text:
How can atoms be influenced by light?
I was watching a SciShow video "Most dangerous Chemicals". It said that a compound called Azidozide azide explodes on being exposed to light. My question is: how does the the atom feel light. Heat can be felt (let us not go into the philosophical question why do we feel warm while standing below the sun) and can excite molecules. How can light excite molecules?

Light can excite atoms a great deal harder than heat. Related: https://chemistry.stackexchange.com/questions/86931/reactions-in-the-presence-of-sunlight/86935#86935

Light can be ‘felt’ because light is transmitted by photons. Depending on how you choose to observe light, it will either manifest itself as waves — for example in interference experiments — or as particles — for example when describing the photoelectric effect.
When described as a wave, the energy of a certain source of light will be determined by its wavelength by equation $(1)$ in which $h$ is the Planck constant and $c$ is the speed of light.
$$E = \frac{hc}\lambda\tag{1}$$
It may not immediately be obvious how this translates to the energy of a photon, but the key is the de Broglie wavelength as given in equation $(2)$.
$$\lambda = \frac hp\tag{2}$$
This equation connects all particle and wave properties of types of matter. (Indeed, due to the wave-particle dualism all particles can theoretically behave as waves and vice-versa; the wave nature of particles as large as entire molecules has been shown in interference experiments.)
When a photon with a certain energy travels through an atom or molecule, it may interact with the electrons present; often this interaction causes the photon to be absorbed and the electron to be excited to a higher energy level. As chemists, we describe this as an electronic transition from one orbital to another. The entire process can also be reversed, i.e. if an electron is excited to a higher energy level by any means, it can relax following certain quantum-mechanical selection rules back down to a lower energy level and emit a photon of a definite wavelength. The latter process can be observed in the flame colour given by sodium atoms: their 3s electron is excited to a 4p level thermally due to the heat of the flame. Upon relaxation, the electron emits a photon of $\pu{589nm}$ wavelength which we perceive as yellow. We could write that in an equation:
$$\ce{4p ->[][$\lambda = \pu{589nm}$] 3s}\tag{3}$$
If the target of a photon happens to be a molecule rather than an atom, we need to consider what type of orbitals we are exciting from and exciting into. Typically, the occupied orbitals will be bonding orbitals while unoccupied orbitals will be antibonding. Thus with the correct photon wavelength, an electron can be excited from a bonding into an antibonding orbital which we can, in layman’s terms, call ‘breaking a bond’. Whether or not the bond actually breaks depends on whether or not the atoms can dissociate apart.
A beautiful reaction to exemplify this is the reaction of chlorine and hydrogen in equimolar mixtures. Under ambient conditions, these two are metastable. If the activation energy is supplied, they will react to form hydrogen chloride.
$$\ce{H2 + Cl2 -> 2 HCl}\tag{4}$$
Activation energy can be supplied not only by a flame but also by light as long as its wavelength is short enough. A flash of blue light will work, a flash of red light will not. What happens? First, a photon excites an electron in a chlorine molecule. It will reach an antibonding, unoccupied orbital, the $4\sigma$ orbital. This causes the $\ce{Cl-Cl}$ bond to break liberating two chlorine radicals.
$$\ce{Cl2 + $h\nu$ -> [Cl2^*]^\ddagger -> 2 Cl^.}\tag{5}$$
These chlorine radicals are then energetic enough to react with hydrogen molecules to give hydrogen radicals and so on — a radical chain reaction.
$$\begin{align}\ce{Cl^. + H2 &-> HCl + H^.}\tag{6}\\
\ce{H^. + Cl2 &-> HCl + Cl^.}\tag{7}\end{align}$$
And this goes on until all the reactive species have been consumed and the radicals recombine. The reaction releases $\pu{92.3kJ/mol}$ and is therefore strongly exothermic, even explosive.
This mechanism of course also works for infrared light which we perceive as heat. However, since the energy of infrared light is very low, few reactions can be initiated by it.

Photons in the UV or visible ranges of the EM spectrum can have sufficient energy to excite electrons. In simplified terms, this usually involves an electron entering an anti-bonding orbital, which may be sufficient to start a reaction. If that reaction yields enough energy to start the next one, you have a chain reaction. Alternatively, those electrons relax back to their ground states, photons will be emitted, and the atom or molecule will give off visible light of specific frequencies. This is called fluorescence. So in a nutshell light gives off energy and that is what will excite your molecules along with the energy from anything else that makes contact with the chemical. Ask if you need clarification.

I think this answer could be improved if it addressed how the absorption of light could result in an explosion per the OP's example, not just the emission of a photon.

Thanks for the suggestion. I'm not so sure what it is but simple warmth of light could be enough to cause it to explode perhaps because of the burning point of the chemical is so low the heat overcomes it and it ignites. And when i mentioned how the energy from contact of outside sources ignites it the friction would be that cause of explosion due to the heat generated. Feel free to correct my thinking.


text:
Why secondary alkyl groups are more prone to shifting than primary?
Why secondary alkyl groups are more prone to shifting than primary? I attach example from Grossman book [p. 20]:

When $\ce{R} = \ce{CH2CH2Ph}$, the coordinated $\ce{Al}$ simply transfers a $\ce{Me}$ group to the carbocation $\ce{C}$ (σ bond nucleophile). The $\ce{O}$ atom then coordinates another equivalent of $\ce{AlMe3}$ before the product is obtained upon workup.

When $\ce{R} = \text{cyclohexyl}$, the $\ce{R}$ group migrates (1,2-alkyl shift) to give a new carbocation. (2° Alkyl groups are more prone to migrate than 1° alkyl groups.) After $\ce{Me}$ transfer to the new carbocation and coordination of another equivalent of $\ce{AlMe3}$, workup gives the product.

This is quoted from "The Art of Writing Reasonable Organic Reaction Mechanisms", Second Edition by Robert B. Grossman, DOI: 10.1007/b97257.

I don't know what Grossman book is, and Google finds way too many results. Please add human-readable references and link through a stable URL, preferably a DOI. See here for more information.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Insertion vs epoxide formation
How can it be explained that in the first reaction mentioned below, a 7-membered ring dominates when the leaving group is nitrogen and an epoxide forms when the leaving group is dimethyl sulfide?
This is quoted from "The Art of Writing Reasonable Organic Reaction Mechanisms", Second Edition by Robert B. Grossman, DOI: 10.1007/b97257.

The $\ce{C}$ in diazomethane is nucleophilic. The product of attack of diazomethane on the carbonyl $\ce{C}$ has a leaving group a to the alkoxide, so either a 1,2 alkyl shift or direct nucleophilic displacement can occur. The insertion product happens to dominate with $\ce{H2\overset{-}{C}-\overset{+}{N}_2}$, but with $\ce{H2\overset{-}{C}-\overset{+}{S}Me2}$ the epoxide dominates.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Loss of aromaticity vs low entropy of activation
Even if reaction has low entropy of formation does that mean it is favorable to loss aromaticity and form highly strained cyclopropane ring? Why this reaction happens? Maybe someone can explain that in more details? Is this equilibrium reaction? I assume cyclopropane ring can be as easily opened with nucleophile as it closed? It does not make any sense for me here. Thank you for comments.
This is quoted from "The Art of Writing Reasonable Organic Reaction Mechanisms", Second Edition by Robert B. Grossman, DOI: 10.1007/b97257.

The reaction proceeds by an $\mathrm{S_N2}$ mechanism. The reaction has a very low entropy of activation, so it proceeds despite the loss of aromaticity. The product is a model of the antitumor agent duocarmycin. DNA reacts with duocarmycin by attacking the $\ce{CH2}$ group of the cyclopropane ring in an $\mathrm{S_N2}$ reaction.

There are several factors we can point to before resorting to entropy of activation.
1)  At the end of the "chain" is displacement of an iodide ion.  Iodine, with its diffuse valence orbitals, forms rather weak bonds with carbon making this displacement thermodynamically and kinetically favorable.  Presumably the nucleophile that re-opens the cyclopropane ring forms a stronger bond with the carbon.
2)  We may no longer have an aromatic ring, but a carbonyl group flanked by two conjugating carbon-carbon bonds still has plenty of resonance stabilization.  We certainly do not lose all that much stabilization.  Moreover...
3)  Upon further review, there is still an aromatic contribution after all.  The cyclopropane ring is well known to stabilize carbocations by hyperconjugation, and with the carbonyl oxygen tending to withdraw electrons we may argue that the quinone-like ring has some carbocation character.  Adding a pair of pi electrons to this ring through hyperconjugation will then give aromatic contributing structures in which the cyclopropane ring is positively charged and the oxygen regains its negative charge.
All told, we don't have to reach for "entropy of activation".  The reaction is better off than it looks just in terms of energy.


text:
How to calculate molarity and the number of molecules for a mixture in a simulation box?
I want to do a simulation of a mixture which consists a total of $512$ molecules of glycine, ammonia, and water molecules mixture in a cubic box of, say, length $\pu{2.6 nm}$; $\pu{2.6 nm}^3 = \pu{17.576 nm3}$ in terms of volume.
I want [1 glycine molecule + $x$ ammonia molecules + $y$ water molecules] which together comes to a total of $512$ molecules and in a box volume of $\pu{17.576 nm3}$ and molarity should be $\pu{15 M}$.
Actually here I am fixing the glycine molecule which is only one and varying only the number of ammonia and water molecules to get different molar concentrations, say, $\pu{5 M / 10 M / 15 M}$ and $\pu{18 M}$, where the total number of molecules is constant, i.e. $512$ molecules only (see the example below). Since a maximum of $\pu{18 M}$ solution can be prepared (since ammonia solubility is $\pu{31 g}$ in $\pu{100 mL}$), so, I am trying to study with varying the $\%$ composition of ammonia and water molecules.
In this case I want to get a concentration of $\pu{15 M}$.
How to calculate the number of $x$ ammonia and $y$ water molecules that need to be taken in order to get $\pu{15 M}$, considering the conversion of number of molecules to mass, amount of substance, etc..
For example:

$1$ molecule of glycine + $52$ molecules of ammonia + $459$ molecules of water is $10 : 90$ mixture (total $512$ molecules);
$1$ molecule of glycine + $104$ molecules of ammonia + $408$ molecules of water is $20 : 80$ mixture (total $512$ molecules).

Then what would be the molarity of the above two mixtures ($1 : 90$, $20 : 80$) considering the same volume ($\pu{17.576 nm}$)?

Do you need molarity specifically? Often with molecular simulations, you will see molality used instead because it is easier to calculate when you dealing with a reasonably countable number of solvent molecules.

