text:
Is it advisable to make vinegar at home using acetic acid?
I've taken only a couple university chemistry courses. In the lab, when we always used acetic acid in powdered form, it struck me how when buying vinegar in a bottle from the store, you must be paying mostly only to transport water. Wouldn't it be more economic to simply mix acetic acid solutions for yourself? Would any of you use acetic acid to make vinegar at home? Are there any dangers or drawbacks?
Edit: Thanks for the informative responses! It seems my memory was mistaken. We clearly must not have been using acetic acid in powdered form.

To turn it into powder means removing all the water, which is energy intensive and expensive.

Acetic acid is liquid at rt so I don't understand what the powder woukd be that you used

Perhaps glacial acetic acid. How could OP use it and not know it's dangerous, is another question...

But glacial acetic acid isn’t a powder either :/

Glacial acetic acid is a significant contact hazard for skin.

Vey concentrated (glacial) acetic acid is a corrosive acid that can severely damage your skin and airways. Acetic acid does come in solid form but has a melting point of 16 to 17degC. Diluting it can be dangerous if not done properly due to dilution of acid releasing a lot of heat. It can spit acid all over you. In all honestly it's just easier buying it as 5% acetic acid (What you get from shops) rather than go through all the dangerous effort to try and dilute pure solid acetic acid.

Thanks. It seems my memory was mistaken. We must not have been using acetic acid in a powder. Very good response!

As someone who was in a lab when a student used a vacuum pump to pull lots of glacial acetic acid into the air...ouch. "Airway damage" hurts like hell the whole time it's going on, and the stuff is not much nicer to your eyes.

That sounds really bad @chipbuster . I've had to work with glacial acetic acid and it's not very pleasant


text:
Why is antimony oxide used as a flame retardant?
What is the mechanism of action of antimony oxide as a flame retardant? Is this to do with radical absorption?

According to FLAME RETARDANT THEORY:

Antimony oxide by itself is not a fire retardant, and the halogens by themselves, mainly bromine and chlorine, are weak fire retardants. However, when they are combined they become synergistic and are the most effective and most widely used flame retardant system for plastics.

...

Two mechanisms exist in the synergistic system.
First is the "free radical capture" process that takes place in the vapor phase. On combustion at a temperature of over 600 [degrees] F, the halogen forms hydrochloric or hydrobromic acid that reacts with the antimony oxide to form antimony trichloride, antimony oxychloride, antimony tribromide, or antimony oxybromide. The flame retarding action takes place in the vapor stage above the burning material. It is thought that "free radicals propagate" the flame. But, the antimony trihalides or antimony oxyhalides act as "free radical traps", and take up free radicals. They inhibit ignition and pyrolysis in the solid, liquid, and vapor phases.
A second process occurs in a solid phase and is the "char process". The antimony oxide promotes the formation of "char" (essentially carbon) on the substrate which reduces volatile gas formation. The barrier between the substrate and the vapor phase reduces the available oxygen to the underlying substrate. The barrier effect is obtained by almost any inert additive. In plastics there is a cross linking with antimony to produce a more stable thermoset polymer. Additionally in the solid phase, the formation of SbCl3 and SbOCl acts as a dehydrating agent that increases charring. Sometimes phosphorous compounds (TCP), magnesium oxide, alumina trihydrate, molybdic oxide, zinc borate, or zinc oxide are used in combination or in place of antimony oxide to reduce costs, to increase char formation, or to reduce smoke. However, the substitution of the other retardants greatly reduces the flame retardance normally rendered by antimony oxide. Testing of the amounts of the halogen and antimony oxide in each formulation is necessary to optimize the flame retardance and lower costs. Alumina trihydrate is not synergistic with halogenated flame retardants. It functions as a flame retardant by the release of its water of hydration and cannot be used in high temperature processes. Zinc borate, molybdic oxide, zinc oxide, and magnesium oxide can be used in conjunction with antimony oxide to augment char formation and decrease smoke. Replacement of the antimony oxide to meet a smoke requirement compromises the flame retardance.

See also Flame retardant mechanism of antimony trioxide

source: Stack Exchange

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text:
Molecular orbital diagram and irreducible representations for dinitrogen
I'm trying to understand how to draw molecular orbital diagrams, since I'd like to use the molecular orbitals to determine the total state of the molecule (i.e. $A_1$, $B_1$ etc.), so I prefer the notation with irreducible representations of orbitals like it's shown in this video for $\ce{H2O}$.
The problem is that I'm not sure how to make molecular orbitals for $\ce{N2}$ and $\ce{N2^+}$.
I know that $\ce{N}$ has atomic orbitals $\ce{1s^{2} 2s^{2} 2p^{3}}$. $\ce{N2}$ is a linear molecule, so it belongs to the $D_{\mathrm{\infty h}}$ symmetry point group, but we can descend in symmetry, so we'll use $D_{\mathrm{2h}}$.
But now I have no idea how to proceed. Should I work with both nitrogen atoms together, i.e. with group orbitals, similarly to the case with two hydrogen atoms in $\ce{H2O}$? Or am I supposed to work with them separately?
So could you please write down one of the diagrams using irreducible representations and explain it step-by-step? I think it would be really useful here, as I couldn't find any source with some general explanation.

Because this is a diatomic molecule, there are no group orbitals. Put another way, the group orbitals are the molecular orbitals. Knowing the nitrogen atomic orbitals (AOs) and their irreducible representation (irrep) labels is enough.
Since we'll work in the $D_{\mathrm{2h}}$ point group, we need its character table:
$$\begin{array}{c|cccccccc|cc} \hline
D_\mathrm{2h} & E & C_2(z) & C_2(y) & C_2(x) & i & \sigma(xy) & \sigma(xz) & \sigma(yz) & & \\ \hline
\mathrm{A_g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\
\mathrm{B_{1g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & xz \\
\mathrm{B_{3g}} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \\
\mathrm{A_u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\
\mathrm{B_{1u}} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\
\mathrm{B_{3u}} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline
\end{array}$$
Using the nitrogen atom's electron configuration $\ce{1s^{2} 2s^{2} 2p^{3}}$ as our minimal basis, no d-orbitals will be present, so we can assign irreps to each AO right away, in part because we assume the principal rotation axis (the one of highest order) is aligned along the z-axis:
$$\begin{array}{cccc}
\hline
\mathrm{s}     & \mathrm{p}_x    & \mathrm{p}_y    & \mathrm{p}_z    \\
\hline
\mathrm{A_{g}} & \mathrm{B_{3u}} & \mathrm{B_{2u}} & \mathrm{B_{1u}} \\
\hline
\end{array}$$
Then, proceed by starting to form the standard MO diagram for a diatomic, but add the irrep labels to each AO:

Note that I haven't spaced the energy levels properly; the $\ce{1s}$ should be much lower than it is relative to the $\ce{2s}$. More on this later. Regardless, because this is a homodiatomic, all AOs will mix at each energy level, even the core $\ce{1s}$ AOs. Then, attempt to form the MOs:

This is probably wrong, but it's a starting point. Again, energy levels will be discussed later. Finally, add symmetry labels to each MO, remembering that

MO symmetry is derived from AO symmetry,
numbering is consecutive within each irrep, not with respect to the set of all MOs, and
lowercase is used to signify that these are MOs; uppercase is for the irreps themselves.

This is our final MO diagram for $\ce{N2}$. To form the diagram for $\ce{N2^+}$, remove an electron from $\mathrm{4a_g}$. This ignores orbital relaxation effects, but for the purposes of working this out on paper, it should be fine.
Now for the matter of relative and absolute energy levels. It is probably possible to get the correct relative ordering of the MO energy levels. Here, I assume that since AO mixes with an identical partner on the other atom, the splitting for $\ce{1s}$ would be the same as $\ce{2s},~\ce{2p_z}$, etc. Since I drew the $\ce{1s}$ too high, the $\mathrm{3a_g}$ is almost certainly too high, and perhaps should even go below what is labeled as $\mathrm{2a_g}$. The way to confirm this, and the only way to get absolute energy levels, is to perform a quantum chemical calculation. Since we've used a minimal basis for the drawing, we'll stick a minimal basis in the calculation. Here is a Psi4 input file:
molecule {
    N 0.0 0.0 0.0
    N 0.0 0.0 1.0975
}

set {
    basis sto-3g
    scf_type direct
    df_scf_guess false
    cubeprop_orbitals [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
}

e, wfn = energy('hf', return_wfn=True)
cubeprop(wfn)

and from its output:
Orbital Energies (a.u.)
-----------------------

Doubly Occupied:

   1Ag   -15.518067     1B1u  -15.516124     2Ag    -1.442840
   2B1u   -0.722491     1B2u   -0.573123     1B3u   -0.573123
   3Ag    -0.539495

Virtual:

   1B2g    0.281319     1B3g    0.281319     3B1u    1.123476

It looks like I might have messed up the diagram, because the ordering isn't what's expected (why isn't $\mathrm{B_{1u}}$ degenerate with the other two p-orbitals?), and the virtual degenerate p-orbitals have gerade symmetry. Time to plot!

In my haste, I forgot an important point. When you make the MOs from AOs, you're making two linear combinations:
\begin{align}
\psi_{s} &= \frac{1}{\sqrt{2}} (\chi_{l} + \chi_{r}) \\
\psi_{a} &= \frac{1}{\sqrt{2}} (\chi_{l} - \chi_{r}),
\end{align}
which means that an antisymmetric combination of two s orbitals will look like a $\mathrm{p}_z$ orbital. This doesn't fully explain everything, but is a good starting point. The lesson up to now is that while drawing the diagrams by hand is a useful exercise, it is not enough to do only that when the goal is to perform a correlated ab initio electronic structure calculation.

Please forgive me for the ugly diagrams, I can try and redraw them on a tablet if desired. I am working on an answer for your other (Molpro) question, but it will take me a little more time.

Why do the cubes look so strange?

Thank you very much! And don't bother with other new diagrams, these are certainly good enough ;)

Could you, please, just briefly mention, in which situations the "group orbitals" are used?

@pH13-YetanotherPhilipp I don't know. Really, the whole thing confuses me, they shouldn't look mixed like that. I'm going to try Q-Chem next.

@Eenoku Group orbitals are used when one of your bases for the MO diagram is just that: a group. For example, in ammonia, the nitrogen atom may be on one side, but the $\ce{H3}$ is taken together as a group on the other side. Same thing would go for a $\ce{ML6}$. The way I think about it is that you can't have more than two sides on an MO diagram...what do you do when you have more than two atoms? Combine ones that have a characteristic symmetry.

