Figuring out the concentration of weak polyprotic acids in presence of strong acid We have a $\pu{0.1 M}$ solution of $\ce{H2S}$ with a $\pu{0.1 M}$ solution of $\ce{HCl}$. We need to figure out the concentration of $\ce{S^2-}$ ions at equilibrium.
$$\ce{H2S <=> HS- + H+}$$
$$\ce{HS- <=> S^2- + H+}$$
Given $K_{\mathrm a_1} = 9.1 \times 10^{-8}, K_{\mathrm a_2} = 1.2\times 10^{-13}$.
I have no idea how to do this problem. I can't relate the $\ce{S^2-}$ concentration to the first equation. I first tried determining the $\rm{HS^-}$ concentration from the first equation, then putting that into the second one to obtain the required concentration, but then I realized that as the second reaction approaches equilibrium, the concentrations of the $\ce{HS-}$ ions will change, which will then throw the first reaction out of whack.
I tried taking the required concentration as $\pu{y M}$ and then tried to establish a relationship, but the only thing I could get was that the concentration of the hydrogen ions will go up by $y$, which is of no use at all.

The first step in solving this kind of problems is to write the reactions that take place in your solution, and the equations of the acidity constants :
$$\ce{H2S + H2O <=> HS- + H3O+}$$
$$K_{\mathrm a_1}=\frac{[\ce{HS-}][\ce{H3O+}]}{[\ce{H2S}]}$$
$$\ce{HS- + H2O <=> S^2- + H3O+}$$
$$K_{\mathrm a_2}=\frac{[\ce{S^2-}][\ce{H3O+}]}{[\ce{HS-}]}$$
Since $\ce{HCl}$ is a strong acid ($\mathrm pK_\mathrm a < 0$), the dissociation equilibrium for it will be totally shifted to right:
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
And the dissociation of water, occuring in any solution:
$$\ce{H2O + H2O <=> H3O+ + HO-}$$
$$K_\mathrm w = [\ce{H3O+}][\ce{HO-}]$$
*The concentration of water is not taken into account because it remains approximately constant as it is much bigger than the concentration of the $\ce{H2S}$ dissolved. It is also right to write the equations as you did above, considering the dissociation of the acid, but it's better to use the hydronium ion, $\ce{H3O+}$, as we know $\ce{H+}$ is quite unstable on its own. (In some books $[\ce{H+}]$ is used to ease writing)
As you saw, you cannot calculate the concentrations at equilibrium (written in square brackets, i.e. $[\ce{HS-}]$) considering only the first reaction (unless you consider the shift of the second reaction doesn't affect significantly the first, as Nicolau said, and so we use the analytical concentration of a compound (in this case $\ce{H2S}$ and $\ce{HCl}$) which means the specific concentration of the solute when the solution is prepared ($\pu{0.1 M}$ for both acids, if these are the concentrations in the mixture of acids solution and not the initial concentration of each acid solution). It is usually denoted $c_{\ce{H2S}}$ and $c_{\ce{HCl}}$ respectively.
Looking at the above reactions, we can write the formula for the analytical concentrations :
$$c_{\ce{H2S}} = [\ce{H2S}] + [\ce{HS-}] + [\ce{S^2-}]$$
and
$$c_{\ce{HCl}} = [\ce{Cl-}] $$
As you might notice no $\ce{H3O+}$ concentration appeared in any of these formulas, because it depends on both $\ce{HCl}$ and $\ce{H2S}$ concentrations.
However, this is not enough to solve our problem, and so the electroneutrality principle is used, stating that the concentration of the negative ions must be equal to the concentration of the positive ones, and in our case may be written as:
$$[\ce{HS-}] + 2[\ce{S^2-}] + [\ce{Cl-}] + [\ce{HO-}] = [\ce{H3O+}]$$
*The concentration of $\ce{S^2-}$ has the coefficient $2$ because of the two negative charges, unlike the other ions, which have only one negative or positive charge and don't need any.
Using the equations of the analytical concentrations in the electroneutrality formula, we get this :
$$c_{\ce{H2S}} - [\ce{H2S}] + [\ce{S^2-}] + c_{\ce{HCl}} + [\ce{HO-}] = [\ce{H3O+}]$$
And, using also the acidity constants and the water constant :
$$c_{\ce{H2S}}-\frac{[\ce{H3O+}][\ce{HS-}]}{K_{\mathrm a_1}}+\frac{K_{\mathrm a_2}[\ce{HS-}]}{[\ce{H3O+}]} + c_{\ce{HCl}} + \frac{K_\mathrm w}{[\ce{H3O+}]} = [\ce{H3O+}]$$
The equation has 2 unknown concentrations ($[\ce{H3O+}]$ and $[\ce{HS-}]$) and you may approximate the concentration of $\ce{H3O+}$ to $c_{\ce{HCl}}$ since the acidity constants for $\ce{H2S}$ are very small:
$$[\ce{H3O+}]=c_{\ce{HCl}}$$
So, the concentration of $\ce{HS-}$ can be calculated:
$$[\ce{HS-}]=\frac{c_{\ce{H2S}} + \frac{K_\mathrm w}{c_{\ce{HCl}}}}{\frac{c_{\ce{HCl}}}{K_{\mathrm a_1}} + \frac{K_{\mathrm a_2}}{c_{\ce{HCl}}}}$$
However, since $\frac{K_\mathrm w}{c_{\ce{HCl}}} << c_{\ce{H2S}}$ and $\frac{K_{\mathrm a_2}}{c_{\ce{HCl}}} << \frac{c_{\ce{HCl}}}{K_{\mathrm a_1}}$, the above equation can be approximated to:
$$[\ce{HS-}] = \frac{c_{\ce{H2S}}K_{\mathrm a_1}}{c_{\ce{HCl}}}$$
The approximations lead us to a simple formula, which could be denoted directly from the first equilibrium, but without understanding exactly why. From this the concentration of $\ce{S^{2-}}$ can be calculated easily, just like you did first. Even though some would just ignore the second equilibrium for $\ce{H2S}$, that is NOT the way to solve analytical chemistry problems. First you consider all of the equilibriums and processes occuring in a solution and then (if possible) approximate, and not the reverse.


Melting post-1982 minted U.S. pennies and aluminum: Why does the slag turn blue? After melting aluminum inside of an iron crucible, I tossed in a handful of U.S. pennies that were minted post-1982. The slag began to burn a brilliant blue. What caused this?

Zinc burns in air forming zinc oxide with a bright blue flame. The bit of copper in US cents will add a greenish color to the flame.

N.B. Fumes of burning zinc and some other metals are toxic. Avoid contact or inhaling the smoke.


Synthesis of branched alkanes The question might seem kind of trivial, although it is even harder than I've thought. How can we synthesise branched alkanes from linear one? For example, I want to obtain 3,4,7,8-tetramethyldecane from n-decane. I've already found Stille coupling, but I guess I'm searching for easier method. I'm just a high school student, studying chemistry by myself as well.

@Radi: Critical changes to original post! I will guide you through a reasonable approach. You have to realize that you have not specified the stereochemistry of 11 -- of which there are 16 possible stereoisomers (8 racemates) -- but I will ignore the issue. You will recognize the symmetry in 3,4,7,8-tetramethyldecane (11), so we will look for a coupling reaction of two 7-carbon entities. (E)-2-Buten-1-ol (crotyl alcohol) (4) and triethylorthopropionate 5 are commercially available. The reaction of 4 and 5 is a Johnson-type Claisen rearrangement (Google it!) which produces racemic esters 7 via transition state 6. Reduction of the double bond of unsaturated esters 7 leads to saturated esters 8. Reduction of the ester group to an alcohol followed by conversion to a bromide gives structure 9. The Wurtz reaction (9 --> 11) is rarely used but here is an opportunity.
Alternatively, esters 8 under the same conditions of the Wurtz reaction produces the α-hydroxyketones 10 (acyloin condensation). While the Clemmensen reduction (10 --> 11) of aliphatic ketones does not work as well as with aromatic ketones, under these conditions you would lose both oxygens. If the ketone is produced (structure 10 less the hydroxyl group), a Wolff-Kishner reduction would lead to the tetradecanes (plural) 11.Read about these name reactions which will be of more value to you than your original proposal. Good luck!


What can be a metal cleanser for aluminum? I want to remove corrosion products in aluminum, but I don't have supply of nitric acid, as suggested by an ASTM method. Which can be a good substitute?

Really, ASTM procedures must be followed as written for certified testing. However, if this is for your edification or for a school project, you might use a commercial cleaner, such as those below, or just one ingredient such as phosphoric acid.

There are a number of commercial aluminum corrosion cleaners, such as Loctite Duro Aluminum Jelly and CRC MaryKate Aluminex Clean and Shine.

If you want to make your own:

Ingredients listed in the Aluminum Jelly Material Safety Data Sheet include phosphoric acid, diethyleneglycol monoethyl ether, nonionic surfactant, isopropyl alcohol , polysaccharide and sulfuric acid. [Similar formulaions may have contained hydrofluoric acid to break the highly-adherent $\ce{Al2O3}$ surface film and corrosion, but due to the hazards of $\ce{HF}$, I believe it is no longer sold in most places, for good reason.]

Aluminex Clean and Shine's MSDS includes ethoxylated tallow alkyl amines, sodium decylglucosides, hydroxypropyl phosphate and potassium iodide.


Hydrocarbons with only 4 carbon atoms How many different organic structures (from the pure theoretical viewpoint) can be drawed with only 4 (exact) carbon atoms and with/without hydrogen? Polycyclic compounds and bridged compounds are also allowed. Please, name them all as well! :)

Remark: I have drawed the chemical graphs in a piece of paper (the skeletons) and I get 37 graphs (including linear with simple, double and triple bonds, and planar and nonplanar compounds), without geometric isomerism (cis/trans). Is that OK?

Final update, all earlier edits incorporated.

Groundrules: Considering compounds with:

only carbon and hydrogen
only 4 bonds to carbon

There are 37 isomers without considering trans isomers; 49 when trans isomers are included.  Also, many of these compounds seem extremely unstable and therefore unlikely to exist.

Note to self: check back in 20 years and see how many of the unlikely ones have been detected.

Names by row:

butane, isobutane
but-1-ene, but-2-ene, 2-methylpropene
buta-1,3-diene, buta-1,2-diene, buta-1,2,3-triene
but-1-yne, but-2-yne, but-1-ene-3-yne, buta-1,3-diyne
methylcyclopropane, 2-methylcyclopropene, 1-methylcyclopropene, methylenecyclopropane, methylenecyclopropene, methlycyclopropadiene
methylcyclopropyne, methlenecyclopropyne
cyclobutane, cyclobutene, cyclobuta-1,2-diene, cyclobuta-1,3-diene, cyclobutatriene, cyclobutatetraene, cyclobutyne, cyclobutenyne, cyclobutadiyne
bicyclobutane, bicyclobut-1(3)-ene, bicyclobut-1(2)-ene, bicyclobuta-1,3-diene, bicyclobuta-1,2-diene
tetrahedrane, tetrahedrene, tetrahedradiene


What is the mechanism of azulene synthesis? In the Organic Syntheses preparation of azulene (http://www.orgsyn.org/demo.aspx?prep=CV7P0015) there are two reactions. First is a clevage of the C-N bond in pyridine and the second one is the annelation of a seven-memebered ring:

I suppose that the first step of the first reaction  is a nucleophilic substitution by pyridine at the 1-chloro-2,4-dinitrobenzene, but I have no idea how the mechanism could continue. Could some one please help me? Thank you.