I would use a slightly different approach than andselisk, based on the volume on your box, and that I don't think you can neglect the volume of your ammonia molecules. Given $V(\text{sol}) = \pu{17.576E-27 m3}$ and $c(\ce{NH3}) = \pu{15.0E3 mol  m-3}$ we can calculate the number of ammonia molecules we need:
\begin{align}
  && c(\ce{NH3}) &= \frac{n(\ce{NH3})}{V(\text{sol})};\\
  && n(\ce{NH3}) &= \frac{N(\ce{NH3})}{N_\mathrm{A}};\\
&\to&
     c(\ce{NH3}) &= \frac{N(\ce{NH3})}{N_\mathrm{A}\cdot V(\text{sol})}\\
&\Longleftrightarrow&
     N(\ce{NH3}) &= c(\ce{NH3})\cdot N_\mathrm{A}\cdot V(\text{sol}) \approx 159
\end{align}
Since you want to limit the number of molecules to $512$, we know that all other molecules have to be water:
$$N(\ce{H2O}) = 512 - N(\ce{Gly}) - N(\ce{NH3}) = 352$$
Lastly, let's check whether the density is anywhere close to what we would expect, we take $M_\mathrm{m}(\ce{Gly}) = \pu{75 u}$, $M_\mathrm{m}(\ce{H2O}) = \pu{18 u}$, and $M_\mathrm{m}(\ce{NH3})= \pu{17 u}$.
\begin{align}
  \rho(\text{sol}) &= \frac{m(\text{sol})}{V(\text{sol})}\\
  &= \frac{\left(
       N(\ce{Gly})M_\mathrm{m}(\ce{Gly})
     + N(\ce{H2O})M_\mathrm{m}(\ce{H2O})
     + N(\ce{NH3})M_\mathrm{m}(\ce{NH3})\right)\frac{m_\mathrm{u}}{\pu{u}}
     }{V(\text{sol})}\\
  \rho(\text{sol}) &\approx \pu{861 kg m-3} = \pu{0.861 g cm-3}
\end{align}
Citing Wikipedia's entry of ammonia I would say you are quite close:

Solvent properties
  Ammonia is miscible with water. [...] The maximum concentration of ammonia in water (a saturated solution) has a density of $\pu{0.880 g cm-3}$ and is often known as '.880 ammonia'. [...]

The Handbook of Chemistry and Physics has a concentration table (91st Edition p. 8-53; currently 98th Edition, also online), and I'll reproduce a couple of values for comparison, at $\pu{20 ^\circ C}$.
\begin{array}{rrr}
  \text{Mass\%} & c / \pu{mol L-1} & \rho / \pu{g cm-3}\\\hline
    5.0   &   2.868    &   0.9770\\
    9.0   &   5.080    &   0.9613\\
   12.0   &   6.695    &   0.9502\\
   18.0   &   9.823    &   0.9294\\
   24.0   &  12.826    &   0.9102\\
   28.0   &  14.764    &   0.8920\\\hline
\end{array}
In conclusion, you are a few water molecules short, or your box is slightly too large, or your solution is not dense enough. I would expect some trade-offs when doing such a study and keeping the named constraints, so I guess you should have a close look at everything and settle for the optimum you can afford.

The last row (28 Mass%) would be well reproduced by 139 ammonia, 471 water in a cubic box of 2.5 nm. (18 Mass%) 93 NH3, 394 H2O, 2.5 pm.

Can u kindly tell what is mu/u in the density equation..?? and one more thing... how to type the rho symbols, and subscripts/superscripts, equations here..??

@D.H.N Have a look here and here for MathJax tutorials. The $m_\mathrm{u}/\pu{u}$ is only the conversion factor from the unified atomic mass unit, to the SI unit (kg). (And when you click the [edit] link, you can have a look at the source code, too.)

@Tyberius I actually really dislike using \to instead of \implies, for me it simply has a completely different meaning. Why are you doing that anyway? Thanks for catching the spelling errors though.

@Martin I tend to change \implies to \to because \implies doesn't show up in the app and I personally consider them as conveying the same idea. I can understand wanting to use them differently and that it might not help much in this case since \pu blocks a lot of the post anyway. I've just figured in cases where it was the one obstruction to reading the post on the app, I would change it and if the poster didn't think it was right it could be quickly reversed.

@Tyberius As you might have noticed, I did not change it back, I just thought it is useless here, since I am using \pu which gets wrangled by that stupid app... oh well....

Using conventional macroscale quantities and SATP, one can express the desired concentration of aqueous ammonia using the amount $n$ and the volume $V$ of the solution:
$$C(\ce{NH3}) = \frac{n(\ce{NH3})}{V(\ce{H2O})} \tag{1}$$
Let's find the ratio between the numbers of molecules of ammonia ($N(\ce{NH3})$) and water ($N(\ce{H2O})$). By definition of a mole ($N_\mathrm{A}$ – Avogadro's number):
$$n(\ce{NH3}) = \frac{N(\ce{NH3})}{N_\mathrm{A}} \tag{2}$$
Volume of water required to achieve given concentration can be found using density $\rho$ and molecular weight $M$:
$$V(\ce{H2O}) = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{n(\ce{H2O}) \cdot M(\ce{H2O})}{\rho (\ce{H2O})} = \frac{N(\ce{H2O}) \cdot M(\ce{H2O})}{\rho (\ce{H2O}) \cdot N_\mathrm{A}} \tag{3}$$
Assembling (2) and (3) in (1):
$$\frac{N(\ce{H2O})}{N(\ce{NH3})} = \frac{\rho (\ce{H2O})}{C(\ce{NH3}) \cdot M(\ce{H2O})} = \frac{\pu{1.0e3 g L-1}}{\pu{15 mol L-1} \cdot \pu{18.015 g mol-1}} = 3.70 \tag{4}$$
Now, knowing the ratio between ammonia and water, total number of molecules in system and assuming that single glycine molecule remains intact, one can count the exact numbers:
$$1 + N(\ce{NH3}) + 3.70 \cdot N(\ce{NH3}) = 512$$
$$\to N(\ce{NH3}) \approx 109 $$
$$\to N(\ce{H2O}) \approx 402 $$
By the way, 402 water molecules would only occupy $\pu{1.20e-26 m^3}$ (formula (3)), assuming this molecular assembly (cluster) behaves similarly to its macro counterpart (which is probably not exactly true). This means that you need a slightly smaller cube with a side of about $\sqrt[3]{\pu{1.20e-26 m^3}} = \pu{2.3 nm}$.
The same procedure can be repeated for other concentrations.

I've looked up the densities of ammonia solutions, and what I calculated is surprisingly close to what you would find in the theoretical setup.


text:
A problem in solubility of salts with multistep hydrolysis of an ion possible

Calculate the solubility of $\ce{MnS}$ in pure water, $K_\mathrm{sp}(\ce{MnS}) = \pu{2.5e-10}$. Assume hydrolysis of $\ce{S^2-}$ ion. $K_1$ and $K_2$ for $\ce{H2S}$ are $10^{-7}$ and $10^{-14}$, respectively.

Let unknown amount of $\ce{MnS}$ be dissolved in $\pu{1 L}$ of water, so $\ce{Mn^2+} = s~\mathrm{moles}$ and $\ce{S^2-} = s~\mathrm{moles}$ where $s$ is the solubility of $\ce{MnS}$ in $\pu{mol L-1}$.
Now let $\ce{S^2-}$ participate in hydrolysis by dissociating its $x$ moles and combining with $\ce{H2O}$ to form $x$ moles of $\ce{HS-}$ and $x$ moles of $\ce{OH-}$, so final moles of $\ce{S^2-}$ is $(s-x)$.
I considered only $K_\mathrm{h1}$ as it is $10^7$ times larger than $K_\mathrm{h2}$, $K_\mathrm{h1}$ is $K_\mathrm{w}/K_\mathrm{a2}$ which is equal to $1$. So I got $s^2 = (s-x)$ and we know that $K_\mathrm{sp} = s(s-x)$. On substituting I got the following equation:
$$K_\mathrm{sp} = s^2 - s^{1/2} \cdot K_\mathrm{sp}^{1/2}$$
I am stuck here. I know that $K_\mathrm{sp} = \pu{2.5e-10}$, so I tried solving for $s$, but could not find the correct answer.
Can you please point out my conceptual mistake?
Sorry, if this question is of homework type, I really tried solving this before posting.

I edited the question a bit according to local guidelines (if you are interested in formatting on Chemistry.SE, check out this page, this page and this one). Have you tried to solve the last equation for $s$, and if so, what did you get and what was the answer?

I think this page might hold your answer : https://socratic.org/questions/what-is-the-solubility-of-mns-in-grams-per-liter-in-a-buffer-solution-that-is-0-

I entered the equation in wolfram, it showed pretty complex solution, the correct answer is $6.3*10^{-4}$

Let's denote all processes of interest which are occuring or might occur in the system:
\begin{align}
\ce{MnS &<=>[$K_\mathrm{sp}$] Mn^2+ + S^2-} && K_\mathrm{sp} = [\ce{Mn^2+}][\ce{S^2-}] \tag{R1} \\
\ce{MnS + H2O &<=>[$K_\mathrm{h1}$] Mn^2+ + HS- + OH-} && K_\mathrm{h1} = [\ce{Mn^2+}] [\ce{HS-}] [\ce{OH-}] \tag{R2} \\
\ce{MnS + 2 H2O &<=>[$K_\mathrm{h2}$] Mn^2+ + H2S + 2 OH-} && K_\mathrm{h2} = [\ce{Mn^2+}][\ce{H2S}][\ce{OH-}]^2 \tag{R3} \\
\ce{H2S &<=>[$K_\mathrm{a1}$] HS- + H+} && K_\mathrm{a1} = \frac{[\ce{HS-}][\ce{H+}]}{[\ce{H2S}]} \tag {R4} \\
\ce{HS- &<=>[$K_\mathrm{a2}$] S^2- + H+} && K_\mathrm{a2} = \frac{[\ce{S^2-}][\ce{H+}]}{[\ce{HS-}]} \tag {R5} \\
\ce{H2S &<=>[K_\mathrm{d}] 2 H+ + S^2-} && K_\mathrm{d} = K_\mathrm{a1} K_\mathrm{a2} \tag{R6}
\end{align}
and also agree on labeling $C_\mathrm{A}$ total concentration of sulfide anions in solution in form of $\ce{S^2-}$, $\ce{HS-}$ and $\ce{H2S}$; $\alpha_2$ – fraction of $\ce{S^2-}$ among those; $s$ – solubility. Then from (R1)
$$K_\mathrm{sp} = [\ce{Mn^2+}] C_\mathrm{A} \alpha_2 \implies s = \sqrt{\frac{K_\mathrm{sp}}{\alpha_2}} \tag{1}$$
$\alpha_2$ can be expressed via known equilibrium constants and $\mathrm{pH}$:
\begin{align}
\alpha_2 &= \frac{[\ce{S^2-}]}{[\ce{S^2-}] + [\ce{HS-}] + [\ce{H2S}]} &&\Bigg| \cdot \frac{[\ce{H+}][\ce{HS-}]}{[\ce{H+}][\ce{HS-}]}\\
&= \frac{K_\mathrm{a2}}{K_\mathrm{a2} + [\ce{H+}] + \frac{[\ce{H2A}][\ce{H+}]}{[\ce{HA-}]}} &&\Bigg| \cdot \frac{K_\mathrm{a1}}{K_\mathrm{a1}} \\
&= \frac{K_\mathrm{a1}K_\mathrm{a2}}{K_\mathrm{a1}K_\mathrm{a2} + K_\mathrm{a1}[\ce{H+}] + [\ce{H+}]^2} \tag{2}
\end{align}
Since hydrolysis might occur partially (R2) or completely (R3), in order to find$[\ce{H+}]$ from
$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} \tag{3}$$
let's estimate whether the equilibrium in (R3) is going to be shifted left or right. Solubility of sulfide anions in the absence of hydrolysis is
$$[\ce{S^2-}] = \sqrt{K_\mathrm{sp}} = \sqrt{\pu{2.5e-10}} = \pu{1.6e-5 M} \tag{4}$$
whereas solubility when $\ce{H2S}$ is forming (from (R6)) would be
$$[\ce{S^2-}] = \sqrt[3]{\frac{K_\mathrm{a1} K_\mathrm{a1}}{4}} = \sqrt[3]{\frac{10^{-7} \cdot 10^{-10}}{4}} = \pu{1.4e-6 M} \tag{5}$$
so that formation of less soluble product would indeed favor (R3). Hence, in order to estimate the increase in $\mathrm{pH}$, it's better to use $K_\mathrm{h2}$:
\begin{align}
K_\mathrm{h2} &= [\ce{Mn^2+}][\ce{OH-}]^2[\ce{H2S}] && \Big| \cdot \frac{[\ce{H+}]^2[\ce{S^2-}]}{[\ce{H+}]^2[\ce{S^2-}]} \\
&= \frac{K_\mathrm{sp}K_\mathrm{w}^2}{K_\mathrm{a1}K_\mathrm{a2}} \tag{6}
\end{align}
On the other hand,
$$K_\mathrm{h2} = [\ce{Mn^2+}][\ce{OH-}]^2[\ce{H2S}] = 4[\ce{OH-}]^4 \tag{7}$$
Equating (4) and (5):
$$[\ce{OH-}] = \sqrt[4]{\frac{K_\mathrm{sp}K_\mathrm{w}^2}{4 K_\mathrm{a1}K_\mathrm{a2}}} = \sqrt[4]{\frac{\pu{2.5e-10} \cdot (10^{-14})^2}{4 \cdot 10^{-7} \cdot 10^{-10}}} = \pu{5e-6 M} \tag{8}$$
Now back to (3):
$$[\ce{H+}] = \frac{10^{-14}}{\pu{5e-6}} = \pu{2e-9 M} \tag{9}$$
Knowing how the $\mathrm{pH}$ of the system has been changed, let's find what fraction of sulfide anions are in solution:
$$\alpha_2 = \frac{10^{-7} \cdot 10^{-10}}{10^{-7} \cdot 10^{-10} + 10^{-7} \cdot \pu{2e-9} + (\pu{2e-9})^2} = 0.0467 \tag{8}$$
Finally, solubility can be determined:
$$s = \sqrt{\frac{\pu{2.5e-10}}{0.0467}} = \pu{7.3e-5 M} \tag{10}$$
e.g. taking hydrolysis into account, solubility of $\ce{MnS}$ is approx. $4.5$ times higher (see equations (4) and (10)).

