible that I don't take enough points in the area of the equivalence point and this makes the pH>7?

Please look at the second graph here and also note that titration end point is not necessarily the same as equivalence point for an indicator.

The PH at equivalence point of titration for a strong acid and strong base should be 7 theoretically specking because, they produce a neutral salt on reaction. This means that the salt wont be hydrolysed or anything like that to raise the PH above 7. But during actual titration
its important that we use an indicator to map the equivalence point. Generally during experiments we cannot precisely map the equivalence point and just stop at the exact instant  when the equivalence point point is reached,you might have added a few extra drops of strong base which raised the Ph of resulting solution.as none can stop at exact moment.
while conducting experiment the Ph transition range of the indicator that you used might not have been near about 7 which could have led to you adding more than required NaOH .

The equivalence point should be at pH 7, if it is higher it means that too much NaOH was added. Seems like you should indeed add more points.
https://www.khanacademy.org/test-prep/mcat/chemical-processes/titrations-and-solubility-equilibria/a/acid-base-titration-curves

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How to make plasma in microwave (without grapes please)?
I am planning to make plasma in microwave but the grape method just seemingly sparks and simply isn't impressive. I read some articles and they don't seem to be doing a great job at explaining. So, if someone could explain how to do it with (preferably without) grapes or something else (I believe it can also be recreated with metal) it would be great. Also if you could link me to some actual scientific papers that would be great.
The articles I read:
https://www.sciencenewsforstudents.org/article/why-microwaving-grapes-makes-plasma-fireballs
https://www.vox.com/xpress/2014/12/3/7326643/grape-plasma-microwave

With some care, you may consider putting a glass slide of microscopy into a not-too thick walled suction flask, removing most (but not all) air and put this into the microwave oven (Karnes, American Journal of Physics 89, 372 (2021), link (open access) + video). With some pump/thaw cycles, further enriching the remaining air by argon (e.g., from a welder, or protecting gas from the chem lab) you might get closer to the emission occasionally seen for argon (on much smaller scale) in the glove box.

Run the following demos with a cup of water in the microwave, and the test object on an inexpensive glass or ceramic plate, to reduce the risk of damage to the microwave itself. Even with care, expect plates to crack, and there is still possibility of damage to the oven. You might also want to remove or cover the light built into the microwave oven to see these more clearly.

Incandescent light-bulbs (lamps) are filled with a variety of gases, or vacuum for some. Try a few different clear lamps and watch the glow and spark discharges from the filament. Note the color of the discharge - argon tends towards a purple, nitrogen is more bluish, usw. Use a hand spectroscope and compare the spectra with those of some common gases.
Even at atmospheric pressure, helium may glow in a home microwave. Stretch out some steel wool and put it into a flask. Flush the flask with helium from a party-balloon tank to displace air, cap it and put it into the oven.
A flame is at least partially ionized. Use one to start the plasma.

Run these test for no more than 15 seconds each, and expect the apparatus to get very hot.


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Is Thermoplastic Rubber Food/Pet Safe? Confusions between TPR and TPE
I'm not sure if this is the correct place to post this or if I should be posting at Crafts or Engineering, but since my primary concern is about the safety of its use for my dogs I will post here.
I have been investigating getting a "lickimat" for our dogs, and because it's something they're going to be constantly licking I wanted to make sure what it was made of was completely safe.
I did some research into all the different types of plastics and found out that plastics:

#2 - High-density polyethylene (HDPE)
#4 - Low-density polyethylene (LDPE)
#5 - Polypropylene (PP)

...appear to be the safest ones.
However, I then stumbled into a product made of "Thermoplastic Rubber (TPR)", so I started researching that.
I found out there is a very similar product in the same family called "TPE", and the problem with this is it seems some companies use the terms interchangeably.
Such as:
https://www.ecovessel.com.au/blog/what-is-bpa-and-why-is-it-bad/

The SCOUT and FROST bottles use a TPR spout that is kid-safe and eco-friendly. TPR stands for for Thermoplastic Rubber (also known as Thermoplastic Elastomer, or TPE). It is a mix of rubber and plastic polymers that is highly resistant to biting, chewing, heat, and bending. Like silicone and stainless steel, TPR is recyclable and does not contain leaching chemicals that would contaminate your water. We chose it as the spout for our Kids’ Straw Bottles for its properties of safety and durability.

Now, whilst TPR is also from the Thermoplastic Elastomers family, the two products differ slightly and are used for different things.
This page explains the differences in them.
Namely, you will see TPE is used for food and medical uses, whilst TPR is primarily used for industrial purposes.
If you check the below table it makes no mention of TPR being used for food or medical devices, whilst for TPE it does:

Then for good measure we get this Amazon answer about TPR's safety with children stating that:

TPR is PVC- and BPA-free, so is very safe for children. For instance, it's widely used for catheters where nylon block copolymers offer a range of softness ideal for patients. Also used in medical devices and toys so is very safe for kids.

So I am a little confused about the whole thing on if TPR is food and pet safe.

PVC stands for polyvinyl-chloride and isn't really safe, Especially if it catches fire. I worked in the plastic injection industry for 12 years.

Material food contact safety rating is not a simple topic. Depending on how a product is made, it may be approved or not. BPA and other plasticizers are only one component.
See the particular manufacturer's NSF, FDA or USDA rating. Can they substantiate the claims for thermoplastic elastomer (TPE) or thermoplastic rubber (TPR) with accreditation? Note that NSF is more globally accepted, while FDA and USDA are used in the United State of America.


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Making a metal oil paint by the use of unsaturated polyester resin
How can we make metal coating or oil paint by the use of unsaturated polyester resin and its suitable solvent, plus some pigments only if really required?


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How can the difference between surface energy and electrostatic repulsion be calculated in electroplating?
When metals are electroplated unto surfaces, there are a myriad of forces at play. To set a clear example, if one were to electroplate a magnet with a diamagnetic metal such as bismuth, there will be a certain amount of repulsion from the substrate from the surface. Hence, how could one check how much distance there will be from the applied metal to the surface?

There will not be any distance: either electroplating happens or does not, resulting in a  non-adherent outcome.

The diamagnetic force is relatively weak. Not sure what happens on long term, but it should stick to the surface at least after the electroplating. For example, a fridge magnet will easily attract a paper clip, but you won't feel the repulsion if you bring a gold coin near it.

I think the question is pretty clear so I don't know why there are so many close votes.

Hope I could know was well, Maiti


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How are acids in aqueous solution able to react again after reacting with water?
I hope to clear up some confusion I have about acid-base reactions in aqueous solution.
Based on the Bronsted-Lowry definition of acids, I know acids donate protons. In water, a strong acid should react completely with water to form hydronium and some anion. For example,
$\ce{HCl + H2O -> H3O+ + Cl-}$
However, my confusion arises when this acidic solution is reacted with another solution. For purposes of demonstration, I’ll use aqueous $\ce{NaOH}$. Usually, the acid-base reaction is
$\ce{HCl + NaOH -> H2O + NaCl}$
How is this possible if the $\ce{HCl}$ already reacted with water to form hydronium and chloride ions? Is it perhaps better to think of it in the Arrhenius sense where the original reaction is just $\ce{HCl -> H+ + Cl-}$ that way $\ce{HCl}$ can still be a reactant in the acid-base reaction without already being “used up”?

In a way, you may think that HCl doesn't exist in the solution and hence doesn't react. H3O+ does.

What you have drawn as a single arrow reaction for HCl + H2O is an equlibrium

See also: https://chem.libretexts.org/Courses/Purdue/Purdue%3A_Chem_26505%3A_Organic_Chemistry_I_(Lipton)/Chapter_8._Acid-Base_Reactions/8.4_Solvent_Effects or https://en.wikipedia.org/wiki/Leveling_effect

The proton transfers may happen in cascade.
For example, pure $\ce{HCl}$ is a gas that dissolves and reacts with water (acting as a base) according to a proton transfer from HCl to $\ce{H2O}$ :  $$\ce{HCl + H2O -> H3O+ + Cl- ... \ (1)}$$ But $\ce{H2O}$ does not hold its proton very strongly. Here is why : if another base like ammonia is added to this solution, a new proton transfer occurs from $\ce{H3O+}$ to $\ce{NH3}$ which is a stronger base than $\ce{H2O}$ $$\ce{H3O+ + NH3 -> NH4+ +  H2O} ... (2). $$ A last proton transfer happens from $\ce{NH4+}$ to $\ce{OH-}$ if some solution of $\ce{NaOH}$ (containing $\ce{OH-}$ ions) is added, according to : $$\ce{NH4+ + OH- -> NH3 + H2O} ...\ (3) $$
This shows that

$\ce{H2O}$ is a base just strong enough to pick a proton from $\ce{HCl}$, according to ($1$).
$\ce{NH3}$ is a stronger base as it is able to pick up a proton form $\ce{H3O+}$ according to $(2)$. Of course $\ce{NH3}$ is also able to pick a proton for from $\ce{HCl}$, although it was not shown in the preceding cascade of events. But if  pure $\ce{HCl}$ and pure $\ce{NH3}$ are mixed, the following proton transfer happens from $\ce{HCl}$ to $\ce{NH3}$  direct : $\ce{NH3 + HCl -> NH4+ + Cl-}$
$\ce{OH-}$ is the strongest base of all, as it is able to pick up a proton from $\ce{NH4+}$ as seen in (${3}$).

