Why does the reduction of a Weinreb amide give an aldehyde instead of an amine? The reduction of a regular amide with $\ce{LiAlH4}$ yields an amine. However, with a Weinreb amide, the product is an aldehyde. How can this be justified? I can't find the mechanism for the Weinreb amide reduction - is it known?

Reaction of acyl carbonyl compounds with strong nucleophiles (hydrides, organometallics) is a very common and useful reaction, although it has some limitations. One of the simplest cases is the reaction of an ester with a strong nucleophile. As one would expect, the strong nucleophile attacks the electrophilic carbonyl carbon, giving a tetrahedral intermediate. In most cases, the tetrahedral intermediate breaks down readily to give a new carbonyl compound. In this case, that carbonyl intermediate will be a ketone (from a Grignard, e.g.) or an aldehyde (from LiAlH4). Both of those functional groups are more reactive than the ester starting material, so the next equivalent of nucleophile will react faster with the aldehyde/ketone, eventually giving a primary (from LiAlH4) or tertiary (from a Grignard) alcohol product.

The situation at the tetrahedral intermediate changes slightly when an amide is the electrophile. As cfcief points out, the less electronegative nitrogen will kick out the more electronegative oxygen (stabilized by aluminum) to give an iminium intermediate. Again, this intermediate is more reactive than the starting material, so a second nucleophilc attack takes place.

The key feature of the Weinreb amide is that after the nucleophilic attack at the carbonyl, the tetrahedral intermediate is stabilized as a five-membered cyclic chelate. Without decomposition of the tetrahedral intermediate, the only active electrophile is more of the Weinreb amide starting material. On workup, the strong nucleophile is quenched, the metal washed out, and the tetrahedral intermediate is allowed to decompose to the aldehyde/ketone product.

Although the N-methoxy-N-methylamide was specifically designed by Weinreb to behave this way (ref), it wasn't until more than 20 years later that mechanistic evidence was observed (ref).


Is it possible for halogens to show oxidation states less than -1? Halogens like $\ce{Cl,Br,I}$ generally show oxidation states between $-1$ and $+7$. However, is it possible that they can show an oxidation state beyond $-1$ (e.g. $-3$)?

I was thinking of some compound in which chlorine is bonded with a less electronegative element and forms more than one bond with such element(s).

Is it theoretically possible?

Yes, they can show oxidation states lower than -1. For example in hyperlithiated compounds, like $\ce{Li3Cl}$, the chlorine would formally have an oxidation state of -3.

However, a word of caution, oxidation states are a tricky thing. They are determined by adopting the hypothetical view that the compound you are looking at consists only of single-atomic ions, e.g. in the case of $\ce{Li3Cl}$ of 3 $\ce{Li+}$ and $\ce{Cl^{3-}}$. The hypothetical charges of these "ions" are then your oxidation states. Clearly, this view has nothing to do with the real bonding situation in the compound and oxidation numbers correlate very poorly with the real electron distribution. It is simply a useful little device for keeping track of redox reactions. So, the -3 oxidation state of chlorine in $\ce{Li3Cl}$ does not mean that chlorine has a 10-electron valence shell. The real bonding situation will be somewhat more complicated.


Balancing Redox Equations - Half Reactions How do I balance this using the half-reaction method: $$\ce{Zn(s) + HCl(aq) -> Zn^{2+} (aq) + H_2 (g)}$$

I would first split it up in its ionic components:
$\ce{Zn(s) + H^+ + Cl^- -> Zn^{2+} (aq) + H_2 (g)}$.

Then we have first the oxidation-reaction: $\ce{Zn(s) -> Zn^{2+} (aq)}$. Balancing this with electrons gives $\ce{Zn(s) -> Zn^{2+} + 2e^-}$.

The reduction-reaction is $\ce{H^+ + Cl^- -> H_2 (g)}$. I need to balance this with $\ce{Cl}$? So that would get me $\ce{H^+ + Cl^- -> H_2 (g) + Cl}$. Then adding hydrogen to the left side with $\ce{H^+}$ gives $\ce{2H^+ + Cl^- -> H_2 (g) + Cl}$.