And yes, quantitative analysis can be tedious at times.

Thanks @andselisk your answer is correct I think, as I made a typo in the question, K2 was actually 10^(-14). Should I correct it?

@drake01 No prob, thank you for letting me know. Sure, go ahead and edit your question, I'll edit my answer at some point today:)

How did you obtain equation (5) and kh2 is net kh right?


text:
How do we identify the molecular formula from the empirical formula of a completely unknown substance?
Suppose you are given a sample of new (undiscovered) hydrocarbon with 600/7% $\ce{C}$ and 100/7% $\ce{H}$. (From combustion or any other method) We can find the empirical formula to be CH. Then how the we figure out whether it will be $\ce{C2H2}$ or $\ce{C4H4}$ or even something else?
Now we have identified the different compounds and can identify it through various reactions, but what would we do if we had something entirety new? How did chemists initially differentiate between these compounds?

If the empirical formula is known, then any method that will determine the molecular weight will do. Of merely historical interest is the use of freezing-point depression, nowadays, one would use mass spectrometry (electron-spray ionisation will leave most molecules intact).
This assumes that it is a reasonably small molecule, but only for those will the empirical formula be known before the molecular one.

Historically, chemists initially differentiated by subjecting this compound to all sorts of conditions, notifying the outcome and if somebody else had previously reported such a compound compare to their published properties to determine whether it was the same compound or not. They did not know what the molecular structure or sum formula was — and they did not need to. Merely knowing ‘this compound reacts just like Berzelius’’ or ‘this one has the same composition but is notably different’ was enough.
While the concept of e.g. tetravalent carbon was also a very old one, it did not mean that chemists had any understanding of what molecular structures looked like. For example, one German chemist from the late 19th or early 20th century preferred drawing ‘sausage depictions’, where a carbon would be a sausage of length 4 while each hydrogen would be a sausage of length 1 and oxygen of length 2. These would then be put together so that each sausage was double-layered. These served as depictions for the chemists’ imaginations, they were not considered an accurate representation of molecules in any way.
Finally, the ‘all sorts of conditions’ approach also allowed some deductions on the chemical nature. For example, if you were able to introduce iodine while something else gets removed and you end up with a known compound (identified previously e.g. as 1-iodobutane) you could infer that there must have been some functionality likely at the 1 position of a four-carbon chain. And so on and so forth.


text:
Why does the carbocation in this scheme react under elimination conditions rather than being attacked by water?
In the final step of the mechanism, elimination occurs over substitution, why is this?

For tertiary carbocations elimination products are generally produced, nothing special here.  On the other hand rearrangement that happens earlier is a bit interesting.

The simple answer is that water is a very weak nucleophile and substitution does happen reversibly, but usually dissociates immediately. After which, the final reaction shown happens.
There is likely a (very) minor product that follows along the substitution pathway with an alcohol final product.

So if water added to the carbocation, the c-o is cleaved mostly as oppose to deprotonating and forming the alcohol?

That is essentially correct

Moreover, for the bottom two examples here: https://prnt.sc/hny81u why is the substitution product formed over elimination?


text:
Is there a method to recover camphene from sublimation at ambient conditions?
Camphene solidifies and sublimates at ambient room temperature below 45°C. I'm aware that a glass funnel is often used to separate mixtures through sublimation of the volatile component (in this case camphene) from another material.
Is it possible to recover sublimated camphene using this or another method?

I think that easy setup should work. Are you afraid that it does not condensate ?

Yes. I'm worried the vapor will not condense on the inside of the glass funnel, and instead become a semi-solid or liquid mess. I think it's completely solid and sublimates at 20°C, and at around 25°C it starts to become semi-solid. At 45°C it's almost completely melted.

Cab you adapt something to be a cold finger? In principle you can use a beker full of water if you can confine the vapours.  Roll something around it

Thanks Alchemista! I think you had the right idea for sure.

Yes the answer confirms this

MIT has a brilliant technique for sublimating and condensing:
https://www.youtube.com/watch?v=dBNELFi5XiY
The sample is placed in a culture dish sitting on a hotplate. The lid is placed on top.  Finally a beaker with ice is placed on top of the dish to condense the sublimate. After a while crystals form on the top of the lid. These pure crystals are periodically scraped with a policeman and collected.


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Stability of extended enolates
I'd expect an extended enolate to be more stable than a normal enolate, when both can form, but my textbook suggests otherwise.
What is the stability of an extended enolate compared with a normal enolate? For example, following an enolate alkylation of the molecule depicted below, which product ($\alpha$- or $\gamma$-alkylated) will form?

It might be beneficial if you could provide a reference to your textbook.

It probably depends on the deprotonation conditions

@orthocresol Thanks! But could you please elaborate? Do you mean bulkiness of the base? How could you favour one product over the other?

I was looking for a post, because it has been discussed here before, and I found it: https://chemistry.stackexchange.com/a/43024/16683

@orthocresol Thanks a lot! The answer makes sense, but it doesn’t explain why alpha-deprotonation is kinetically favoured. Do you have any ideas?

With LDA for example you can form a six-membered TS (Ireland model) with a $\ce{C=O...Li}$ interaction. I've heard that the model is a simplification of the truth, but the bottom line is that there's probably some kind of involvement of the carbonyl group.

Yes, it’s almost a Duplikate of that question except one is about cyclic compounds the other about linear ones …


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Absorption spectrum of Hydrogen molecule in UV region 200 nm - 300 nm (not Atomic Hydrogen)
Can somebody provide Absorption spectrum of Hydrogen molecule? I have searched days for absorption spectrum of Hydrogen molecule in 200 nm - 300 nm UV region and All I got is absorption spectrum of atomic hydrogen. I do find little about absorption spectrum of hydrogen molecule in the region 90 nm to 180 nm in some astrophysics research papers.

Although now an old book  Herzberg, 'Spectra of Diatomic Molecules' gives details of the energy levels of numerous excited states (up to $\approx$ 100000 cm$^{-1}$) and references to the original papers. The ground state dissociation energy is $\approx $ 4.476 eV.

Being a homonuclear diatomic molecule, molecular hydrogen does not posses a pure rotational of vibrational (electric dipole) spectrum. The lowest excited singlet state is the B state (for historical reasons there is no A state in H$_2$) and transitions from this state to the X ground state are known as the Lyman band which starts around 120 nm, thus outside the range you are interested in. In principle, electric quadrupole transitions may be observed but these are several orders of magnitude weaker than electric dipole transitions. In addition, the dissociation energy of molecular hydrogen lies around 36000 cm$^{-1}$, so that only radiation longer than about 277 nm can result in bound-to-bound transitions. However, I expect that the Franck-Condon factors to these highly excited vibrational states are very poor.
So there reason that you cannot find a spectrum is that there is none.


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Finding the percent void of the crystalline structure of NaCl

Find the percent of void in the crystal structure of $\ce{NaCl}$.

I am trying to calculate the percent void of the packing structure of $\ce{NaCl}$. Let's denote radius of anion as $r$, radius of cation as $R$. What I know
$$\text{side length} = 2R + 2r$$
$$R + r = \pu{2.50 Å}$$
The ratio of cation radius to the anion radius is $0.600$. The equation to find void space:
$$100 \times \frac{\text{total volume} - \text{volume of spheres}}{\text{total volume}}$$
Yes, this is a homework question, but I'm stuck and I just need some help to get started.

Homework is fine, you just have to explain what you have tried and what specifically is giving you trouble. For example, you need to find the void space. It looks like you have the formula for it, so are you having trouble obtaining the total volume, volume of spheres, both?

For the assignment I can't use decimals.  I have not figured out how to set up the equation without solving for both of the variables.  Would I just call one R and the other 0.6R?

Yes of course! The absolute value of R+r is irrelevant, the percentage of voids is the same in any arbitrarily scaled model or in actual NaCl.

Ok, I tried that I ended up with 18%. Is that right?


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Chemical reaction to produce 5 liter of gas
I'm looking for a chemical reaction in order produce about 5 liter of harmless gas.
The goal is to start with something that takes as little volume as possible to start with, and reacts to fill up a bag, and also not producing too much heat to not harm the bag.
Every answer is appreciated!

I'd go with carbon dioxide $\ce{CO2}$ as a rather convenient gas which is non-flammable and doesn't require special equipment or exothermic reaction to synthesize. Common ways of obtaining $\ce{CO2}$ gas are:

Dry ice + ambient temperature above $\pu{−78.5 ^\circ C}$;
Carbonate + acid. These reactions rarely have significant thermal effect. One of the best mixture would be sodium bicarbonate + citric acid + water. Assuming dissolving of citric acid is endothermic and takes some time, prepare the solution of citric acid beforehand.