Also, for the 2nd reaction, is it an electrophilic substitution at the cyclopentadieneyl anion, or does it have an  entirely different mechanism?

The @orthocresol link to the Zincke reaction is a good start to the mechanism. You are correct about the first step. Dimethylamine adds to the electrophilic pyridinium salt 1 as illustrated in 1 --> 5.  The reaction is driven by loss of the poor nucleophile (decent leaving group), DNP-NH2. The (Z)-isomer 5 is in equilibrium with the (E)-isomer 6 but the reactions leading to azulene 12 require a cisoid conformation to allow for cyclization.
Cyclopentadienide anion, which is delocalized but for convenience is shown as a localized anion, displaces dimethylamine which is protonated by the methanol in the reaction mixture (see your link to Organic Synthesis). Fulvene 8 must generate the geometry shown to complete the remaining steps to azulene (9 --> 12). ADDENDUM: An added thought. The transformation of 8 ---> 10 may be a 4n+2 disrotatory thermal electrocyclization that forms the new sigma bond of 10 with cis hydrogens.


Hamiltonian for a two electron system What is the hamiltonian for an atom with two electrons and a nuclear charge $Z$?

The Hamiltonian always takes the general form: $\hat H=\hat T + \hat V $

The kinetic energy of each of the electrons needs to be taken into account so: $\hat T=-\frac{\hbar^2}{2\mu}(\nabla_1^2+\nabla_2^2)$

The potential operator is where I am finding trouble. There are two terms for the attraction towards the nucleus - one for each electron: $-\frac{Ze^2}{4\pi\epsilon_0r_1}-\frac{Ze^2}{4\pi\epsilon_0r_2}$. However, the potential energy involved in the electron electron repulsion needs to be taken into account. My question is: are there two terms for this? To clarify do I need to calculate the increase in potential of electron 1 caused by electron 2 AND the increase in potential energy of electron 2 caused by the presence of electron 1? Or would adding: $\frac{e^2}{4\pi\epsilon_0r_{12}}$ take into account all of the potential energy involved in the electron electron interaction?

The electron-electron repulsion potential is an interaction potential, therefore it has one term per interaction. If you have two electrons, you will have only one term, representing the interaction between them. The number of potential terms will be, in your case:

Attraction between electron 1 and the nucleus (one interaction)
Attraction between electron 2 and the nucleus (one interaction)
Repulsion between electron 1 and electron 2 (one interaction)

In total, you will have three interactions and therefore only three potential terms (two for electron-nucleus and one for electron-electon interactions).

In general, an interaction potential that only depends on the distance of two particles can be written as
$$
U = \sum_{i<j} u(|\vec{r}_i - \vec{r}_j|),
$$
where $u$ represents the interaction between only two particles. You see that for two electrons you will have only one term, since $i<j$. For three electrons you will have three terms (interaction between 1 and 2, between 1 and 3 and between 2 and 3) and so on.

Maybe your confusion comes from the fact that, sometimes, this interaction potential is written as
$$
U = \frac{1}{2}\sum_{i\neq j} u(|\vec{r}_i - \vec{r}_j|)
$$
In this case, for two electrons, you will have two terms. However there is the factor $1/2$ which cancels out the double counting, since
$$u(|\vec{r}_i - \vec{r}_j|)=u(|\vec{r}_j - \vec{r}_i|).$$


Angular Momentum of an orbital and orbit While pursuing forth in Atomic Structure, I encountered the following concepts...

According to Bohr, the angular momentum of an orbit remains quantized, i.e $\displaystyle mvr = n\hbar .$

But he wasn't able to expalin the reason that why is it so.

But later Louis de Broglie gave his hypothesis:

Every particle in the universe has dual nature, one wave nature an other particle nature.

Focusing on wave nature of particle he gave an equation, known as de Broglie equation or de Broglie wavelength as

$$\displaystyle \lambda = \frac {h}{mv}.$$

This was proved years later by Davisson and Germer's Diffraction Experiment.

Now, we can also prove the quantization of angular momentum of an orbit as:

For any orbit the number of waves produced by an electron in one complete revolution is $\displaystyle \frac {2\pi r}{\lambda}.$

Substituting the values of $r,v$ given by Bohr, then

$$\displaystyle \frac {2\pi r}{\lambda} = \frac {2\pi r \cdot mv}{h} = \frac { n^2 \hbar^2 \cdot mkZe^2}{\hbar \cdot mkZe^2 \cdot n\hbar} = n$$

As $\displaystyle r =\frac { n^2 \hbar^2 }{ mkZe^2}  $ and $\displaystyle v = \frac {kZe^2}{ n\hbar}.$

Now , as $ \displaystyle \frac {2\pi r}{\lambda}= n ,$ From this we can prove that angualr momentum of an orbit is quantized.

Now according to Quantum wave mechanical model, the angular orbital momentum of an orbital is : $ \displaystyle \sqrt {l(l+1)} \hbar .$

Now which one is actually valid, because in many of the books both of the momentums are refered only by the term angular momentum.  But both are not same.

My question might sound silly, for such topic... But I would like to know which one is to be refered when we encounter angular momentum.

Bohr Model came before the quantum theory and it was a success for explaining the non- continuous emission spectrum of excited hydrogen atoms. However, Bohr model was not sufficient for explaining many other phenomena related to the interaction of matter with light, such as the non-continuous  emission spectrum of multi-electronic atoms,  Zeeman Effect,... Therefore, the quantum theory is considered as a more complete theory.
The quantification of the angular orbital momentum should be linked to the second quantum number $l$ not $n$.

Up till now, we use Bohr model in chemistry textbooks in spite of its limitation. This is because it's simple and doesn't require complex mathematical tools as quantum theory.


shifting of equilibrium Does it mean that if one side of a reaction is favoured, the reaction will shift to the other side? As I was reading up on the van't Hoff equation, it says that an increase in temperature of an endothermic reaction will favour the products. It also showed a graph where lnK increases with temperature (T). But I thought what would happen in this case is the reactants will be favoured and then the equilibrium will shift to the right to de-stress the equilibrium?

So in summary, I'm just quite confused about what it means by favouring a side of an equilibrium.

In general, a equilibrium will shift in a direction that opposes the stress that was placed on the system. For example, if you add heat to an endothermic reaction, you will receive more products, as this will lower the temperature of the solution. When you have the dissociation of an acid, the acid will dissociate more (act like an acid) when you make the solution more basic. An equilibrium will shift to minimize any change in a system, so when evaluating what will happen after a system is stressed, look to see which side of the equilibrium relives that stress, and that side will be favored.


What is p defined as for the expansion and compression of an ideal gas system in the equation ΔU = ΔH - pΔV? $$\Delta U = \Delta H - p\,\Delta V$$

Consider a gaseous system in a container with a frictionless piston, and a pressure outside the container.

Is the pressure outside the container assumed to always be constant?
Is the pressure inside the container assumed to always be constant? I assume no looking at the graph of Boyle's Law.
For expansion of the gas, work is done by the system (i.e. the gas in the container) on the surroundings, so is $p$ the pressure of the surroundings as work is done against the pressure of the surroundings?

This doesn't seem to be the case in several examples in my lecture notes:
e.g. calculate the work done against pressure of 1 bar at 298 K when 1 mol of zinc is dissolved in aq. HCl.

Here, work is being done against the pressure of the surroundings, so I'd expect $p$ = pressure of surroundings. However, the notes say 'consider $p(\Delta V) = (\Delta n)RT$' and proceeds to input the number of moles of hydrogen gas. So here $p$ is taken as pressure of the hydrogen gas produced.

For compression of the gas, work is done by the surroundings on the system, so is $p$ the pressure of the gas in the container as work is done against this pressure?

In the following I limit discussion to the particular expression $$ΔU = ΔH - pΔV$$
which follows (spoiler alert) from $$H\equiv U+pV$$ when the pressure $p$ of the system (the container) is  identical at the beginning and end of the process.

  Is the pressure outside the container assumed to always be constant?

For an isobaric process such as the stated process involving zinc, by definition yes. Generally speaking the expansion work in the 1st Law is written  $$dw = -p_{ext}dV$$ where $p_{ext}$ is the pressure of the surroundings against which the system does work. If $p_{ext}$ is constant work is still against or by the pressure of the surroundings and the integral can be evaluated as $$w=-p_{ext}\Delta V $$ For a reversible process the pressure of the system and surroundings are equal and we evaluate $$w=-\int_{Vi}^{Vf} pdV$$ where $p$ is the pressure of the system.

  Is the pressure inside the container assumed to always be constant? I assume no looking at the graph of Boyle's Law.

In the case of a reversible process against constant external pressure, yes.

BTW Boyle's law ($p \propto V^{-1}$) does not consider changes in the number of particles during a reaction.

  For expansion of the gas, work is done by the system (i.e. the gas in the container) on the surroundings, so is p the pressure of the surroundings as work is done against the pressure of the surroundings?

The pressure of the system (container) and surroundings are assumed to be equal (balanced) for a reversible process. However, if the pressure of the system is greater than that of the surroundings (the latter assumed constant) then the work is still $$w=-p_{ext}\Delta V$$

  This doesn't seem to be the case in several examples in my lecture notes: e.g. calculate the work done against pressure of 1 bar at 298 K when 1 mol of zinc is dissolved in aq. HCl

$pV$ work involves a change in volume. Here the volume change is due to formation of gas which causes the pressure in the container to increase by an infinitesimal amount $\mathrm{d}p$. The gas pushes against the piston, increasing the volume and returning the system to mechanical equilibrium when the internal pressure equals the external pressure $p$.

  For compression of the gas, work is done by the surroundings on the system, so is p the pressure of the gas in the container as work is done against this pressure?

Assuming no chemical processes, compression requires the surrounding pressure to be higher than that of the system. It suffices that the difference be infinitesimal in magnitude, resulting in a reversible process. During such a process, the pressure $p$  of the system may vary during the entire process, remaining always equal (to within an infinitesimal amount) to the applied pressure.


Lone pair on carbon in CO2? CO2 has sp hybridization and no lone pairs on carbon.

For some reason calculation with VSEPR gave sp2 hybridization and lone pair on carbon.

The carbon atom in $CO_2$ has zero lone pairs of nonbonding electrons.

[By the way, the carbon has $sp$ hybridization, as it requires the bonding orbitals to have the same lobe sign on each side to make two identical $\sigma$ bonds. The $2p_z$ orbital alone is not sufficient to make two identical bonds, because the molecule would then lose the reflection symmetry it ought to have.]

Let's construct the Lewis structure ourselves:

Carbon has four valence electrons. We accept that.
Each oxygen has six valence electrons. We also accept that.

Therefore, the total number of valence electrons is $4+6+6=16$. Carbon is the larger atom, so this is a more usual structure with the largest atom in the center. The skeletal structure so far is:

$\text{O}-\text{C}-\text{O}$

We place three lone pairs of electrons on each oxygen next, to account for $2+2$ bonding and $2\times3+2\times3=12$ (currently) nonbonding electrons, for a total of $16$ as we needed.

Lastly, we use one lone pair from each oxygen to make a $\pi$ bond. This only redistributes electrons, without adding any extra or losing any. That results in:

$:\ddot{\text{O}}=\text{C}=\ddot{\text{O}}:$

And are the formal charges (FCs) minimized?