However, I feel like I made a mistake somewhere. Any help please?

You are right for the half reaction for oxidation is:
$$\ce{Zn(s) \rightarrow Zn^{2+} + 2e^-}$$
Now, you have to write properly the  half reaction for reduction:
 $$\ce{2H^+(\mathrm{aq}) + 2e^- \rightarrow H_2(g)}$$

This means, that the electrons given by zinc are taken by ion hydrogen.
So, if we add the two half reactions:
$$\ce{Zn(s) + 2H^+(\mathrm{aq}) \rightarrow Zn^{2+}(\mathrm{aq}) +  H_2 (g)}$$
Notice, that ion chloride doesn't change its oxidation state. Its role is to keep the medium neutral electrically. So, we'll add two ions chloride to the two members of the equation:
$$\ce{Zn(s) + 2(H^+_{\mathrm{aq}} + Cl^{-}_{\mathrm{aq}}) \rightarrow (Zn^{2+}_{\mathrm{aq}} +2 Cl^{-}_{\mathrm{aq}}) + H_2(g)}$$
Remember, redox reaction are reactions of electrons exchange between an oxidant and an reducer.


Internal Energy Changes in Adiabatic Reactor If we consider adiabatic reactor (one which doesn't exchange energy and matter with surroundings) with some reaction in liquid phase, energy balance for such reactor is written as heat of reaction being equal to time derivative of enthalpy at any point in time. Since reaction is in liquid phase during reaction there are negligible changes in volume and pressure of reactor so that change in enthalpy is approximately equal to change in internal energy. Problem here is that how can internal energy (or enthalpy in this case) change in adiabatic reactor since it doesn't exchange energy with surroundings? Internal energy is a sum of all microscopic kinetic energy and potential energy of interaction of all bonds in the system. During chemical reaction interatomic potential energy is converted to microscopic kinetic energy (or vice versa) because of which temperature changes, but change in each type of energy must be equal due to energy conservation. Since this is the case internal energy doesn't change after reaction in adiabatic reactor. If so, why does equation of energy balance say it does change as temperature of system is changed during reaction?

You omitted the energenics associated with breaking chemical bonds and making new chemical bonds.  If this contributes a negative change to the internal energy (exothermic reaction), then the temperature would have to rise to offset this effect so that the overall change in internal energy would be zero.  If it contributes a positive change to the internal energy (endothermic reaction), then the temperature would have to drop to offset this effect so that the overall change in internal energy would be zero.  Saying this in a different way, when you have a chemical reaction occurring, the internal energy can change even if the temperature is held constant.


Orbital angular momentum For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?

Also, in the orbital approximation (ie the solutions obtained after ignoring electronic repulsions in the Hamiltonian) of the many electron atom( say helium) how does one go about finding total orbital angular momentum?

For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?

Well, you can't. The different components of the angular momentum operator $\hat{\mathbf{L}}$, i.e. $\hat{L}_{x}$, $\hat{L}_{y}$, and $\hat{L}_{z}$, do not commute, so you cannot simultaneously determine all of them. But the squared angular momentum operator, which gives you the squared length of the angular momentum vector commutes with the components of $\hat{\mathbf{L}}$, so you can simultaneously determine the length of the angular momentum and one of its components. Which component you choose is immaterial, usually $\hat{L}_{z}$ is chosen. What you get out of those two operators can be pictured like this: you know the length of the angular momentum vector and its z-component; the other two components are undetermined but because you know the length they can only take on values that lie on a circle centered around the ${L}_{z}$ axis, whose offset is determined by the value of ${L}_{z}$ and whose radius is determined by $| \sqrt{\mathbf{L}^{2}}|$. For more details have a look at this Wikipedia article or any general physical chemistry or quantum mechanics textbook.

  Also, in the orbital approximation (ie the solutions obtained after
  ignoring electronic repulsions in the Hamiltonian) of the many
  electron atom( say helium) how does one go about finding total orbital
  angular momentum?