Of course, $\ce{OH-}$ is also able to pick up a proton from  $\ce{H3O+}$ and $\ce{HCl}$
Just before the end, I would like to state that some teachers are reluctant to accept the reaction $({1})$, because nobody is sure that $\ce{H3O+}$ does exist. It could be admitted that the reaction is more complicated and does happen according to $\ce{HCl + 2 H2O -> H5O+ + Cl-}$ or even $\ce{HCl + n H2O -> H_{2n+1}O+ + Cl-}$. So, in order to simplify their teaching, they would recommend to simply write down the equation of $\ce{HCl}$ with water as : $\ce{HCl -> H_{aq}^+ + Cl-}$ And indeed the reaction of $1$ mole $\ce{HCl}$ with $2$ moles $\ce{H2O}$ is more exothermic that the reaction with $1$ mole $\ce{H2O}$.


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what is the difference between valency and oxidation number?
I know that oxidation number is the charge any species has in its molecular or ion form. And valency is the number of electrons a species contain in its valence shell, that always tends to be used in bond formation. But is there any relation between the oxidation number and those valence electrons?

Better relearn from the start - in particular oxidation number has little to do with actual charges.

There is no simple and general relationship between valency and oxidation number. Look at the example of Oxygen. It depends of experimental evidences.
Oxygen has always valency $2$. But it has an oxidation number equal to $-2$ in the vast majority of its compounds, like water $\ce{H2O}$ or $\ce{CO2}$. It can also be at oxidation number $-1$ in $\ce{H2O2}$, or at oxidation number zero in $\ce{O2}$. It can be at $+1$ in the molecule $\ce{O2F2}$, and at $+2$ in $\ce{OF2}$


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Why does boron form only covalent compounds?
I got two options,

The size of $\ce{B^3+}$ ions, is very small and the ion has high charge density. Thus, following Fajan's rule, boron forms covalent compounds.
The sum of the first three ionisation enthalpies of boron is too high, thereby restricting the formation of $\ce{B^3+}$ ions. Thus, boron is forced to form covalent bonds with other compounds.

Which one is a better explanation?

Some borides can be considered metallic.

Another thing is what you consider a covalent compound...

No: https://chemistry.stackexchange.com/questions/143805/does-boron-form-compounds-without-covalent-bonds?r=SearchResults

To say that boron forms only covalent compounds is an oversimplification. It can be involved in metallic or even predominantly ionic bonding, especially when combined with elecyropositive elements like magnesium. Compounds such as magnesium diboride ($\ce{MgB2}$), discussed in the referenced answer, or calcium hexaboride ($\ce{CaB6}$) are best described as having covalent bonds between boron atoms, but mostly ionic bonds between the boron (which is assembled into an electron-deficient macromolecular structure) and the metal. Transition metals are in general less electropositive, and similar borides of these metals have more metallic bonding between boron and the metal.

Magnesium diboride by Ben Mills

Calcium hexaboride by J. Etorneau et al.

Another explanation.
It costs something to remove one election from an neutral atom, whatever the nature of this atom. It costs more to remove a second electron from an atom that has already lost one electron. But it is feasible. As a consequence, removing a third electron form an atom that has already lost two electrons may be too much, if there is another possibility like covalences. This is the reason why Boron makes $\ce{BF3}$ or $\ce{BCl3}$. The real question is : Why does Boron chloride not make dimers like $\ce{B2H6}$ or Aluminum chloride, which is made of covalent $\ce{Al2Cl6}$ ?


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How to compare magnetic moment of chromium(III) and cobalt(II) ions?
Is there any way of comparing the magnetic moments of $\ce{Co^2+}$ and $\ce{Cr^3+}$ gas phase ions by just going through their electronic configuration?Or is it just experimental data?

What are the complexes you are considering? Are the ions in water, or are they chelated or something else? You can make a good guess of the magnetic moment but they are not always right.

Both of them are in gaseous form this is the only information provided.

For spin only magnetic moment, you can just do $u_{spin} = \sqrt{n(n+2)}$ where n is the number of unpaired electron. Experimental data will differ though because this is based on spin only I believe.

If they are in gaseous form, then yes, the use the formula in the above comment, and take n as the total number of unpaired electrons.

But in this particular example they both are having same number of unpaired electrons so how can I further think about it?

Could you post the actual question that you have. This might depend on ligands and other conditions..

The actual question is the comparison of magnetic moment of these two ions provided that both of them are in gaseous state. No other information is provided other than this.

Hmm I think what I wrote before is actually wrong. Because in the gaseous state, without any ligand, the contribution of the orbital angular momentum (l) would not be cancelled, so you cannot use the spin only formula

I think the implicit question here is:
Given the same S and L, what is the value of J for less than half-filled vs more than half-filled magnetic ions?
The answer is Hund rule #3:

For atoms with less than half-filled shells, the level with the lowest
value of J lies lowest in energy. When the shell is more than half
full, the opposite rule holds (highest J lies lowest in energy).

So, for Cr$^{3+}$ the following applies:

S(Cr$^{3+}$)=3/2
L(Cr$^{3+}$)=3
J(Cr$^{3+}$)=L-S (since less than half filled, per Hund's rule #3)=3-3/2=3/2

Whereas for Co$^{2+}$ the following applies:

S(Co$^{2+}$)=3/2
L(Co$^{2+}$)=3
J(Co$^{2+}$)=L+S (since more than half filled, per Hund's rule #3)=3+3/2=9/2

To calculate the magnetic moment one would also need to calculate $g_J$. Higher values of $J$ result in smaller values of $g_J$, like in the case of lanthanides, but overall larger $J$ still result in higher magnetic moments. To visualize this general principle one can check the case of lanthanides:

So, to explicitly answer the question, while these are just approximated models and the experiment is always preferrable, yes, just by looking at the electronic configurations and applying Hund's rules you can have a good deal of information about the relative magnetic moment of these two ions.

In the equation for the magnetic moment, don't you have a Lande g-factor in front of $\sqrt{J(J+1)}$ ? As the g factor contains a term with J(J+1) at the numerator and 2J(J+1) for the denominator, so for higher J values the g-factor should be lower (?). So without doing the full calculations, is it really possible to say that the higher J means higher magnetic moment?

As an example, here are the g·(sqrt(J·(J+1)) values for lanthanides. It is fairly clear that the less than half-filled have overall smaller values: https://www.radiochemistry.org/periodictable/la_series/L8.html

Is the Lande formula applicable to d-block as well?


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Can ions of a pure substance be obtained in only gaseous form?
I was studying mass spectrometry where I learned that the sample is first vaporized before ionization. And ions are formed in gaseous from in an ionization chamber.
Why is there need of vapourizing our sample first? Can't we directly ionize the solid sample without vapourizing it?

It depends on what is being analyzed. Possibly sputtering. could be used.

Maybe look up techniques like LIBS (laser induced breakdown spectroscopy), being used on Mars for zapping rocks to determine elemental composition.

Once the molecules are in the gaseous phase (well, vacuum), the molecules are isolated from neighboring molecules and ions which would allow to "return" to the neuter, uncharged state. What could be alternatives to accelerate them than as a charged item in an electric field and separate them by electromagnets (in case you use a sector field MS)?  Even the MALDI-(TOF)-MS rely on an ionization of the desorbed molecules.

The point of ionization is that you can readily separate charged species based on charge to mass ratio. Separation is much more difficult for neutral species (but certainly not impossible). So for MS to work you need to both bring the species into the gas phase and ionize them. You seem to wonder whether you might ionize first and then bring the ions to the gas phase, or maybe do both things simultaneously. Is that right? Otherwise it seems unclear what you are asking.

Ok i got the points. Thanx much. But as ionization energy is defined as energy required to remive an electron from an isolated gaseous atom. Then all the metals or element in solid form in periodic table are first vapourized and then their ionization energies are determined experimentally?. Like ionization energy of Na metal is determined in it's vapour form??

Consider that ions, by definition, have an electric charge. If you have ions in a matrix, e.g. table salt, $\ce{NaCl}$, crystals, the negative and positive charges balance. and the ions can get cozy. But pure ions, e.g. $\ce{Na+}$ or $\ce{Cl-}$, do not find each other attractive enough for a really solid relationship.
As an exercise, calculate the repulsive force of ions in condensed matter, e.g. a hypothetical $\ce{Na+}$ crystal sans balancing $\ce{Cl-}$ ions... you might be surprised at its strength.
That said, a small number of ions can be trapped in a crystal, as point defects. Electromagnetic radiation spectroscopy can be used, though since they're surrounded by other species, the light spectra are not those of the pure ion. As for mass spectroscopy, it would be hard to get an ion trapped in a solid to move in a magnetic or electrostatic field, though not impossible. Perhaps you might explore this in heated ionic crystals with point defects?


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Finding the rate of a reaction given experimental data for reaction kinetics
Here is the question:

For the reaction:
$$\ce{2 A + 2 B  →  C + D}$$
The following data was obtained from three experiments:
\begin{array}{c|ccc}\hline
\bf{Experiment} &\textbf{[A] (mol/L)} &\textbf{[B] (mol/L)} &\textbf{Rate of Formation of C(mol min/L )} \\ \hline
1 & 0.60 &0.15 &6.3 \times10^{-3} \\
2 & 0.20 & 0.60 & 2.8 \times 10^{-3}\\
3 & 0.20 &0.15& 7.0 \times10^{-4} \\ \hline
\end{array}
a) What is the rate equation for the reaction?
b) What is the numerical value of the rate constant $k$?
c) Propose a reaction mechanism for this reaction, identify any intermediates, and the rate-determining step.

What I have done:

However, I don't know if this is correct and how to proceed forward :/
Edit:
I fixed my rate law however When determining the reaction mechanism, I am having a hard time cause I am not sure of the possible mechanism that indicates a slow step, how would I do this part?

I am unable to understand what your efforts stand for. Could you please clarify?