Carbon owns 4 valence electrons and came in with 4 valence electrons, so $FC = 0$.
Oxygen owns 6 valence electrons and came in with 6 valence electrons, so $FC = 0$ as well.

This must be the actual Lewis structure, and it also is in the correct image you have placed in your question.


Steric hindrance of an unbonded electron pair This question is half inspired by this question, and half inspired by the structure of molecules like hydrazine and hydrogen peroxide.

When I was looking at the aforementioned molecules, I started trying to reason as to why the lowest energy conformation of these molecules are what they are.

$\hspace{34ex}$

In hydrazine, the lowest energy conformation is not the one in which the unbonded electron pairs are anti-periplanar. This makes me think that hydrogen have a greater steric hindrance than unbonded electron pairs do.

$\hspace{34ex}$

In hydrogen peroxide, however, the lowest energy conformation is not the one in which the hydrogen are anti-periplanar. This contradicts what I reasoned form the structure of hydrazine.

Being confused by this, I tried a different approach—what about the cyclohexyl carbanion? Surely using $\Delta G = -RT \ln{K_\mathrm{eq}}$ the relative steric hindrance of an unbonded electron pair and a hydrogen atom could be determined. What I became unsure about, however, was how rapid inversion might affect this, and additionally whether or not there even is even any data regarding the major and minor confirmations of the cyclohexyl carbanion (because I am unsure if it even exists in such a way that this can even be measured).

So am I flawed in my reasonings? Does the equilibrium data for the cyclohexyl carbanion exist?

Since I just covered this material in depth in my physical organic chemistry class, I figured I'd write up an answer.

Electronegative elements lower the energies of all molecular orbitals to which they contribute, of which their low-lying $\sigma ^{*}$ orbitals are of particular importance. Electronegative atom lone pair MOs, which have little bonding character and are also (relatively) high-lying in energy, are very close in energy and will donate their electrons to the low lying empty orbitals, producing a stabilizing interaction. Molecular conformations are in part shaped by these interactions, which is known as the donor-acceptor effect.

Hydrogen peroxide is a classic example of this effect. In addition to sterics predicting that the hydrogens be anti-periplanar, this conformation also minimizes the molecule's net dipole (another stabilizing effect), yet its most stable conformation adopts anti-clinal geometry. This can essentially be thought of as a compromise between between sterics and dipole minimization, and the donor-acceptor effect:

$\hspace{4.3cm}$

The $\sigma ^{*} (\ce{O-H})$ is an excellent acceptor, and oxygen's highest energy unbonded pair is one of the strongest donating. The lower energy lone pair occupies a $\sigma\text{-out}$ orbital, and the higher energy lone pair occupies a pure $\mathrm{p}$ orbital. The mixing of this orbital with the $\sigma ^{*} (\ce{O-H})$ orbital favors a $90º$ dihedral angle, but due to steric and dipole effects, the actual angle is $\approx 120º$.

$\hspace{5.3cm}$

In hydrazine, the $- 2.5 \ \mathrm{kcal \ mol^{-1}}^{[1]}$ preference for the two $\mathrm{n}(\ce{N})$ orbitals being syn-clinal over anti-periplanar is also explained by the donor-acceptor effect. When the $\mathrm{n}(\ce{N})$ are syn-clinal, they are each eclipsed by the $\sigma ^{*} (\ce{N-H})$ orbitals, and orbital mixing occurs. The effects of this are further seen in the $\ce{N-N}$ bond length difference between the two conformations: $1.448\ \mathrm{Å}$ in the syn structure and $1.489\ \mathrm{Å}$ in the anti structure. The shortening of the bond length in the syn structure is a direct result of the multiple bond character introduced by the $\mathrm{n}(\ce{N}) \rightarrow \sigma ^{*} (\ce{N-H})$ interactions$^{[1]}$.

$^{[1]}$ Wilcox, C.; Bauer, S. Journal of Molecular Structure: THEOCHEM 2003, 625 (1-3), 1–8.


Why does a larger DNA gel box require a higher voltage to move DNA at the same speed? If I run a small agarose DNA gel and use a voltage of 90V, the DNA migrates the length of the gel in about an hour. If I use a medium size gel I have to use a voltage of 140V to move the same DNA the same distance (the equivalent of the small gel) in the same amount of time. For a large gel it takes about 170V. What is it exactly that requires a higher voltage to maintain the same speed of DNA movement?

I think this is more of an electrochemistry question than a biology questions so I put it here.

It's actually a physics question. The speed of the particle in the gel is proportional to the force that acts on it. The force that acts on the charged particle is proportional to the strength of the electric field $E$. The strength of the electric field is proportional to the voltage $U$ and inverse proportional to the length of the gel $d$.

$$E = \frac{U}{d}$$

So if you want to have the same strength of the electric field you have to double the voltage if you double the length of the gel.


Mathematically, how does the exchange integral for a closed-shell system reduce to zero? I am looking at the Hartree-Fock method of approximation for a closed-shell two-electron system.

I have the basis functions

$$
\chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(s_1) \\
\chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(s_2)
$$

where $\vec{x}_i$ is the combined spatial and spin coordinates. The opposite spins are illustrated by the $\alpha$ and $\beta$ labels.

The above gives the total wave function to be the following Slater determinant

$$\Psi(\vec{r}_1,\vec{r}_2,s_1,s_2) =
\begin{vmatrix}
\psi_1(\vec{r}_1)\alpha(s_1) & \psi_1(\vec{r}_1)\beta(s_1) \\
\psi_1(\vec{r}_2)\alpha(s_2) & \psi_1(\vec{r}_2)\beta(s_2)
\end{vmatrix}
$$

Let's bypass a lot of math, and just look at the exchange integral, where the spin and spatial parts have been separated:

$$
\int ds_1 \alpha^*(s_1)\beta(s_1) \cdot \int ds_2 \beta^*(s_2)\alpha(s_2) \cdot \int d\vec{r}_1 \int d\vec{r}_2 \psi_1^*(\vec{r}_1) \psi_1^*(\vec{r}_2) \dfrac{1}{r_{12}} \psi_1(\vec{r}_2) \psi_1(\vec{r}_1)
$$

So the spin integrals do not survive in this case, due to the closed shells, but I am not sure if I see why. Is there some orthonormality for the spin states at play here?

Let's do the math as explicit as possible. So, the exchange integral is defined as follows,
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
:=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\iint\limits_{-\infty}^{+\infty}
\bar{\chi}_1(\vec{x}_1) \bar{\chi}_2(\vec{x}_2)
r_{12}^{-1}
\chi_2(\vec{x}_1) \chi_1(\vec{x}_2)
\mathrm{d} \vec{r}_1 \mathrm{d} \vec{r}_2 \, ,
$$
where I used a more appropriate label for the spin coordinate ($m_{s}$) as well as distinguished between summation over discrete variables $m_{s1}$ and $m_{s2}$ and integration over continuous $\vec{r}_1$ and $\vec{r}_2$ ones.

Then, for the restricted spin orbitals,
$$
  \chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(m_{s1}) \, , \\
  \chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(m_{s2}) \, ,
$$
where $\alpha$ and $\beta$ are the so-called "spin up" and "spin down" spin functions defined as follows,
$$
    \alpha(m_{s}) =
    \begin{cases}
        1, & m_{s} = +1/2 \\
        0, & m_{s} = -1/2
    \end{cases} \, ,
    \quad
    \beta(m_{s}) =
    \begin{cases}
        0, & m_{s} = +1/2 \\
        1, & m_{s} = -1/2
    \end{cases} \, ,
$$
we get then
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle \\
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\iint\limits_{-\infty}^{+\infty}
\bar{\psi}_1(\vec{r}_1) \alpha(m_{s1})
\bar{\psi}_1(\vec{r}_2) \beta(m_{s2})
r_{12}^{-1}
\psi_1(\vec{r}_1) \beta(m_{s1})
\psi_1(\vec{r}_2) \alpha(m_{s2}) \\
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2})
\iint\limits_{-\infty}^{+\infty}
\bar{\psi}_1(\vec{r}_1)
\bar{\psi}_1(\vec{r}_2)
r_{12}^{-1}
\psi_1(\vec{r}_1)
\psi_1(\vec{r}_2) \, ,
$$
where we used the fact that spin functions $\alpha$ and $\beta$ are real-valued, so that $\bar{\alpha} = \alpha$ and $\bar{\beta} = \beta$.

At this point we could concentrate our attention exclusively on the coefficient in front of the double integral above, since we could show that it is equal to zero, so that the whole expression vanishes irregardless of the value of the integral. So, the coefficient can be written as follows,
$$
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2}) \\
=
\alpha(-1/2) \beta(-1/2) \beta(-1/2) \alpha(-1/2)
+
\alpha(-1/2) \beta(+1/2) \beta(-1/2) \alpha(+1/2) \\
+
\alpha(+1/2) \beta(-1/2) \beta(+1/2) \alpha(-1/2)
+
\alpha(+1/2) \beta(+1/2) \beta(+1/2) \alpha(+1/2) \\
= 0 + 0 + 0 + 0 \\
= 0 \, ,
$$
where all the terms vanish due to at least either $\alpha(-1/2) = 0$ or $\beta(+1/2) = 0$, so that we indeed get
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
=
0 \, .
$$

It can be noted that there is a shortcut in proving that the coefficient above is zero that uses the defining properties of spin functions. First, we could rearrange spin functions as follows,
$$
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2})
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s1})
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\beta(m_{s2}) \alpha(m_{s2}) \, .
$$
Secondly we note that by the very definition of $\alpha$ and $\beta$ presented above,
$$
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \alpha(m_{s}) \beta(m_{s})
  =
  \alpha(-1/2) \beta(-1/2) + \alpha(+1/2) \beta(+1/2)
  =
  0 \cdot 1 + 1 \cdot 0
  =
  0 \, ,
$$
which is already enough to establish that the whole exchange integral vanishes. The same is, of course, true for the second factor as well,
$$
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \beta(m_{s}) \alpha(m_{s})
  =
  \beta(-1/2) \alpha(-1/2)
  +
  \beta(+1/2) \alpha(+1/2)
  =
  1 \cdot 0 + 0 \cdot 1
  =
  0 \, .
$$
It is also quite customary to use Dirac bracket notation for such expressions over spin orbitals, so that one could write these findings concisely as
$$
  \langle \alpha \lvert \beta \rangle
  =
  \langle \beta \lvert \alpha \rangle
  =
  0 \, ,
$$
where $\langle \alpha \lvert \beta \rangle$, for instance, is defined as follows,
$$
  \langle \alpha \lvert \beta \rangle
  :=
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \bar{\alpha}(m_{s}) \beta(m_{s}) \, .
$$
In general, for two spin functions $\gamma_1$ and $\gamma_2$, expression $\langle \gamma_1 \lvert \gamma_2 \rangle$ defined as follows,
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  :=
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \bar{\gamma_1}(m_{s}) \gamma_2(m_{s}) \, ,
$$
is zero if the functions correspond to different spin states and one otherwise which can be expressed as follows,
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  =
  \delta_{\gamma_1 \gamma_2} \, .
$$

Finally, I would like to make a short comment on "integrating" over discrete spin coordinates rather then summing up over it. One could indeed define
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  :=
  \int \bar{\gamma_1}(m_s) \gamma_2(m_s) \mathrm{d} m_s \, ,
$$
and then think of integration over discrete $m_s$ variable being reduced to summation. But the integration here is just a "symbolic shorthand" since, strictly speaking, we have a summation over discrete $m_s$ variable from the get go. We could just symbolically write it down as an integration as well, if we think it looks more cute for some reason.