$\ce{O2}$ and $\ce{H2}$ are rather dangerous in terms of flammability, also common synthetic methods require exothermic reactions.
$\ce{N2}$ is inert, but not easy to come by; the most effective way of obtaining nitrogen would be thermal decomposition of azides $\ce{M^IN3}$ and ammonium nitrite $\ce{NH4NO2}$.

There used to be little ‘bomb bags’ available. Sealed bags containing sodium hydrogencarbonate and a smaller internal bag with acetic acid. You broke the inner bag and after some time the pressure built up inside would make the bag explode.

@Jan Interesting. Must be local though; when I was a kid calcium carbide and some water in a tightly closed bottle used to be a trend:)

@andselisk damn, I don't think acetylene in shoddy sealed containers is a good thing to play with as a child hahaha

@Eskir Definitely not! I was referring to different time and place, namely in USSR and 1990s Russia it was totally normal for kids to "borrow" some calcium carbide from a construction site and go for blast fishing.


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Does 1-ethylidene-2-(propan-2-ylidene)cyclobutane show geometrical isomerism?

I think it should be yes, but my book says no.
I think the two geometrical isomers are
Z-isomer

E-isomer

Which book? Please state the source properly.

The book is wrong unless one of the two does not exist. For the (hypothetical? ) structures you are right.

Well, there would be sever sterical clash for Z, so it's practically non-existant.

The two isomers are distinct substances, differing only in the orientation of a double bond, so they must be geometrical isomers. However, as a practical matter, they may interconvert under mild conditions and/or one may be much less stable than the other.


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Is steric inhibition of resonance or steric inhibition of protonation dominant in o-toluidine?
I wish to know which effect out of steric inhibition of resonance (SIR) or steric inhibition of protonation (SIP) is dominant over the other when comparing basicities of o-toluidine and aniline:

Case 1, using only SIR effect
We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. Hence, we can say o-toluidine is more basic than aniline (as nitrogen's lone pairs are in resonance with benzene).
Case 2, using only SIP effect
If we consider that methyl's bulkiness shields the lone pair of nitrogen from protonation in o-toluidine and no such inhibition or shielding happens in aniline, we can conclude that aniline is more basic than o-toluidine.
Both these cases are contradictory. Can anyone please resolve the discrepancy?

The reason for the decreased basicity is essentially the ortho-effect as outlined here: Ortho-effect in substituted aromatic acids and bases.
Your question on the other hand asks about a seeming contradiction, but I believe there is a small misconception on your part.

We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. [...]

I don't see how resonance could be inhibited, as that essentially requires that the lone pair orbital of nitrogen would have to rotate significantly towards an in-plane arrangement. The methyl group is simply not big enough for that to happen, and if it were, the lone pair would more likely rotate toward the more shielded side, further reducing acidity.
The highest occupied molecular orbitals (HOMO) of the molecules show this quite nicely, as they are very, very similar.

On the DF-B97D3/def2-SVP level of theory the RMSD of the core structure (excluding the hydrogen and the methyl group) is negligible with only 0.03 Å.
From this I would conclude that there is no steric inhibition of resonance.
On the other hand, steric inhibition of protonation is given in 2-methyl aniline, where it is obviously missing in aniline.
Additionally there is a stabilising non-covalent interaction between the methyl and amine group, which can be analysed with NCIPLOT, and then displayed. The green blob is weakly attractive, red is strongly repulsive, blue would indicate strongly attractive interactions.

TL;DR: The dominant effect is steric inhibition of protonation, as steric inhibition of resonance is negligible (if possible at all).

Nitrogen’s lone pair can be in a number of different orbital types depending on the geometry of the protons around it. The most important are $\mathrm{sp^3}$ and pure $\mathrm p$. The interconversion between the two can lead to the substituents switching sides; therefore, this process is known as nitrogen inversion.
Nitrogen inversion means that the lone pair is accessible from both sides, so you need to block both sides sterically if you want to inhibit protonation.
To inhibit resonance sterically, introducing an inhibitor on one side is enough since the group must be perfectly in plane for resonance to be significant.
Thus, aniline is less basic than 2-methylaniline.

But I have found out that pKb value of 2-methylaniline and aniline are 9.56 and 9.4 respectively, making aniline more basic. sources:https://en.wikipedia.org/wiki/Aniline and http://www.microkat.gr/msds/o-toluidine.html

@ShrishShankar Interesting. Although Wikipedia does not quote a $\mathrm pK_\mathrm a$ value for 2-methylaniline or ortho-toluidine, I have confirmed the prediction elsewhere. I am at a loss how to explain that, since also the inductive effect of the methyl group should make 2-methylaniline more basic. We may argue that the effect is extremely small, though; there may be a countereffect I can’t think of right now.

There are two methyl groups needed to really block conjugation .

@Mithoron one methyl group is more than enough

@ShrishShankar Then you should revise your knowledge, as Martin's answer to question of which this one was marked as duplicate, proves otherwise.

@ShrishShankar Because resonance isn't really inhibited in  2-methylaniline, it's acid-base properties are connected with so-called ortho effect.

To inhibit resonance sterically you have to include a huge™ moiety, because you have to essentially rotate the lone pair into the plane of the aromatic ring. "Inhibition of protonation" (as I read it: hindering protonation) is rather easy, just put a propeller right next to it, blocking one of the entry paths.

@Martin-マーチン If we consider pyramidal inversion in combination with one side of the nitrogen atom being sterically blocked (so proton approach is stopped)... It does seem that one would need more than one bulky group to effectively stop protonation.

From what Martin says, the delocalization of the electron pair is almost unaffected in the two species being compared. However... In aniline, pyramidal inversion should be less significant than in 2- methylaniline. Reason: In aniline, the lone pair is more or less always oriented with the p orbitals of the benzene ring. Nitrogen is between sp3 and sp2. The same is not the case for 2- methylaniline, because of the presence of a methyl group. So, while Resonance effect is slightly lesser in 2- methylaniline, pyramidal inversion makes the location of the lone pair uncertain. Two factors at play.

Just thinking. I'm not really very sure about this.


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Elements in the lanthanide and Actinide series
How many elements are there in lanthanide and Actinide series? My big doubt is does the 57th element Lanthanum and 89th element Actinium include in it?
In some periodic table the entire 57-71 is given outside the box but in some of them only 58-71 is given outside the box.

The arrangement of elements in periodic table is a subject of debate. There is no particular arrangement which is the best. This is the reason of confusion.
However, Concise Inorganic Chemistry by J.D. Lee mentions:

These four elements [Sc,Y,La,Ac] are sometimes grouped with the 14 lanthanides and called collectively the 'rare-earths'. This is a misnomer because the scandium group are d-block elements and the lanthanides are f-block elements.
Pg 677, Concise Inorganic Chemistry, J.D. Lee, Blackwell Science Ltd, 1996.

So you can consider that the lanthanide and actinide group each contains 14 elements and Lanthanum and Actinium are d-block elements.

Couldn't it be a 15 in Lanthanide series and 15 in actinide series? Because chemical property is the period function of the atomic number.


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Why is the bond angle of N-N-N in hydrogen azide not 180 degrees?
Chen and Wang (2009) have reported an $\ce {N-N-N}$ bond angle in the hydrazoic acid molecule of $171.5$ degrees using B3LYP/6-311G** model. The wikipedia page for the molecule also reports a similar bond angle of $171$ degrees. However, this seems to go against predictions by VSEPR theory for the case of two regions of electron density (which gives a bond angle prediction of $180$ degrees). It is known that VSEPR theory is limited in its theoretical predictions. However, I would like to understand what accounts for the discrepancy in the two.
Reference
Chen, F., & Wang, F. (2009). Electronic Structure of the Azide Group in 3′-Azido-3′- deoxythymidine (AZT) Compared to Small Azide Compounds. Molecules, 14, 2656-2668. doi:10.3390/molecules14072656

Also relevant paper on the recent SCXRD investigation (DOI: 10.1021/ja2027053) suggests somewhat higher average $\ce{N-N-N}$ angle 172.8°. Besides, all 3 N atoms and both bonds in $\ce{N3}$ are quite inequivalent, so there is no wonder the angle is not 180°.

VSEPR doesn't predict anything.

related https://chemistry.stackexchange.com/questions/19904/what-is-the-hybridization-of-the-nitrogen-in-the-azide-ion

https://acs.figshare.com/articles/dataset/Molecular_Structure_of_Hydrazoic_Acid_with_Hydrogen_Bonded_Tetramers_in_Nearly_Planar_Layers/2624300/1     Molecular Structure of Hydrazoic Acid with Hydrogen-Bonded Tetramers in Nearly Planar Layers


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Mechanism of hydrogen evolution on metal-oxide electrodes
I am looking into the mechanism of different electrocatalytic reaction in acidic media on metal-oxide electrodes.
Reduction of on such electrodes generally happens with the help of electrons from the conducting band of the electrode, which is mainly d bands of the metal. That made me wonder about the mechanism of hydrogen formation: I assumed hydrogen is formed from protons chemisorbed on the surface. However, protons are likely to attach to surface oxygen ions, not metal ions, therefore it seems difficult to receive electrons from the conducting band.
Different of water-related electrocatalysis seems very extensively studied, including many atomistic simulations, however, I haven't found much on hydrogen evolution on oxide electrodes. Is there an established mechanism for hydrogen formation?

I don't think there's established mechanism, there may be postulated one, but heterogenic reactions are notoriously difficult as far as finding mechanism is concerned.


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Have I inadvertely diethyl ether peroxides?
About 1 week ago, I used diethyl ether to wash some solid products.
I dumped it in my temporary waste container, which also contains water and some other solvents like DCM.
I returned today, and discovered that my waste container is full of yellow to brown crystals, and possibly some oil of similar color.
Needless to say I didnt stick around for long to investigate or even touch anything, as I realized that I might have turned my fume hood into a minefield (ether peroxides are friction explosive).
I wilk of course E&HS at my faculty, but it's saturday so no one is around.
Is it plausible that I've made dangerous peroxides? Has anyone else done the very stupid thing which is to store diethyl ether alongside with water in an open container for 1 week?

Diethyl ether left in an open container will be completely gone in a matter of hours. Peroxides are scary, but I wouldn't expect them here.

Additional comment.  Try not to contaminate organic solvents waste with water based one as it can pose severe corrosion problems at a big and  further stages of the disposal process. Moreover the treatment of halogenated waste is more expensive and is not economic to add too much not alogenated mass to it.

@Alchimista It depends on the waste solvent removal policies of the institute. When I did my bachelor’s, we separated organic, halogen-organic and aquaeous. When I arrived at my master’s, halogen-organic and organic were no longer separated because people were too stupid to do that (that was the reason they gave!). When I did my PhD at a different institute, everything was poured into a single container. Now as a postdoc on the other side of the planet, all three are separated again.

It's only halogen containing org. solvents and non-halogen containing org. solvents here. No rules about getting water in either one of them, although you're advised to use your head and not do things like pouring saline water into the halogen container (dissolved NaCl is technically 'halogen containing'...)


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Can suspensions be separated by centrifuge or by separatory funnel?
I want to know if we could separate suspensions whose particle sizes are more than $\pu{100 nm}$ like mud by a normal centrifuge?
Could we use a separatory funnel for these situations?

I mean, filtering might work. What is the range of sizes of particles?

Also, suspensions eventually settle

@JavaScriptCoder After watching my colleague separate some iron nanoparticles I can tell you that ‘eventually’ can mean an incredibly long timeframe …

@Jan yes, but wouldn’t mud be different than iron nanoparticles?