The image you posted has a very low resolution so it is hard to see what you wrote there. Could you perhaps post an image with a good resolution, or even better, type it in with MathJax ?

Your method is correct. Try getting the exponent of $[A]$ again. It should be +2, not -2.

As I said in the comments, your image is quite low resolution. But, if I read it correctly, you have got the exponents in the rate law wrong.
Solving this type of problem is simple, you just need to check if the rate law makes intuitive sense. The data given in the question contains various instantaneous rates (or initial rates maybe) with concentrations at that point.
First, look at the data:
\begin{array}{c|ccc}\hline
\bf{Experiment} &\textbf{[A] (mol/L)} &\textbf{[B] (mol/L)} &\textbf{Rate of Formation of C(mol min/L )} \\ \hline
1 & 0.60 &0.15 &6.3 \times10^{-3} \\
2 & 0.20 & 0.60 & 2.8 \times 10^{-3}\\
3 & 0.20 &0.15& 7.0 \times10^{-4} \\ \hline
\end{array}
From experiment 1 & 3., the rate increases by 9 times when you increase $\text{[A]}$ by 3 times. Clearly, the exponent of $\text{[A]}$ in the rate equation is $2$.
Now from experiments 2 & 3, the rate increases by $4$ times when you increase $\text{[B]}$ by 4 times. The exponent of $\text{[B]}$ in rate eqn. is 1.
So your rate law should be $$r=k\mathrm{[A]^2[B]}$$
I think in your calculation you switched the numerator and denominator somewhere so you got negative exponents.
Then you can use the rate equation with the data given to find the rate constant.
Proposing the reaction mechanism is a bit difficult, remember that most elementary reaction steps are bimolecular. Trimolecular steps are rare. Any step where more than 3 species collide is almost impossible, so make sure your mechanism doesn't have anything like that.
Edit: For writing a plausible mechanism, in general keep in mind that the rate law will have terms for all of the reactant species that are present in and before the slow (rate limiting) step.
An example mechanism for the reaction could be something like the following:
$$\ce{A +B-> X_1}$$
$$\ce{X_1 +A \overset{slow}{->} C + X_2}$$
$$\ce{X_2 + B -> D}$$
$\ce{X_1}$ and $\ce{X_2}$ are intermediates.
To get the rate law, count the number of reagent molecules (not the intermediates) for the slow step and all steps before that. So in the slow step you have one molecule of A, in the previous step you have one molecule of A and one of B. So, your rate law should be $\mathrm{r=k[A]^2[B]}$ which is the same as you got from experiment, so this is a possible mehcanism (this is not a proof that the mechanism is right!)
Note that the above way of getting the rate law from mechanism is simplistic, so there might be some unusual reactions where it fails.

Thanks so much for the help! I fixed my rate law the way you have explained it! When determining the reaction mechanism, I am having a hard time cause I am not sure of the possible mechanism that indicates a slow step, how would I do this part?

@Mia See the edit.

Thank you so much for the clarification!


text:
What is the purpose of HFCs (Hydrofluorocarbons) numbering system?
I read that for hydrofluorocarbons, they're named HFC-xyz, where
x = number of carbon - 1
y = number of hydrogen + 1
z = number of fluorine
I'm curious why they were named like this.
Is it better than simply putting numbers of carbon, hydrogen atoms into x and y?

Apart from the numbering system help distinguish between various isomers (see @DrMoishePippik's answer), it also helps grouping compounds into various series:

000 series: methane-based compounds

100 series: ethane-based compounds

200 series: propane-based compounds

300 series: cyclic organic compounds

400 series: zeotropes

500 series: azeotropes

600 series: organic compounds

700 series: inorganic compounds

1000 series: unsaturated organic compounds

For example: HCFC-22 is in the "000 series" of refrigerants, meaning it is a methane-based compound. R-134a is in the "100 series" of refrigerants, meaning it is an ethane-based compound.
Source: https://www.achrnews.com/articles/108910-ice-breaker-refrigerant-numbering-system-explained

Does it suggest that, in order to distinguish "hydrofluorocarbons with 7 carbons" from "inorganic compounds", they gave up assigning numbers according to the amount of carbon atoms?
If this is the case, then why would they add 1 to the amound of hydrogen atoms?

@Wang (1) for a HFC to be in 700 series, number of carbon should be 8 (as x = no. of carbon - 1) and yes (2) The numbering system is done in such a way that you can also do the vice-versa (obtaining the formula from the number). Add 90 to the number to obtain a 3-digit number and each digit represent number of carbon, hydrogen and fluorine atoms and a certain calculation for chlorine atoms. Because of this simplicity and versatility in getting both the formula/number if one of the either is given, the addition and subtraction is done. Consider it a "trick".

(1) I meant if they use the amount of atoms as number, 7xx would have 7 carbon atoms.
(2) Wouldn't it be better and more simple if they just set the number as the formula? Then we won't need any tricks. This is exactly where my question is. What merit does the system have, that they would give up simplicity for it?

Some "hydrofluorocarbons" used in refrigeration and other applications actually are fluorocarbons, sans hydrogen. To make numbering unambiguous, even tetrafluoromethane, $\ce{CF4}$, which is "R-14" in ASHRAE parlance, would be 114 in that HFC nomenclature.
Manufacturers use proprietary and somewhat arbitrary nomenclature. For example, isomers,must be distinguished, as in 1,1,2,2-tetrachloro-1-fluoroethane, Freon R-121 vs. 1,1,1,2-tetrachloro-1-fluoroethane, Freon R-121a. Compounds with other substituents, e.g. chlorine, as in chlorotrifluoromethane, R-13 and bromochlorodifluoromethane, $\ce{CBrClF2}$, known Halon 1211 or Freon 12B1, need to be assigned names, too.

Does it mean that the name does not have a certain purpose?
I was wondering why they would give up the easy thought, which is assign numbers according to number of atoms.

@Wang I imagine that having fairly simple names aids in the use of the correct chemical when they are being used by technicians skilled in other areas. Also, imagine the size of the label on a car's air conditioning refill port if it had to say "2,3,3,3-Tetrafluoropropene" instead of "HFO-1234yf".

I wonder why it couldn't be "HFO-1324yf"，with 3 for the real amount of carbon atoms, and 2 for the real amount of hydrogen atoms. What merit does the "carbon-1, hydrogen+1, fluorine" really has, compared with simple " carbon, hydrogen, fluorine".

@Wang, There are interesting comments on haloalkane nomenclature at https://www.chemeurope.com/en/encyclopedia/Haloalkane.html . Again, because these are large-scale commercial products, a company may choose it's own naming system, rather than IUPAC's, both for convenience and to hide proprietary information.


text:
Constraints on pyrotechnic explosives that won't detonate spontaneously at 500 °C when landing on Venus?
The Space Exploration SE question Premature detonation of explosive bolts when landing on hot Venus? explains the need for and widespread use of pyrotechnic fasteners and cable cutters for Mars landers and elsewhere in spaceflight, and points out that Venus' atmosphere is already 75 °C and 1 atmosphere at an altitude of 50 km and roughly 500 °C and a dense 90 bar supercritical fluid at the surface.
So anything on the outside of a lander is going to get hot during the long descent to the surface in this thick soup.
Questions: What are the chemical and physical constrains on reliable, easily to electrically detonate explosives for pyrotechnic devices that will prevent them from spontaneously detonating at 500 °C when landing on Venus? Are there any realistic, potential candidates?
For example; do we need reactions that are less exothermic? Or a higher barrier between initial and final states? Or a lower density so that below some critical temperature an occasional molecule "exploding" doesn't trigger the next one so easily?

These examples are on Mars, not Venus. Click for larger

left: From How were Perseverance's cables "cut" after touching down? in Space Exploration SE right from this answer there. source: NASA/JPL-Caltech and source: NASA/JPL-Caltech

Out of curiosity, how did the first landing on Venus manage the feat? I ask because it is not clear that what works well on the thin Martian atmosphere would be adaptable for the “thick soup” Venusian atmosphere.

@EdV Good question! There is an answer posted on my linked question. Those pyrotechnic devices were kept internal to the craft, which was "chilled" before atmospheric entry and insulated thereby allowing its electronics to work as well, for a few hours. That was a long time ago. In the future Venus landers and rovers will use some combination of (perhaps) refrigeration and high band gap/temperature semiconductors My question here is purely about the chemistry of potential 500 °C pyrotechnics, whether or not they will actually be used.

@EdV Just fyi; per that answer, they used special cords on the parachute that held it partially closed so it would fall very fast until reaching 200 °C when the parachute opened fully to avoid crashing. The fast fall was to shorten the time to the surface so that more of its limited "chill time" would happen after landing.

Ah, thanks for the information! I was thinking some sort of insulated internal “shotgun” mechanism, maybe with small shaped charges, could blow the bolts off or out. Seems like that is essentially what was done. So how about using two substances that are thermally stable, but explode on contact? Ram them together at the right locations and take out the bolts.

@EdV Interesting thought, if it has a chance of working please consider describing it in an answer post!

I suspect NH4NO3 based explosive may work. Per the Texas City disaster, just heating NH4NO3 may not result in an explosion (complex product range as a function temperature, which may be favorable in your situation). Also, N2O dissolved in fuel is an idea where N2O is more electrostatically given to explode than just by heating (but does result in extremely high temperatures on detonating, vaporizing metal).

Because oil wells reach high temperature at depth, there are special explosives, e.g. https://www.freepatentsonline.com/y2010/0006193.html . As you mention, wide-band-gap semiconductors such as SiC first received use in well loggers. Interesting - "As above, so below."

The question is rather which explosive does not decompose/sublimate and possibly not even melt at 500 °C?