In exact same way the exchange integral can also be symbolically defined as follows,
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
:=
\iint
\bar{\chi}_1(\vec{x}_1) \bar{\chi}_2(\vec{x}_2)
r_{12}^{-1}
\chi_2(\vec{x}_1) \chi_1(\vec{x}_2)
\mathrm{d} \vec{x}_1 \mathrm{d} \vec{x}_2 \, ,
$$
where the "integration" is informally done over joint spin-spatial coordinates of the electrons. And again, strictly speaking, a summation is done over the spin coordinates and an integration over spatial ones, as it is written explicitly in the very beginning of the answer.


What chemicals are mixed together to cause chemoluminescence? My brother is supposed to do a science experiment at school for an open house. We found an interesting experiment in which you pour several chemicals together, causing the resulting concoction to glow brightly. You can make several colors in this way. Then you can dump them together, to demonstrate how white light is made up of several colors mixed together. Several questions.
1. Which chemicals are used?
2. How do you get the colors?
3. Where can you buy the chemicals you need?

If you want different colours, you might want to give the peroxyoxalate reaction, known from glow sticks, a try.

Typically, you will need three components:

a diaryl oxalate
hydrogen peroxide (to oxidize the diaryl oxalate, probably to a 1,2-dioxetanedione)
a fluorescent dye, which determines the final colour.

Wilhelm J. Baader has examined the reaction in great detail, and you'll find a lot of references on his website.


Geometry of tetrabromidomanganate(II) What is geometry and hybridization of $\ce{[MnBr4]^2-}$ for $\ce{Mn}$ ($Z = 25$)?

I am not sure if it is tetrahedral or square planar.
I assumed $\ce{Br}$ is a weak ligand and got its geometry to be tetrahedral.
Am I right?

I am a 12th grader and I only know the basics of coordination chemistry.

A square planar geometry is somewhat constrained, if you wish. There need to be compelling electronic reasons to move four coordinating anions (especially large ones like bromide) closer together than tetrahedral — which is the case in square planar, cf the smaller bond angles. Usually, this geometry is associated with $\mathrm{d^8}$ ions but it can happen for other compounds, too.

Another thing to consider is the field split. Its size is influenced by the metal’s charge. Higher charge typically means greater field split. Square planar complexes typically require a large field split to stabilise their low-spin state.

Manganese(II) is a $\mathrm{d^5}$ ion. Since it only has a low charge, it is very likely to be in the high-spin state. The high-spin state of a $\mathrm{d^5}$ ion does not stabilise any geometry whatsoever. Thus, no reasons exist to go square planar and the tetrabromidocomplex must be tetrahedral.


Precedence of 1,2 carbocation rearrangement How can one choose which group has more shifting tendency in 1,2 carbocation rearrangement? The obvious order is via the stability of the carbocation of the group. But, phenylic groups have high shifting tendency--and a phenylic carbocation is unstable.

Is there a way of predicting these? I was thinking that it will be related to the delocalization in the triangular intermediate formed.

Update: I'm talking about comparing "migratory aptitude" of the $R$ groups in a system similar to $>C(+)-C(R_1R_2R_3)$

The obvious order is via the stability of the carbocation of the group.

I think you meant the migrating groups stability?
This is not what textbooks say.
Reactions are ruled by the $ΔG$ of the activated complex.
This may be close  to educts, or to products, or some halfway
state or a real intermediate minimum is on the reaction path.
So, in some cases the model of a separate carbenium ion
(in a ionic pair maybe) is useful, sometimes you need to
think of a three-center bond state.


Spectrometry in Chemistry & Flame Test Lab Question 1:
You currently see two colors that are two similar (dark yellow and orange) in a flame test lab. And also you have a hard time differentiating the colors, what is an alternative to distinguish the two compounds?

My thoughts:
I thought that it was because if you use a wooden splint to dip it into the chemicals, you accidentally combust it causing the inaccuracy and also the source of the flame is blue making it hard to see, thus you use a metal wire and dip that and put it just so slightly on top of the flame to not mix the blue color of the butane flame and the flame that came with the chemicals on the wire. Also to use a digital spectrometer for more accuracy

Am i right about my answer?

Question2:
Let's say that you are using reusable metal loops instead of wooden splints, and you were doing two flame tests of the same chemical and you get different colors, what is the error and the reason?

My thoughts:
I was really confused, maybe because for one of them you placed it too far from the flame source and for the other you placed it too close to the flame source?

... two colors that are two similar (dark yellow and orange) in a flame test ...

Do you know what the two compounds might be?

   The yellow colour typically indicates sodium, the orange flame colour calcium.

In order to distinguish the two compounds, you have several options:

Use a different method

   You can try to precipitate the calcium ions in the form of an insoluble salt, such as its oxalate.

Refine your flame test

   View the flames though a piece of cobalt glass! This filters of the yellow flame colour typical for sodium. Note that the flame colours for the other metals will look different now!


Radioactivity and Kinetics of disintegration Well I've just started on kinetics and there are a few things I am absolutely confused about!

First of all, is a neutron actually made up of a proton and electron? If yes, then how is this possible?
In K-electron capture where the electron of the K-Shell is captured by the nucleus, doesn't the species become positively charged or ionized as it is losing an electron?
When metastable technetium disintegrates into normal technetium and a gamma particle, what is the change brought about? Isn't the technetium on both sides of the equation basically the same?

No, as hey answered, neutrons are made of two down quarks and an up quark.  Protons are made of two up quarks and a down quark.  It is the conversion of a down quark into an up quark that results in the neutron --> proton conversion.  Read here for more information about this reaction.
The confusion with this probably comes from the fact that net reaction  in a neutron to proton conversion is $n^{0} \rightarrow p^{+} + e^{-} + \overline{\nu_e}$, so it looks like a neutron is falling apart into three sub-particles.
No, the atom will not become ionized from this event itself.  Consider what is happening in this reaction:
$p^{+} + e^{-} \rightarrow n^{0} + \nu_e$
Yes, an electron is removed from the atom, but at the same time a proton is as well.  We should know that the atoms remains neutral because of conservation of charge. If the atom has no net charge at the beginning of the reaction, and no charged particles are ejected, it must be neutral at the end of the reaction as well.
Jon Custer's response is spot on here.  Gamma ray emission represents the movement of the nucleus from a metastable state to its ground state via the release of electromagnetic radiation.


Bond length in ICl2- and ICl2+ Why is the $\ce{I-Cl}$ bond length in $\ce{ICl2-}$ (charge of -1) is longer than in $\ce{ICl2+}$ (charge of +1) ?

Here are a few explanations that I have in mind. Please critique.

I in $\ce{ICl2-}$ is $sp^3d$ hybridized. $\ce{I}$ in $\ce{ICl+}$ is $sp^3$ hybridized. Since additional promotion energy is required to form the $sp^3d$ hybrid compared to the $sp^3$ hybrid, the bond lengths in $\ce{ICl2-}$ is longer.
For $\ce{ICl2-}$, if we let $\ce{I}$ be $sp^2$ hybridized such that the 3 lone pairs lie 120 apart in a plane, and then let the 2 $\ce{Cl}$ atoms and the $\ce{I}$ atom lie in a straight line, then we can form a 3-centred, 4 electron system. This is weaker than the 2-centred, 2 electron system in the $\ce{I-Cl}$ bond in $\ce{ICl+}$.
As $\ce{I}$ in $\ce{ICl2+}$ has a higher effective nuclear charge, its valence orbitals are smaller and more compact, hence the $\ce{I-Cl}$ covalent bond is stronger as the electrostatic attraction between the nuclei and the overlap density is stronger.

The iodine atom in $\ce{ICl2}$ is what is termed hypercoordinated (or hypervalent). Basically, this means that the 3 atoms combine to produce 3 molecular orbitals (the 3-center 4-electron model). There is no need to invoke d-orbitals.

In neutral $\ce{ICl2}$ there are three electrons spread across the two $\ce{I-Cl}$ bondss. Placing those electrons in our molecular orbitals, 2 go into the bonding orbital and one into the nonbonding orbital.

In $\ce{ICl2^{+}}$ we have removed the one nonbonding electron.  As a consequence there will be less electron-electron repulsion and only the bonding molecular orbital is occupied, because of this we would expect shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2}$.

In $\ce{ICl2^{-}}$ we have added another electron and it will go into the nonbonding orbital.  This will increase the electron-electron repulsion in the molecule without increasing bonding, because of this we would expect longer bond lengths in $\ce{ICl2^{-}}$ compared to $\ce{ICl2}$.

These arguments tell us that we would have shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2^{-}}$.


Extracting herbs with grape juice I tried searching my question on google and did not find anything. I am looking to extract herbs with pure 100%
grape juice http://www.lakewoodjuices.com/product_detail/id-32/ ( it is pasteurized ) instead of using the old school
method of extracting with a 50% alcohol solution.My theory as of now would be that it would be more beneficial and it
would ferment or extract the same plus the extra beneficial minerals of the juice. I am wonder what your opinion on
this is. Do you think it would extract the same and be just as beneficial as a tincture? Would it be more beneficial?
How long should I let it sit for? should I let it sit at room temp or in cooler temp say, 45 degree F? Should I do neutral water dilution with the grape juice? how much, 50% neutral water 50% grape juice?
Thank you so much for your time and knowledge.

The point of using alcohol to extract things from herbs is that alcohol solubilises things that are not soluble in water. So using grape juice as an alternative is pretty pointless since it is just water with a lot of other water-soluble things in it (like sugars).

So using grape juice is a pretty poor alternative to using alcohol.

Grape juice will ferment under some circumstances, converting some of the sugars into alcohol. But this won't work well unless the conditions are controlled carefully and won't work at all if it is cold (45 F is cold). If you don't do what winemakers do, there is no telling whether it will work or whether it will work without producing nasty results. And pasteurised grape juice won't ferment at all if you don't add yeast and keep it clean. If you don't keep it clean you might well get some nasty strain of natural bug rather than a good wine yeast which will have unpredictable results.

Moreover, the best level of alcohol you are going to get even as an experienced winemaker is 10-15%. This is nothing like as useful as pure or 50% alcohol for solubilising or extracting things that are not water soluble. And grape juice with no alcohol isn't going to be a much better solvent than water (and most likely will be worse as it is already full of other dissolved things).

The old school recipes you are trying to avoid exist because people have tried these experiments before and found what works. Trying to invent new recipes that ignore the "old school" experience is a pretty futile task: what you propose isn't in the old school recipes because it doesn't work.


Does boron form compounds without covalent bonds? I have read that boron, due to the very high sum of its first three ionization energies, it is not able to form its +3 ions, and thus it generally forms only covalent compounds. But in a popular Indian engineering entrance exam: JEE MAINS-2014 19th April paper, this question was asked:

Which of these statements is not true ?

In aqueous solution, the $\ce{Tl+}$ ion is much more stable than Tl (III)

$\ce{LiAlH4}$ is a versatile reducing agent in organic synthesis.

$\ce{NO+}$ is not isoelectronic with $\ce{O2}$

$\ce{B}$ is always covalent in its compounds.

The answer given was 4.
I get that there might be an exception but I'm not able to find it anywhere. If there exists an ionic compound of boron, could someone please mention that to me?