What about a trigonal bipyramidal complex, where there may be 3 distinct groups ($\ce{AX3Y2}$)? The whole thing is $D_{\mathrm{3h}}$. Form $\ce{Y2}$ in $D_{\mathrm{3h}}$ (MO diagram not necessary), form $\ce{X3}$ in $D_{\mathrm{3h}}$ (MO diagram not necessary), then have each group on a side of an MO diagram to help form $\ce{X3Y2}$, then finally make an MO diagram with $\ce{X3Y2}$ on one side and the central atom on the other. The $\ce{X3}$, $\ce{Y2}$, and $\ce{X3Y2}$ are all group (intermediate) orbitals.

This is a correctly ordered molecular orbital diagram with $D_{2h}$ irreducible representations and energies included for every orbital.
The energy values and orbital ordering were get with a help of Psi4 script from pentavalentcarbon's answer.

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text:
Phenol preparation from Grignard reagent
Reaction: Phenyl magnesium bromide in presence of excess oxygen, on acidification yields phenol, as below.

My instructor told me that the above reaction occurs, and if the reagent is in excess then phenol is the end product. However, I'm not able to figure out the mechanism.
I've tried and got stuck at this point:

I wish to know the complete mechanism of this reaction.

https://synapse.embibe.com/learn/chemistry/concept-preparation-of-phenols-from-grignard-reagent-16311.

In practice this is not how lab chemists  make phenols from aryl halides as the yield not reliable. The route is to make the aryl boronate from the Grignard and oxidise that.

@Waylander, could you please post that as an answer with the reaction/mechanism?

I doubt that the peroxide compound is the real intermediate. In reality the Grignard probably reacts directly with water to give the phenol.

As requested by the OP - the route that lab chemists use to form phenols from aryl halides (not aryl fluorides) is via formation of the aryl boronic acid or boronate. These may be formed by either metallation of the aryl halides to form the aryl lithium or Grignard and reaction with a trialkyl borate $\ce{B(OR)3}$, or by Pd catalysed reaction with pinacol borane (Suzuki-Miyaura reaction). These boronates or boronic acids may be isolated, but in practice an oxidative workup, most often with basic peroxide, yields the phenol in good to excellent yield.
(read more links: 1, 2, 3)

image from US7678158B2.

Thanks to @Avatar Shiny, I found the mechanism: (a link has been posted in the comments section)
For the sake of completeness, I add the mechanism here as an answer -

Why have an anion attack a negatively charged oxygen? The other oxygen of the hydroperoxy anion is the electrophile.

Does the oxygen lone pair or the electrons of the O−O bond attacks BrMg? Also what w/u means? Thanks in advance.


text:
How to know when Sn2 or Sn2' reaction will take place?
The following reaction has the product as shown, as the major product $S_N2'$(allylic nucleophilic substitution) path is followed:

However, if instead of using $\ce{H3C-O-}$, we use $\ce{H3C-OH}$ we get (both $S_N2'$ and $S_N2$ paths are followed):
It seems to me that when we use $\ce{H3C-O-}$, only $S_N2'$ reaction takes place but on using $\ce{H3C-OH}$, both $S_N2'$ and $S_N2$ are equally taking place. What is the reason?

The second set of reactions is probably $S_{\mathrm{N}}1$. In general, it's quite difficult to tell between 2 and 2'. It's going to be fairly dependent on the specific conformation of the substrates, the nature of the solvent, and a long list of other factors.

The reason is the Pearson's SHAB (Soft Hard Acid Base) principle. In the first case , when you're using $\ce {CH_3O^-} $, it is a very hard base, and therefore a softer nucleophile. So it prefers to attack at the softer electrophilic centre. In the compound the $\ce {C} $ centre with $\ce {Cl} $ is the hard acid centre, whereas the $\beta $ positioned carbon to it i.e. the Carbon with double bond is the soft acid centre. That's why, the major product is $\ce {S_N2' } $.  But in the next case, $\ce {CH_3OH} $ is not a very hard base , nor a very soft base. So , both the reactions are possible.


text:
Is the application of Le Chatelier's principle wrong in this case?
In the reaction:
$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$
Suppose the system has 3 moles of $\ce{N2}$, 1 mole of $\ce{H2}$ and 1 mole of $\ce{NH3}$ at equilibrium. $Q_\chi=K_\chi=8.33$ and $Q_p=K_p$.
Now I add 5 moles of $\ce{N2}$ in the container by keeping (i) pressure and (ii) volume constant (temperature constant in both).
According to Le Chatelier's principle:

If pressure is constant: equilibrium shifts to backward direction.
If volume is constant: equilibrium shifts to forward direction.

But if I look at the equilibrium in terms of mole fraction, $Q_\chi$ will be the same in both the cases and hence equilibrium should shift in the same direction i.e. in backward direction.
Why do they predict different equilibrium states?
[$K_\chi$ is equilibrium constant in terms of mole fraction, $K_p$ is equilibrium constant in terms of partial pressures].

Try doing the calculation using the fact that $K_x$ actually is not constant at a given temperature.  It depends on total pressure as well as temperature.  Assuming you maintain ideal gas behavior $K_p$, and only $K_p$, is constant when temperature is fixed.  In terms of that equilibrium constant you should find the reaction following Le Chatelier's Principle both of your cases.

Le Chatelier's principle says that, for a system at equilibrium, a change in temperature, pressure or concentration of ONE of the components will tend to shift to counteract the change.
Consider first case ii, constant volume. If additional nitrogen is introduced into this volume, the concentration (pressure) of nitrogen will increase, while all other concentrations (partial pressures) remain the same. By Le Chatelier's principle, this single change tends to drive the reaction toward a higher concentration of NH3 (forward, not backward).
Keeping total pressure constant (case i) requires expanding the volume by a factor of two (3+1+1 moles increases to 3+1+1+5 moles). This changes ALL of the concentrations (pressures).
The concentration of N2 increases by a factor of (8/2)/3 = 4/3. This is 8 moles in twice the volume vs 3 moles in the original volume, thus favoring the forward direction. The effect of H2 is a factor of 8 in the backward direction (1/2 cubed) or 1/8 forward and the effect of decreased NH3 is a factor of 4 in the forward direction (1/2 squared). The overall effect will be 4/3 x 1/8 x 4 = 16/24 = 2/3 in the forward direction; i.e., the addition of extra N2 and volume expansion now favors the reverse direction, compared to the original situation direction.
Keeping the total pressure constant violates Le Chatelier's principle by changing more than one variable (concentrations of reactants and product). However, it is just a complication that can be handled by looking at the equilibrium constant; it's nice that the temperature remains the same.

In the constant volume case, the concentration of N2 does not increase by a factor of six. Before, there were 3 moles in a given volume, now there are 8 moles in twice the volume, so a ratio of 3 to 4, not 1 to 6. Also, adding the effects does not make sense, you have to multiply them.

@Karsten Theis: You are quite right! Thank you. I think I must have been using wrong numbers. But the multiplication rather than addition of effects is one for me to remember.

For this reaction:
$\Delta n=-2$
______________________________________________________________
(i) Adding an inert gas at constant temperature and pressure implies an increase in total volume, so:
$V_1<V_2$
Also at constant temperature and pressure, we have:
$Q_c=K_c*\left(\frac{V_1}{V_2}\right)^{\Delta n}$
Let:
$\Upsilon=\frac{V_1}{V_2}$
$Q_c=K_c*\Upsilon^{-2}$
$\ce{V1 < V2 -> \Upsilon < 1->Q_c > K_c}$
So equilibrium shifts to the left.
______________________________________________________________
(ii) Adding an inert gas at constant temperature and volume implies an increase in total pressure, so:
$P_1<P_2$
Also at constant temperature and volume, we have:
$Q_p=K_p*\left(\frac{P_2}{P_1}\right)^{\Delta n}$
Let:
$\rho=\frac{P_2}{P_1}$
$Q_p=K_p*\rho^{-2}$
$\ce{P1 < P2 ->\rho > 1->Q_c < K_c}$
So equilibrium shifts to the right.


text:
Resemblance of molecular orbital formed by mixing of s and p orbitals to hybridised sp orbital
See the following two images.

I think that the two orbitals should not resemble each other because s and p orbitals overlap in different fashions in the two cases.
In the case of sp hybrid orbital, both s and p orbitals are centred at the same point. As a result, s orbital does constructive interference with one lobe and destructive interference with another lobe of p orbital.
But in the case of molecular orbital, s and p orbitals are centred at different nuclei (and those nuclei can't be at the same point). Then how come the resultant molecular orbital resembles the sp hybrid orbital?
The right lobe of the molecular orbital should get enlarged due to constructive interference of s and p orbitals, but why does the left lobe get diminished? Shouldn't it remain as it is in the p-orbital?

It's important to realize that both of these drawings are highly stylized renderings of the electron density that are intended to provide a qualitative understanding. They don't accurately represent the actual shape of the orbitals. For example, $p$ orbitals are much rounder and taller relative to their length than we usually draw them. They are drawn elongated so that they don't cover each other up if you are drawing more than one on an atom. So the fact that the drawings of the $sp$ hybrid and the $p+s$ $\sigma$ bond look similar should only be taken to mean that they both have a bias of density to one side. You can't really conclude much else from the drawings.
The Orbitron website is a great resource for seeing the actual shape of the electron density of different types of atomic orbitals, which you can compare to the drawings.
MolCalc produces similar images of molecular-level orbitals. The bonding orbital of HF is probably the closest you can get to the $p+s$ $\sigma$ bond orbital in your drawing.

The center of the positive nuclei is the same.
Imagine a line between the two nuclei in the sigma bonding MO.
The nuclei in the sp hybrid orbital is to be found somewhere on this line.
This is the negative center of the orbital and naturally the spot where the positive nuclei will be. In the first picture there are two nuclei and their positive center (together) will be there too.
Hope I could help. English is not my native language.


text:
Thermodynamic stability of benzene derivatives
Q1. How can we comment on the thermodynamic stability of the following benzene derivatives?

Q2. What does thermodynamic stability mean, in general? What does it mean to compare the thermodynamic stability of more than compound?
Q3. How do we compare the thermodynamic stability of benzene derivatives? Is there a specific rule, or way to go about it? Would MOT (frost circles, to be specific) help?
I looked at the extent of resonance, and other such stabilizing phenomena/interactions. However, that only tells the extent of stabilisation of the compound due to those phenomena, not as a whole.
Edit: I wish to predict the order of stability using knowledge of chemistry, and not check enthalpy of formation data and then establish the order.