@Poutnik why would melting have to be explicitly excluded? For the purposes of this question it only needs to work at hight temperature, perhaps melting could be required to produce the explosive. What I've asked about here is how to avoid the problem of spontaneous explosion at that 500 °C.

possibly does not mean have to. I had in mind eventual phase change may affect the functionality of explosive device. But thermal  degradation along high T safety are the major concerns.

@Poutnik yes, it could activate it for example, or melt two precursors that then mixed https://en.wikipedia.org/wiki/Binary_explosive

Sure, binary explosives are safer way.  But generally, even if binary, energy rich molecules do not like high temperatures.

it would have to be based on solid salts, like barium nitrate + aluminium powder

@FrancisL. I think that if you explain why in an answer it might be well-received. The question asks about constraints and it seems you've given that some careful consideration.

How small could a thermobaric explosive be made? Basically, use a burst of oxygen gas to get flammable particulates suspended in a very small compustion chamber and then use a spark plug to ignite, sending the bolt-cutting ‘piston’ to do its job. Basically just a flour mill explosion in miniature.

@EdV that's excellent, a mini-space-grain silo explosive bolt/frangible nut I love it! Better file the patent ASAP (seriously!) and/or post an answer.

When I was writing the comment, I had grain silo first and then changed it to flour mill. But it really does not answer your actual question about pyrotechnics and constraints, etc. I don’t much care about the rep stuff, so if someone else wants to develop (or elaborate) and post an actual answer, it is fine with me. By the way, I have one (now expired, I think) patent that got me nothing but interaction with a lawyer. I will pass on doing that again.

@EdV yes I know the experience of which you speak and would personally "run away" if ever asked again to write something up as a patent, but this is a really cool (or in this case hot) idea no matter what.

Ionic compounds as a general rule have greater bond strength than covalently bound molecules. This makes them more heat resistant, therefore it seems likely that a binary explosive consisting of an oxidizer salt and a fuel would be the way to go.
Since you don't want the explosive to change phases during the mission, since it could leak or unmix, you'd want a salt with melting temperature above 500 °C, the same goes for fuel.
One possibility would be barium nitrate and powdered aluminium mixture.

The trouble could be safety of their mixture, not to be trigerred prematurely, as 500 °C is a lot.

I don't doubt your answer, is it possible though to add some reference about reactions between barium nitrate and powdered aluminium for completeness? Thanks!

I can only provide a link to a video with someone testing it out https://youtu.be/BB4Nj7ZkUKw


text:
bromobenzene and NaOH reaction at high temperature and pressure

Find the product of the following reaction

Note that the carbon marked as a star is a special labelled carbon which may be considered as a $\ce{C-14}$ isotope
The given answer is (c)
Why would the product in option (b) form ?
I know that it proceeds via SN2 mechanism but is there some sort of benzyne reaction where we would get the product in option (b) ?

It is similar to dow's process, which takes place via Sn Benzyne mechanism

The point is, as mentioned in the comment by @Jayadithya, that given the reaction conditions, the explicit «marking» a carbon atom as $\ce{^{13}C}$ (instead of the typical $\ce{^{12}C}$ / $\ce{^{13}C}$ according to the general isotopic distribution for this element) does not introduce a significant site selectivity for the reaction of the intermediate benzyne:

Thus, one anticipates an equal molar ratio of products formed along pathway a), and b).  It is not the kinetic H-D isotopic effect you may observe e.g., with the $S_N2$ reactions


text:
Which is more reactive towards electrophilic aromatic substitution? Benzene or naphthalene?
My attempt:
The compound which would be more electron rich would also be more reactive to electrophilic aromatic substitution.
In benzene 6 pi electrons are being shared between 6 carbons while in napthlene 12 pi electrons are being shared between 10 carbons so it is more electron dense.
So, napthlene should be more reactive.
I am having doubts because I had read something about aromaticity being shared is not a good thing.Perhaps while reading about anthracene. I exactly can't remember.
Edit: As pointed out in the comments, I have made an error in counting the number of pi electrons in napthlene.
What I wanted to ask was: What effect does one ring have on the other ring? Do they increase each other's electron density or decrease each other's electron density?

I think you need to recount the number of pi electrons being shared in naphthalene

In naphthalene 10 pi electrons are shared by 10 carbon atoms. However the distribution is not as even as in benzene.

Naphthalene is more reactive towards electrophilic substitution reactions than benzene.
On a quick glance you might think that as 10 pi electrons are delocalized on 10 carbon atoms in case of naphthalene, it should have resonance energy per bond similar to that of benzene and thus making both equally active towards electrophiles. But in practise it is observed that naphthalene is more active towards electrophiles.
This is because the delocalization in case of naphthalene is not as efficient as in benzene. Unlike in case of benzene all pi electrons of naphthalene are not equally delocalized on the 10 carbons. This fact becomes quite evident after one draws all the major canonical structures of naphthalene.

The two structures on the left
have one discrete benzene ring each, but may also be viewed as
10-pi-electron annulenes having a bridging single bond. The structure
on the right has two benzene rings which share a common double bond.
From heats of hydrogenation or combustion, the resonance energy of
naphthalene is calculated to be $\pu{61 kcal/mol}$, $\pu{11 kcal/mol}$ less than
that of two benzene rings ($2 \times 36)$. As expected from an average of the
three resonance contributors, the carbon-carbon bonds in naphthalene
show variation in length, suggesting some localization of the double
bonds. The $\ce{C^1{–}C^2}$ bond is $\pu{1.36 Å}$ long, whereas the $\ce{C^2{–}C^3}$ bond length is
$\pu{1.42 Å}$. This contrasts with the structure of benzene, in which all the $\ce{C–C}$ bonds have a common length, $\pu{1.39 Å}$.

As one can see, the 1-2 bond is a double bond more times than not. Similarly, the 2-3 bond is a single bond more times than not. This shows that pi electrons are not equally delocalized in naphthalene and thus causing the 1-2 bond order to be near 1.67 and the 2-3 bond order to be near 1.33, contrasting to equal bond order of 1.5 among all bonds in case of benzene.
This partial localization of pi electrons increases the nucleophilicity of naphthalene and thus makes it more active towards electrophiles than benzene.

Reference:
(1) Reactions of Fused Benzene Rings
https://chem.libretexts.org/@go/page/1206
(accessed Jun 13, 2021).

Thank you. Wouldn't it increase nucleophilicity as well as electrophilicity of napthlene? The electrophiles can react with bonds with more bond order and nucleophiles can react with bonds with less bond order.

@satyamkumarjha Yes you are right, naphthalene has higher nucleophilicity as well as higher electrophilicity than benzene.


text:
Does pressure affect pH?
Since pH is the function of $\ce{H+}$ ion concentration, the change in volume will change the equilibrium according to Le Chatelier's principle and affect the pH. Is it correct, or are there any additional factors?

Yes, pH is affected by pressure.

But the effect is many orders weaker then uncertainty of pH prediction or measurement.

Yes, pressure does affect pH of a solution.
In basic solutions, pH increases with increasing pressure, while in acidic solutions, pH decreases with increasing pressure.

(It may seem obvious but just in case:) Apart from the effect mentioned by MaxW, the "Yes" answer may be particularly true if your solution is exposed to acidic vapours. For instance if you have CO2 around, more of it will dissolve into your solution with an increasing pressure (see Henry's law) making the solution more acidic.

@mranvick It would not be pH change based solely due pressure, but because of pressure dependent acid concentration.

@Poutnik, yes indeed. But since it results in an overall pH modification I thought it could be worth mentioning it. The question being quite laconic, I thought it may be worth mentioning various sources of pH variation with pressure, as they may be encountered in real-life situation by the asker. It was a more generic attempt to emphasize the importance of a pH pressure dependence caused by other chemical phenomena that might be affected by pressure e.g. the dissolution of a gaseous acid (CO2).

In basic solutions, the pH value increases with increasing pressure (e.g. NH3 in water) and in acidic solutions it decreases (e.g. SO2 in water). If a reference measurement is made under normal pressure conditions, the measurement result will be correspondingly higher or lower.

Please could you indicate a source (citation) to support your claim?

From Albaha University:
Please look at the following below links which may help you in your analysis:
https://derangedphysiology.com/main/core-topics-intensive-care/arterial-blood-gas-interpretation/Chapter%201.1.4/neutrality-and-influence-temperature-and-pressure-ph
https://www.sciencedirect.com/science/article/pii/S0304420316301943
http://scrippsscholars.ucsd.edu/adickson/content/effects-pressure-ph-tris-buffer-synthetic-seawater-0
http://jes.ecsdl.org/content/112/3/350.abstract
Thanks

@Krish These links need to be converted into the proper citations and added to the answer. Also, what are "normal pressure conditions", exactly?


text:
What is role of copper powder, calcium chloride and cuprous chloride in the SN1 reaction of hydrochloric acid with propargylic alcohol?
For the synthesis of a 2H-cromene from a phenol I need to prepare 3-chloro-2-methyl-but-1-yne from the corresponding alcohol 2-methylbut-3-yn-2-ol. This reaction could be seen as a simple $S_\text N1$ substitution.
Plausible mechanism:

However it requires a catalytic amount of copper powder, and almost half equivalent of $\ce{CaCl2}$ and $\ce{CuCl}$, in addition to concentrated $\ce{HCl}$ at 0°C temperature. The authors say that these additives are necessary to avoid the formation of a series of subproducts but they don't talk about mechanistic explanation about the role of these additives.
Possible subproducts:

In my opinion, the formation of subproducts derived from the eneyne, obtained from dehydration, is suppressed using low temperature.
So the additives act probably on the carbocation. We may suppose the formation of a copper acetilyde that makes the resonance structure of carbocation less important. This is corroborated by the use of concentrated HCl during the work-up. But I don't understand how it is possible to form a copper acetilyde in acid media (usually it requires the presence of a base that deprotonate the alkyne).
The autors says that the formation of subproducts is more evident for propargylic alcohol that have a low solubilty in water (this is not the case of 2-methylbut-3-yn-2-ol).Regarding the presence of $\ce{CaCl2}$, I know that this salt is used in the distillation of some halocompounds because it interacts in some way with alcohol. So maybe $\ce{CaCl2}$ helps the solubilization of alcohol.
Can you please explain the role of each additives?
Reference:
G. F. Hennion, A. P. Boisselle, J. Org. Chem. 1961, 26, 725–727.