@JavaScriptCoder Sure, but your ‘all’-implying sentence needed a partial rebuttal ;)

If you are indeed dealing with a proper suspension (not a colloid and not a solution) then centrifugation should always work to separate the components. It may be that you need a very high rotation speed or a long time, but you will manage to get the solid particles to the bottom of the tube and the clear liquid above.
A separatory funnel is a bad idea for a number of reasons. For one, you don’t know how long it will take for the components to separate. For two, you will probably have difficulties removing the solid portions; they might clog the opening or they might attach together not intending to go through at all. Finally, you will always get some liquid together with the solid fraction in a separatory funnel while you should arrive at a much better dryness with centrifugation.


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How to tell 'orthokinetic' apart from 'perikinetic' based on their prefixes?
During a test in colloid and surface chemistry, students were asked to distinguish between two terms: orthokinetic and perikinetic coagulation.

The rate of aggregation is in general determined by the frequency of
  collisions and the probability of cohesion during collision.

If the collisions are caused by Brownian motion, the process is called perikinetic aggregation;
[If the collisions are caused] by hydrodynamic motions (e.g. convection or sedimentation), one may speak of orthokinetic aggregation.

from  Everett, D. H. 'Manual of Symbols and Terminology for Physicochemical Quantities and Units, Appendix II: Definitions, Terminology and Symbols in Colloid and Surface Chemistry'. Pure and Applied Chemistry, 2009, 31 (4), 577–638, p 610. doi: 10.1351/pac197231040577

Perhaps unsurprisigly, I got the terms mixed up, and used definitions the wrong way round. The problem was that nothing too obvious popped out when considering the prefixes and the respective processes.

Question: To which meaning or association are the prefixes peri- and ortho- referring? The definitions are mostly clear. It is just that I would rather avoid memorisation, and would prefer to be able to always recall which is which based on the prefixes alone.

This means other meanings and etymology of peri- and ortho- are welcome, but will be considered off-topic on their own, unless a clear connection to the definitions of orthokinetic and perikinetic is provided. A short answer with an authoritative reference is very acceptable.

Haven't found a good source, but my guess is the difference is related to undirected vs directed motion. Peri- means near or around, so perikinetic aggregation is just due to any motion around the cluster. Ortho- can mean straight, which to me would suggest orthokinetic aggregation is caused by processes directed at the cluster.


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Formal charges on polyatomic ions
Why is it that sometimes a polyatomic ion's Lewis structure is adjusted to remove the formal charge on the central ion, but sometimes it is not?
For example, in the $\ce{I3-}$ ion (triiodide), the Lewis structure is the following:

This gives a single formal negative charge to the central iodine. In my textbook, it says to remove formal charges from the central atom whenever possible, so couldn't we form a double bond with one iodine and therefore move the formal charge off of the central atom like this?

Comparatively, in $\ce{IO4-}$ (periodate), the Lewis structure is:

There, the single formal negative charge is on one of the oxygens instead of on the central ion, making it a resonance structure.
So basically what I'm asking is why is there a formal charge on the triiodide ion's central atom but not on other ions' central atoms? Like why is it okay that the triiodide ion has a formal charge on the central atom when it could be moved off to create a resonance structure, such as in $\ce{IO4-}$?

Ouch, all those structures hurt my eyes D=

Yeah, sorry :( I had to create the second two using Google drawings because I couldn't find an appropriate picture. Getting all of the electrons to line up well is hard! Haha sorry about that

They hurt because they are so wrong, not because of the way you drew them (I have seen much worse attempts at drawing structures; yours’ are actually legible, rather well-spaced, etc.)

First off, in neither triiodide nor periodate have the structures you drew. They all assume octet expansion on iodine which has been disproven. It was traditionally explained by the ‘participation’ of iodine’s empty 5d orbitals but these are energetically too far removed and do not take part in bonding to any noticable extent.
Second off, it is wrong to give a formal charge to the central iodine of triiodide. In fact, it is the two outer iodines that carry the negative charge. The structure should be given as follows with a four-electron-three-centre bond between the two iodines:
$$\ce{I-I\bond{...}\overset{-}{I} <-> \overset{-}{I}\bond{...}I-I}\tag{1}$$
Because iodine adheres to the octet rule, there is no contribution by any double-bonded structure. That would create a superoctet structure.
In periodate, there is no question that oxygen is far more electronegative than iodine and that iodine’s oxidation state of $\mathrm{+VII}$ does not go well with a formal or partial negative charge. Thus, it is only logical for the outer oxygen atoms to bear the negative charge. A better way to draw it would be in adherence to the octet rule, i.e. four $\ce{O-}$ substituents and a central iodine with a $3+$ formal charge.
There is nothing inherently wrong with a formal charge on the central atom, though. Take for example tetrahydridoborate $\ce{BH4-}$, the addition product of borane and hydride. Here, boron should only have three valence electrons but if bonded to four hydrogens it must have four. Thus, it carries a formal negative charge. (In reality, the hydrogens carry the negative charge, though, since they are again more electronegative than boron.)


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Does dinitratomethane exist?
In 2016 synthesis and structure of tetranitratoethane $\ce{C2H2N4O12}$ were reported [1]. Currently (as of 2017-Q4) this is the only dinitrato-carbon compound with the well-established crystal structure and composition:

This article also mentions dinitratomethane (reference number updated from 6 to 2):

While nitrate esters such as nitroglycerin (NG) and pentaerythritoltetranitrate (PETN) (Figure 2A and B) are well known,  only very few examples of germinal C-nitrato compounds are known. The simplest one, dinitratomethane (Figure 2C) is a liquid which is obtained from nitration of 1,3,5-trioxane in a $\ce{HNO3/H2SO4}$ mixture [2].

Figure 2. Structural formula of nitroglycerin (A), PETN (B), and dinitratomethane (C).

However, I wasn't able neither to get the full-text of original paper from 1938, nor any more recent proofs of existence for dinitratomethane. Google Scholar returns only a couple of entries one of which is [1], and another one is a PhD thesis. There are also zero Ngrams for "dinitratomethane", which is unusual. ChemSpider also lists this compound as nitrooxymethyl nitrate (CAS 38483-28-2).
Can anyone approve or disapprove the existence of dinitratomethane $\ce{CH2N2O6}$ (feel free to utilize computational chemistry too)?
References

Fischer, D.; Klapötke, T. M.; Stierstorfer, J. Chem. Comm. 2016, 52 (5), 916–918. DOI: 10.1039/C5CC09010E.
G. Travagli, Gazz. Chim. Ital., 1938, 68, 718.

"dinitratomethane" sounds off to me, I'm afraid authors weren't too good in nomenclature-fu. Better names can be Methylene (or methanediyl) nitrate, or nitroxy instead of nitrato. Nitrato is used for inorganic nomenclature.

For some reason I’m not surprised at all to see Klapōtke in that author list. (Same for Stierstorfer)

After some more research I'm pretty much convinced in its existence. As Mithoron noticed in the comments, the name dinitratomethane is somewhat confusing, and is indeed not preferred.
I suppose the most accurate IUPAC-compatible one would be (nitrooxy)methyl nitrate. It also turned out that in the literature (mostly on regulated and hazardous chemicals) this compound is generally known as methylene glycol dinitrate.

The second volume of a classic handbook "Chemistry and Technology of Explosives" [1, p. 155] gives the most extensive description I discovered (reference numbers are updated):

Methylene glycol dinitrate, or "nitromethylene glycol", is an oily liquid having a boiling point of $\pu{75-77 ^\circ C}$ at $\pu{20 mm Hg}$.
It is a strong explosive, capable of dissolving collodion cotton. However it is not stable, being readily hydrolysed and therefore it is of no practical significance.
A method of obtaining this explosive, by the nitration of moist
  paraformaldehyde was patented by Travagli and Torboli [2]. They utilised the well-known property of formaldehyde of reacting like methylene glycol in the presence of water.
According to Travagli [3,4] methylene glycol dinitrate is obtained as follows:
  paraformaldehyde is dissolved in sulphuric acid and added drop-wise at $\pu{3-5 ^\circ C}$ into a mixture of nitric and sulphuric acid. The resulting oil is separated and washed with water to give a final yield of $35\%$.

The recent handbook of liquid explosives from 2015 [5, p. 227] contains virtually the same information without any updates, which indicates there simply hasn't been much of an interest in this compound due to its instability and lack of possible applications.
References

Tadeusz Urbanski. Chemistry and Technology of Explosives, 1st ed., Pergamon Press GmbH, 1965; Vol. 2.
G. O. Travagli and Torboli, Ital. Pat. 338080 1935.
G. O. Travagli, Gazz. Chim. Ital. 1938, 68, 718.
G. O. Travagli, Chimica ed Ind. 1938, 20, 389.
Jiping, L. Liquid explosives; Springer Berlin Heidelberg: New York, NY, 2015. ISBN 978-3-662-45847-1.

Admit it: you knew this all along and just wanted to get two hats ;p

@Jan No, I didn't know it at a time. The fact I might get two hats was a good enough carrot to do more research:)


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Home test for magnesium carbonate
People that do rock climbing and weightlifting usually use magnesium carbonate on their hands to decrease slippery hands.
I would like to know how can someone that buys MgCO3 can really be sure it's magnesium carbonate and not another thing.
Is there a simple test people could do at home?

Magnesium carbonate is dirt-cheap and I can hardly imagine why anyone would substitute it with anything else. Are you trying to proof it's mostly MgCO3, or discover any third-party (organic/inorganic) components in there?

@andselisk, I'm trying to discover if it's mostly MgCO3. I imagine the manufacturer could add calcium carbonate (CaCO3) or something else into it without the customer knowing it.

Flame test can tell you something. The title is misleading as for identifying X is one thing. Assuring that only X is in there is a whole different story.

If all you want to know is the relative amounts of calcium and magnesium carnbonates disolve it in vinegar (it will fizz). Now add lye until the pH is about neutral (use pH test strips) in order to neutralize any excess acid. Obtain a test kit for determining water hardness that will give both the magnesium and calcium hardness values for a sample. Dilute with DI (or nearly DI i.e. RO) water to be within range of the kit Be sure you understand what units the kit is reporting in. Usually it is units related to the equivalence of the calcium and magnesium so that if calcium harness turns out to be 1/5 magnesium hardness that tells you the original mix has 1/5 of a mole of CaCO3 for each mole of MgCO3.


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Charge separation and stability in resonating structures
I know that when there are like charges on a resonating structure, the farther apart they are, more stable is the structure due to reduction in repulsive forces.Is the same true for opposite charges? which structure is more stable -
1)with positive charge on one C and negative on adjacent C
2)with positive charge on one C and negative charge on C separated by 4 carbons between them?

I haven't seen this before but i think a hydrogen will shift from the positively charged carbon to the negative carbon... Thus neutralising both charges

In general, the stability of resonating structure is of important within the hybrid of resonance. From a purely electrostatic point of view distance improves stability. A very long dicarboxyloc acid should have just a pKa value, to give you an idea.  As far as I know the importance of the zwitterions you described should be negligible anyway....

A carbon with a negative charge adjacent to carbon with a positive charge will form a pi bond.