Boron can form ions but there is some fine print.  You won't get monatomic cations like the metals below it.  Instead, ionic boron structures are formed from clusters where the ionic bonding is driven by the molecular orbital structures in these clusters, not by electronegativity (cf. This answer).
Such clusters are internally held together by covalent bonds between the boron atoms, so in this sense boron is still forming covalent bonds.  The ionic bonds would be with atoms of other elements outside the boron cluster.  Since the valence shells of a neutral boron atom are less than half filled the clusters will likely have low-energy, bonding orbitals that require electrons from outside atoms.  Thus the boron clusters will be anionic and the ionic bonds will be most likely formed with electropositive metals.  As suggested in the comments, magnesium diboride, $\ce{MgB2}$, is one of the most widely studied compounds containing such boron clusters. It has drawn much research interest because of its relatively high critical temperature (39 K) for superconductivity, which may be related to the impact of ionic magnesium-boron bonding on the eletronic interactions that lead to superconduction.
Magnesium diboride has a layered structure in which magnesium layers alternate with boron layers.  The latter are covalently bonded into a hexagonal honeycomb, resembling a carbon layer in graphite.  However, in the boron layers each atom supplies only three electrons per atom instead of four, so the layers may act as electron-accepting structures to form macro-anion having the formula $\ce{B^-}$.  An ionic model for the diboride would then have the empirical formula $\ce{Mg^{2+}(B^-)2}$.  Here I discuss two references I have examined, in which the bonding is examined and the results may be compared with this model.
De la Mora et al. [1] compare magnesium diboride with other $\ce{MeB2}$ diborides using early transition metals and aluminum (the latter might also be regarded as having early-transition-metal character, as there is no $d$ block separating this element from magnesium).  They find that while all the diborides have significant ionic character, this ionicity is enhanced in the magnesium compound.  Thereby the magnesium compound has increased electrical-conduction anisotropy as the valence electrons are strongly localized towards the boron layers.  Zirconium diboride, with less ionic bonding and less electron localization, is also superconducting, but its critical temperature according to this reference is only 5.5 K versus 39 K for the magnesium compound.  The authors also suggest that an isoelectronic, even more strongly ionic $\ce{Li(BC)}$ compound may offer an even further enhancement in superconductivity.
Nishibori et al. [2] found that at room temperature, magnesium is essentially fully ionized to $\ce{Mg^{2+}}$ while the boron remains neutral; the negative charge is associated with the interstitial regions as if to constitute metallic bonds.  This still represents two-thirds of the theoretical charge separation for an ionic model and in that sense, the bonding between magnesium and boron may be deemed predominantly ionic.  At 15 K the electrons become more localized so the boron now has a significant negative charge and the percentage of theoretical charge separation exceeds 80%.
Thus both references agree that in magnesium diboride, the combination of an electropositive electron source with a favorable molecular structure for electron acceptance leads to strongly ionic bonding between the magnesium and the boron.  This applies especially in the low-temperature superconducting state, even as the boron-boron bonding within the boron layers themselves remains covalent.
References
1.
Pablo de la Mora, Miguel Castro and Gustavo Tavizonb,
"Comparative study of the electronic structure of alkaline-earth  borides (MeB2; Me=Mg, Al, Zr, Nb, and Ta) and their normal-state conductivity", Journal of Solid State Chemistry 169 (2002) 168–175, https://doi.org/10.1016/S0022-4596(02)00045-2.
2.
Eiji Nishibori, Masaki Takata, Makoto Sakata, Hiroshi Tanaka, Takahiro Muranaka and Jun Akimitsu, "Bonding Nature in MgB2", Journal of the Physical Society of Japan 70:8 (2001), 2252-2254,
https://doi.org/10.1143/JPSJ.70.2252.


To calculate total vapour pressure of a solution Question:
1 mole each of A and B were mixed to obtain an ideal solution of which one mole went to the vapour phase at equilibrium.
If $P°_\mathrm{A} = \pu{100 torr}$ and $P°_\mathrm{B} = \pu{900 torr}$, then $P_\mathrm{S}$ would be?

For the above question, I first calculated vapour pressure of A and B which came out to be $\pu{50 torr}$ and $\pu{450 torr}$ respectively (using $P°_\mathrm{A} × X_\mathrm{A}$ where $X_\mathrm{A}$ is mole fraction of A) and then I calculated vapour pressure for A and B in vapour phase which came out to be $0.1$ and $0.9$ respectively, so the amount remaining in the liquid phase will be $\pu{0.9 mole}$ for A and $\pu{0.1 mole}$ for B, according to which the $P_\mathrm{S}$ should be $\pu{180 torr}$, but the answer provided is $\pu{300 torr}$, am I missing something? (also see the image below for better understanding of my approach).
Any help is appreciated! Thanks in advance!

The starting point to solve the problem is Raoult's law. For each component we can write that the vapour pressure (related by Dalton's law to the total pressure) is equal to the product of the mole fraction in the solution and the vapour pressure of the pure component:
$$y_iP_S = \chi _i P^{\circ} _i$$
You can exploit the fact that the molar amounts in the problem are particularly tidy:
\begin{align}
  n_{1l} + n_{1g} &=1 \\
  n_{2l} + n_{2g} &=1 \\
  n_{1g} + n_{2g} &=1
\end{align}
So that for the first component:
\begin{align}
  y_1 &= n_{1g} \\
  \chi _1 &= n_{1l}  \\
  \rightarrow n_{1g}P_S &= n_{1l}P^{\circ}_1 \\
  \rightarrow n_{1g}P_S &= (1-n_{1g})P^{\circ}_1  \\
  \rightarrow P_S &= \frac{(1-n_{1g})P^{\circ}_1}{n_{1g}}
    = \frac{n_{2g}P^{\circ}_1}{(1-n_{2g})}
    \tag{1}\label{eq:firstcomponent}
\end{align}
For the second component we can similarly write
\begin{align}
  y_2 &= n_{2g} \\
  \chi_2 &= n_{2l}  \\
  \rightarrow n_{2g}P_S &= n_{2l}P^{\circ}_2 \\
  \rightarrow n_{2g}P_S &=(1-n_{2g})P^{\circ}_2  \\
  \rightarrow P_S &=\frac{(1-n_{2g})P^{\circ}_2}{n_{2g}}
    \tag{2}\label{eq:secondcomponent}
\end{align}
Equating the expressions for $P_S$ in \eqref{eq:firstcomponent} and \eqref{eq:secondcomponent} and rearranging terms we can write a quadratic equation in $n_{2g}$ whose solution can be inserted back into the expression for $P_S$.


how to modify a transport device with an internal combustion engine with different atmospheres How to modify a transport device with an internal combustion engine with:

the methane-nitrogen atmosphere
carbon dioxide atmosphere
the atmosphere of argon and chlorine.

Which substances will react? How will they feed? How effective will it be?

Each of the cases will strongly depend on the non-atmospheric natural
resources available on each planet.

For example in an atmosphere of $\ce{O2}$ + inert gases, it would be desirable to have deposits of hydrocarbons or $\ce{H2}$ gas to be mined and pumped into your fuel tank to burn with the atmospheric $\ce{O2}$.  In each case proposed below it is assumed that the non-atmospheric reagents would be available as deposits on the planet or that they could be synthesized in a way that they are acting as batteries, storing the energy that went into their synthesis.  That said, here are some possible simplistic solutions for the different scenarios.

Scenario 1: $\ce{CH4}$ + $\ce{N2}$ atmosphere:

Here you could simply reverse the typical internal combustion engine process used in Earth-vehicles. Rather than filling your tank with hydrocarbons to be mixed and burned with atmospheric $\ce{O2}$, you would fill your tank with (probably liquid to keep the volume small) $\ce{O2}$ to be mixed and burned with atmospheric $\ce{CH4}$.  Just as with conventional internal combustion engines on Earth, the reaction products would be $\ce{H2O}$, $\ce{CO2}$, and heat.

Scenario 2: $\ce{CO2}$ atmosphere:

As $\ce{CO2}$ is already thoroughly oxidized, you would need to fill your tank with a good reducing agent.  If there were deposits of $\ce{H2}$ gas trapped within the planet, then you would mine the $\ce{H2}$ and fill your tank with pressurized $\ce{H2}$ gas (the resources required to form and store liquid $\ce{H2}$ just wouldn't be worth the trouble, though there are other ways to concentrate $\ce{H2}$ then liberate it when needed).  The $\ce{H2}$ would then be mixed and burned with $\ce{CO2}$, ideally giving off $\ce{CH4}$, $\ce{H2O}$ and heat as products.

Scenario 3: $\ce{Ar}$ + $\ce{Cl2}$ atmosphere:

First to address is the composition of the vehicle, or at least the parts that would come into contact with the $\ce{Cl2}$ and other reactants and products, I propose a $\ce{SiO2}$ based coating for all components that would be in such contact.  For a gas-phase ignition scenario, I propose a diesel type of concept (ignition heat from compression of the mixed gases) to avoid the need for electrical ignition, metal wires, etc.

One possibility is to use the $\ce{Cl2}$ analogously to how $\ce{O2}$ is used in internal combustion engines, namely as the oxidant.  A hydrocarbon, $\ce{CH4}$ to keep it simple, would act as the reductant.  This reaction would ideally produce $\ce{CCl4}$, $\ce{HCl}$ and heat.  The atmosphere is already toxic and corrosive so the $\ce{HCl}$ exhaust is probably OK.

In the same manner as the $\ce{CO2}$ atmosphere, $\ce{H2}$ is another possibility.  In this case, the ideal reaction product is just $\ce{HCl}$.

One final possibility is worth considering here.  In all other scenarios I've only proposed fluids as fuel for the combustion.  But $\ce{Cl2}$ reacts very exothermically with $\ce{Fe}$ metal.  And the ideal product, $\ce{FeCl3}$, is also a solid.  So introducing a fine iron dust in some carrier gas could actually have the advantage of ease of collecting the exhaust do be disposed of at a dedicated treatment center.  That way we wouldn't pollute the air for the $\ce{Cl2}$-breathing natives.

Note that most of the proposed fuels here are composed of some of the most abundant elements in the universe.


What determines "polarizability" of an element? In this question, the answer given discusses that fluorine has a good orbital overlap with hydrogen and also that fluorine is not very polarizable. However, is it always true that strong bonding is inversely related to polarizability?

Fluorine is not very polarizable because it is small. Its electrons, therefore, are all close together. A polarized atom has shoved all its electrons to one side. Since they are close together in Fluorine, the negative-negative repulsion is too large for that. In another halide such as Iodine, whose valence electrons are much farther from each other, much better polarization can be achieved.

More polarizability does NOT always mean better or weak bonding. Generally, polarizable atoms like to bond to each other, and non-polarizable atoms like to bond to other non-polarizable atoms. For more info, see:

http://www.adichemistry.com/inorganic/cochem/hsab/hard-soft-acid-base-theory.html


Equivalent of Szabo and Ostlund book for DFT Szabo and Ostlund book Modern quantum chemistry [1] is extremely useful to understand Hartree-Fock and post-Hartree-Fock methods. Not only it explains the theory behind such methods, but it is also oriented towards a practical implementation. Following this book I was able to implement an Hartree-Fock code from scratch in order to perform Born-Oppenheimer molecular dynamics.