Q1, you cannot. Q2. It doesn't mean anything. Q3. You could only do that for specific types of interconversion reactions. || For your edit: there is only order with respect to the same variables. What you are looking for is a rule of thumb for some wishy-washy hand-wavy guessing game organic chemists love to play...

Wikipedia defines thermal stability as .

Thermodynamic stability occurs when a system is in its lowest energy state, or chemical equilibrium with its environment. This may be a dynamic equilibrium, where individual atoms or molecules change form, but their overall number in a particular form is conserved.

For the particular problem in hand the lowest energy state definition is more relevant for us .
Enthalpy of formation can be used for judging the energy states.
A  negative enthalpy change implies that the product has released heat thus lowering the amount of energy present in the product
This implies.

More negative the enthalpy of formation, more stable the compound.

Enthalpies of formation

Phenol
$\ce{ -> -165.0 kJ/mol}$.
Hydroquinone
$\ce{-> -371.1 ± 1.3 kJ/mol}$
P-Nitrophenol
$\ce{-> -207.1 ± 1.1 kJ/mol}$

All data has been taken from http://webbook.nist.gov , for solid state.

The order can be calculated now as
Hydroquinone $>$ paranitrophenol $>$ phenol

Great! How can I PREDICT the order of enthalpy of formation, though?

@schrodinger_16 That's a question by itself, but I would go about seeing the number of bonds and how electronegative the constituent atoms are.

Why, though? How does that help? Is it a general approach?

Creating a bond releases energy , more electronegative the atoms greater the lattice enthalpy . It's a very general approach. Read chemical bonding again it's all there .

Thermodynamic stability of compounds can be determined by obviously enthalpy of formation ($\Delta H_{_\mathrm f}$) of individual compounds. The enthalpy of formation will be lesser if the compound is formed from its constituent elements enjoys some greater stability.
Among the above given compounds, considering any extra stability of the compounds, their thermodynamic stability can be guessed.
For phenol, when 1 mole of it is formed there can be slightly weak intermolecular Hydrogen bonding($\ce{C6H5-OH...OH-C6H5)}$. The hydrogen bonding is slightly weak because the lone pair of oxygen atom is delocalised with the benzene ring thus creating a $\delta$+ charge on the $\ce O$ atom, due to which the dipole dipole interaction between $\ce H$ of one $\ce{OH}$ and $\ce O$ of other phenol molecule is weakened. Thus, phenol enjoys only weak intermolecular hydrogen bonding.
For 4-nitrophenol, there is stronger intermolecular hydrogen bonding, which can exist between two phenol molecules as:

Due to enhanced no. of hydrogen bonds, the compound enjoys more stability than ordinary phenol and therefore its enthalpy of formation is more than phenol.
For hydroquinone, the number of hydrogen bonding is even more. One molecule can have a hydrogen bond with four other molecules through four hydrogen bonds as shown:

Thus the molecule of hydroquinone enjoys much more stability due to large number of hydrogen bonding, making its thermodynamic stability even more than p-nitro phenol.
Thus the stability order will be $$\text{hydroquinone > 4-nitrophenol > phenol}$$

That is alright, but isn't that just the extent of hydrogen bonding?

How are you using that to compare thermodynamic stability on an absolute scale?

Extent of Hydrogen bonding can give us only the RELATIVE STABILITY  between them. I am only saying that if you have such compounds you can JUST COMPARE their stability based on these observations. But It will not help to produce those numbers as said in another answer. The best way to logically deduce might be  what I have written.  But I myself dont really know what the values of$\ce { $\Delta $ H_f} $ are. I tried to logically figure out what might be their orders.


text:
Why are allyl halides reactive towards both SN1 and Sn2 reactions?
I know that allyl halides are reactive towards SN1 reactions due to the stabilization of carbocation due to resonance but why are they reactive towards SN2 too?

I actually read that allyl halides are more reactive towards SN2  than primary alkyl halides, whom I thought were most reactive.

Can you add what you just said above to your question?

In a reaction, SN2 as well as SN1 reactions take place. It only matters that which reaction pathway gives the major product. The answer to your question depends on the type of base used, the substrate and the overall stability of the product. Allyl halides are also reactive to SN2' reactions, i.e allylic nucleophilic substitution reactions. There are a lot of possibilities.
Keep in mind that the electronegativity order of hybridized carbons is $\ce{sp > sp2 > sp3}$. So in an allylic halide, such as $\ce{CH2=CH-CH2Br}$, the alpha carbon is much more partially positive than the alpha carbon in $\ce{CH3-CH2-CH2Br}$. Therefore, the attacking nucleophile will be attracted to the alpha carbon more in the allylic halide compound than in the primary halide. Hence, allylic halides tend to follow SN2 pathways more than primary halides.

Can you explain why are they more reactive than primary halides in SN2 reactions?

@KirtiAgrawal I have edited my question with the answer. I hoped it helped you! But next time please give one example at least and describe your problem instead of making your question so broad. This will help us answer you much better!


text:
Which out of the two has greater bond angle? Water or diethylether.Why?
My theory:
The ethyl groups in diethylether being larger in size when compared to hydrogens in water, should be repelled more by the lone pairs on the central atom oxygen. Hence, the ethyl groups should come closer, thereby decreasing the bond angle.
Moreover, the electronegativity difference between the central atom and the peripheral atoms is more in case of water i.e., $\ce{O-H}$ electronegativity difference is more than that in $\ce{C-O}$ as per Pauling’s scale. Hence, the oxygen atom should pull the bond pair closer, thereby allowing to open up the $\ce{H-O-H}$ bond increasing the bond angle.
Hence bond angle of water should be greater than that of diethylether. Am I right?
PS: I know that in reality dimethylether has greater bond angle than water.

You are mostly correct with your reasoning , and the $\ce{C-O-C}$ bond angle was indeed expected to be lesser than that of $\ce{H-O-H}$ bond angle of water . But it's all theoretical. Experimentally the opposite was observed and was explained on the basis of steric hindrance or steric repulsion of the two bulky $\ce{C2H5}$ groups.
In case of $\ce{(C2H5)2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two ethyl groups. They repel each other and as a result , they do not let the lone pairs of Oxygen compress the bond angle more .
$\ce{H-O-H}$ bond angle is found to be $104.5^\circ$ whereas the $\ce{C-O-C}$ bond angle in diethyl ether is found to be roughly close to $110.5^\circ$. It can be attributed to the steric repulsion that trumps the bond pair-bond pair and lone pair-lone pair repulsion in this case.


text:
How does iodide replace bromide when the latter is more basic?
The following reaction follows the $S_N2$ pathway: $$\ce{CH3CH2CH2-Br + I- (presence of DMSO) -> CH3CH2CH2I + Br-}$$ I found this reaction in my textbook and according to me this reaction should not even take place. $\ce{Br-}$ is a stronger base than $\ce{I-}$ and keeping in mind, stronger the base, lesser is the leaving ability, why will $\ce{I-}$ replace $\ce{Br-}$ in the first place?

As you suggest "$\ce{Br−}$  is a stronger base than $\ce{I−}$ and keeping in mind, stronger the base, lesser is the leavability, $\ce{I−}$ should not replace $\ce{Br−}$ "
I actually agree with the logic and according to me, infact, the reverse reaction, i.e.
$$\ce{CH3CH2CH2I + Br- (presence of DMSO) -> CH3CH2CH2Br + I-}$$
should be much more favourable. Because we already know, that in the aprotic solvents, the nucleophilicity of halides is  $\ce{F− > Cl− > Br− > I−}$, so $\ce{I−}$ should be a better leaving group and $\ce{Br−}$ should be a better nucleophile .

This reaction would be similar to Finkelstein reaction. Here instead of using acetone, you are using $\ce{NaI}$ in DMSO.
We usually use $\ce{R-Cl}$ in a non polar solvent in Finkelstein reaction, not only for increasing nucleophilicty of the attacking reagent, but also to precipitate the $\ce{NaCl}$ formed in the reaction.

The increasing order of solubility in non polar solvent for sodium halides is:
  $\ce{NaF < NaCl < NaBr < NaI}$

The order is similar for potassium halides. This order is due to polarisation. In non polar solvents, $\ce{KI}$ is soluble but $\ce{KCl}$ is not.
This drives the reaction forward to completeness.
The difference in solubility in $\ce{NaI}$ and $\ce{NaBr}$ (or the potassium derivatives) is not very appreciable, so the reaction will take place in the forward direction, but will not give a very good percentage yield of product.


text:
Are intermolecular forces a type of chemical bond?
My chemistry teacher told me that chemical bonds are of two types: intramolecular and intermolecular. He said that intermolecular forces come under the category of intermolecular chemical bond.
I have never read such statement anywhere. Nor can I find anything on the Internet that would support this statement.
My understanding is that chemical bond is a force that holds atoms together in a chemical species. Since intermolecular forces do not hold atoms together, they should not be termed as chemical bond.
So, are intermolecular forces a type of chemical bond?

The IUPAC definition of "chemical bond" is:

When forces acting between two atoms or groups of atoms lead to the formation of a stable independent molecular entity, a chemical bond is considered to exist between these atoms or groups. The principal characteristic of a bond in a molecule is the existence of a region between the nuclei of constant potential contours that allows the potential energy to improve substantially by atomic contraction at the expense of only a small increase in kinetic energy. Not only directed covalent bonds characteristic of organic compounds, but also bonds such as those existing between sodium cations and chloride anions in a crystal of sodium chloride or the bonds binding aluminium to six molecules of water in its environment, and even weak bonds that link two molecules of $\ce{O_2}$ into $\ce{O_4}$, are to be attributed to chemical bonds.

So the answer is "yes" in some cases.
See also the IUPAC definition of "molecule":

... must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state

So, for example, a water-water dimer, held together by hydrogen bonding, has a monomer-monomer potential energy surface that is deep enough to confine at least one vibrational state, and it would be appropriate to refer to the hydrogen bond as a chemical bond.

But if we ignore the last sentence for a moment, the whole definition indicates that intermolecular forces should not be considered as chemical bond. Phrases such as 'stable independent molecular entity' and 'existence of a region between nuclei...' make an image that chemical bonding is an intramolecular phenomenon. Anyway, the last sentence sidelines all this and makes it clear that intermolecular forces are indeed chemical bond.