I thought about this as well, since e.g. the most stable canonical for the sulfate anion is at least yiilidic(formally doubly positively charged sulfur directly connected to formally singly negatively charged oxygens).

Let's take the example of 1,3-butadiene:

In order to talk about the energies and stabilities, we need to look at the molecular orbitals of the compound.
Look at this diagram from ChemLibreTexts

The two filled ground states, both have $\ce{\pi}$ bonds between $\ce{C_1}$ and $\ce{C_2}$, and $\ce{C_3}$ and $\ce{C_4}$ suggesting that separation of charges is not favorable.
The charged resonating structures with charges on $\ce{C_1}$ and $\ce{C_4}$ corresponds to $\ce{\Psi_3}$ in the MO-diagrams where the $\ce{\pi}$ bond is present between $\ce{C_2}$ and $\ce{C_3}$.
This observation questions which structure $\ce{\Psi_4}$ corresponds to. Since no $\ce{\pi}$ bonds are present, I reckon the corresponding resonating structure would be one with charges on each of the four carbons:

Conclusion: At least for the case of but-1,3-diene, structures with both charges placed on the $\ce{\alpha}$-Carbon (to the $\ce{\pi}$ bond) are more stable.
I also want to briefly talk about the MO-diagram of 1,3,5-hexatriene:

Here too, keeping the charges $\ce{\alpha}$ to a $\ce{\pi}$ bond seems more important than the separation between the charges. I wonder if the "transition" charged states even exist under normal conditions.
Reference
Francis A. Carey, Richard J. Sundberg - Advanced Organic Chemistry Part A. Structure and Mechanisms-Springer (2007)

Criteria to determine stability of resonating structures in decreasing priority are listed below :

Neutral resonating structures with complete octets are most stable
Ionic resonating structures with complete octets of all atoms is the next criteria to determine stability of resonating structures
Ionic structures in which carbocations are present (with incomplete octets) having unlike charges closer, is ranked at the third priority to determine stability.
Finally, anions at more electronegative element are more stable.

So, according to the information you have provided, the resonating structure which corresponds to the first case is more stable (following the third point)

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text:
Is selenocysteine C3H6NO2Se or C3H7NO2Se?
I'm a beginner in chemistry, and recently I am interested in selenocysteine.
While most websites (Wikipedia, ChemSpider) say it is $\ce{C3H7NO2Se}$, PubChem says it is $\ce{C3H6NO2Se}$.
Which one is correct? Or are both reasonable?

I think the only difference is whether the selenium is protonated or not. So the only difference is where you might encounter each form.

Selenocysteine can be either of the following: $\ce{C3H8NO2Se+, C3H7NO2Se, C3H6NO2Se-}$ or $\ce{C3H5NO2Se^2-}$. This is because the neutral form of selenocysteine is both a base and a diprotic acid: the amino group can be protonated and both the carboxylic group and the selenol can be deprotonated.[1]

The above graph taken from the chemicalize.com property calculation site[2] shows the distribution of the different forms of selenocysteine in aqueous solution. The underlying data has been calculated by chemicalize.com and is not experimental data. For the cell, neutral pH is most important. We can see that the most common form by far at neutral pH is $\ce{C3H7NO2Se}$, but a small quantity is present as $\ce{C3H6NO2Se-}$. Once selenocysteine has been incorporated into a protein, only the selenol function remains potentially acidic in aqueous solution; however, it’s $\mathrm pK_\mathrm a$ value is higher than neutral pH by three logarithmic units. Thus, we can assume that it will not be deprotonated notably.
The actual $\mathrm pK_\mathrm a$ values have been measured by Huber and Criddle.[3] These show that chemicalize.com’s predictions are rather bad. They determine the $\mathrm pK_\mathrm a$ values to be $2.01,5.24$ and $9.96$ for the carboxylic acid, the selenol and the amino group, respectively. Using these values, we expect selenocysteine to be present in the monoanionic form $\ce{C3H6NO2Se-}$ at cellular pH. A distribution graph based on these values (provided by Canadianer) can be seen below.

PubChem’s suggestion of $\ce{C3H6NO2Se}$ is wrong. If you add up valencies, you will notice that it is only possible as a radical structure. It would be acceptable if a negative charge had been included (which I don’t see on the site).
On the other hand while the chemical formula on Wikipedia or ChemSpider is correct and gives a neutral compound with all spins paired, this formula does not correspond to the species predominantly present in strongly buffered aqueous solutions.

Notes:
[1]: This way of writing it assumes the uncharged neutral form. In aqueous solution, the molecule is expected to adopt a zwitterionic neutral form with the carboxylate being deprotonated and the amino group being protonated. From this form, the carboxylate is the base while the amino group is one of the two acidic functions.
[2]: Unfortunately, I am not aware of a way to link directly to the calculation. However, this calculation is free if you register, since the molecule contains less than twelve heavy atoms.
[3]: R.E.Huber, R.S.Criddle, Arch. Biochem. Biophys. 1967, 122, 164–173. DOI: 10.1016/0003-9861(67)90136-1.

The legal notice of  chemicalize.com is hilarious. They are not responsible for the accuracy of the data, don't allow you to use third-party tools to work with their service, have their own micro-transaction credit system, and on top of that: no refunds. Wow. Great answer by the way:)

Not sure why our comments were deleted, but thanks for looking into experimental pKa. You mentioned that you were looking for a graph similar to that in your post but with the experimental data. You can use this one if you like; I tried to infringe on chemicalize's style as much as possible. PS: I realize this minor detail isn't the crux of your answer in any way, I just had some free time on my hands.

@Canadianer  Comments are supposed to be deleted after they served their purpose. I deleted mine and probably flagged yours’ for deletion, the usual Stack Exchange way. Thanks for the graph which I included.


text:
Calculating the pH of a ammonium hydrogen sulfate solution
I am attempting to complete a question in which I need to use an ICE table to calculate the pH of a $\pu{2.61 mol L-1}$ $\ce{NH4HSO4}$ (ammonium hydrogen sulfate) solution.
When $\ce{NH4HSO4}$ dissociates I know that it splits into $\ce{NH4+}$ and $\ce{HSO4-}$ ions.
According to my textbook, $\ce{HSO4-}$ is actually an acid and not a base, so I set up my equilibrium expression like such:
$$\ce{NH4+ + HSO4- + 2 H2O <=> 2 H3O+ + SO4^2- + NH3}$$
$\ce{NH4+}$ has a $K_\mathrm{a}$ of $\pu{5.56e-10}$. $\ce{HSO4-}$ has a $K_\mathrm{a}$ of $\pu{1.2e-2}$.
According to my ICE table:
$\ce{NH4+}$ and $\ce{HSO4-}$ have starting concentrations of $\pu{2.61 mol L-1}$.
Both lose $x~\pu{mol L-1}$, and have equilibrium concentrations of $2.61 - x$.
$\ce{H3O+}$ has a concentration of $2x$ at equilibrium. $\ce{SO4^2-}$ and $\ce{NH3}$ each have a equilibrium concentrations of $x$.
To calculate the final answer I am using the $K_\mathrm{a}$ of $\ce{NH4+}$ and plugging in the values from my ICE table for the equilibrium concentrations of the products and reactant.
Since $K_\mathrm{a}$ of $\ce{NH4+}$ is so small, I am disregarding the changes and substituting $2.61$ instead of $2.61 - x$.
The final answer I am getting is not the same as the answer in the back of the book which is approximately $\mathrm{pH} = 0.76$.
What is wrong with my process?

The answer in the textbook looks way off. Are you sure it's pH = 0.76?

I mean, there is no way for me to be sure that the answer is correct because I don't even know if the process is right. But the textbook does say 0.76

The hydrogen sulphate ion is a much stronger acid than that of the ammonium ion. Thus, almost every hydronium ion in the solution comes from the dissociation of hydrogen sulphate ions.
This said, the assumption,

"$\ce{H3O+}$ has a concentration of $2x$ at equilibrium. $\ce{SO4^2-}$ and $\ce{NH3}$ each have a equilibrium concentrations of $x$"

is wrong.
When you solve the quadratic equation
$$K_\mathrm{a} = \frac{[\ce{H3O+}] [\ce{SO4^2-}]}{[\ce{HSO4^-}]} = \frac{x^2}{c - x}$$
you will get a pH near your text book's value.


text:
Product of primary amine and acid anhydride
How does an acid anhydride react with a primary amine? I thought it'll be some sort of condensation reaction, so I made the product by removing a water molecule and got a cyclic product. However, the product given is an open chain molecule (major).

How does the reaction occur? How does the ring open? Could someone please give a detailed mechanism?
P. S. What is the general reaction for primary amines and acid anhydrides?

It should be very clear that a primary amine is a nucleophile and an acid anhydride an electrophile. In the series of carbonyl and carboxyl derivatives, an acid anhydride is classified as quite reactive: more so than aldehydes or ketones, much more than esters or amides but less than acid chlorides. Thus, we should all agree that the first step is the nucleophilic attack on the $\ce{C=O}$ π system via a tetrahedral intermediate to give N-methyl succinic monoamide.

Scheme 1: Mechanism of the monoamide formation.
With that, we can consider whether further reactions will take place. Where we originally had an anime and an acid anhydride, we now have a carboxylate (the $\ce{C=O}$ bond that wasn’t attacked) and a secondary amide. The secondary amide still has a lone pair we might consider to be nucleophilic — but that lone pair is taking part in amide resonance and is thus not available for nucleophilic attack (at least not well). And to remove the amide’s proton, we would need a strong base which we don't have. Thus, the amide side of the equation is far less reactive.
What about the carboxylate? Thus highly symmetric structure is also very unreactive towards nucleophilic attack in the same way as an amide. Due to the negatively charged oxygen donating one of its lone pairs to the π system, the carbon atom is much less electrophilic than in an acid anhydride. Thus, in this case also there is no desire to react.
Since we have two functional groups of low reactivity, the reaction ends here. However, with enough heat (supplying activation energy), it is possible for the sluggish intramolecular reaction between the the amide and the carboxylate to occur. In principle, the mechanism is the same except that you must transfer a few protons and remove water.

Scheme 2: potential onwards reaction to succinimide under harsh conditions.

How do you explain the simultaneous formation of N-methylsuccinamic acid?

@user28968 Well, with enough heat, some of the amide may attack the carboxylate anyway, but it is going to be a slow process.

That's particularly the reaction I need the exact mechanism for. I'll be very grateful if you could help me with the same

@user28968 Wait … maybe I’be got my nomenclature wrong but the major product according to your question is the one whose formation my answer describes. I thought that was the one you needed? (In any case I can’t draw until tomorrow.)


text:
Reaction of Diols with HX hydrohalic acids, Competing SN2 Reactions
Suppose 1,5-pentanediol were exposed to something like hydroiodic acid. Would the alkoxonium end of the diol be exposed to a nucleophilic attack by the remaining hydroxide at the other end for a ring formation? Or would the newly-free iodide ion be more competitive for substitution? Thanks!
edit: It was pointed out to me "newly-free iodide" was quite confusing. I meant it as though HI dissociated, the Hydrogen ion attacked one end of the diol, and the iodide from the HI being "newly-free". Now that one end of the diol has a good leaving group (H2O), is it more susceptible to be attacked by the iodide, or by the other end (OH) of the diol.