My take as to the likely role for a catalytic amount of copper powder, $\ce{CaCl2, HCl},$ and $\ce{CuCl}$ is as follows:
$\ce{ Cu + Cu(II) <=> 2 Cu(I) }$
where the $\ce{Cl-}$ presence forms a soluble complex to keep the reaction reversible:
$\ce{ CuCl + Cl-  <=>  CuCl2- }$
The first equation is an example of a working so-called redox couple with an associated presence of solvated electrons $\ce{(e-)}$ in the alcohol medium. Further, in the presence of H+ (from the $\ce{HCl}$), we have:
$\ce{ H+ + e- <=> H^.}$
The formed hydrogen radical can further attack an organic complex resulting in either hydrogen or water with the introduction of an associated organic radical:
$\ce{ RH + H^. <=> ^.R + H2}$
Or:
$\ce{ ROH + ^.H <=> ^.R + H2O}$
At low temperatures, the above reaction is moved to the right. A minor formation reaction is also possible:
$\ce{ R^. + ^.R -> R2 (g)}$
where the above reaction could account for any reported associated formation of a metal acetylide.
Lastly, I suspect, presence of $\ce{CaCl2}$ adds $\ce{Cl-}$ and also removes water and further engages with the alcohol creating an addition compound.
I trust this helps.


text:
Kinetic energy of liquid
Does intermolecular force of attraction have anything to do with molecules having more energy in their liquid state than their solid state? If not, what is responsible for greater kinetic energy in liquids?

A new degree of freedom occurs in liquids : translation. In the solid state, the only degrees of freedom are rotation and vibration.

@Maurice  With unusual exceptions, solids have vibrational freedom only.

@theorist  On the contrary, the translational kinetic energy of a substance is a function of temperature not state. At a phase transition, for example, the solid and liquid phases, being at the same T, have the same translational kinetic energy. Obviously, in the solid the average displacement of each molecule is much smaller, but there's definitely translation happening. More here: https://chemistry.stackexchange.com/questions/88803/kinetic-energy-of-molecules-in-liquid-state/89011#89011

@Maurice - see comment above

@Andrew I don't see that. Indeed, back in Jan. I questioned the ans. you linked, and never rec'd an adequate reply. Physically, it doesn't makes sense to me that particles in solids have translational freedom (w/ the exception of free electrons in metals). And the basic models we have for the heat capacities of solids treat them as either collections of harmonic oscillators (Einstein), or as having collective vibrational modes (Debye); i.e., no translational motion is included. See also my ans. here: https://chemistry.stackexchange.com/questions/129290/what-exactly-is-temperature/129366#129366

@Andrew And from a quick search of the literature:  "A monoatomic molecule in the gas or liquid phase moves with free motion. Such a movement of a monomolecule is called a translational mode (Fig. 1(i)). By contrast, bonded atoms in a molecule or a solid phase vibrate with other atoms coherently, which is called the vibrational mode." [ https://pubs.rsc.org/en/content/articlelanding/2020/sc/d0sc02605k#cit1 ]

@theorist - it does get into semantics a bit, but a vibrational mode in which the center of mass is in motion technically is or contains a translational mode. In a typical solid, these can be modeled as harmonic oscillators in which the restoring force is intermolecular rather than intramolecular. If they become coupled across a larger scale, the result is a phonon, but that coupling isn't required for translation to be happening.

@Andrew Sure, but by that way of thinking rotational motion is also translational, because the atomic centers are are moving through space as the particle rotates. But no one calls that translational. By the same token, no one (at least that I know of) refers to vib. motion as a translational DOF just b/c it includes particle motion. The point is that all modes involve motion, and thus to call all modes translational because of it makes the distinction between translation, rotation, and vibration meaningless. And it is important/useful to distinguish these b/c they are qualitatively different.

@theorist -The key point is the movement of the center of mass of the molecule. Movement of atoms in a rotation is not translation if the center of mass remains stationary. The normal vibrational and rotational modes of a molecule are clearly defined as having a stationary center of mass. Any movement of the COM requires activating a translational mode in addition to the vibration.

@Andrew In liquids (or anytime the molecule is subject to external forces), the CM can move as a result of rotations. So you're saying anytime there is vibration or rotation in which the CM moves, that should be considered to have a translational component. Questions: (1) I offered generally-accepted approaches in which the motion of solids is treated as purely vibrational. Can you give me examples of generally-accepted approaches in which the vibrational DOF's in solids are broken up into translational and non-translational components? I've never seen this done.  What I'm really asking is...

...for you to show that your anzatz, if you will, is also generally-accepted within the field (as I've done), rather than an outlier view.  (2) Can you explain why your anzatz is useful, i.e., why it would give a better model of the properties of solids than the standard ones, in which the partition functions are vibrational only? (3) You wrote "the solid and liquid phases, being at the same T, have the same translational kinetic energy". That would only be the case if the translational states are equally accessible in both.  Is that the case, and how would you demonstrate this?

Let us continue this discussion in chat.

It is a common misconception that molecules in a liquid state categorically have more kinetic energy than molecules in a solid state. This is only true when the liquid is at a higher temperature than the solid, simply because translational kinetic energy is proportional to temperature.
In a molecular solid, the individual molecules cannot move very far, but they can still move, and if their centers of mass are moving at all, they have translational kinetic energy. The kinetic energy in these small motions is proportional to the temperature, just as is the case with the larger motions of molecules in a liquid.
So if a solid and liquid are in equilibrium at the same temperature (the melting point), both phases will have the same amount of kinetic energy (assuming we are not at such a low temperature that quantum effects come into play).
You are correct, however, that molecules in the liquid state have more energy than in the solid state and that intermolecular forces play a role. The difference in energy, though, is in the potential energy rather than the kinetic energy. When bonds form, whether intramolecular or intermolecular, there is a decrease in potential energy. In order to break those bonds, we have to increase the potential energy. When we add energy to convert a solid to a liquid (or a liquid to a gas), that energy goes to increasing the potential energy rather than the kinetic energy.

There's a form of inverse relationship between the strength of intermolecular force of attraction between molecules of a substance and the degree of kinetic energy of the substance. The higher the strength of intermolecular force between the molecules of a substance the lower the kinetic energy(K.E.). The molecules of solid have greater intermolecular force that exist between them compare to the molecules of liquid, and as such they have less kinetic energy (the energy of a body in motion).
The the link below explains the concept of kinetic energy
https://www.khanacademy.org/science/biology/energy-and-enzymes/the-laws-of-thermodynamics/a/types-of-energy


text:
What are chemicals that will create a similar reaction to baking soda and vinegar for my book?
I'm writing a book based on a dream I had, but after some research I can't find too much on a specific reaction I want. The main character mixes a clear liquid and yellow powder which hisses and bubbles creating a lot of yellow gas. The gas was slightly toxic, but he was exposed for around a minute and nothing super dangerous happened.
Are their any chemicals that make this reaction and includes a powder and liquid? The colors don't have to match, but I would prefer if the gas had some visible color so it's easier for me to describe.

This might fit the bill. You're not doing it IRL, so the safety concerns are minimal. You might employ some literary freedom for the liquid part: ethanol is used here, but only very sparingly. You could also stick with vinegar and baking soda and just add some turmeric spice to the baking soda - it'll be a yellow mixture and maybe (?) produce a gas. Try experimenting.

The reaction of ammonium dichromate described by Todd Minehart is spectacular. But it uses a substance which is cancerigen, And so it is forbidden to buy it in many countries. On the other hand, it produces a huge amount of green ashes, whose volume is  much greater than the original volume of ammonium dichromate. It produces y gas which is colorless, nitrogen.

The number of colored gas is not huge. Chlorine $\ce{Cl2}$ is pale green. But it is toxic : it was used during World War I as a poison gas. $\ce{NO2}$ is brown, but it is corrosive and toxic. Iodine $\ce{I2}$ gives violet vapors, but it must be a little heated. And this gaz is also corrosive. So your dream is probably impossible to realize. Sorry for being negative !

Some pretty interesting ideas in chemistry have been preceded by dreams. A good mystery novel could be the beginning of a career.
Well, the colored gases are listed on Wikipedia (also colors of some solids):

Considering first your specification of slightly toxic, I would aim for a small quantity, which suggests that we need intense color. Iodine is one of the best: very purple, very visible. It needs a bit of heat to make it gaseous, but not much: a few crystals dropped on a very warm surface will generate a lot of purple. Not really hissing; perhaps this isn't exactly what you want. You could drop a few iodine crystals into a dish or cup of hot water...
Then there is NO$_2$. While the picture says dark brown, I think deep orange red is more accurate. It's toxic, but you can see it and smell it, so it may not be a terrible hazard because you can get away from it - again, in small amounts. Not like the explosion in Lebanon (Ref 2). I was in the lab when we made some NO$_2$ - accidentally. A clear liquid (a 5-gallon bucket of 25% HNO$_3$) was treated with what was thought to be sodium nitrate, but the 50# bag of solid was only partially filled, and was torn so it read SODIUM NITR&)E, or something similar. It was yellowish - thought to be somewhat aged - but it was really sodium nitrite, NaNO$_2$. About 5 pounds got dropped into the bucket. That was followed by a large POOF - not really an explosion, but suddenly a HUGE red cloud of NO$_2$ appeared. Fortunately, there was another exit to the lab, and no one was even slightly injured. It took a few hours to ventilate the lab. Now, on a small scale, any clear acid will react with NaNO$_2$ to produce orange-red NO$_2$. It's heavier than air, so it can be fairly easily contained in a beaker if your liquid (acid) is only a small amount at the bottom. So it could be dangerous, but can be handled safely.