@AumkaarPranavShukla for IUPAC, once the force is enough to confine at least one vibrational state, the definition of "molecule" is met.  For example, a water dimer, held together by hydrogen bonding, has monomer-monomer vibrational states, so it is considered a stable independent molecular entity.


text:
Why does reducing pressure cause the equilibrium to shift towards the side with less moles?
Take the example:
$$\ce{N2(g) + 3H2(g) <-> 2NH3(g)}$$
I understand that if I increase the pressure of the system, it'll shift towards the $\ce{NH3}$ side. This is because of the reaction rate increasing (increased collisions) more for the forward reaction than the reverse reaction. What I don't understand is why a decrease in pressure of the system would shift the equilibrium to $\ce{N2(g) + 3H2(g)}$. There is overall less collisions, but there's still more moles of reactants to collide and less molecules for $\ce{NH3}$ to collide and decompose.

When you decrease the pressure, the reaction will try to keep the pressure up i.e try to increase the pressure. This is only possible if the reverse reaction takes place which is increasing the number of moles so that more collisions can occur thereby increasing pressure.

Actually, the shift of reaction towards left on decreasing pressure (and towards right on increasing pressure) is due to Le Chatelier's Principle, which states that if a change is brought in the equilibrium conditions of a reaction, the reaction will proceed in such a manner that it counteracts the change.
In case of increasing pressure the reaction shifts to right due to lesser number of moles on right. And according to gas equation, lesser moles means lesser pressure. The opposite happens when the pressure is decreased.

So, it is essentially less collisions = less pressure and vice versa. Initially when the pressure is decreased, does this mean that the reverse reaction occurs faster than the forward reaction until equilibrium is reestablished?

@CipherBot Yes, definitely. It no longer remains in equilibrium, until a new one is reestablished. Also in new equilibrium, the concentration of the reactant would be greater than that in old one.


text:
Wieland's method for triphenylamine bromination
I'm looking for a Wieland's paper on triphenylamine bromination. I've found other papers referencing it like this but I don't find the actual paper.
Does someone know where could I find it? Or whats Wieland's method all about?

From the reference you supplied the Wieland paper is Chem. Berichte 40 (1907) 4278
The bromination method (3 eq of Bromine in chloroform slowly added to a chloroform soln of triphenylamine) in the experimental section of the reference you supplied appears to be Wieland's method - it is referenced as such

Thanks ! Side question: Do you know what is (or where I can find) the mecanism for this reaction ?

Straightforward electrophilic substitution by Br+. After the first Br goes in that phenyl is less activated towards a second substitution so the 2nd phenyl reacts preferentially, then the third.


text:
Deciding the aromaticity of two napthalene-like rings
Why is the first compound aromatic and the second compound non-aromatic?
In the first compound, the left side of the ring is aromatic but the right side isn't.
Second compound satisfies all the conditions of aromaticity but why is it non-aromatic?

I think there's something to do with planarity. I don't think the second compound exists in the form you drew it in. Did you think about that factor?

Duplicate of: https://chemistry.stackexchange.com/questions/75688/is-1-2-dihydronaphthalene-aromatic and https://chemistry.stackexchange.com/questions/86734/aromatic-compounds


text:
Why is 15,16-dihydropyrene aromatic?
Why is 15,16 dihydropyrene aromatic?

Here the middle two carbons are $\ce{sp^3}$ hybridized. Then how is it aromatic? Do we only consider the conditions for aromaticity to apply only at periphery in case of fused rings?

How is toluene aromatic when it has a saturated carbon too?  You need to focus on the conjugated atoms, everything else is just a ligand -- even if it's located in the interior.

https://chemistry.stackexchange.com/a/15521/9961

https://chemistry.stackexchange.com/questions/86975/is-this-compound-aromatic

While doing calculations for Huckel's rule, in polycyclic compound only peripheral pi electrons are considered for aromaticity.
So, 7 double bonds in the periphery i.e. 14 pi electrons.
14=4(3)+2
Huckel's rule satisfied. Aromaticity achieved.
Why are only peripheral pi electrons considered?
The reason is that only these elections are participating in delocalization for conjugation. If multiple bonds occur in the centre like in pyrene these central bonds will not be localized. Thus, only peripheral pi electrons are considered in conjugation.

Therefore, first 2 compounds in this image are aromatic and last one is weakly Anti aromatic or Non aromatic.
We can't commit compound to be aromatic on ONLY Huckels rule with this compound must be

Cyclic
Planar
Complete cyclic conjuction
Huckel's rule

Here in 3rd compound,
Due to crowding of Interior Hydrogen compound loses its planer nature. Thus, not called to be Aromatic.
(I have attached 3d structure of cyclotetradeca-1,3,5,7,9,11,13-heptaene (3rd compound)
Please refer it.)

@AniruddhaDeb I would disagree there, AFAIK it is not so clear cut. [14]Annulene has structures which are considered to be aromatic. "... the C2h structure has an aromaticity index of 1.71 and the D2 structure an index of 1.70. Since benzene with an aromaticity index of 1.75 is the prototype of an aromatic system, these two structures must be called aromatic". The other structures are usually taken to be weakly aromatic. Thus, I wouldn't go out and say that [14]Annulene is known to be non-aromatic, it's probably somewhere in the middle.


text:
Electrophilic or nucleophilic addition?
In this reaction:
$$\ce{H2C=O + H2O -> H2C-(OH)2}$$
My textbook says it's a nucleophilic addition reaction.
 But in this reaction:
$$\ce{H2C=CH2 + HBr -> H3C-CH2Br}$$
is an electrophilic addition reaction.
How do we know when a reaction is via an electrophilic or nucleophilic addition?

Additon to aldehydes or ketones are always Nucleophilic. I have rarely seen any electrophilic addition to aldehydes or ketones. But addition to C=C , i.e. alkenes are mostly electrophilic.

I agree. The C=O bond is strongly polarised making the C receptive to nucleophiles

Why can't we say "making the O receptive to electrophiles ". @Waylander Please help me out and thank you so much for your time!

You can say that. There is a well-established procedure for making carbonyl groups more reactive by premixing them with Lewis acids e.g. BF3 before adding a nucleophile to attack the C

Learning organic chemistry is about learning key modes of reactivity and being about to identify key features of reactants that may trigger relevant modes.
Relevant examples here are:

Nature of orbitals: Carbonyl groups have a low lying $\sigma^{*}$ orbital that is quite susceptible to nucleophilic attack. Alkenes, on the other hand, do not. What alkenes do have is a lot of electron density, so they are susceptible to protonation to form cations.
Acidity. Water is very mildly acidic while hydrogen bromide is quite acidic. Therefore different modes of reactivity should be considered.

The real question here is: Why are you trying to compare these two reactions at all? They're using different reagents with very different functional groups. Naturally, very different mechanisms were involved between the two reactions.

I came across these two questions in my textbook and I just wasn't happy with the answer my teacher gave me. Thank you for your time!

As Soumik Das correctly pointed out , "Additon to aldehydes or ketones are always Nucleophilic."
In your case , this can be viewed as ,
$$\ce{H2C=O}$$ has a carbon attached to oxygen , which is fairly electronegative . As a result , a partial positive $(\delta^+)$ charge gets created on the carbon.As a result , the lone pairs on the oxygen atom of $\ce{H2O}$ get attracted towards the slightly positive carbon and carries out nucleophilic addition reaction.
I hope you can carry it from here now.

So does a partial negative charge develop on the oxygen and could attract the H+ from the water molecule, making it an electrophilic addition reaction.

@raajsuriya it could definitely take place, but the step here which ultimately leads to the product formation of the more stable product is the nucleophilic addition.Please upvote and accept the answer if you found it helpful.

Both products are the same, aren't they? @Tanuj

I gave it an upvote alright but I can't accept the answer because it doesn't really help or answer my question. Thank you so much for your time though. @Tanuj

@raajsuriya both products are the same but the product is formed through a transition state which is more stable (in which activation energy is less) and that is nucleophilic addition and not eletrophilic addition.


text:
Diffusion rate of gases
What is the effect of temperature on diffusion rate of gases? Do the diffusion rates of gases decrease with an increase in temperature? Why?
I just read it somewhere that the diffusion rate of gases decreases when temperature increases. So i just wanted to know if its true.

Temperature is directly proportional to diffusion rate.
Diffusion, being the dispersion of molecules throughout a space, is affected by the speed of molecules and the density of the space through which the molecules are being dispersed.
An increase in temperature increases the speed at which molecules move at. This is because the average kinetic energy of the gas molecules is proportional to the absolute temperature of the gas. Increased speed will yield faster and more even diffusion.
Average Kinetic Energy of gas molecules = $\frac{3}{2}RT$.
Therefore, decrease in temperature will cause molecules to slow down (due to reduction in average kinetic energy). Hence rate of diffusion decreases.
I hope this helps!

@MollyCool I want to have a discussion with you on this . Kindly join https://chat.stackexchange.com/rooms/73715/gas-rate

No worries. Will join now @Tanuj

This should be a comment but I don't have enough reputation to comment yet.
Diffusion coefficient, as defined by Fick's equation
$$
D_\mathrm {AB} =\frac{\text{flux of mass diffusing } (\pu{kg m^-2 s^-1})}{\text{concentration gradient  }(\pu{kg m^-4})} =  -\frac{J_\mathrm A}{\mathrm dC_\mathrm A/\mathrm dx}
$$
increases with temperature as pointed out by @MollyCool.
Your doubt might come from the fact that gas viscosity increases weakly with increasing temperature. And you may know that diffusivity is inversely proportional to viscosity. Note that the increase of kinetic energy of gas molecules has a much larger effect than the weak increase of viscosity and, overall, gas diffusivity will increase with temperature


text:
Why are nitro compounds with at least one alpha-H soluble in NaOH?
Are all the nitro-compounds having at least one alpha-$\ce{H}$ soluble in $\ce{NaOH}$? And what does solubility has to do with alpha-$\ce{H}$?
I was reading about tautomerism today and got stuck on this thing. Is it due to its conversion to the enol form? Does the keto form also show this? What could be the possible reasons?

You should provide more details to your question , show some effort , and ask us where you got stuck and whats your doubt. Please note this is not a homework site !

I was reading tautomerism today and got stuck on this thing. Is it due to its conversion to enol form? They also wrote that enol form is acidic. How is that possible when enol form contains one -OH group?

@kirtiagarwal a compound containing $-OH$ group isn't necessarily basic . See it this way. If some strong base in the solution is able to abstract $H^+$ from the $-OH$ of enol , the enol would be termed as an acid , not a base.

@kirtiagarwal Add more details for more people to have a look at this question.