Unclear as to the meaning of "newly-free iodide". Did you mean 5-iod0pentane-1-ol? Depending on conditions, the cyclic ether (tetrahydropyran) would be susceptible to HI cleavage with formation of 1,5-diiodopentane.

Thank you so much for the help/feedback! I did my best to clarify in the edit.

Basically, you’re asking whether the product will be 1,5-diiodopentane or tetrahydropyrane (oxane), aren’t you?

Bob:
Here is the likely process. The alcohol 1 is protonated to provide intermediate 2 which undergoes SN2 displacement. The closure of 3 to  tetrahydropyran 6 is unlikely under the reaction conditions. Rather a second displacement via 4 would lead to diiodide 5. If the tetrahydropyran did form, ethers are cleaved by HI. The conversion of 3 to 6 is normally accomplished in the presence of base (NaH/THF).


text:
Calculation of equivalent weight

When a solid element is reacted with chlorine, a gaseous chloride of vapor density $68.75$ is formed. If this reaction is performed at constant temperature and pressure, the volume of the system reduces by $\frac{1}{3}$. What is the equivalent weight of the solid element?

I found out the molar mass of the gas $68.75 \times 2$. But don't know how to proceed from here?

I would express equivalent weight $M_\mathrm{eq}(\ce{E})$ of unknown element $\ce{E}$ using the number of equivalents $x$ and corresponding molecular weights $M$:
\begin{align}
M(\ce{ECl_x}) &= x \cdot M_\mathrm{eq}(\ce{E}) + x \cdot M(\ce{Cl}) \tag{1} \\ \to \quad M_\mathrm{eq}(\ce{E}) &= \frac{M(\ce{ECl_x}) - x \cdot M(\ce{Cl})}{x} \tag{1a}
\end{align}
To find $x$, let's apply to the balanced reaction, assuming it is complete and there is no chlorine left:
$$\ce{E (s) + x/2 Cl2 (g) -> ECl_x (g)}$$
The amounts $n$ of gaseous products are bound by stoichiometry:
$$n(\ce{Cl2}) = \frac{x}{2}n(\ce{ECl_x}) \tag{2}$$
At constant temperature and pressure $n_i V_i = \mathrm{const}$. The problem says the volume of the system reduces by $1/3$, which means:
\begin{align}
V(\ce{Cl2}) &= \frac{3}{2}V(\ce{ECl_x}) \tag{3} \\
\to \quad n(\ce{Cl2}) &= \frac{3}{2}n(\ce{ECl_x}) \tag{3a}
\end{align}
Now, comparing/equating (2) and (3a) it's easy to see that $x = 3$:
$$\frac{x}{2} = \frac{3}{2} \quad \to \quad x = 3$$
You already correctly determined molecular weight from vapor density $\rho_\mathrm{v}$:
$$M(\ce{ECl_x}) = \rho_\mathrm{v} \cdot M(\ce{H2}) = 68.75 \cdot \pu{2.016 g mol-1} = \pu{138.60 g mol-1} \tag{4}$$
All that's left now is to do the math with (1a):
$$M_\mathrm{eq}(\ce{E}) = \frac{\pu{138.60 g mol-1} - 3 \cdot \pu{35.45 g mol-1}}{\pu{3 equiv}} = \pu{10.75 g mol-1 equiv-1}$$
I'm not sure this makes any sense from the chemistry prospective as this would suggest that element $\ce{E}$ is sulfur and so the gaseous product is $\ce{SCl3}$. I would've expected the compound to be something like $\ce{BCl3}$, but in this case the vapor density should've been $58.12$.

I just made some rough calculations. And my answer (equivalent weight of the solid) comes out to be 10.33g.
As you have not mentioned the correct answer and since my answer has not matched with that of my previous answerer, so I am just providing an image of my drafting work and I am not going to type the full answer.

Since you decided not to type the answer properly, I decided not to spend time trying to decipher the handwritten draft; but the equivalent weight of an element (from the header) is supposed to be given in $\pu{g mol-1 equiv-1}$, not just in grams.

Actually I am a student and not quite used to the SE script writing.....so, it would have taken much time and by the way I am not sure of the answer.

It's not scripting language, it's two markup languages which are relatively easy to get used to, especially if you used sites like GitHub before. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Many people here are also students, some of them have jobs and families, and they don't use this as an excuse.

Ok..I agree.... But can you please"decipher" my solution once... If I've done any thing wrong

I posted an answer below on how I'd approach this problem. Compare yours with that and you'll get the same result. If you are unsure/in doubt or have another question, post it as, well, another question.

This link takes you to a question asking form related to ionic equilibrium

This link is for the Ask-a-new-question form.

Why have you sent that link??

sigh Because you said you are unsure about your answer, and the best way to sort out your doubts is to ask a separate question.


text:
Why are not all metals included in the reactivity series
Why aren't all metals included in the reactivity series? For example why isn't molybdenum included? Now we know that hydrogen displaces those metals from their oxides that are less reactive than iron. But how do I know whether hydrogen would also displace molybdenum if it's not there in the reactivity series?

What is "reactivity series"? You mean galvanic series?

@Karl I mean reactivity series ...https://en.m.wikipedia.org/wiki/Reactivity_series

Oh god. That is a stupid, unscientific try to make chemistry more palatable for students (and insulting them on the way) by removing the numbers. No reason why molybdenum should be missing, the authors were just lazy. Check the galvanic series or https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page) .

Both thermodynamic and kinetic reactivity orders differ a lot in various scenarios and with respect to each other. Particularly,  mutual displacement of metals in oxides differ a lot wrt metal ions in water solutions.

It's a good question: "reactivity" may vary depending on concentration, temperature etc. but should generally closely approximate the electromotive (galvanic) series.
That said, as @Karl mentions, reactivity series I've seen usually are incomplete, often leaving out transition metals with multiple valences such as Mo and W. To fill in the missing species, look at a good table of the electromotive series.
The omission of so many metals is downright shocking, revolting. One would think a current table would have included more metals; these reactivity series seem like works in transition.

Thank you , I was really confused at first when I was asked to find whether Hydrogen would displace Molybdenum form it’s oxide or not

I'll say it does vary with concentration, temperature, the solvent ... ;-)

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text:
do organic- compounds can be synthesized from inorganic compounds?illustrare with example
Do you think that organic compound can be synthesized from inorganic compounds? Illustrate with example.

Yes, of course! Friedrich Wöhler, a German chemist, synthesized Urea, a clearly organic molecule since it is produced in every animal and human body, in 1828.
He used different sets of inorganic substances: a combination of cyanic acid and ammonia, a combination of silver cyanate and ammonium chloride, a combination of lead cyanate and ammonia and a combination of mercury cyanate and cyanatic ammonia (which is again cyanic acid with ammonia).
Ever before humanity was sure that organic substances could only be formed by god. Wöhler proved they were wrong.

Perhaps not really by God. But by means of a living sort of spirit / property (vis vitalis) peculiar to living organisms.

Hermann Kolbe (1818-1884) ended the concept of vitalism with his synthesis of acetic acid from iron pyrite and carbon (1845).

Correction: Lampadius (1796) made CS2 from FeS2 and carbon. Kolbe used CS2 and chlorine to make acetic acid. Inorganic ---> organic.


text:
What pH indicators are commonly used for non-polar medium?
Compounds, commonly used as pH-indicators, are soluble in aqueous media and polar solvents (DMSO, DMF, NMP etc.) and are functional pretty much across all these solvents. However, practically none these indicators are soluble in non-polar solvents (hexane, chloroform, dioxane etc.).
I briefly searched for the suitable available systems and so far there is only an article about oxoporphyrinogen derivatives used as pH-indicators in $\ce{CH2Cl2}$ [1]. Are there any other common substances that are typically used for the visualization of acid-base equilibrium in solvents with low polarity?
Reference

Shundo, A.; Ishihara, S.; Labuta, J.; Onuma, Y.; Sakai, H.; Abe, M.; Ariga, K.; Hill, J. P. Chem. Commun. 2013, 49 (61), 6870–6872. DOI: 10.1039/C3CC42859A.

pH does not really make sense outside of aquaeous solutions, so~

I disagree with @Jan's comment a bit. The notational definition of pH isn't specific to water. Relative activity of a proton is equally well/poorly defined in non-aqueous solvents. Comparing them is difficult because of different reference points. There are proposed solutions (e.g., a unified acidity scale), and this has been realised experimentally for HPLC liquid phases. That said, I don't see a straight-forward way of comparing pH across solvents via a colour indicator. If this is not a problem, five titration indicators are apparently listed here. (not a sci paper)

(ref-s for previous comment: doi: 10.1002/anie.201000252, doi: 10.1021/ac504692m)

@Jan, pH+ is quite sensible for Lewis and for Bronsted-Lowery acids-bases in non-aqueous media.

@LinearChristmas Of course, scales and definitions exist; see for example the DMSO $\mathrm pK_\mathrm a$ values in the Bordwell data. But the thing is that when one sees ‘pH’, the layman assumes 7 being neutral and the scale essentially going from 1 to 14 which is only true in water. Of course, protons are still acidic outside aquaeous solution and certain solvents can accept them in acid-base reactions (or donate their own). I think this shouldn’t be called ‘pH’ though.

True, pH+ is associated with aqueous solutions in most people's minds. BTW, 7 is neutral only at ~300 K. At boiling, 373 K, neutral's ~6.14.


text:
Prop-2-yn-1-ol NMR spectrum differences
Looking at prop-2-yn-1-ol spectrum that was taken from:
Spectrometric identification of organic compounds. Robert M. Silverstein and G. Clayton Bassler. J. Chem. Educ., 1962, 39 (11), p 546. DOI: 10.1021/ed039p546.
Why OH peak is so sharp? And why it is located around 1 ppm? The book does not specify the solvent used.

At the same time if one takes a look at 1H NMR spectrum of the same compound taken on 90 MHz machine in CDCl3. Source: Spectral Database AIST with SDBS No. 74HSP-00-252

Why the OH peak shifted so much? But it is still as sharp as in other spectrum.  Please someone explain such pattern.

There is no way that top spectrum comes from the reference indicated. It is a FT spectrum; FT NMR was not developed commercially until several years after that article was published.

@long Well spotted!

The chemical shift of protons in polar groups can vary a lot between solvents of different polarity, because the average electron density around the proton changes. Hydrogen bonding etc. A nice overview can be gotten from the article NMR Chemical Shifts of Trace Impurities: Common Laboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the Organometallic Chemist.
For example water shifts from 4.79ppm in water($D_2O$) down to 0.4ppm in d6-benzene. Or the OH group of ethanol from 0.5 in benzene to 1.3 in chloroform to 4.6(!) in DMSO. The methyl and methene groups otoh just shift by 0.3 ppm max. between solvents.
The signals in your spectra are sharp, because there is either a fast exchange, meaning the peak comes at the average CS of both positions of the proton, or no exchange. Slow exchange (either statistically or because the reaction is slow) leads to broadening.
(Slow = exchange happens on the NMR timescale, a millisecond plus/minus a few decades, depending on your $T_{2}$ etc.)

The chemical shift and linewidth of the hydroxy signal varies considerably, as has been discussed. Exchange rates will influence both linewidth and position, but in the case of the sharp signal shown you most likely have very fast exchange, otherwise coupling to the -OH signal would be apparent.
The shift of -OH signals will vary considerably, and is influenced by solvent, temperature and concentration. More specifically, the shift will be influenced by conditions that affect H-bonding. H-bonding leads to greater downfield shifts. Usually, you see that as you increase the concentration of the alcohol, the shift of the -OH peak moves downfield.
So, all else being equal (same solvent, same temperature), it is likely that the bottom spectrum is recorded at a higher concentration than the top spectrum.