Ignoring relativistic effects the Hamilton operator of a system consisting of $M$ nuclei and $N$ electrons, which are considered to be point charges takes on the form

\begin{align}
  \hat{H} &= \underbrace{ \sum_{j=1}^{M} - \frac{ \hbar^{2} }{ 2 M_{j} } \mathbf{\nabla}_{ \mathbf{R}_{j} }^{2} }_{ = \, \hat{T}_{\mathrm{K}} } + \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{M} \frac{ e^{2} Z_{i} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{R}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{K--K}} }  \\
  & \quad \underbrace{- \sum_{i=1}^{N} \sum_{j=1}^{M} \frac{ e^{2} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{e--K}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{N} \frac{ e^{2} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{r}_{j} | } }_{ = \, \hat{V}_{\text{e--e}} }
\end{align}

where $ \mathbf{R}_{j} $, $Z_{j}$, and $M_{j}$ are the coordinate, the charge number, and the mass of the $j^{\mathrm{th}}$ nucleus, respectively and $ \mathbf{r}_{i} $ is the coordinate of the $i^{\mathrm{th}}$ electron and $m_{\mathrm{e}}$ is the electron mass.
Furthermore $e$ is the elementary charge und $\epsilon_{0}$ is the vacuum permittivity.
The different terms of this Hamilton operator can be explained as follows: $\hat{T}_{\mathrm{e}}$ and $\hat{T}_{\mathrm{K}}$ are the terms representing the kinetic energy of the electrons and the nuclei, respectively, $\hat{V}_{\text{K--K}}$ describes the coulomb interaction between the nuclei, $\hat{V}_{\text{e--e}}$ is the interelectronic repulsion, and $\hat{V}_{\text{e--K}}$ represents the coulomb interaction between the electrons and the nuclei.
Using the Born-Oppenheimer approximation you can seperate the electronic degrees of freedom from the ones of the nuclei and treat the electronic problem independently and for the many-electron atom you also only have one nucleus to consider so the terms $\hat{T}_{\mathrm{K}}$ and $\hat{V}_{\text{K--K}}$ drop out of the Hamiltonian. If you now also ignore the electron-electron interaction you are basically left with a Hamilton operator describing two independent electrons moving in the electrostatic field of the nucleus:

\begin{align}
  \hat{H} &= \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \sum_{i=1}^{N}  \frac{- e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} | } }_{ = \, \hat{V}_{\text{e--K}} }
\end{align}

For this you can easily seperate the $N$-electron problem into $N$ one-electron problems

\begin{align}
  \hat{H} &= \sum_{i=1}^{N} \hat{h}_{i} \qquad \Rightarrow \qquad \Psi(\mathbf{r}_{1}, \ldots, \mathbf{r}_{N}) = \psi(\mathbf{r}_{1}) \cdots \psi(\mathbf{r}_{N})
\end{align}

where each one-electron Hamiltonian $\hat{h}_{i} = - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} - \frac{e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} |}$ and one-electron eigenfunction $\psi_{n}(\mathbf{r}_{i})$ basically satisfy a Schroedinger equation for the hydrogen-like atom

\begin{align}
  \hat{h}_{i} \psi_{n}(\mathbf{r}_{i}) = \varepsilon_{n} \psi_{n}(\mathbf{r}_{i}) \qquad \text{with} \qquad \varepsilon_{n} = -\frac{m_{\mathrm{e}} e^{4} Z^{2} }{ 32 \pi^{2} \epsilon_{0}^{2} \hbar^{2}} \frac{1}{n^{2}} \qquad n = 1, 2, ...
\end{align}

Using spherical coordinates, $x, y, z \to r, \theta, \phi$, you can rewrite the kinetic energy operator $- \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2}$ in terms of $\hat{\mathbf{L}}^{2}$ so that the Hamilton operator becomes

\begin{align}
  \hat{h}_{i} = \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} r^{2}} \frac{ \partial }{ \partial r } \biggl( r^{2} \frac{ \partial }{ \partial r } \biggr) + \frac{ 1 }{ 2 m_{\mathrm{e}} r^{2} } \hat{\mathbf{L}}^{2} - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r }
\end{align}