I think that it is only short chain nitro groups that are aq NaOH soluble

Might help lie here? Google: Evans pKa

https://chemistry.stackexchange.com/questions/44420/reaction-of-nitro-compounds-with-strong-alkali

The compounds containing nitro groups when dissolved in basic medium, the following things can happen.The base can take the $\ce{$\alpha$-H}$ of the nitro-group, which will create a negative charge on that $\ce{$\alpha$-C}$, which can further delocalise with the Nitro-group (similar to Nitro-Acinitro tautomerism).
So, the whole compound will be in the state of a Resonance Hybrid where there will be partial double bond character between $\ce{$\alpha$-C, N }$ and two $\ce O$ atoms. and also there will be a $\delta$- charge on the $\ce{$\alpha$-C }$ atom as well as on the $O$ atoms. , and $\delta$+ charge on the $N$ atoms.Due to such salt like compound formation, it becomes soluble. This $\delta$- charge created on the Carbon atom is stable if the side chain is short, but starts to get destabilised as the side chain grows longer. So, I think the solubilities of the Nitro-compounds decrease as the side chain grows longer, and the compound with $10-12$ C atoms may not be at all soluble in base because of such large Hydrophobic part and destabilisation of the $\delta$- charge.

Didn't I put you a link to a page about proper formatting of posts on Chem.SE? Even when edit is started you automatically can see some tips, like using enters instead of breaklines...


text:
Sodium and potassium produce a strange liquid, why the liquid explode when it gets contact with water?
I saw an experiment sodium and potassium mixed together to produce a liquid alloy  but when the liquid get contact with water then some kind of explosion occur.
Why the explosion occur when the liquid  made contact with water?
Same explosion will occur if it would get contact with other form of liquid?

This site provides an explanation about the mechanism of the explosion.

...causes an explosion because the metal dissolves, generating an
extreme amount of heat and transferring electrons to the water. The
dissolution step also generates steam and forms hydroxide ions and
hydrogen, which can be ignited, making the process even more
energetic.
Some researchers have puzzled over how the process can occur so
quickly, though. They recognized that the steam and hydrogen generated
early on in the reaction should form a buffer layer over the metal
surface and impede water from continuing to react.
...within a fraction of a millisecond of making contact with water,
the Na/K droplets form numerous spikes that protrude into the water.
Molecular dynamics analysis indicated that nearly instantaneous
transfer of electrons from the spikes to the water rapidly generates
positively charged alkali ions, which vigorously repel and cause a
so-called Coulomb explosion. It is the speedy manner in which that
process propagates and generates reactive metal surfaces that triggers
the overall explosion.
The researchers have figured out many of the key aspects that enable
this highly exothermic reaction to become explosive, rather than
self-quench.

It does not generally reacts with any other liquids. Hence they are stored in mineral oil.

It is highly reactive with water and is stored usually under hexane or
other hydrocarbons, or under an inert gas (usually dry nitrogen or
argon) if high purity and low levels of oxidation are
required.(Wikipedia)


text:
Boiling point trend in group 13
My Theory:
Since atomic mass increases down the group, the van der Waal's forces should also operate to a greater extent, thereby making it difficult to change the phase of the substance. Hence, boiling point should increase down the group from boron to thallium.
But in my book, the answer given is that the forces of attraction decrease down the group because of the increase in size of atoms down the group. Hence, the boiling point decreases down the group.
I'm confused. Even though the size of the atom increases, shouldn't the boiling point follow the same trend of increasing because atomic mass increases?
 Moreover, due to poor shielding of d-electrons the increase in size down the group is very minimal.

Experimental data: (source) shows a regular decrease in boiling point down the group
$$
\begin{array}{|c|c|}\hline
\text{Element}&\text{Boiling point (K)}\\\hline
\ce{B}&\pu{4200K}\\\hline
\ce{Al}&\pu{2743K}\\\hline
\ce{Ga}&\pu{2673K}\\\hline
\ce{In}&\pu{2345K}\\\hline
\ce{Th}&\pu{1746K}\\\hline
\end{array}
$$

EDIT:
Can we explain the above trend with metallic bonding present in these elements? Boron has its highly networked structure so let us leave it alone. As we move down the group, size of the element increases. This increase in size, decreases the strength of metallic bonding as the same number of valence electrons now have to hold a larger cation. Metallic bonding is viable in liquid state too. So could this be the reason?

How are the Van der Waal forces related to atomic size?

Down the group, with increasing amount of electrons, there can be more correlated motion and hence there is a stronger interaction between the molecules right? @JonCuster

I spent like an hour the only useful things I got was that boron a network  structure that's why it has a high boiling point and a shady answer on Quora which said that boron is small in size so more attraction. I would ignore the latter.

Oh! Thanks for your efforts. I know boron compounds being electron deficient exist as polymers but don’t quite get how that will influence boiling point of atomic boron. (Which is about 4200K)

London forces are the forces that you should always consider in the end. They're like the slow pawns on the chessboard, their effect is negligible when the fast rooks and the industrious queens are around!
In this case, our queens are the different structures of these elements. Let me first tell you the fact that even the melting point of boron ($\pu{2349K}$) is more than the boiling points of thallium and indium!
So, first off, why is the melting point of boron higher than that of all other group thirteen metals? From this Wikipedia page, its α-rhombohedral allotrope has B-12 icosahedral units arranged in a crystalline structure, and connected by strong covalent bonds[a]. As a comparison, the other metals in their solid state have unit cells whose atoms are bonded by only weak metallic bonding. The larger the atoms get, the weaker the metallic bonding gets, and hence, the melting point decreases down the group.[b]
Coming to the trend in boiling points now. Part of the difference in the trend has to arise from the excessively large melting point of boron. You see, thallium boils off even when boron has yet to melt, so that is a contributing factor in the excessively large boiling point of boron over the others.
According to this paper[1], for liquid thallium,

The coordination number changes from 11.5 atoms at $\pu{350^\circ C}$ to 10.9 atoms at $\pu{700^\circ C}$

This is said to be due to:

creation of vacant spaces about an atom. Although in liquid thallium the interatomic distances remain unchanged, some of the atoms are moving farther apart within the first coordination shell.

Another paper[2] however shows that, while CN of liquid indium decreases with increasing temperature, their inter-atomic separation also decreases. We can interpret this to be a "reinforcement" that helps increase the boiling point of indium over thallium.
With indium and thallium cleared, notice that the boiling points of indium, gallium, and aluminum are almost identical. Sadly, I've been unable to find any research papers dealing with these three elements in liquid state near their boiling point. For example, this paper[3] deals with liquid aluminium, but only upto $\pu{1273K}$.
We do have some data for boron. You'll be surprised how strong its covalent bonds are. In fact, this paper[4] states that, even at as absurdly high temperatures as $\pu{2600K}$, though the "icosahedra are destroyed", the "atoms still form an open packing with sixfold coordination". Moreover, this paper[5] observed that the density of liquid boron, at temperatures above $\pu{2360K}$, is higher than that of solid boron, which is unlike aluminum and indium (though for gallium the density at melting point is still slightly higher). Another paper[6] has measured the properties of liquid boron, but that sadly only at around $\pu{2400K}$.

I have checked a dozen best-matching research papers, and this was all the relevant data I could pull out. This isn't the complete answer, but again, without any proper data in the first place, there really cannot be a proper qualitative analysis either. So, this answer is really all there is.

References:

Temperature Dependence of the Structure and Transport Properties of Liquid Thallium, N. C. Halder and C. N. J. Wagner, The Journal of Chemical Physics 45, 482 (1966); https://doi.org/10.1063/1.1727593
Temperature Dependence of the Structure of Liquid Indium, H. Ocken and C. N. J. Wagner, Phys. Rev. 149, 122 – Published 9 September 1966
The structures of liquid mercury and liquid aluminum, P. J. Black and J. A. Cundall, Acta Cryst. (1965). 19, 807-814 https://doi.org/10.1107/S0365110X65004383
Structural and electronic properties of liquid boron from a molecular-dynamics simulation; N. Vast, S. Bernard, and G. Zerah; Phys. Rev. B 52, 4123 – Published 1 August 1995
Millot, F., Rifflet, J.C., Sarou-Kanian, V. et al. International Journal of Thermophysics (2002) 23: 1185. https://doi.org/10.1023/A:1019836102776
Structure of Liquid Boron; S. Krishnan, S. Ansell, J. J. Felten, K. J. Volin, and D. L. Price; Phys. Rev. Lett. 81, 586 – Published 20 July 1998; https://doi.org/10.1103/PhysRevLett.81.586

[a]: actually "the icosahedra are connected via a combination of two-center and three-center bonds" as claimed in [6], but suffice to say that they are almost as strong as the covalent bonds, if not more
[b]: except a few exceptions

How does inter-atomic separation decrease in indium but not in thallium when CN decreases?

@MollyCooL Good question! Unfortunately, the linked paper does not explain why this is so. However, they do note that this is also the case with tin (as detailed by Furukawa et. al) - the inter-atomic separation of tin remains constant upto $\pu{250^\circ C}$ over its melting point.

So, is it all experimental data and no common reason to explain trends like in other groups?

@MollyCooL I would say that the other groups are lucky to have good-looking, easy-to-digest explanations. Unfortunately, as the anomalies (like allotropes) start entering our picture, it gets blurred and tough to justify by theoretical claims.


text:
Is borazine aromatic?

I came across this structure which has got pi bonds only in its canonical forms (which are also unstable compared to the Ist structure), is it aromatic?
Also, is it compulsory for an aromatic compound to contain a carbon ring? Can't a conjugate system without carbon be aromatic (as in this case)?

The modern definition of aromaticity from deep theoreticians is that the π-system needs to support aromatic ring currents. Borazine can support it, so it is technically aromatic.
Aromatic systems that do not contain carbon are not really all that common, but they are known. Pentazole was detected, for example. The problem is, that for aromaticity to be a thing we need a π-system, and π-systems effectively restrict us to BCNOS elements, since other elements are not eager to form π-bonds. Aromatic systems are usually 5- or 6-membered rings. Add in the fact that rings with a lot of N/O atoms often happily go kaboom after a funny look, and it's not surprising we have very few actual examples here. Still, some people insist on doing unreasonable things with results from as far back as the 50s.

Borazine is slightly controversial, I think the general consensus seems to be aromatic, but not very much. e.g. NICS values borazine -2.1, benzene -11.5.

Yes, Borazine is aromatic.
According to Hückel’s rules, a compound is aromatic if it satisfies three conditions:

Planarity
Complete delocalisation of π-electrons
Presence of $(4n+2)$ electrons where $n$ can be $1, 2, 3, \dots$.