Since I already work with software packages based on density functional theory (Gaussian09 for calculations in vacuum and QuantumESPRESSO for solid state systems), I would like to implement a simple DFT code in order to better understand what is going on inside the "black box". However, I would like perform this task using a plane wave basis set: this is somewhat more involved than using Gaussian basis sets since it involves periodic boundary conditions and pseudopotentials. In addition, I will have to take care of fast Fourier transforms everywhere and of the approximation of the exchange-correlation functional.

Supposing that I will be happy with the local density approximation, there is a book out there similar to Szabo's book for DFT? Notice that I am not interested in the basics of DFT but I am searching for a good book explaining in details the practical implementation using PW.

Here some comments on the books I already considered:

Martins [2]: This book contains a lot of informations and is oriented to solid-state electronic structure, therefore it explains plane waves and pseudopotentials. However I think that it works well as a reference but not as first reading to understand in details the concepts.

Giustino [3]: Very nice book with the goal of teaching DFT to undergrad students. For this reason is too simplistic and does not go into technical details.

[1] A. Szabo and N. S. Ostrlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Doever Publications, 1996.

[2] R. M. Martins, Electronic Structure: Basic theory and practical methods, Cambridge University Press, 2004.

[3] F. Giustino, Materials modelling using density functional theory, Oxford University Press, 2014.

I can recommend the book of Yang and Parr, although I am not sure it deals  with plane wave bases at all. In principle, a DFT code is not really different from a HF code. In fact, you can use exactly identical codes, just replacing the Coulomb and exchange integrals by numerical integration of some functional. E.g., if you choose $X\alpha$-LDA, than your exchange functional is
$$K = C_\alpha \left(\frac{3}{\pi}\right)^\frac{1}{3} \int \rho(\mathbf{r})^\frac{4}{3} \mathrm{d}\mathbf{r}$$
You can evaluate your density numerically, choose any numerical integration scheme and obtain the exchange energy. That works even, if you still calculate the Coulomb integrals the HF way. However, you wouldn't save much time since you still have the $\mathcal{O}(N^4)$ step in the Coulomb integrals, where $N$ is the number of basis functions. But once you integrate the Coulomb integrals numerically, you have basically a working DFT code.

There is a number of papers about choosing the correct grids and integration schemes for numerical integration. Once, I read one that compared the radial grids quite well, but sadly I don't remember the reference. But the buzz words would be e.g. Becke grid, (Treutler?-)Ahlrichs grid or simply Gauss quadrature for the radial integration (most of the grids are a form of Gauss quadrature, though). For the angular integration there is e.g. the Lebedev grid, but its calculation is a little more involved.

My suggestion, based on teaching experience, is to write first a HF code in a Gaussian basis. Then you pimp it with a numerical integration scheme and something like $X\alpha$. In the next step, you replace your Coulomb integrals. Lastly, you can switch from a Gaussian basis to the basis of plane waves. The last step only changes your matrix elements but not the algorithm (HF/DFT) itself. Since you first implemented a HF code for a Gaussian basis, you can always check your algorithm before going into uncharted waters.


Fractional Bond Order - How to Visualize? Consider nitric oxide, which has a real bond order of 2.5, contrary to what this UC Davis rendering might imply:

What's the best way to visualize a molecule with a fractional bond order? What exactly is "half a bond"? The electron spends half its time participating in bonding and the other half of the time as a lone electron on nitrogen?

Maybe some examples will help.  In the picture below, I've drawn some resonance structures for butadiene and acrolein.  What do they tell us?  They suggest that the central "single" bond should have some double bond character, it's more than a single bond, but less than a double bond.  By the same token, the "double"

bonds at the ends of the molecules should both have some single bond character.  These are examples of fractional bond order.  The real world consequences of fractional bond order is a higher barrier to rotation about the central "single" bond in these two molecules than we might have expected, and a lower barrier to rotation in the pi bonds than we might have expected.


What is an orbital boundary surface? Could anybody explain the following statement:

"The boundary surface is a common way to represent atomic orbitals, incorporating the volume in which there is about a 90 percent probability of finding the electron at any given time.

I don't understand what does the author means starting from "incorporating the volume.."

It means that the boundary surface is a closed surface or set of closed surfaces, such that there is a 90% chance that an electron is interior of the closed surface and a 10% chance that the electron is exterior of the closed surface.

A better way of specifying how an orbital is represented is "equal probability boundary surface" which further specifies that each point on the surface must have any equal electron probablity density, and the surface must be a closed surface or set of closed surfaces with a 90% (or other percent) chance of containing the electron at a given time.

Without the "equal probability" statement there are infinite different boundary surfaces for each orbital (infinite different ways to enclose 90% of the probability density).


Atomic mass number and mass number in atomic symbol In chemistry classes or even many different branches of chemistry textbooks, I am told that the atomic symbol

$$\Large ^A_Z\ce{X}$$

includes the following notations:

$A$ – "atomic mass" or "mass number";
$Z$ – "atomic number".

But, because I want to make myself to remember easily, I would prefer to call $A$ "Atomic mass number" instead of simply "mass number". I think many say that I am making a big deal and unnecessary things. But as a non-native learner, if I can confirm that using the term "atomic mass number", it would make me easier to remember the term.

The IUPAC preference for A in $\ce{^A_ZX}$ is mass number. The term would also apply to a molecular fragment, for instance in a mass spec. So depending on the context atomic mass number or molecular mass number might be used to be more specific.

The problem with you arbitrarily defining A in $\ce{^A_ZX}$ as atomic mass number is that various references might just use mass number. So depending on context you'll need to be able to decide which possibility is the correct one. Of course in spite of IUPAC definitions there are no doubt multiple variations in the past and current literature.


Do all the precipitates undergo an equilibrium with its respective ions, why can't they dissociate fully? I’m really confused about this. If we consider $\ce{Ca3(PO4)2}$, it can be seen it undergoes an equilibrium. But why it’s an equilibrium?

This is the equilibrium it undergoes $$\ce{Ca3(PO4)2_{(s)}<=> 3Ca^2+_{(aq)} + 2 PO4_{(aq)}^3-}$$

By the way I’m having a doubt of why some are using backward arrow for a dissociation of a particular precipitate while some are using a solid forward arrow for the same reaction. For an example let’s take $\ce{AgCl(s)}$

Now this is a place where a backward arrow is used,
$$\ce{AgCl_{(s)}<=> Ag+_{(aq)} + Cl_{(aq)}-}$$

But I have seen in some books where there is a solid forward arrow instead of back-ward arrow for the dissociation of $\ce{AgCl(s)}$.

In the two examples you mentioned, we have an equilibrium between the solid and the small quantity of the solid that is soluble in water and completely dissociated:
$$\ce{Ca3(PO4)2_{(s)}<=> 3Ca^2+_{(aq)}  + 2 PO4^{3-}_{(aq)} }$$
$$\ce{AgCl_{(s)}<=> Ag^+(aq)_{(aq)}  + Cl^{-}_{(aq)} }$$

I think it's wrong to write a solid forward reaction.
It's an heterogeneous equilibrium between the solid state (the precipitate) and the aqueous state (ions constituents of the precipitate). The constant of this equilibrium is the solubility product $K_{\mathrm{sp}}$ of the precipitate.


A contradiction, probably from confusion All things in nature tells us that high entropy is easier to happen.
But then why does it takes energy to break bonds, in order to achieve high entropy.

Any thing in gaseous state has higher potential energy than it is in solid state. But that creates less stability.

Processes in nature that lead to an increase in the entropy of the system are in general favourable, but entropy alone is not the deciding factor!

If we were to follow a full proof we would have to consider Carnot engines and such but for now the result that we are interested in is called the Clausius statement. It is the Clausius statement that really defines the second law and the direction of spontaneity. The Clausius statement leads us to define two parameters depending on the experimental conditions: either the Helmholtz free energy $A$ (when working at constant volume) or the Gibbs free energy $G$ (under conditions of constant pressure). In practice it is much easier to work with $G$ since we can expose our reaction mixture to the atmosphere and call it constant pressure.

It is these thermodynamic potentials $G$ and $A$ that determine the overall thermodynamic picture of spontaneity of a given process.

If we work with the Gibbs potential, after some mathematics we can easily express by the following equation:

\begin{equation}
G=H-TS
\end{equation}
Where $H$ is the enthalpy of the system, $T$ is the temperature and $S$ is the entropy. For a given process it is the Gibbs energy that determines the spontaneity of the reaction. Crudely put, the starting material will have a certain $G$ value and the products will have a certain $G$ value, if the change in $G$ over the course of the reaction is negative then we say it is spontaneous. If it is positive it is not spontaneous.

We can clearly see that the condition for spontaneity is not solely dependant on the entropy alone rather it is also a function of the enthalpy too. Further the entropic contribution to the $G$ is dependent on the temperature indicating that there will be a certain threshold temperature for an endergonic process.

An endergonic process is characterised by having a positive enthalpy change, requiring energy to occur. If the "amount" of energy the process requires is very large then the change in enthalpy will also be large and as we can see this makes the change in $G$ for the process become less favourable.

Moving on, we can now see that a given process will have a certain energy barrier... an activation energy if you like. Whether a process is thermodynamically spontaneous or not one has to consider the Kinetics too! A simple yet ubiquitous model for Kinetics is the Arrhenius equation which in general says the rate of reaction decreases for processes with a higher energy of activation.

Turning the later part of your question. The most stable phase (state of matter) is dependent on the conditions of the reaction. The most stable phase for the given combination of temperature and pressure will prevail in the reaction mixture. In practice this leads to the state with the lowest value of $G$.

Solids actually liberate energy when they come together, especially ionic solids! This energy is called the Lattice energy and it is relative to all the reactants in their gaseous state (as ions). (look at Born-Lande equation for more info)! So although gaseous species will have a higher entropy due to all the rotational vibrational freedom they have ... the state of matter is dictated by $G$.

Without the full mathematical argument behind what I have said I feel my answer will be incomplete, but I hope it helps you :)


Information depth in x-ray photoelectron spectroscopy (XPS) What is 'information depth' in XPS and what does determine it? Also how does it depend on the 'take-off angle'?

Do you understand the basics of XPS?  The Wikipedia article is pretty clear. In summary, a (usually monochromatic) x-ray source irradiates a sample.  The x-ray can be absorbed by electrons in the sample, and can provide enough energy for a core electron to be set free with an energy equal to the incident x-ray energy minus its binding energy. Collecting the electrons and measuring their energy gives a plot of counts vs electron energy.  Lower electron energy indicates an electron coming from deeper in the atomic shell. Given knowledge of atomic binding levels, you can associate specific peaks in the spectrum with elements (with chemical effects on binding energies giving additional information).

So - what is the "information depth"? A slightly odd term, I've generally heard/used 'detection depth'.  Two questions to ponder - (1) how far will x-rays of a few keV penetrate into a solid (if an atom doesn't see any x-rays, it won't emit any electrons), and (2) how far can a ~100eV-1000eV electron travel in a solid before being scattered (and thus losing the useful info).  Not surprisingly, (2) dominates for XPS, and is often called the 'escape depth'.

As for the 'take-off angle' - again, the interesting part is where your electron detector is placed relative to the sample, and how far the electrons have to travel through the sample to head out in the direction of the detector - the 'escape depth' will vary with exit angle (take-off angle) since the amount of material the electron has to go through is a simple trig function of the exit angle.

I hope that helps.