where $r$ is the distance between nucleus and electron. The operator $\hat{\mathbf{L}}^{2}$ commutes with the Hamilton operator $\hat{h}_{i}$ because $\hat{\mathbf{L}}^{2}$ commutes with itself and does not involve the variable
$r$. Likewise, the operator $\hat{L}_{z}$ commutes with $\hat{h}_{i}$ because it
commutes with $\hat{\mathbf{L}}^{2}$ and also does not involve the variable $r$.
Thus the operators $\hat{\mathbf{L}}^{2}$, $\hat{L}_{z}$, and $\hat{h}_{i}$ have simultaneous eigenfunctions:

\begin{align}
  \hat{h}_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \varepsilon_{n} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \\
  \hat{\mathbf{L}}^{2} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \hbar^{2} \ell_{i} (\ell_{i} + 1) \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad \ell = 0, 1, 2,..., n-1 \\
  \hat{L}_{z} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \hbar m_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad m_{i} = -\ell_{i}, -\ell_{i} + 1, \ldots , \ell_{i} - 1, \ell_{i}
\end{align}

Thus, you have a well defined angular momentum for each electron in the system.


Can sulfuric acid itself be electrolysed? We all could know by now that water can be electrolysed by adding some sulfuric acid.:)

But is it possible that the sulfuric acid itself becomes electrolysed when the concentration of the acid in the water gets higher ?

Would it give sulfur dioxide or sulfur trioxide ?

After some searching, I found this 1956 Thesis(Reference), which describes the production of peroxydisulfuric by electrolyzing concentrated sulfuric acid. The yield varied according to the concentration of the acid, temperature, current, and the nature of the electrode. In the Thesis, the chemist noted that discharged $\ce{HSO4-}$ united together to form peroxydisulfuric acid ($\ce{H2S2O8}$). It further stated that  oxidation of sulfate ions by anodic oxygen was responsible for the peroxydisulfuric acid formation. Also, hydroxyl ion at anode played an important role in the acid formation.

The reaction mechanism was confirmed by electrolyzing sulfuric acid and ammonium sulfate at various concentration at $\pu{16\!- \! 20 ^\circ C}$ using platinum electrode and the reaction occurred in accordance to the theory. It was also confirmed that amount of peroxidic oxygen increased with increase in ammonium sulfate. The specific gravity of sulfuric acid and the current density was varied and accordingly the highest yield of 78.4% was obtained. It was also noted that increasing the current density increases the yield up to a certain point as  peroxydisulfuric acid was found to convert into peroxymonosulfuric acid. At that moment, a small trace of hydrogen peroxide was noted.

So, the chemical species formed during electrolysis are $\ce{HSO4-}$, $\ce{SO4^2-}$, peroxide ion, protons, trace of hydrogen peroxide, and other side products.

Reference:

Sin-Chou Fan, “The electrolytic production of peroxydisulfuric acid using periodically reversed direct current and alternating current superimposed on direct current,” MS Thesis, Virginia Polytechnic Institute, 1956.


How many total absorptions would appear on proton -NMR spectrum for this molecule? I count 7 equivalent absorptions total, but answer says 8. I don't see where this eighth absorption comes from?

Heres how I'm counting starting on right side of molecule:

1 is Carbox acid H

2 is the 1 H just left of the double bonded Oxygen

3 is the 3 H's of the methyl group next to the lone H

4 and 5 are the 2 sets of equivalent H's in the ring structure

6 is is the 2 H's to left of ring

7 is the 2 methlys on the isoproplyl (these are equivalent so same signal)

How do they get 8?

To answer this question properly one must understand that the field of the instrument is important. At 400 MHz, ibuprofen displays 7 distinct carbon-bound proton signals. Below is the spectrum from the SDBS website. Although the two methyls of the isobutyl chain are recorded [G] as a "singlet" at δ 0.889, they are diastereotopic and inherently different no matter how far they are removed from the asymmetry on the other side of the ring. If this spectrum were recorded on a 1,020 MHz 1H NMR, perhaps separation of the two methyl signals would occur. That the two methyls are different, imagine that one of them is a perdeutero methyl group. This compound is a diastereomer of the compound having the other methyl group as the perdeutero group. So there are actually 9 total distinct types of hydrogens (including the carboxyl hydrogen) that, given the right spectrometer, could appear in a spectrum of ibuprofen. It is a matter of resolution. [As a case in point, see the 1H NMR of cholesterol at the SDBS website and note that the two methyl groups at the end of the side chain have different chemical shifts even though the are separated from the nearest center of asymmetry by three methylene groups.]
Acknowledgment: http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi


Can equilibrium be reached in a non-reversible chemical reaction? Can chemical equilibrium only be achieved in a reversible process?