Borazine is aromatic as there are 3 nitrogen atoms in the structure of borazine that donate two electrons each. Boron doesn't donate any, but it has an empty p-orbital available through which the electrons can delocalize. Hence, in total it has 6 π-electrons. Though the delocalisation is lesser than that in benzene, it still is aromatic.
FunFact: Borazine is also called inorganic benzene because of its resemblance to benzene by structure.

Strictly speaking, Hückel's rules is derived from hydrocarbons, and the generalisation to other compounds is neither straight forward, nor necessary correct. Just because it a has been applied before, it does not mean it is founded in rigorous theory.

Oh! Alright. Noted! So aromatic ring currents is the only theory to confirm aromaticity for a $\pi$ system? (Systems that don’t have carbon)


text:
Why is the speed of ideal gases mass dependent while the kinetic energy is not?
The rms speed of an ideal gas is $v_{\text{rms}}$ = $\sqrt{\frac{3RT}{M} }$ and the kinetic energy is $E_\text{k} = \frac32RT$. From this, it is concluded that the speed is mass dependent, while the kinetic energy isn't.
This doesn't make sense to me. I know the speed and kinetic energy for things other than ideal gases are surely mass dependent. The heavier an object is, the slower it is and will have a bigger kinetic energy. Why isn't this true for ideal gases?

As the mass increases, the velocity decreases, yes that is true. But how does the kinetic energy remain the same?
The increment in the mass is balanced by the decrement in the velocity. In other words, the velocity changes as the mass changes in such a manner that the kinetic energy of the molecules of the gas will remain to be same no matter what. It's something like the multiplication of two variables always giving the same constant value. Keeping in mind that this is true in an ideal environment.

I think that the answer should be pointed out that this assumes ideal gas behavior.

I added that last bit. Thank you @MaxW. Since his question was "Why isn't this true for ideal gases?" I assumed I was writing for ideal conditions.

I didn't claim that the kinetic energy doesn't change but that is is independent on the gas mass. as the equation suggests Ek = 1.5RT the only thing that influences the kinetic energy is the temperature.

@ILoveIL Yes, only temperature influences kinetic energy.  What is the issue?

@ILoveIL Two different gases in two separate vessels at the same temperature will have same kinetic energy but not the same Vrms.  The difference in masses compensates for different Vrms values.  Could you reframe your question again? So that I can understand your problem and help you solve it.

We start with two samples of pure gas, one containing molecules of A (the larger molecule) and one containing molecules of B (the smaller molecule) with the same average speed (it will turn out that gas A has the higher temperature). If you mix these samples, their kinetic energy (per degree of translation freedom) will be different at first. However, they will quickly achieve thermal equilibrium. This means that the energy transfer in collisions between molecules of A and molecules of B average out. In a collision between a bat (molecule A) and a baseball (molecule B) coming at each other at the same speed, there is a kinetic energy transfer from bat to ball (the bat slows own a bit, the ball speeds up - if we assume an elastic collision).
Here is a simulation that shows one diatomic molecule and many monoatomic molecules. The diatomic is a bit slower on average.

Source: http://physics.weber.edu/schroeder/md/

imo, the confusion has nothing  to do with ideal gases at all. For any object, the velocity and kinetic energy are NOT always dependent on mass.. It depends on what is kept constant in the comparison. If I accelerate a proton and electron through a p.d, their k.e is independent of mass whereas velocity is not. If I put a proton and an electron in a g-field, their velocities are independent of mass, whereas k.e is not.

The answer can be
$$Speed \propto1/√m$$
And$$ K.E. = 1/2mv^2$$.
Finally m will cut, and therefore K.E. does not depend on mass.

This directly follows from the formula. I believe the OP was asking for an intuitive approach rather than a formula-based approach.

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


text:
Non-ideal velocity distribution of gases
A recent question about the velocity distribution of ideal gas molecules got me wondering. How much would the velocity distribution vary for non-ideal gases?
I was wondering about a mixture of ethane and helium. The ethane molecule would have vibrational modes that the helium atom wouldn't have. Given that the helium velocity was "ideal," how non-ideal could/would the ethane velocity be?
Some other molecule been experimentally analyzed like this? I guess that a molecule with even more atoms would have even more vibrational/rotational modes.

Vibrational modes alone would not make the velocity distribution non-ideal.

@IvanNeretin In ethane, the forces of attraction will obviously be more than that for He. Wouldn’t this make it a non-ideal situation?

A non-ideal situation? Yes. But a non-ideal velocity distribution? Why?

@IvanNeretin I’m confused. By “ideal” velocity distribution OP means Maxwell’s velocity distribution right? If it is so, that was derived for ideal gases and there will be deviations for non ideal situation.

According to Wikipedia: >"The Maxwell–Boltzmann distribution applies to the classical ideal gas, which is an idealization of real gases. In real gases, there are various effects (e.g., van der Waals interactions, vortical flow, relativistic speed limits, and quantum exchange interactions) that can make their speed distribution different from the Maxwell–Boltzmann form."

To begin with, ideal gas molecules don't have size and don't collide with each other, hence they definitely would not establish Maxwell’s velocity distribution. (I don't care if Wikipedia says otherwise.) Sure, it was derived from certain approximations, but these were not what we normally call an ideal gas. As for the rest, well, OK, there would be some marginal difference for the molecules which are currently under the influence of (read: in a collision or a close fly-by with) other molecules, but these form only a tiny fraction of all molecules at any given moment.

Boltzmann's statistics (which do not exist) are used in calculating the velocity distribution,  however, if low pressure or high temperature are assumed then Bose-Einstein/Fermi-Dirac statistics revert to Boltzmann.  The (translational) partition function is calculated and from this the velocity distribution.  If your two species do not interact with one another then the two distribution are independent and will only differ due to mass.  Since equilibrium  exist,  if one type of molecule interacts  to form dimers eg. as $\ce{A_2 <=> 2A}$  then we might consider them as two species.


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Why boiling point of water is much higher that expected from the diagram?
Explain why the boiling point of water is much higher than expected based on the size of the molecule. Has it got to do something with the intermolecular forces ? Please explain.

Hi lizzy,  welcome to Chem.SE! We require you to show your efforts on this question. Why do you think so.?Please add this to your question. Thanks! “Homework questions must demonstrate some effort to understand the underlying concepts.” This is our policy.

If you still don’t have an idea, I can give you a clue.

A water molecule has a small size, and its high boiling point is indeed anomalous with the general trend of boiling points of molecules of similar size as water molecule.
Now the high boiling point of water molecules can be explained on the basis of $\ce{H}$ bonding that takes place among the water molecules in liquid state. A water molecule can form a maximum of $4$ $\ce{H}$ bonds. This increases the effective size of water molecules too, which in turn increases the van der Waals forces operating in water molecules.
The two reasons for high boiling point of water are -

Formation of $\ce{H}$ bonds, which requires more energy for breaking the 'extra' bonds.
Increase in van der Waals forces, as a result of which we need to supply more temperature for water to boil, or change from liquid state to vapour state.

Because it can form hydrogen bonds between its molecules.
The greater the forces of attraction, the higher the boiling point, or the greater the polarity, the higher the boiling point. Water molecules in liquid state have the ability to form hydrogen bonds with each other. These hydrogen bonds are some of the strongest of all intermolecular forces, so a large amount of energy is needed to break these interactions −  this is the main reason why water has such a high relative boiling point.


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Conversion from t-butanol to alkene

How can I accomplish the above conversion?
My thoughts:

One way of introducing the double bond, would be using the appropriate Grignard Reagent after oxidation of t-butanol (somehow)
I think that initial conversion to carbonyl compound is necessary - only then can we follow up with nucleophilic addition.

P.S. The question was in the organometallics section of my assignment, possibly hinting that reagents apart from the above species may be used. However, won't it be great if we could design a mechanism using only the starting species and commonly available reagents?

Do we have to use the same species given , or some other species could be used ?

I suppose you may use some other species as well, but won't it be great if we could propose a mechanism using the given species and commonly available reagents?

t-butanol + allyltrimethylsilane + Lewis acid
This is called the Hosoni-Sakurai Reaction (see J. Org. Chem. 2006 71 8516)
Can only find an image for the more common addition to carbonyl compounds but there are examples of additions to alcohols (see the JoC reference)

One last thing, could you post the relevant image here?

Done, though it is an image of the addition to carbonyls

@Waylander, The first step is the Nucleophilic addition to Carbonyls . But if we have tertiary alcohols, How can there be any addition? Don't we need substitution? And wouldn't that be difficult with a normal Lewis acid ?

See the JoC reference for addition to alcohols. I could not find an image of the addition to alcohols.

A dehydration reaction would be your best bet, you could use $\ce{KHSO4}$ or $\ce{TsOH}$ or $\ce{H3PO4}$. The idea would be to turn the $\ce{-OH}$ into water, making it a better leaving group.

What after turning OH into a good leaving group? A carbocation is formed, what next?

Depending on what reagent you use; you'll have a negative ion - or you can use water as a base to remove the hydrogen from the carbon next to the one with the positive charge. The breaking of the C-H bond will form the C-C double bond.

Yield will really be nothing. Major would be 3 Methylpropene.

What you could do is convert the t-butanol into a tertiary bromide by heating it with concentrated sulfuric acid and then treating it again with hydrobromic acid to get the tertiary bromide. After that, convert into a Grignard by oxidative addition of a Magnesium atom. Then you can just react it with ethylene oxide, and after working it up one more time, we can again heat it under concentrated sulfuric acid to eliminate the hydroxyl and form the double bond.

I think you should correct your answer, from using ethylene oxide to using 1,3 propylene oxide instead. Hence the downvote.

tertiary Grignards are lousy nucleophiles


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Is the equilibrium constant in the expression based on pressure or concentration?
$$\Delta G^\circ=-RT\ln K$$
I have always assumed it to be $K_p$ because $\Delta G^\circ$ represents standard free energy change and the standard state for a gas is 1 bar. So I thought we must express the activities of gases as ratios of partial pressures to $\pu{1 bar}$. But while solving sums, I found my answer to match only if I consider $K$ as $K_c$. Where am I wrong? Or am I missing something trivial?

Your question is indeed natural to arise, but I think it's way complicated than it seems. Read more here.

Sure thing! I don’t understand a word in that link.