Which element has a greater second electron affinity, sulfur or oxygen? I found a question asking the above which states that sulfur has a lesser second electron affinity than oxygen. But since the inter-electronic repulsion in sulfur is lesser, shouldn't it be willing to take up another electron more easily than an oxygen ion with a -1 charge?

The most important fact here is that the electron added to $\ce{S-}$ goes into an $n=3$  orbital ($3p$), whereas the one added to $\ce{O-}$ goes into an $n=2$ orbital ($2p$).

The larger $n=3$ orbitals are further from the nucleus and hence experience a weaker attraction to the nucleus and are much higher in energy.  This dependence with quantum number scales approximately $E\sim - n^{-2}$ so is very important [1].

The dependance of orbital energy also scales quadratically with effective nuclear charge, however effective nuclear charges falls significantly compared to the true nuclear charge due to shielding and repulsion from other electrons.  A common first order approximation for the effective nuclear charge is given by Slater's rules [2]

Combining these two approximations we can calculate very approximate orbital energies.

$$
\begin{align}E_{2p}(\ce{O^2+}) & {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
& {} = - R_\infty h c \frac{(8-4.15)^2}{2^2} \\
& {} = - 3.7 \text{ Ry} \\
E_{3p}(\ce{S^2+}) & {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
& {} = - R_\infty h c \frac{(16-11.25)^2}{3^2} \\
& {} = - 2.5 \text{ Ry}
\end{align}
$$

Where 1 Ry is the electronic energy of a hydrogen atom.

As you can see, to first order, the energy of the new electron is much lower when added to $\ce{O-}$ than when added to $\ce{S-}$, leading to a more favourable electron affinity for $\ce{O-}$ than $\ce{S-}$, driven by the fact that the orbital occupied has a lower principal quantum number $n$

[1] - See the energy of a hydrogenic ion - Wikipedia.

[2] - Slater's rules - Wikipedia


How does the "kreosan" Coca-Cola rocket work? A YouTube channel called "kreosan" (made by Russian-speaking people who I believe live in Ukraine) posted a video showing rockets made from bottles of Coca-Cola.

Here's what I've figured out from the video.

First step: pour out some Coca-Cola to make room in the bottle to add gas.
Second step: add butane (looks like a canister for refueling butane lighters)
Third step: this is unclear.  He adds a gas from a large container that appears to say "GAS-1".  The YouTube description says "Coca Cola + propane = turbo ROCKET" so I guess it's propane.

Unknown: Does the rocket require both butane and propane?  I think the second
   step was just to show that you don't get very much gas out of a
   little butane canister, and all you really need is the propane from
   the last step.

Final step: invert the bottle, and some reaction pushes Coca-Cola out the mouth of the bottle with enough force that the whole bottle flies quickly as a rocket.

What is going on here?  How does inverting the bottle trigger the reaction?  Is this like the "Mentos and Diet Coke" thing where the trick was to get the carbon dioxide dissolved in the drink to come out quickly?

My understanding is that the surface of Mentos candy provides a bunch of nucleation sites for bubbles, causing a lot of carbon dioxide to come out of solution quickly.

So, does the "kreosan" rocket fly on just carbon dioxide or is there some other reaction going on here?

Here's a paper written about propane forming small ice crystals when sprayed into a cloud of water vapor: James R. Hicks and Gabor Vali, J. Appl. Meteor., 1973, 12, 1025–1034.

I'm no expert, but I'm guessing the propane forms small ice crystals as it passes through the Coca-Cola, and when flipped, the Coca-Cola warms up the liquid propane, which causes it to expand as it travels upwards. This provides an immense surface of nucleating sites, similar to a Mentos reaction, but with millions of microscopic Mentos (this means larger surface area). The pressure waves caused by the violent nucleation further drives the reaction, as the liquid propane expands further and freezes more ice crystals and so on. This all happens quite fast, and with one easy way out, everything shoots out the mouth of the bottle.

Of course this was written by a Materials Engineering undergrad, so don't take this explanation too seriously.


Why is this carbocation more stable? For comparing the stability of different compounds that undergo resonance, I was taught to first check for equivalent resonance, then the number of atoms on which the delocalization was taking place and so on(e.g phenolate ion vs acetate ion)

But comparing benzyl carbocation with $\ce{CH2=CH-CH2+}$, even though $\ce{CH2=CH-CH2+}$ has equivalent resonating structures(and hence a greater extent of delocalization of electrons), it is less stable than benzyl carbocation. Why?

What all the comments boil down to is we are comparing different things:

Allyl vs benzyl cations:  Both contain onlybhydrogen and carbon, with the positive charge formally distributed over two carbon atoms in allyl versus four carbon atoms.  Putting the positive charge on more carbon atoms makes the benzyl cation more stable.

Acetate vs phenolate anions:  Here, unlike the hydrocarbon actions mentioned above, oxygen enters the picture.  The negative charge in the acetate ion is distributed over two oxygen atoms, versus phenolate having its negative charge on one oxygen plus multiple carbons.  Having the second oxygen atom wins out because of the superior electronegaytivity of oxygen, which is a nonfactor for the cation case mentioned above.

Note that the "rules" above are not absolute laws.  You have to consider other factors such as aromatic or antiaromatic coupling that involves the charge; e.g. cyclopropenyl cation is much more stable than benzyl cation because of the positive charge being inherent to an aromatic ring, even though cyclopropenyl has only three instead of four  carbons to take up the charge.


How do I select a sub-freezing temperature for my experiment that will allow the formation of frost, but not supercooled liquid water (rime) I once conducted a set of experiments in which I had to form ice by direct vapor deposition (not condensation of liquid water followed by freezing, aka riming) onto a Peltier-cooled glass-coated copper rod.

The experiments were conducted outside at night within a small wind tunnel, with ambient air temperatures ranging from -2 oC to -25 oC and the glass substrate was cooled from 1 oC to 10 oC below the ambient air temperature.  Occasionally supercooled water droplets would condense, followed by freezing (forming rime ice) rather than direct deposition of ice crystals from the vapor phase.  These events would ruin my time-consuming experiments.

The only relevant parameter under my control was the temperature of the glass-coated copper rod.  The only relevant parameters I could measure during the experiment were the air temperature and humidity. I also introduced trace-levels (single to hundreds of ppb) of several C1 to C4 alcohols in order to later measure their incorporation into ice during ice deposition from the vapor phase.  However, the low levels of these compounds did not affect whether riming or direct vapor-ice deposition would occur, as based on a series of 'blank' experiments without the trace gases.

If I were having trouble with liquid water condensation, in what way would I want to alter the glass substrate temperature so that only the direct deposition of solid ice could occur?

Surprisingly, at a glance, I had to run my experiments with the glass substrate at higher temperatures in order to ensure ice would form and not liquid water.  When you think in terms of the difference between the saturation vapor pressure of liquid water and that of ice at the same temperature, this solution is obvious.  The temperature has to be above the dew point but remain below the frost point, which is always higher than the dew point at temperatures below 0 oC.  The reason for this is that at any given temperature, liquid water has a higher vapor pressure than ice.

As a real-world example of the ice-growth mechanisms observed in my trace-gas uptake experiments, what was a problem for my experimental conditions actually has important implications for ice-particle growth in clouds, in which context it’s called the “Wegener-Bergeron-Findeisen Process”.  Here, when there is a mixture of ice and supercooled water droplets, water vapor condenses onto the ice particles so that the humidity with respect to liquid water decreases enough for the liquid water to further evaporate.  Thus there is a net transfer of water mass from liquid droplets to solid particles and the solid particles may then grow large enough to precipitate (as in fall to the ground in this context).


Deprotonated form of phenolphthalein The wikipedia page states clearly that the coloured form of phenolphthalein is the one with two protons removed. What is the evidence that phenolphthalein loses both protons in basic solution? And does the only once deprotonated form ever appear?

The Wikipedia article has a nice picture of the stages of deprotonation of phenolphthalein:

https://en.wikipedia.org/wiki/Phenolphthalein

However, there is a big jump here between H2In and In2-. In fact, smaller steps, showing just one deprotonation, are very reasonable. Whoever heard of two protons coming off at the very same time?

http://www.ch.ic.ac.uk/vchemlib/course/indi/indicator.html

Of course, a fuller explanation is needed.

Thymol Blue has a structure siilar to phenolphthalein in that it has two phenolic groups. They deprotonate at different pHs and thymol blue has two color changes.(https://en.wikipedia.org/wiki/Thymol_blue):

The uk reference also lists the pH for a color change of various indicators and also the range over which the color change occurs.

And litmus. Litmus is a water-soluble mixture of different dyes extracted from lichens. The litmus mixture has the CAS number 1393-92-6 and contains 10 to 15 different dyes. All of the chemical components of litmus are likely to be the same as those of the related mixture known as orcein, but in different proportions. In contrast with orcein, the principal constituent of litmus has an average molecular mass of 3300.
https://en.wikipedia.org/wiki/Litmus

In the list of indicators, all have a range of color change of 1.6 pH units, even thymol blue, which has two phenolic groups. Except phenolphthalein, which changes with a broader range of 1.8 pH units. Then there's litmus, with a color change range of 3.8!

The range for litmus can be ascribed to its being an extract of many similar dyes, a natural product, with slight differences between individual compounds. Thymol blue has a sulfonate moiety which will ionize more easily than the carboxyl in phenolphthalein. It seems that the colorless form is singly ionized, being a single quinone moiety and two aromatic rings, without extended conjugation.

In thymol blue, the second deprotonation gives a phenolate which more certainly conjugates with the quinone, establishing a larger conjugated system, thus highly colored. (Yeah, the unconjugated neutral compound is red, how do I explain that - I guess it's just the starting color of a pretty aromatic compound.)

Now in phenolphthalein, the first deprotonation yields an ion, shown in the uk reference. (Although they say it is pink, I am going to disregard that. In addition, I'm going to postulate that the mono-ion would be better described as an ester with a phenolate group, not as a quinone-carboxylate.) This first-deprotonated ion resembles the colorless form of thymol blue: a quinonoid structure with two phenyl rings attached to a central carbon. I might guess/conclude that this form could be colorless and that the second deprotonated form is the pink/fuchsia compound. The (slightly) larger range for the color transformation could well come from the lower tendency of the carboxylate (compared to the sulfonate in thymol blue)  to dissociate after the first proton is drawn off; as long as the central carbon is tetrahedral, the molecule has one phenol and one phenolate. The central carbon goes planar in phenolphthalein AFTER the second proton comes off, whereas the second proton comes off in thymol blue from an already planar structure.

To restate: in phenolphthalein, the central carbon is tetrahedral and the two phenolic groups have very similar pKa; after both phenolic protons are removed, the carboxyl detaches and the extended resonant system is formed (pink/fuchsia). In thymol blue, the first proton removal is easy because the sulfonate is a more stable ionic group than carboxyl, so this proton is more acidic than phenol, but the second proton is just part of a substituted phenol.

Just for reference, the pKa of phenol is 9.95 (https://en.wikipedia.org/wiki/Phenol).


Derivation of formula used in calculating molecular mass from a mixture In a solution to a problem, to calculate Molecular mass of unknown gas from a mixture containing $\ce{O2}$ in $80~\%$, the author used the formula :

  $\frac{1}{\sqrt{M_\text{mix}}}=\frac{X_{\ce{O2}}}{\sqrt{M_{\ce{O2}}}}+\frac{X_\text{gas}}{\sqrt{M_\text{gas}}}$.