While all the other gases could also be handled safely, they do seem more hazardous to me - but in a book you can always specify the precautions you want to emphasize.
Ref 1. https://en.wikipedia.org/wiki/Color_of_chemicals
Ref 2. https://www.bbc.com/news/explainers-53664064

(+1) Nice answer, but iodine is now problematic, as per the DEA in the USA, because people making illicit drugs use it. Both nitrogen dioxide and bromine vapor are way too toxic, so no good options.

@Ed V: Perhaps the plot depends on not-so-good options, like finding stocks of illegal or hazardous chemicals in the basement of the school/hospital/courthouse.


text:
Why d and s orbitals overlap even when there is considerable difference in their energies?
In my grade 11 chemistry NCERT textbook, its written that hybridization happens when orbitals of almost same energy overlaps.
But S orbital and D orbital have significant difference in their energies. But still they overlap .I would be grateful to know why it happens then.

4s and 3d are particularly close in energy values this is a major factor in transition metals which frequently (In the context of VBT) undergo hybridisation with these orbitals. What you might be mistaking about must be the energy levels of s and d of same shells which do have significant energy gaps.

In multielectronic atoms we have a relatively large difference between $s$ and $d$ orbitals when they have the same $n$ quantum number, or in terms of the actual quantum mechanics when they have the same total number of nodes in the wavefunction (this being what we label as $n-1$). But in the case of transition metals the $s$ orbital that mixes in with the $d$ orbitals has one higher $n$ or equivalently one more total node -- for instance, with iron it would be $\color{blue}{4}s$ (three total nodes) mixing with $\color{blue}{3}d$ (two total nodes). The opposing differences between angular momentum and between total nodal count leave the relevant orbitals close together in energy.

ok i am satisfied with ur answer, thanks


text:
Is bond dissociation enthalpy same for homolytic and heterolytic cleavage?
In David R Klein's Organic Chemistry Book, Chapter of Radicals, one of the 'skillbuilder' questions is to identify the weakest C-H bond in the following molecule:

The solution was to split each carbon-hydrogen bond homolytically one at a time and find which radical generated was most stable.
However, the question asked 'weakest C-H bond' without mentioning what kind of bond splitting is happening, so does this mean bond dissociation enthalpy for homolytic and heterolytic split of bond is same?

Well, if it was chapter about radicals, then it should be clear that it's about homolytic - hardly any ambiguity.

I don't think one can assume so unless it's explicitly stated.

related https://chemistry.stackexchange.com/questions/133669/heterolytic-vs-homolytic-cleavage-of-the-c%e2%88%92h-bond-in-cyclopentadiene

Well, if context wasn't enough then you should have check definition.

That question contains discussion of the post but really is not asking about the same question I have here.

Dissociation of a bond in thermochemistry & thermodynamics includes the products. Physically there is never a doubt. While speaking perhaps there might be a doubt, in cases. But as said in other comments, it shouldn't be the current case.

The question is for the weakest CH bond. Heterolytic bond cleavage is in almost all cases a more energy demanding process than homolytic cleavage; whether or not it is stated, it is irrelevant.


text:
Diffusion coefficient for acid base reaction
I am trying to model an acid-base neutralization reaction.  There are two adjacent slabs - one acid and one base. Each slab is the same thickness and I'm only looking at one-dimensional diffusion/reaction.  At time $t = 0$, the acid slab has a concentration of cA0 throughout and the base slab has a concentration of cB0 throughout.  At $t>0$ diffusion occurs and reaction occurs at an interface between the two slabs.  In general cA0 is greater than cB0 and therefore A penetrates into the B slab; in other words the reaction plane is a moving boundary.
I have an analytical expression for the concentration profiles that appears to work very well based on comparisons with MatLab simulations.  However, it was developed assuming a constant value for the diffusion coefficient, D, for both acid and base. But, in reality, the diffusion coefficients are different.  Is there a weighted or average or calculated diffusion coefficient that I could use that would apply to both acid and base?  I've tried just using an average, but now the results of the analytical expression are noticeably different than the MatLab simulation.  Maybe an average isn't the right way to do this.
Also, which diffusion coefficient for each species should I use?  Let's say acid is HCL and base NaOH. Should I start with a coefficient for hydrogen ion and one for hydroxide ion?  Or should it be a D for the HCL molecule and the NaOH molecule?  Or.......

Are you serious about HCl and NaOH molecules in water solutions ? // Generally, Ds for particular ions will differ significantly for thick solutions very different to water. Their values are highly empirical.

In solution HCl and NaOH don't exist as molecules but only as ions, $H^+, OH^-$ so you need to find diff'n coefficients for these. You will also need to include chemical reaction in your equations for the equilibrium to form water molecules.  The rate constants for this are known via T jump relaxation experiments, have a look in a phys chem text book.

Well, I didn't mean molecules exactly.  If charge separation is to be prevented doesn't that mean that even though $H^+$ has a significantly higher value of D than $Cl^-$, it will be slowed down and $Cl^-$ will speed up.  Any comments on my question about possibly using an "average" or "weighted" diffusivity for both reactants?

@porphyrin.  Strictly speaking I should include the equilibrium with water.  It means that $OH^-$ will never be exactly zero, which is what my models assume at the reaction interface (i.e. instantaneouus reaction where A and B cannot coexist). But ultimately I am trying to determine the time for neutralization starting with an $OH^-$ concentration of $10^{-2}$.  So five orders of magnitude reduction is "effectively" complete reaction.  Any thoughts on use of an "average" or "weighted" diffusivity?

You could use an average value, but would it not be better to have equations for the diffusion of each species coupled with the $H^+ + OH^- = H_2O$ equilibrium and solve at each time and spatial step? Alternatively, as an approximation  assume  $D_H$ is  v large and equilib instantaneous so the process is effectively determined by relatively slowly diffusing OH$^-$

I am having difficulty determining what values for the diffusion coefficients I should use.  Any guidance is appreciated.


text:
Industrial recycling of mirrors
I learned recently that mirror is mostly manufactured by pouring thin metal coating (i.e alumunium) on top of a substrate surface (i.e glass),
Due to the nature of this composition it cannot be easily disposed. I searched everywhere on how to recycle mirror and it always end up suggesting to recycle it physically (i.e artwork props etc.).
Assuming on industrial level mirror waste, how its getting recycled? I found no machine in market to do so.
My first thought will be separating the two in heat since both have different melting point, however not sure.
Any input is greatly appreciated!

I did not know the mirror can't be disposed on the glass recycling bins. I suspect that the whole story is economic in reason. The reflective coating is so thin that in some mirror can be even scratched by nails. Mirrors probably ruins the glass makers recipes (which already accommodate for a wide range of glasses, just think on the huge gamma of amber brown for beer or the dark for wine... I am really surprised and curious to see the real reason for the lack of mirror recycling companies / techniques, even more as I recall alumina and aluminosilicates are obviously in the feed.

As @Alchimista states, the coating is nanometers, or at most a micron or two thick. It is not made by pouring aluminum, but by evaporating it in a vacuum. It is not difficult to remove such a thin coating by immersing the mirror for just a few seconds in $\ce{NaOH}$ solution; this is part of the process of re-aluminizing (is that re-aluminiumizing in GB?) telescope mirrors.
However, to protect the thin layer of metal on back-surface (daily use) mirrors, a layer of lacquer is applied, which might be more difficult to remove chemically.
That said, many recyclers reduce glass to cullet to be melted into generic glass items, such as glass-fiber insulation, or used to make other material, such as concrete and ceramics. Mirrors probably can be recycled as cullet.

For illustration about the described (first surface) telescope mirrors, which undergo intervals of getting this fine coating removed, and reapplied again a small scale, and large scale example. The (modern bright, second surface) mirrors,  lacquer-backed (e.g., in the bathroom) are double coated, first Sn to improve adhesion, then topped by the fine layer of silver reduced in situ (example). The classic almagam mirror is darker and less frequently seen.

Thank you for the useful information. But then how to strip off the coating completely in lacquer backed mirror? Leaving it out to pure glass which is much more convenient to handle/recycle @Buttonwood

@gardin If the glass is of soda-lime type, then at the high melting temperature (vicinity of 1000 C), all organic material may combust. Perhaps the bit of Cu renders the glass unsuitable for the production of optical lenses (turquoise color, as reported here), yet there are multiple other uses of recycled glass in addition to bottles and windows (ref).

To summarize, mirrors (even for the case of lacquer double-backed mirror), could be recycled by putting them into certain solution for awhile to strip off the outermost coating, and then melting to flush the remaining? End result will be pure glass? Am i missing something? Again thanks a lot @Buttonwood


text:
Which compound of these following have electrovalent, covalent, coordinate as well as hydrogen bond?