Neither. It is exactly what is given there $K$, the standard equilibrium constant, and the equation given is its definition,
$$\Delta G^\circ=-RT\ln K.\tag{1}\label{def}$$
A general equilibrium constant for any reaction
$$\ce{\nu_A A + \nu_B B + \nu_C C + \cdots
   <=>\nu_{A'} A' + \nu_{B'} B' + \nu_{C'} C' + \cdots  }$$
is given by
$$K_x = \frac{ x_{\ce{A'}}^{\ce{\nu_{A'}}}\cdot x_{\ce{B'}}^{\ce{\nu_{B'}}}\cdot
               x_{\ce{C'}}^{\ce{\nu_{C'}}}\cdots}{
               x_{\ce{A}}^{\ce{\nu_A}}\cdot x_{\ce{B}}^{\ce{\nu_B}}\cdot
               x_{\ce{C}}^{\ce{\nu_C}}\cdots},$$
or more general, gathering all components in $\mathbb{B}=\{\ce{A, B, C, \dots, A', B', C', \dots }\}$
$$K_x = \prod_{B~\in~\mathbb{B}} x_B^{\nu_B}.\tag2$$
If you insert that into \eqref{def}, then $x$ becomes the chemical potential of the component. You can derive the link to the activity of the component (see here), and you can link the activity to concentrations, pressures, etc.
Given the right boundary conditions, e.g. ideal dilution, $K$ essentially becomes $K_c$, etc.
$$\mu(i) \propto a(i) \propto c(i) \implies K \propto K_c\tag3$$
Obviously, this is a very sloppy generalisation, and you should study thermodynamics in more detail to fully understand the differences between the quantities used, and when to use which.

x is never equal to chemical potential.  I think you meant activity (or perhaps fugacity).  Chemical potential is partial molar free energy, and has units of energy per mile.

"ideal dilution" Do you mean infinite dilution i.e. the Kohlaurcsh law limit of ionic conductivity in electrochemistry? (that's a very vague connection though)

Although the OP has accepted the answer, I don't see how this answers the question. It does not tell when to use which type of equilibrium constant.

@ApoorvPotnis Martin does say that "you should study thermodynamics in more detail to fully understand the differences between the quantities used, and when to use which" (my emphasis) i.e. without knowing the exact specific conditions it is difficult to predict if it is $K_\mathrm{c}$ or $K_\mathrm{p}$ or a hybrid of both.

@GaurangTandon Then how are we expected to use the equilibrium constants in problems that we are expected to solve?

Yep, I wonder the same...  @Martin Can you help here?

I agree @ApoorvPotnis I accepted the answer for the aforesaid reasons. Is there something like for an equilibrium with only gases, we should use $\pu{K_p}$  and for other equilibria we use $\pu{K_c}$?

With gas phase reactions the equation for equilibrium is $\Delta G^\mathrm{o} =-RT\ln(K_p)$ where because the equilibrium constant $K_p$ may be defined in terms of the free energies in their standard states (as you mention), i.e. at 1 bar (or atm.) it must be independent of pressure.
In terms of the partial pressures if the reaction has species $M_i$ the general  reaction is $\ce{v_1M_1 + v_2M_2<=>v_3M_3 + v_4M_4}$ then, with $p_i$ being the partial pressure of species $i$, $K_p = \prod p_i^{v_i}$ where the stoichiometric coefficients are $v_i$, and are positive for products and negative for reactants, thus for the reaction $\ce{A_2<=>2A}$, $K_p= p_A^2/p_{A_2}$.
The equilibrium constant can also be defined in terms of concentration since for a perfect gas the partial pressure of the $i^{\text{ th}}$ species is $p_i=n_iRT/V=c_iRT$. The new equilibrium constant ($K_c = \prod c_i^{v_i}$) when written in terms of $K_p$ becomes  $\displaystyle K_c=K_p(RT)^{-\Sigma v_i}$ which is often written as $\displaystyle K_c=K_p(RT)^{-\Delta n}$ where $\Delta n= \Sigma v_i$. Thus for the reaction $\ce{2A + 3B<=> 4C}$, $\Delta n\equiv \Sigma v_i= -2-3+4=-1$ and so $\displaystyle K_c=RTK_p$
(Incidentally, the temperature dependence if the equilibrium constant $K_p$ is given by the Van't Hoff equation, sometimes called the Van't Hoff Isochore).


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How to name alicyclic compound with hydroxymethyl substituent?
What is the IUPAC name of the given compound?

Is it cyclohexane dimethanol (or) 1,2-dihydroxymethyl cyclohexane?

The principal characteristic group of the compound that is given in the question is $\ce{-OH}$, which is expressed as suffix (‘-ol’). Thus, the senior parent structure of this compound is methanol. The problem in this case is that the compound contains two separate methanol structures.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical senior parent structures that are linked by di- or polyvalent groups can be named by multiplicative nomenclature if certain conditions are met (otherwise the usual substitutive nomenclature is used).

P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES
P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
  (1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
  (2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
  (3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.
When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

P-15.3.1.3 Multiplicative name formation
  Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
  When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:
(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: omitted (The locant 1 is the only available locant in methanol.)

(b) the name of the linking multiplicative substituent atom or group;

Here: cyclohexane-1,2-diyl

(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: methanol
Therefore, the preferred IUPAC name is (cyclohexane-1,2-diyl)dimethanol.

By way of comparison, the (correct, but not preferred) name based on usual substitutive nomenclature is [2-(hydroxymethyl)cyclohexyl]methanol

Can you please tell what the rulebook means by "substitutive nomenclature"? Thanks!


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What's the reaction mechanism for the oxidation of primary alcohols (e.g. ethanol) to carboxylic acids using acidified KMnO4?
I've seen mechanisms for $\ce{MnO4-}$ in base reacting with primary and secondary alcohols to oxidise them, but what about $\ce{MnO4-}$ in acid?
All the ones I've found have either been referring to $\ce{MnO4-}$ in base or just showed the redox reaction occurring. I've been trying to look for one for a while now, but I still haven't found a satisfying one (i.e. one that goes beyond showing the reaction in redox equation form).
I hope anyone who knows anything about it can answer me on that, and as thoroughly as you can. Thanks.

related https://chemistry.stackexchange.com/questions/4143/reaction-mechanism-for-oxidation-of-primary-alcohol-to-carboxylic-acid https://chemistry.stackexchange.com/questions/49542/mechanism-of-the-oxidation-of-alcohols-with-kmno4

also https://chemistry.stackexchange.com/questions/9001/difference-in-the-functioning-of-alkaline-and-acidified-kmno4 https://chemistry.stackexchange.com/questions/43283/mechanism-for-oxidation-of-primary-alcohols-to-carboxylic-acids but I don't see proper duplicate.

I was looking for a mechanism specifically about acidified KMnO4. I can't seem to find one very easily, not even here on Stack Exchange. I hope someone has an answer for it.

Well, it looks like doesn't it? And it's not particularly strange, redox mechanisms are notoriously obscure and often debatable. Even if you find it there's high probability it's more or less wrong.


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Why is helium used in diving apparatus?
In my textbook it said that helium is used in oxygen tanks (diving apparatus) by deep sea divers.
When I searched the internet for the reason it said that :

A mix of helium and oxygen is used in diving apparatus by deep sea divers, due to low solubility of helium in blood even at high pressure.

My doubts:

Why to mix helium when we can just fill the tanks with pure oxygen?
Why can't other inert gas be used in place of helium?

Pure oxygen is problematic: https://en.wikipedia.org/wiki/Hyperoxia

There's https://chemistry.stackexchange.com/questions/61406/why-would-breathing-pure-oxygen-be-a-bad-idea already, so this might be qualified as dupe of this and another(?) or too broad.

https://chemistry.stackexchange.com/questions/8791/xenon-and-the-human-body

As TAR86 said pure oxygen is problematic.
A little something from the Wikipedia page on oxygen toxicity

Severe cases can result in cell damage and death, with effects most often seen in the central nervous system, lungs and eyes. Oxygen toxicity is a concern for underwater divers, those on high concentrations of supplemental oxygen (particularly premature babies), and those undergoing hyperbaric oxygen therapy.

Turns out nitrogen is also toxic at high pressure see nitrogen narcosis.
Helium is used primarily because it's cheaper as compared to other alternatives and is not narcotic.
Other gases are too heavy and they diffuse into the bloodstream.
For more information see this PDF on exotic diving gases.

Fun Fact
Argon is used but can you guess why?

 Used to inflate scuba suits, helps in insulating them.

It is not that helium is cheaper than anything. It is just that there are no alternatives. All other gases start dissolving in blood, causing weird effects.

Indeed, helium is not 'cheap' and only getting more expensive.

@JonCuster cheaper than neon at least(which is the only other somewhat safe option).

Yeah, the price (and supply) of neon has also gotten out of hand, but at least it can be distilled out of air. Helium comes from (a small number of) natural gas fields, and is the only gas that is worthwhile to ship internationally.


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Electrochemistry molar conductivity

Molar conductivity increases with decrease in concentration. This is because the total volume, $V$, of solution containing one mole of electrolyte also increases. It has been found that decrease in $\kappa$ on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, $\Lambda_m$ can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration
approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol $\Lambda^\circ_m$. The variation in $\Lambda_m$ with concentration is different (Fig. 3.6) for strong and weak electrolytes.

Can anyone explain what does this sentence mean : "it has been found that decrease in k on dilution of a solution is more than compensated by increase in its volume"
Molar conductivity : $\kappa \times V$ ($\kappa$ is specific conductance), $V$ is volume of solution containing 1 mole of electrolyte.
I understand that molar conductivity increase with decrease in concentration. But at the same time this sentence is contradicting that because it says the decrease in "$\kappa$"(specific conductance) is more than increase in volume Which should decrease molar conductivity. I am confused.

I think you missed the word "compensated".

The decrease in specific conductivity is more than compensated by increase in its volume.

This simply answers your question. The reduction in molar conductivity due to the decrease in specific conductivity is  overridden by the increase in volume.
It is saying that the latter overrides the former, and not just overrides, but overrides by a lot.


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Reduction of methylene blue
Why does methylene blue (which is initially blue) reduce to form a colorless solution?
I was told by my teacher that it has to do with the presence of an additional double bond on methylene blue, but I can't figure out or find any source on the internet that claims so.

The color change results from the reversible oxidation-reduction reaction of the methylene blue indicator.  In alkaline solutions, glucose is oxidized to D-gluconic acid or alpha-D-gluconolactone.
$\ce{HOCH2(CHOH)4CHO + 3OH- -> HOCH2(CHOH)4CO2 + 2H2O + 2e-}$
In the course of this reaction, methylene blue is reduced from the blue (oxidized) form to the colorless (reduced) form.