+1 Also you don't see the chloroform peak in the lower spectrum. That being said, why is it invisible? 99.99% CDCl3? Bit strange, that.

Alcohol + chloroform don't make a fast exchange. I wager the sample also contained a rather large amount of water, with which the exchange takes place.

@Karl - No suggestion that alcohol exchanges with chloroform. All untreated commercially available CDCl3 contains at least trace levels of water. Apologies if that was not clear. The amount of water does not have to be large. In both cases, the -OH peak appears to give very good integration. And as you say, the sharpness of the peaks indicate either fast or no exchange. No exchange is not an option for a simple primary alcohol.

@Karl - AINST-SDBS remove solvent signals from the spectrum prior to public release. This is stated as per their experimental conditions "Solvent, reference, and impurity peaks were removed prior to open to the public[sic]".


text:
Irreducible representations and system states connection
Let's say I have the molecule $\ce{N2+}$. Its symmetry point group is clearly $D_{\infty\mathrm h}$. But I'm confused by its irreducible representations. I know there are 8 of them, but as I understand it, they're all representations of the same group, just with different bases.
But today I found out, that they're not completely "equal", i.e. that they mark different states of the system ($\mathrm{A_g}$ is the ground state, $\mathrm{B_{1u}}$ is one of the excited states etc.).
What I'm completely missing is - how are irreducible representations linked with the system states?
Please, try some "low-level" explanation, I'm a beginner both in quantum chemistry and group theory.

The point group $D_{\infty h}$ has actually infinitely many IRREPs, the 8 IRREPs you are referring to  are from $D_{2h}$ which is the largest abelian subgroup of $D_{\infty h}$.

Agreed that Eenoku is referring to $D_{2h}$. However, $D_{2h}$ is not the largest abelian subgroup of $D_{\infty h}$. There are infinitely many larger abelian subgroups of $D_{\infty h}$, namely the $C_n$  (n>4) and $S_{2n}$ groups (n>2). FWIW, I believe it is the largest non-cyclic, abelian subgroup though.

How are irreducible representations linked with the system states?

In short, the irreducible representation (IRREP) tells you about the symmetry of the electronic state or orbital.
Okay, this seems like a quite obvious answer. Let me try to use some different words. The term "symmetry" has two slightly different meanings here: symmetry of the molecule and symmetry of the wave function.
The symmetry of the molecule, which is the same as the symmetry of the electron density, is always either symmetric or not symmetric with respect to a certain operation. This is because the electron density is positive at all points in space. This is what we use to determine the point group of the system.
On the other hand, the symmetry of the wave function can be symmetric or anti-symmetric. The symmetry operations are the same as for the molecule, but the wave function has nodes and therefore different signs. Therefore a symmetry operation can keep the sign or switch it. This is encoded by the characters in the character table.
Example
\begin{array}{c|cccccccc|cc}
\boldsymbol{C_S} & E & \sigma_\mathrm{h} \\ \hline
A^\prime & 1 & 1 \\
A^{\prime\prime} & 1 & -1 \\
\end{array}
The $C_S$ point group has 2 symmetry operations ($E$ and $\sigma_h$) and accordingly 2 IRREPs ($A^\prime$ and $A^{\prime\prime}$). The total symmetric IRREP ($A^\prime$) has the character $+1$ for both operations. This can for example be some $s$ orbital. The other IRREP has character $+1$ for $E$ and $-1$ for $\sigma_h$. This could for example be a $p_z$ orbital where $\sigma_h$ is the $xy$ plane. (Note that $p_x$ and $p_y$ would be of IRREP $A^\prime$ here, since their symmetry plane is not included in the point group discussed here.)
The same can be applied to the wave function of an electronic state. @tobiuchiha already explained in his answer how to go from the symmetry of the orbitals to the symmetry of the state (strictly speaking configuration, but the symmetry will be the same anyway).

The symmetry of a state is the MO (molecular orbital) product of its occupied molecular orbitals. Let's take the example of water molecule. We can see the symmetry label of each of the MO in the diagram. Each MO is assigned an irreducible symmetry label (associated with the point group of molecule) which can be deduced from the symmetry of MOs in space (or from the LCAOs). The wavefunction associated with the MOs also have the same symmetry species. Now, this arrangement of electrons represents the ground electronic state of water, so the symmetry label of the ground state of water can be obtained as
$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2=(A_1)^2\bigotimes (B_2)^2\bigotimes (A_1)^2\bigotimes (B_1)^2 =A_1$
Follow the same procedure to obtain ground state of $N_2^+$. Let me know if you have any questions.

Thank you very much for your answer! I have just one question - how can you multiply two irreducible representations? Up today I understood it just as a row in a character table...

@Eenoku This type of multiplication is called a 'direct product'. It's a general mathematical concept used in group theory and other branches. The following link describes it in the context of molecular symmetry https://www.staff.ncl.ac.uk/j.p.goss/symmetry/Products.html


text:
Does DIBAL-H reduce carboxylic acids?

Diisobutylaluminium hydride(DIBAL-H) reduces only carboxylic acid, ester or both?

The answer given was that it only reduces ester.
Is it correct that under any condition carboxylic acids are not reduced by DIBAL-H?

Bulky groups are there for selectivity, so while it might do that, it's not supposed to.

Yes I think it does. Like BH3, DIBAL-H is an electrophilic reducing agent. In other words, it is best able to reduce electron-rich compounds like amides, acids and esters. I have seen various sources saying that too. However, it is best known for its reduction of the ester to the aldehyde and not the alcohol.

Conditions matter a lot in this question. Are you using a full equivalent or two equivalents? Your answer might be very different with 10 equivalents...

Does diisobutylaluminium hydride (DIBAL-H) reduce only carboxylic acids? Or  only carboxylic acid esters? Or reduce both both types of compounds?
The answer is yes, it does reduce both. Not only them, it also reduces nitriles to aldehydes, and is a more selective reagent than lithium aluminum hydride (LAH) in the reduction of nitriles (Ref.1). About reduction of carboxylic acid, Miller et al. (Ref.1) found that DIBAL-H can be used to reduce benzoic acid to give 72% yields of benzyl alcohol in 1959. In same publication, they have also shown that esters can be reduced to corresponding alcohols as well.
In reduction of nitriles to aldehydes, following results were achieved and compared with the result obtained by Amundsen and Nelson (Ref.2) using $\ce{LiAlH4}$:

The summery of reduction of carboxylic acids and their esters are illustrated in following scheme, where two different groups (Ref.1 & Ref.3) achieved two different results for ester reduction under different conditions:

References:

A. E. G. Miller, J. W. Biss, and L. H. Schwartzman, “Reductions with Dialkylaluminum Hydrides,” J. Org. Chem. 1959, 24(5), 627–630 (DOI: 10.1021/jo01087a013).
L. H. Amundsen, L. S. Nelson, “Reduction of Nitriles to Primary Amines with Lithium Aluminum Hydride,” J. Am. Chem. Soc. 1951, 73(1), 242–244 (DOI: 10.1021/ja01145a082).
L. I. Zakharkin, I. M. Khorlina, “Reduction of esters of carboxylic acids directly into aldehydes with diisobutylaluminum hydride,” Tetrahedron Letters 1962, (14), 619-620 (https://doi.org/10.1016/S0040-4039(00)70918-X).

Sir, please tell that does DIBAL-H reduce aldehydes and ketones to alcohols? If it does, then why aldehyde formed in reduction of esters is not reduced further to alcohols ?

@An_Elephant: Look here the mechanism for ester to aldehyde reaction and description given.

My book (Cengage for JEE Adv) doesn't mention anything about carb acids, but I was able to find this website that says DiBAL-H reduces carb acids to aldehydes and alcohols, but requires 2-3 equivalents, at a high temperature.


text:
What is the purpose of CaO in Oakwood reaction
In the Oakwood reaction, i.e.
$$\ce{RCOOH ->[NaOH + CaO] RH + Na2CO3},$$
is there any role or importance of $\ce{CaO}$, because it doesn't appear to be anywhere in the product.

Sodium hydroxide plus calcium oxide forms soda lime. Its use purely handling convenience, it has a high surface area for reaction and is far less hygroscopic than solid sodium hydroxide powder.

Sodium hydroxide plus calcium oxide forms soda lime. Its use is mainly down to handling covenience as it is far less hygroscopic than powdered sodium hydroxide.

Well, now your comment is obsolete, also it would be good if answer was longer then comment ;)


text:
How does CaO react with NaOH?
$\ce{CaO}$ being a basic oxide should not react with $\ce{NaOH}$.
Then how come decarboxylation takes place using soda lime? I am confused.

Sodalime when used in decarboxylation, the reaction does not take place between NaOH and CaO. In Oakwood reaction, Carboxyl group is the one which gets reduced by NaOH being base.

The decarboxylation using soda lime does not involve any reaction between $\ce{NaOH}$ and $\ce{CaO}$.
It is in fact the $\ce{NaOH}$ which reacts with the carboxylic acid to form the alkane, alongwith the formation of $\ce{Na_2CO_3}$.
$\ce{CaO}$ is simply used to make the reagent easier to handle. $\ce{NaOH}$ is highly hygroscopic and easily forms a concentrated sodium hydroxide solution when exposed to air. Sodalime does not absorb moisture that easily.


text:
Mechanism of Oxymercuration-Demercuration reaction
I am trying to write a mechanism for OMDM reaction.

I could write the mechanism for oxymercuration part of reaction but I couldn't write the mechanism for remaining part (demercuration) of the reaction. Could anybody help me to proceed forward.

Here is the mechanism I wrote out for you

Why does the hydride attack the mercury?! wouldn't it attack the carbon bonded to the mercury?

Actually, the oxymercuration involves the formation of a cyclic carbocation. This is a non traditional carbocation, but the final reaction becomes non-stereospecific, as you will later find out.

First, the mercuric acetate breaks apart into ions, the acetate anion, and the mercuroacetic cation, which you have correctly depicted.
In the next step, the mercuroacetic cation gets added on to the double bond, which forms a cyclic cation like this:

(Don't bother about the red arrow for now, we'll come to it in the next step.)
Now, water adds on to the carbocation formed from the other side, and this process follows Markownikoff's rule. Now, one $\ce{H+}$ leaves, and a result, we get exactly what you got in your last step. This is, as you called it, the end of the oxymercuration step.
Now, for the demercuration part, we can use any reducing agent that will be able to replace the mercuroacetic cation with $\ce{H}$. (E.g. $\ce{NaBH4}$). This essentially demercurates the product, forming an alcohol.

The entire process is shown here:
.
Now, as it is written here, the addition of the mercuroacetic cation followed by water will be anti addition, but when the reducing agent is used, the hydrogen can be added from either side. As a result, we get a non-stereospecific product. In short, this can also be written as addition of just $\ce{H}$ and $\ce{OH}$ without rearrangements, which followed Markownikoff's rule. In fact, the formation of the cyclic carbocation prevents rearrangements, hence enabling chemists to get desired products without worrying about random stabilisations of the carbocations.