Which compound have electrovalent, covalent, coordinate as well as
hydrogen bond ?
A) $\ce{[Cu(H2O)4]SO4.H2O}$
B) $\ce{[Zn(H2O)6]SO4.H2O}$
C) $\ce{[Fe(H2O)6]SO4.H2O}$
D) $\ce{[Fe(H2O)6]Cl3}$

Answer given:

 Option (A)

My approach
All the compounds contain electrovalent, covalent as well as coordinate bonds hence the deciding factor is hydrogen bonding. (D) clearly would not have hydrogen bonding. I know that in (A) which is $\ce{CuSO_4.5H_2O}$ the fifth $\ce{H_2O}$ which is something that was taught to me earlier.
My Doubt:
Yes I get the answer to the given mcq problem above but I was not able to understand the case of (B) and (C). I think that the $\ce{H_2O}$ inside the coordination sphere are bonded through dative bond and the $\ce{SO_4}$ outside by ionic bonds but I am not able to understand the nature of bond formed by "$\ce{.H_2O}$" present outside the main coordination sphere given in the complexes above. Would they not form hydrogen bond? The answer scheme doesn't think so. So if they don't what is the nature of bond formed by these "$\ce{.H_2O"}$ !?

Edited the question. Please let me know if any further clarifications are required.

All compounds have electrovalent bonds between the metal and the anion (sulfate or chloride). All have covalent bonds between O and H in the water molecules. All have coordinate bonds with water around the metal. And A, B, and C have hydrogen bonds between the inner and the outer water molecules.


text:
Could Gaussian or Spartan software determine stereoisomers from conformational landscape of a molecule alone?
I read that, on some papers, conformers were generated by stereoconfiguration of the molecules and from there, they evaluated stability of each stereoisomer in terms of free energy differences between two stereoisomers. Why people illustrated the energy profile of each stereoisomers from their fixed stereoconfiguration, not from global landscape of conformers (conformational distribution) of the molecule and sorting out the conformers by giving plane of symmetry or other methods defining and distinguishing stereoisomers of the molecule? A molecule is always allowed to rotate freely on 3D space and the results of its ability to rotate into a mirror image of its origin is the chirality we are interested in.
Is it possible to tell Gaussian or Spartan to assign stereoconfiguration of each stereoisomers by solely from its generated conformers (conformational distribution) alone without any preconfiguration of its stereoconfiguration? Or I missed a theory that I can’t follow up the general methods on deducting 3D structure of a stereoisomer?

I really don't understand what you are asking for. A molecule is allowed to rotate in a 3D space, but it will never rotate into a mirror image—that's just not possible. Conversion between stereoisomers requires an inversion, which cannot happen via any bond rotation that you might conceive. Besides, sampling the full conformational space is very expensive, particularly with QM methods.

@ShoubhikRMaiti This figure said it’s possible to do so https://www.nature.com/articles/s41557-018-0043-6/figures/1

Yes, but as you can see from the images, there are some which are easily interconvertible and some which aren't. Generating a global landscape of conformers is very costly and time consuming. It is far easier to generate possible conformers and then generate stereoisomers from them. (AFAIK Gaussian does not have any option to deal with stereoisomers, at least not automatically. I have never used Spartan so can't comment on that.)

Because Jmol may assign stereogenic centers, it is likely other programs (perhaps including Gaussian) are equally capable to perform this task.  For example, on a hypothetical bromofluoromethanol:

you select the carbon atom in question, and type on Jmol's console the commands
set labelalignment right;
set labeloffset -20 0;
label %[chirality];

to obtain the requested CIP-label (further examples and details here).
For the figure in question:

(credit to Canfield et al.)
the e.g., line about the interconversion of atropisomers has to be read altogether with the energy scale / time scale provided below the same figure, too.  For 1,1'-biphenyls with little or none steric hindrance at 2,2',6,6'-position, it is indeed plausible that enough heating might suffice to interconvert the $\Lambda$ into the $\Delta$ atropisomer (or, P into M)

Assuming typical reaction conditions, steric hindrance for larger molecules like e.g., BINAP, or BINOL establishes an energetic barrier this high that you observe them in either the P, or M configuration, even if there is some conformational flexibility left.  The barrier is this high that these molecules are reliably used as chiral ligands in chiral syntheses.

(credit)
If you have access to a molecule kit, I recommend you build these molecules by your own.  Experience the steric hindrance the hydrogen atoms not explicitly drawn impose if one attempts to pass for the other enantiomer; this requires to break the $\ce{C-C}$ bond.
Reference:
Canfield, Peter J. et al., A new fundamental type of conformational isomerism. Nat. Chem. 10 (2018), 615–624; doi: 10.1038/s41557-018-0043-6; the author's preprint.


text:
Bond Polarity vs bond length's effect on reactivity of haloalkanes
My textbook says:

Carbon acquires partial positive charge whereas halogen acquires partial negative charge. Halogen becomes nucleophilic in character, which can be replaced by another nucleophile on the basis of bond polarity, reactivity of alkyl halides decreases in the following order.
$\ce{R-F > R-Cl > R-Br > R-I}$.

My question: Shouldn't $\ce{R-I}$ be the most reactive since it is the least polar and I can be replaced by more electronegative nucleophiles? Also it's bond length is also the longest and it should have lower bond energy hence higher reactivity. So why does my book say otherwise?

The order you provided describes the decreasing bond strength not reactivity. Bond strength of R-X decreases from Fluorine to Iodine as the orbitals become more diffuse and effectiveness of overlap decreases. Hence, due to the ineffective overlap, the covalent bond strength is weaker
I believe the reactivity of alkyl halides should be R-I > R-Br > R-Cl > R-F. This is because of the decreasing bond strength of the R-X bond which thus slows down the rate of Sn1 and Sn2 reactions.
Why does the strength of R-X bond affect the reactivity of halogenoalkane?
This is because, the slow step (i.e. rate determining step) of the Sn1 and Sn2 nucleophilic substitution mechanism involves the breaking of the R-X bond. Either to form a carbocation (for Sn1) or the tetrahedral compound (for Sn2). Thus, the stronger the bond, more energy would be required to break it, making it less reactive.
However, there is a contrasting effect: the polarity of the R-X bond decreases down the group. The R-I bond is less polar than the R-F bond. Hence it may seem as though the nucleophile would be more strongly attracted to the electron deficient carbon. But based on experimental results, it has shown that the bond strength is in fact the main factor that determines the reactivity of the halogenoalkane towards Sn2 reactions.
The magnitude of the partial positive charge of the carbon in the C-X bond does not affect the reactivity. The main factor that affects reactivity is bond strength
Note that we do not need to consider this point about bond polarity of the C-X bond when we are looking at Sn1 mechanism, because the rate determining step of Sn1 reactions does not involve the attack by a nucleophile.


text:
Doubt in a question on cleavage of ethers (Solomons Practice problem 11.18)
I have reffered to the mechanism shown in the text which is:

But I am not able to figure out how this took place:

Please help me with it. Though it is a homework question but I have tried my best to solve it and so I request you not to report this as homework question.
Thank you.

Relative (thermodynamic) stability of carbacations, as a second hint.

I also agree that this product is more stable; but here, Sn2 takes place so how can rearrangement take place? (Or correct me if it's not Sn2 as it is not mentioned in the book.)

It might depend on the reaction conditions; my initial thought involved a Sn1 mechanism. Frankly, I'm hesitant for @Waylander's suggest of a leaving MeCl; this because right now I do not recall an example (so it is my lacuna ...).

This is all driven by the steric hindrance around the protonated methyl ether.
First step - protonation of the oxygen, clear enough?
Second step - attack of chloride ion: the oxygen of the protonated species is in a hindered environment so the nucleophilic attack of Cl- on the back face is blocked. Instead you get attack on the Me group to give MeCl (which leaves as its a gas) and the secondary alcohol.
Third step - there is xs HCl present so the secondary alcohol protonates. Water leaves to give a secondary cation. There follows a hydride shift to give the more stable tertiary cation which is captured by Cl- to give the prodouct observed.

I fully agree with Waylander's suggest to let the strongest Lewis acid ($\ce{H+}$) and base (the alcohol) react with each other to yield an early equilibrium.

Contrasting to Waylander, I speculate methanol could then leave.  Because of the presence of water (as a proton shuttle), the intermediate secondary carbacation however would give way to the more stable tertiary carbacation then awaiting ($S_N1$ mechanism) the addition of $\ce{Cl-}$.
This suggested mechanism however lacks the liberation of $\ce{MeCl}$ postulated by Waylander; the low boiling point were favorable to leave the system and thus advance the reaction forward.  Equally, the intermediate secondary carbacation drawn by mine could rearrange to yield a 1-chloro-2-methylcyclohexane, or / and 2-methyl cyclohexanol for that in comparison to other cyclohexanes, the formation of both 5- and 6-membered cyclohexanes is favorable.

It might depend on the reaction conditions if there is evidence for one, or the other, or for both occurring simultaneously.

If you see the Step 2 written in the explanation I attached, you can clearly see that no carbocation is forming and the reaction is Sn2.

Wikipedia argues there are two plausible mechanisms, either SN1, or SN2, which may compete with each other. The examples on www.masterorganicchemistry.com condense this at one point stating "The Mechanism For The Cleavage Of Unsymmetrical Ethers Is Hard To Generalize" and later "SN1 and SN2 is a continuum. You’ll likely have a mixture of SN2 and SN1 pathways operating." about secondary ethers.  So it is still up to the experiment and its specific conditions.


text:
Does release of nicotine from tobacco require combustion, or an open flame?
I am interested in the combustion argument Phillip Morris is making regarding its iqos product. They claim there's no combustion in their product but how do you take nicotine out of something without a combustion process? Enough heat would still create a chemical reaction... I'm asking based on a lawsuit they have going on. Is it possible to say there's zero combustion in a product that heats a material to extract a chemical(nicotine) in its vapor smoke?