I have only seen equilibrium in processes that are reversible so I was wondering if it were possible in reactions that are not reversible.

In any chemical reaction there is a generally a potential energy barrier between reactants and products. How easily this is crossed from reactants to products or vice versa depends primarily on the temperature. This is because to react, molecules have to gain energy that is more that the average to be able to cross the barrier. (This is called the activation energy). The energy is accumulated randomly from solvent in solution and by collisions in gas phase, which is why some chemical reactions can be slow. Barriers are often high compared to thermal energy so it can take ages to get enough energy by chance to cross a barrier. The chance of having a certain energy E is determined by the Boltzmann distribution $exp(-E/k_bT)$ where $k_B$ is the Boltzmann constant, T temperature. As Boltzmann distribution also depends on temperature reactions are generally observed to be faster at higher temperatures.
If a reaction is called 'reversible' then one expects to see reactants and products in the mixture at the same time at amounts depending on the equilibrium constant. The product is reverting to reactant and reactant to product, assuming that all product is not continuously being removed, say by precipitation or otherwise.  In an 'irreversible' reaction the barrier from products to reactants is so high that in normal experience enough of the product never returns to reactant to be measurable. The effect can be very dramatic (orders of magnitude) with a relatively small change in activation energy because of the exponential dependance of the Boltzmann distribution.  All this means that there is no fundamental difference in 'reversible' and 'irreversible' reactions, just a matter of time-scale. However, it is obviously practically useful to distinguish these cases and so we can look at equilibrium constants to guide us. Note (a) That thermodynamics does not inform us about time-scales but only the relative free energy of reactants and products. (b) The words ' reversible' when used as 'reversible process' has a special meaning in thermodynamics and needs to be used with care. (c) I have excluded the special case of 'barrierless' reactions.


What is a "ynoate" in the context of alkynes? In the literature, I have read the term, "ynoate" refer to a sort of alkyne. Searching through the IUPAC gold book and Google has failed to return an explicit definition.

ynoate refers to a derivative of an alkyne acid. e.g. ethyl-3-butynoate  structure here

The simplest of these is propiolic acid  https://en.wikipedia.org/wiki/Propiolic_acid


Solubility in different pressure conditions I am working on my scholarship exam practice. I believe this exam assumes high school + first year university knowledge. And I'm not quite sure what I did wrong. Could you please have a look?

  The solubility of oxygen in 1.0 L water is 28 mL at 25 °C and 1.0 atm.
  How much oxygen can be dissolved in 1.0 L of water at 25 °C and 4.0 atm?

The answer remains 28 mL but IMO I thought when pressure goes up 4 times while other factors are constant, the solubility would do that as well. Please help on where I missed and what topics I should focus on specifically.

As noted by Wikipedia,

  "Henry's Law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid."

So you are absolutely right. 4 times the pressure of oxygen should result in 4 times as much oxygen being dissolved.

The answer key for the problem must be wrong.

  User Delta_G made an interesting observation/correction. The problem is asking about the volume of gas that dissolves, not the concentration of oxygen in the water. So 28 ml at 4 atm of oxygen gas contains four times as much oxygen as 28 ml at 1 atm of the gas. So in either case 28 ml of the gas phase dissolves into the water.


Ideal gas equation derivation proportionality While deriving the Ideal gas Equation using Boyle's, Charles', and Avogadro Laws,
how are each of the laws, which have different proportionality constants and different quantities as constant combined?
(eg: Boyle's  law which assumes the temperature and moles of gas to be constant, but in Charles' law temperature changes. How is it possible?)