Shaking the flask dissolves $\ce{O2}$ in the solution, which oxidizes the indicator back to the blue (oxidized) form.

Okay, but does the presence of extra double bond has anything to do with colour change?

Can you expand on how the presence of double bond leads to colour change?

Sure ! The double bond changes its position in the product , which is due to tautomerisation. Now , since tautomers  have different physical properties , they have different color. One reason for this is , the lone pairs of Nitrogen which were earlier undergoing conjugation at one side , now undergo conjugation from both sides , resulting in decreasing the electron density on N . Since , it becomes harder for N to donate electrons to the solvent , it now shows different spectral behaviour and hence different color .

Is the same reason applicable for the colour change on methyl orange due to breaking of double bond between the nitrogen ?

Yes !  You're correct.


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What temperature does oxygen have to be at to bond with carbon monoxide?
I was wondering how I would go about calculating the required temperature of oxygen to bond one oxygen particle with carbon monoxide (CO). I know that it depends on some other factors such as air pressure, but that's the extent of my knowledge. I would be really grateful if someone could explain what environment I would need to combine the two.


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Does hyperconjugation stabilize a carbocation more than a free radical?

Which of the following statements is correct about methyl group?

It stabilizes both carbocation and free radical equally.
It stabilizes a free radical more than a carbocation.
It stabilizes a carbocation more than a free radical.
None of these.

I thought the answer would be (1). Keeping in mind these structures: The methyl group can stabilize the carbocation and the free radical group attached to it equally through hyperconjugation. However, the answer given is (3). How is this possible? Why is the carbocation more stable than the free radical? Are the structures I drew wrong?

Can you give an example where stabilization by radical formation is seen.

@AvatarShiny You'll have to scroll a bit and you will see free radical stabilization due to hyperconjugation. https://hubpages.com/education/Thorough-Chemistry-of-Hyperconjugation

related https://chemistry.stackexchange.com/questions/64875/why-are-radical-intermediates-are-more-stable-on-tertiary-carbons

@raajsuriya You can't draw a resonance structure involving all three hydrogen atoms on an adjacent methyl. They can't all be oriented properly for hyperconjugation at the same time. Let's just say that looking briefly at the linked web page, I'm not impressed.

@Zhe The link was not important I googled that quickly for Avtar Shiny. Do you know why carbocations are stabilized more than free radicals through hyperconjugation? Or is this something I should memorize? Thank you for your time

First, why do you think that's true. Second, what do you mean by "stabilized more?"

@Zhe I have edited my question. Maybe this will be clearer.

Well, it certainly can't be equal since their different. But I don't understand how you plan to compare stability of a radical against stability of a cation. They're different and have different modes of reactivity...

I only came across this question, I'm just as confused as well. @Zhe

I think the question is not so difficult (at leats if we accept some ideas). Suppose those structures don't have hyperconjugation, and then, they have. So, you need a difference of energy for each process and compare it. Possibly, it will be enough to say that carbocation atracts more electron density, so, it will be more stabilized..

Related: https://chemistry.stackexchange.com/a/31729/31775

A carbonation is deficient in 2 electrons whereas a free radical is only 1 electron deficient. Thus any type of stabilisation working on a carbonation will be much more effective as it is more electron deficient. Hence, hyperconjugation or inductive electron donation stabilises a carbocation more than it does a free radical.


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Liquid to solid crystal when exposed to air
Im working on a prototype weapon based on the needler from the Halo series. And yes while I do realize that this game series is fantasy, I still find it fun to try and recreate working versions of what is in the game.
I have a concept worked out to keep the crystalline ammunition in a liquid tank and then use compressed air to fire. The liquid will be pushed through perforated holes to for the crystal ammunition. But I don't know of any chemical or compound that turns from liquid to solid, preferably crystalline when exposed to air. If someone could help me with such a substance I would greatly appreciate it.

Science fiction actually; more importantly http://halo.wikia.com/wiki/Type-33_Guided_Munitions_Launcher says nothing about liquid, needles are supposedly as ready to be fired as any ammo. They're just very advanced smart munitions, a bit similar to airburst round but rather penetrating.


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Why vinegar + water to clean stains?
The advice for cleaning salt stains (for example from shoes) with a vinegar + water solution is fairly ubiquitous on the internet. Is there a chemical explanation for why the vinegar is important? Why not just water?
Edit: I don't actually know that vinegar helps clean salt stains. There might be actually be no chemical reason for the advice to be valid.

I see this has been downvoted. This is my first question here - I'm happy to improve the question with some guidance, or even delete if this is totally inappropriate for the site. If that's the case, a suggestion for another site would be greatly appreciated.

How do we know that the statement that vinegar works better with salt (NaCl) stains than normal water is actually a fact? Has anyone performed a scientific experiment and found that vinegar has truly the effect stated (i am not convinced)? I think this is the first question to ask. If you can provide real data that vinegar helps (and not because ppl on the internet say so) then we can start looking into why. Also, by salt i guess you mean NaCl. If it was carbonates (eg. limescale) then use of vinegar would make perfect sense.

@AMM I don't know that it helps. I suppose a valid answer would be that according to the laws of chemistry, vinegar shouldn't help clean salt stains.

What i am trying to say is that you cant know for sure without doing an experiment. This is what scientific method is all about. May vinegar helps and maybe it doesnt until you try. You cant say it shouldnt because maybe for example there is carbonate contamination in the salt they use for the street or maybe etc. etc.

The acid in the vinegar helps to a) dissolve substances or b) makes the substance easier to remove. The water is neutral (more or less) so won't really dissolve any stain or such

How? It appears that salts used to melt ice (e.g. Sodium and Calcium Chloride) are already highly soluble in water.


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Photochemistry and steady state approximation
I am unsure whether to ask this in physics or chemistry as there is overlap but I am taking this in a chemistry module.
System A undergoes photophysical pathways:
$$\ce{A + hv -> A\mathrm{*}}$$
$$\ce{A\mathrm{*} -> A + hv   \tag{k1}}$$
$$\ce{A\mathrm{*} + B -> A + B\mathrm{*} \tag{k2}}$$
$$\ce{B\mathrm{*} -> B + hv      \tag{k3}}$$
$$\ce{B\mathrm{*} -> C           \tag{k4}}$$
It wants me to show that the quantum yield of energy transfer is equal to the quantum yield of fluorescene B and the quantum yield of photochemical conversion from B to C.
I am quite stuck on this problem. I assume the steady state approximation needs to be employed in order to determine quantum yields for the processes asked for.
I am wondering what the general strategy would be to answer this question?

I'd begin by drawing a tree with the possible reaction paths. Which 'quantum yields' are associated with each branch of the tree? Think about which steps are reversible and which steps aren't.
Extra hint: Double funnel.

I think the big question here is which processes do you expect to be reversible? And if the answer is none, then it's just the forward rate, the whole way through...

The processes are absorb light into A, and then A* has only two pathways, i.e. fluoresces and reacts with B to form B*.
The B* has only two pathways, it  fluoresces and converts to C. All B* molecules must therefore follow one of these paths, so the sum of yields of B* (paths fluorescence + forming C) must be the same as the yield to form B* in the first place, thus you don't need to do a calculation.
In general if you have to find a yield of any process x this is, $\displaystyle \phi_x = \frac{\text{rate of process x}}{\text{rate of absorption}} \equiv \frac{\text{rate const x} }{\text{sum of all other rate constants from x}}$.
In your example $\displaystyle \phi_{ET}= \frac{k_{ET}[B]}{k_f^A+k_{ET}[B]}$ where superscript $A$ refers to fluorescence from $A^*$.
To work all this out write down $d[A^*]/dt = +k_{abs}[A]- k_f^A[A^*]-k_{ET}[A^*][B]$ and use steady state.

So then how would I show that Φenergy transfer = Φf(B) + ΦC? Would I need to work out the equations for both the  Φf(B) and  ΦC and then show they are mathematically equal to  Φenergy transfer. Also what specifically is the Φenergy transfer referring to? So for eaxmple you can show that the  Φf(B) =  ΦET + k3/(k3 +k4)

No the argument  is in para 2 and I reword it as follows. There are only two pathways for B* to go to, one is B* fluorescence to  ground state B and the other is to C. These yields must  add to be 1 as there is nothing else B* can do according to your scheme.  As B* is produced from A* it follows that the yield you want can only be that of the amount of B* produced (since the sum of decays of B* is 1) and this is the yield of energy transfer to form B*.


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Why are limes alkaline, but lemons are acidic? They taste about the same
So, this is kinda a dumb question, but I'm curious. Both are citrus fruits and very related to each other and taste sour which corresponds to both being acidic. Why then are limes said to be alkaline, but lemons are acidic?

Limes have calcium oxides which in water forms calcium hydroxide, which is basic due to the hydroxyl ions. Lemons have citric acid (component of vitamin C). I never tasted the former though.

lime: pH 2.40; and lemon: pH 2.30.  source

Doesn’t lime refer to limewater ie., $Ca(OH)_2$?

@avatar just to clarify that page defines pH of the citrus fruit "lime", and not of its other popular usage, which is the inorganic oxides and carbonates of calcium

The word "lime" has two meanings and they have essentially no relationship between them.
The "lime” that is inorganic oxide quicklime, CaO, reacts with water to give a relatively strong-base hydroxide::
$$ \ce{ CaO + H2O -> Ca(OH)2 }$$
$$ \ce{ Ca(OH)2 <=> Ca^2+ + 2OH-}$$
Due to the presence of these hydroxyl ions, the solution is alkaline.
"Limes" that are the green, hybrid citrus fruits are acidic with citric acid, just like lemons.  Both have juice with a pH in the range of 2-3, for limes as in fruit a typical pH level is about 2.8.  Hence the tart flavor for which both fruits are known.  Like all citrus fruits, lemons and limes are a rich source of ascorbic acid, vitamin C.

The OP was obviously talking about the fruit, so why the chemistry lesson on the mineral "lime"?

@James Because the OP was confused. Presumably he read online that "lime" is alkaline ("Why then are limes said to be alkaline?"). This isn't true for the fruit known as limes. But the alkalinity of the unrelated substance known as "lime" is very commonly talked about in gardening. It is incredibly common to see the alkalinity of "lime" discussed online. This answer is almost certainly the correct one the OP was looking for: Limes are as acidic as lemons, but the unrelated substance commonly referred to as 'lime' used in gardening is alkaline.