From what I've just read here, nicotine is released from tobacco at only 150 Celsius. Combustion occurs at 750 Celsius. The heated-tobacco products (iQOS is one of them) appear to target 350 Celsius, which is when pyrolysis occurs but is far lower than the combustion temperature.

Pure nicotine has a well-defined boiling point and can be vaporized just by heating it up. No chemical reaction or combustion is involved in that.

Hope my edits retained the intended original sense of the question. Feel free to roll back to the previous version otherwise. Also, please add references (or links to sources) where possible.


text:
Analysing reaction with chair conformation of substituted cyclohexane
Why doesn't Iodine liberate in the following reaction?

(A) Inadequate ring size
(B) Two Bromines are in equatorial positions
(C) Steric hindrance of t-butyl group towards attacking $\ce{I-}$

Initially, I thought it would simply undergo a simple finkelstein kind of reaction and since two iodines in vicinal carbons aren't stable, it would lose $\ce{I2}$ to give a double bond according to the following mechanism:

But my book said that $\ce{I2}$ would not be liberated.
I had a thought that this might be related to the chair conformation and the bromines being at equatorial positions since the options provided point towards this,
so I felt that's what the exercise wants me to do but I can't get to the reasoning.
For reference, the answer given in my book is:

(B) Two Bromines are in equatorial positions

To loose $\ce{I2}$, the oxidation reaction $\ce{2I- -> I2 + 2e-}$  should happen. But there is no oxidizer in this reaction.

@MathewMahindaratne Oh okay but the reason given in my book says that its because the two bromines are in equatorial positions. Maybe the exercise wanted me to look at this factor rather than that? Can you help me reason this out?

@MathewMahindaratne The OP most probably is talking about homolytic clevage of the vicinal iodines to form iodine and an alkene and not direct oxidation of the iodide ion.

@PrajwalTiwari See the edit I made. Is that what you meant to say? If it's not than feel free to rollback.

How do you know whether the bromine substituents, or the iodine ones, are equatorial? The stereochemistry isn’t marked, so what enables you to make that assumption? I actually think there is a good reason to say so, but you haven’t provided it.

@orthocresol maybe that's just poor question setting in the exercise. But the options provided point towards this factor only. I'm sorry I didn't mention this. I have added this to the post as well.

Yes, that's a pretty bad question; because there's absolutely no reason why the bromines in the starting material should be equatorial. I think there's a good argument that the iodines in your proposed intermediate will end up equatorial. Of course, we're assuming that your intermediate is actually an intermediate; but it doesn't seem unrealistic to imagine that SN2 with a good nucleophile and good leaving group is faster than any elimination.

So assuming they are equatorial, why wouldn't iodine liberate?

Why should we assume that the halogens are in equatorial position?

As @Orthocresol pointed out there is probably enough reason (it should be still mentioned as any other conformer may still exist even if unstable) for both the halogens to be in equatorial position. The t-butyl group attached provides massive steric repulsion to both the halides so they are locked in equatorial position. This is a good proof of tert-butyl's massive steric hindrance which results in cyclohexane ring ditching it's usual chair conformer for a twist-boat conformation for relief.

OK, but why is the configuration of the $\ce{C-Br}$ bonds relevant to the question?

The reaction mentioned here is the well-known finkelstein reaction which prefers $S_{N}2$ mechanism (mainly due to solvent effects). As we know the rate of $S_{N}2$ reaction depends upon it's transition state's stability, whose stability reduces as steric inhibition increases. ($\ce{I^{-}}$ is not exactly the smallest nucleophile out there.)

The problem doesn't end here. The mechanism depends upon nucleophilic attack on $\ce{C-X}$ bond as a back attack. Now we can't expect the to put Br in axial position because of high repulsion of tert-butyl group and in equatorial position the tert-butyl group again interferes directly in approach of nucleophile. As a result this reaction cannot proceed and no $\ce{I_2}$ is liberated


text:
Why is the rate of solvolysis of t-butyl chloride lesser than that of 1-chloro-1-methylcyclohexane?
I came across a question regarding relative rate of solvolysis. According to the answer key:

My analysis:
As in all the three cases, the Cl atom is bounded to a tertiary carbon, hence, all of them are ${S_N1}$ reactions. So, the rate of solvolysis will depend on the stability of the carbocation formed. Upon considering hyperconjugation, t-Bu cation is coming out to be more stable than the tertiary cycloalkyl carbocations.
But, according to the given answer, t-BuCl has the least solvolysis rate as compared to the others. Could anyone please explain why is it so?

@NisargBhavsar Even I thought the same initially, and so to confirm, I checked if solvolysis/hydrolysis can involve elimination - In Wikipedia, for hydrolysis, it is mentioned that "The term is used broadly for substitution, elimination, and solvation reactions in which water is the nucleophile". So I think even if we consider E1 instead of ${S_N1}$, we still need to check the carbocation stability to compare. Am I going wrong somewhere?

You are right. Check the answer I posted.

According to the definition given by The IUPAC Compendium of Chemical Terminology: The Gold Book[1], solvolysis is defined as:

Generally, reaction with a solvent, or with a lyonium ion or lyate ion
involving the rupture of one or more bonds in the reacting solute.
More specifically the term is used for substitution elimination and
fragmentation reactions in which a solvent species is the nucleophile.

It clearly states that elimination occurring due to a solvent is also considered solvolysis and thus elimination in case of t-butyl is also considered.
This means you are right in your reasoning of t-butyl being the most susceptible to solvolysis among the given choices.
I believe this was an error from the question setter. They must have been referring to the substitution reaction due to the solvent but ended up creating a poorly worded question.

Reference:
(1) The IUPAC Compendium of Chemical Terminology: The Gold Book; Gold, V., Ed.; International Union of Pure and Applied Chemistry (IUPAC): Research Triangle Park, NC, 2019.


text:
What are the common procedures for viewing suspensions of volatile organic compounds under an optical microscope?
I am looking to measure a mixture of triethylamine and water in a micro-Raman instrument. This involves placing the sample under an optical microscope. The "default" sample preparation is to place the sample on a microscope slide.
However, that is not an acceptable solution for my use case. Triethylamine is a volatile substance with a foul odor. A thin film of the substance on a glass slide would quickly evaporate, making the measurement pointless and introducing toxic vapors into my university's shared facility (there is no forced ventilation, like a fume hood, available).
Are there sample containers or procedures for viewing volatile substances under an optical microscope?

There are microscope slides which have a mild depression in them. You can add more liquid and cover it. Those slides are called slides with reaction wells. It takes much longer to evaporate now if the wells are covered. BTW, triethylamine and water is not that volatile! Try neat alcohol and it will evaporate in less than a minute.

The point is that the micro Raman set up should not give you any advantage, at least not for what you've described


text:
Am I accidentaly creating dangerous silver nitride?
I am doing some experiments with alternative film chemistry and I am starting to get concerned about safety. One experiment involves converting a small amount of metallic silver to silver chloride and then soaking the film in an weak ammonia solution (< 10% ammonia hydroxide), to dissolve the silver chloride, additionally there may be small amounts of silver oxide and silver metal left on the film. I read somewhere that in experiments involving ammonia and silver compounds that the dangerous explosive silver nitride could be produced. Should I be worried about it? would I be producing it in dangerous quantities? or at all? There is probably only a few grams of silver on the entire film strip.
What are some sensible precautions I could take here?

See this question and links therein.

Wikipedia states specifically that silver nitride is slowly formed from solutions of silver oxide or silver nitrate in aqueous ammonia; it does not mention other silver salts as potential precursors.
In case the nitride does form, you can look for it, and if you suspect anything you can easily dissolve it in dilute ammonia solution.  From the WP article:

Silver nitride may appear as black crystals, grains, crusts, or mirrorlike deposits on container walls. Suspected deposits may be dissolved by adding dilute ammonia or concentrated ammonium carbonate solution, removing the explosion hazard.[1][2]

Cited References
1.
John L. Ennis and Edward S. Shanley (1991). "On Hazardous Silver Compounds". J. Chem. Educ. 68 (1): A6. doi:10.1021/ed068pA6.
2.
Salt Lake Metals Inc., "Silver oxide". Retrieved February 11, 2010.

The ammonium hydroxide would convert the silver chloride to ammonium chloride and a precipitate of silver probably silver oxide also. On a small scale it shouldn't be dangerous , maybe have an ice bath handy just in case and always wear appropriate ppe

No ! $\ce{AgCl}$ gets dissolved in ammonia solution producing the soluble ion $\ce{[Ag(NH3)2]^+}$ ; silver oxide is not precipitated. Silver nitride is dangerous ; it is only formed when dissolving silver oxide $\ce{Ag2O}$  (not silver chloride) in concentrated ammonia solution (> 1 M).

Maurice ,,  in a %10 solution with water? Maybe if left over time, he's dissolving silver chloride not silver oxide.

Read "Silver nitride" in Wikipedia !

Maurice he's not dissolving silver oxide or silver nitrate, he's dissolving silver chloride in aqueous ammonium hydroxide, it's simple Redox chemistry, you should read the reference your citing it states the reaction requirements.

@maurice I have done so. I want a backup plan, hence my answer.

@ Oacar Lanzi. What do you mean with a backup plan ?

Maurice, if the Wikipedia source doesn't list other silver salts as precursors why do you think that silver chloride and ammonium hydroxide in aqueous environment will form the nitride? It specifically says silver nitrate or silver oxide dissolved in concentrated ammonia solution will form the nitride.

@Maurice %10 is not a concentrated solution either.

@Technetium just want to be clear that there is a high chance of a small amount of silver oxide being present as the previous step was a mild oxidizer

@BathroomFilmLab, either way the nitride ion would not have a high potential of stabilising in that environment anyway??