The exact problem is:

  Pure $\ce{O_2}$ diffuses through an aperture in $224\mathrm{s}$, where as mixture of $\ce{O_2}$ and another gas containing $80~\%\ \ce{O_2}$ diffuses from the same in $234\mathrm{s}$. The molecular mass of gas will be?

Solution by author:

  $$\begin{align}\frac{t_\text{mix}}{t_{\ce{O2}}} &= \frac{r_{\ce{O2}}}{r_\text{mix}} = \sqrt{\frac{M_\text{mix}}{32}}\\[1em]
\frac{234}{224} &= \sqrt{\frac{M_\text{mix}}{32}}\\[0.5em]
M_\text{mix} &= 34.92\\[1em]
\Longrightarrow \qquad \frac{1}{\sqrt{M_\text{mix}}} &= \frac{X_\text{gas}}{\sqrt{M_\text{gas}}} + \frac{X_{\ce{O2}}}{\sqrt{M_{\ce{O2}}}}\\[1em]
\Longrightarrow \qquad \frac{1}{\sqrt{34.92}} &= \frac{0.2}{\sqrt{M_\text{gas}}} + \frac{0.8}{\sqrt{32}}\\[0.5em]
M_\text{gas} &= 51.5\end{align}$$

I wish to know why this works here?

Hint: $$v_{rms}=\sqrt{\frac{3RT}M}$$

And weighted average of speeds for mixture.

EDIT:

$$ v_{mix} ^{rms}= x_1 v_1^{rms}+ x_2 v_2^{rms}$$

This will complete the derivation of the formula you gave.


Tertiary carbanion in more alkylated alkene Saytzeff rule says that more alkylated alkene is more stable and the reason is hyper conjugation.
Going by this, 2,3 Dimethyl but-2-ene should be most stable alkene due to 12-alpha hydrogen.
But on drawing hyper conjugative structures of the molecule and a less alkylated alkene, say 2-Methyl propene; there was formation of a tertiary carbanion in 2,3 Dimethyl but-2-ene! and not in 2-Methyl propene. Instead in this molecule there is a primary carbanion which is much more stable than tertiary. I know that there will be 12 hyper conjugative structures in 2,3 Dimethyl but-2-ene rather than 6 structures but isn’t a tertiary carbanion poorly stable and make overall molecule less stable than 2-Methyl propene?

Can anyone please prove me wrong?

Thanks in advance

There are two types of description of hyperconjugation: (1) Valence bond approach & (2) Molecular orbital approach. Your question is pertaining to valence bond approach as you are interested in the stability of the canonical forms.

In Valence bond approach, hyperconjugation is said to happen due to conjugation of C—H bond electron pair with adjacent double bond. We call this phenomenon 'hyper'-conjugation because there is no real negative charge that is getting delocalized. Rather, the partial negative charge which is accumulated on the carbon due to electronegativity difference with hydrogen is getting delocalized.

In a real carbanion, the carbon has ~100% share of the negative charge. However, in this case, the carbon has only partial share of the negative charge. To be precise, about 7%, because 7% is the ionic character of a C—H bond (Calculated using Hannay-Smith's Relationship for Calculating Percentage Ionic Character [1]).

In case of conjugation of real carbanion, the ~100% negative charge is getting delocalized, and you can give importance to the stability of canonical forms to detrmine whether that conjugation is relevant. However, when ~7% negative charge is getting delocalized, it is more important to give importance to the number of such conjugations, than the relevance of each conjugation based on stability of canonical forms, because anyway, ~7% negative charge won't cause much destabilization even in a tertiary carbanion!

In Molecular orbital approach, hyperconjugation is said to happen due to overlap of bonding σ oribitals of C—H bond with adjacent antibonding π* orbital (in case of double bond). Owing to energy & orientation mismatch, the overlap is minimal and thus stabilization arising from hyperconjugation becomes significant only when the number of hyperconjugations is greater. So, this explains why 2,3-dimethyl-but-2-ene assumes much greater stabilization from hyperconjugation compared to 2-methyl-propene.


Experience with PTFE tubing & glass connectors? I'm wondering which PTFE tubing size to get to connect to normal 10mm OD glass connectors. For a silicon hose I would take an ID of 8-9mm as it is flexible but I think PTFE is quite rigid.
Anyone has some experience with PTFE tubing?

In short: Do not attempt a direct connection of the PFTE hose with the glass.
Instead (assuming the tubing shall transfer gases with little pressure only):

Because to the (compared to silicon) lesser flexibility of the tubings made from PFTE, you should consider to use glassware with threaded joints with detachable connectors. This a theme already seen with condensers, e.g.

(source)
but may be equally applied to impingers

(from here)
or overhead exhausts:

(from here)
These connectors may be replaced with ones made mostly out of PFTE, too, e.g.

(source)
Thus you may connect the tubing to the connector at the bench (gentle warming of the tubing with a heat gun may help), secure this end with the clip, and then mount the connected hose to the reactor.  It is faster and more secure (should you need a bit of force, damages are to the tubing and the connector, but not to the reactor) than the direct mount of the tubing to the glass.

The other side of the PFTE tubing will connect with the metal ferrules to your gas supply, e.g.

(screen photo at 1:04 min from this video)
Depending on the dimensions, the fittings by Swagelok are sold in sets of 10; a typical mechanical workshop of your university may have them already in the shelf. The less pricey Bowden connections used for 3D printers (e.g., here) are designed for compressed air.

As for any equipment working with gas: monitor the gas pressure in your reactor and use a safety release / pressure valve.


What is this SMILES notation: C[Si](C)C |^1:1| When converting InCHI to canonical SMILES using Indigo, I will occasionally get an output SMILES string like this:

C[Si](C)C |^1:1|

What does the "|^1:1|" part mean? I can't find this explained anywhere in either SMILES grammar documentation, nor Indigo documentation, although other chemical software seem to accept it as a valid SMILES string. Any ideas on the meaning of this annotation?

The SMILES syntax in question is neither mentioned in the Daylight Theory Manual, nor in the OpenSMILES specification.

It is not part of the canonical SMILES syntax, but an extension, developed  by the ChemAxon company.

According to their documentation, extensions are tagged by leading and trailing | (pipe) symbols. ^1:, followed by an atom index, indicates a

  monovalent radical center:


Why amine salts are soluble in water? Amine salts are soluble in water but insoluble in organic solvents like ether. Why is it so? I guess there is something related to polar nature of amine salts. But I am not sure. Can someone elaborate?

Amine salts are:

charged
polar
hydrogen-bonding (unless they're quaternary)

There some cross-over between those categories, but all will increase solubility in water, a polar solvent, and decrease solubility in a nonpolar solvent, like ether.

That said, if you put very long oily carbon chains on an amine salt, you will decrease the solubility in water, and increase the solubility in nonpolar organic solvents.


What is this glassware? I found this in a second hand store, and I think it has to do with freeze drying. Is there any validity to this assumption? What is this called? Also, the little tuberculina at the bottom are stuffed with fiberglass I believe.

It looks to me like tubes used to distribute gas flow.

If you work with substances that would react with the oxygen in air, you prevent that by working under an inert gas such as nitrogen, i.e. you fill all the flasks containing your substance with nitrogen gas, keeping up a steady flow of the nitrogen in order to prevent any from air leaking in.

The nitrogen is usually provided through a valve in the fume hood, and then distributed through the tubes you have there. The left end in your picture is connected to the valve, and on the ends on the right side you can attach tubes leading to your reaction flasks. The fibreglass is there to prevent air from getting into the nitrogen distributing system while there's nothing attached to the tube.

The ends of the glass tubes look weird to me though, so I might be entirely wrong.


Can I test for lead with household chemicals? I picked up a metal fence loop cap today and took it home just because I liked the shape.

(^ not mine)
I cleaned it, but some of the paint is chipping off. I'm only vaguely thinking about lead, and I'm not going to chew on it, so I don't want to buy a test kit. Is there a way I could test for the presence of lead ions with common/ideally household chemicals? I'd be willing to wait until I'm at my school lab again if I needed HCl or something.

General remark: get a bit of lead (metal or salt - whatever you look for) to have a positive control for your experiments.

More non-chemical tests for lead (metal):

lead is soft: you can cut/scrape it quite easily with a normal knife.
sheets of lead are easily bent, and they stay so.
For bigger pieces: Hammering on it would leave a rather big impression of the hammer. The sound would be dull.
it usually looks grayish, not silvery-metallic, but the cut will be silvery.

If it is about the paint:

what colour is it?
white lead: $\ce{2 PbCO3 \cdot Pb(OH)2}$ I'm more familiar with that in the context of paintings or painting of wooden things (window frames)
$\ce{PbSO4}$ also white. I think it has also been used for things like window frames.
side thought: lead paints for wood may have been popular due to the fact that the toxicity to microorganisms helps very much with the anti-fouling effect. Anti-corrosion paints for metal need different properties.
lead yellow: $\ce{PbCrO4}$ Wikipedia says that was the original pigment for school buses.
$\ce{2PbO \cdot PbO2}$(Mennige) red anti-corrosion pigment that was at least here in Germany the most important lead based priming paint. Only difficulty with recognizing by color: lots of non-toxic replacements are produced with similar colour to have the familiar look...
Very sensitive but but not very distinctive test$^*$ for $\ce{Pb^2+}$: the sulfide is black and very insoluble. Kitchen chemistry: put some egg on and even the surface of the metal should turn black. Possibly some heat is needed (I didn't test, though). Otherwise, use acidic $\ce{Na2S}$ solution from the school lab. If the $\ce{Na2S}$ solution doesn't give a black precipitate, there could in theory still be lead in the paint which is not accessible by the sulfide due to the binding material of the paint. In that case, however, the toxicity will be less of a problem, too (i.e. unless you dissolve the paint somehow, burn it, ...)
$^*$ OTOH it is highly improbable that there are substantial amounts of silver in the paint...
reductive borax bead should be gray, oxidative colourless.
Isolating Pb is quite easy, maybe you can do that in your school lab: precipitate $\ce{PbCl2}$ with cold HCl, filter. The dissolve the chloride in hot water. You can then precipitate PbS (black) or $\ce{PbCrO4}$ (yellow).  Looking at the crystals under the microscope can help further. Qualitative inorganic textbooks often have microscope photos of different precipitates.


How do I calculate weight/number average relative molecular mass for polymers? I have to estimate the weight/number average relative molecular mass of following diagram:

I know the equations:

I just can't interpret the diagram. Do I use the chain length as mass of the individual polymer?

I would then get $\pu{M_w = 52.305}$
and  $\pu{M_n = 68.227}$

Is this alright? I really appreciate any help.

They perhaps want you to make the calculations relatively, which means that even if you assume by looking at the first graph that the mass of the molecular chain length of $1$ (taken relatively and not as per scale) is $x$ units and thus, the mass of $10$ (again, taken relatively and not as per scale) such chains add up to $10x$ units and finally calculate $M_n$ to be $\approx 68.2273x$ it isn't helping us to get a meaningful answer.

A thing that should be noted is that even if you calculate it by following my example, when you'll finally calculate $\displaystyle \frac{M_w}{M_n}$ which (perhaps) the questioner wants you to calculate it won't make difference because this result is independent of $x$.

It appears that the questioner hasn't explicitly stated the mass of a polymer chain or that of the monomer units. They even haven't stated that all the polymers are considered identical with respect; it is highly probable that they want you to assume these things so as not to make things overly complicated.