I understand your concern because this is the way it is taught in South Asian schools and colleges and even general chemistry texts taught elsewhere are silent about it. They state all the individual gas laws using a proportionality sign "$\propto$" and magically all gas laws are combined and that becomes $R$- the universal gas constant. This is done in a single paragraph.
This is not the proper way how the ideal gas law was derived. Out of curiosity, I checked the original paper " Mémoire on the Motive Power of Heat" where the ideal gas law appeared in 1834!!. It is in French, but one can see the equations. Not once he used any proportionality sign but there is a lot of calculus there.
https://gallica.bnf.fr/ark:/12148/bpt6k3414331n/f34.image.r=M%C3%A9moire%20sur%20la%20puissance%20motrice%20de%20la%20chaleur
You should scroll the original paper and feel the length of it to appreciate that the ideal gas law was not derived as shown in your school / college textbooks.
You would see a formal derivation of the ideal gas law in higher classes using the kinetic theory of gases.


The shapes of molecules According to this question/answer, isolated atoms are spherical – and don't have, for example, p-orbital lobes sticking out. The reason is that each complete set of p-, d- and f-orbitals taken together is spherically symmetrical – and the overall wavefunction will be a superposition of all possible states. That superposition will be spherically symmetric, so the atom will be, too. (Sorry for paraphrasing, simplifying etc!).

So my question is: apart from the orbitals involved in making e.g. sigma and pi bonds, is 'the rest' of the atom spherically symmetrical? I have done this rough picture.

You will notice that down the left, the p-orbitals are shown (even though in an isolated atom they would not exist like that – I've kept them so they can show potential 'bonding sites'). But on the right, in the 'finished' molecules, all but the orbitals involved in bonding disappear, into a spherically symmetric shape as before.

I realise there are several to many simplifications here, but this is for a general audience, and I just want to give a sense of molecular orbitals without getting anything horribly wrong. (In every artwork I find online, people only seem to focus on the orbitals involved, as if they are separate from the rest of their respective atoms.)

I'd really appreciate any help and thoughts. Thank you.

Given an external electromagnetic field, the orbitals do split and even an otherwise isolated atom is no longer spherically symmetrical. For a mostly magnetic field, this is called the Zeeman effect, and for an essentially electrostatic field it is the Stark effect.

Of course, this begs the definition of an "isolated" atom. As @Karl mentions, how do you measure the shape without interacting with the atom? Any incident photon, for example, would be "letting the cat out of the box."


Why is Diiron nonacarbonyl so exceptional? My textbook(NCERT) says:

  With exception of $\ce{Fe2(CO)9}$, all other metal carbonyls are soluble in hydrocarbon solvents.

Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed. states:

  The most striking exception among the common metal carbonyls is nonacarbonyl diiron(0), which has a very low vapour pressure and is insoluble in solvents with which it does not react.

Other metal carbonyls are well soluble in hydrocarbon solvents but not Diiron nonacarbonyl. Why? What makes Diiron nonacarbonyl so exceptional? Is it because it has low vapor pressure?

There are no explanations regarding the insolubility of diiron nonacarbonyl. Is there any experiment conducted to determine the reason for its insolubility or is it just mere observation?

Update

@Orthocresol has told me to draw comparison of diiron nonacarbonyl with $\ce{[Mn2(CO)10]}$ and $\ce{[Co2(CO)8]}$ because of their similar molecular weight. Based on this fact, I went on to research further on this topic and from various handbooks and research notes I delved upon, I have drawn the following solubility comparison table. [Note: Only binuclear metal carbonyls were considered of form $\ce{[M2(CO)_x]}$ because they have same structure and thus should have same physical properties].

\begin{array}{c|c}
\mathbf{Metal~carbonyl} & \mathbf{Solubility} \\\hline
\ce{Mn2(CO)10} & \mathrm{ether, other~organic~solvents}\\
\ce{Tc2(CO)10} & \mathrm{ether,acetone} \\
\ce{Fe2(CO)9} & \mathrm{Insoluble~in~benzene,ether,petrol.Only~soluble~in~THF}\\
\ce{Rh2(CO)8} & \mathrm{organic~solvents}\\
\ce{Ir2(CO)8} & \mathrm{Ether,CCl_4}\\
\ce{Co2(CO)8} & \mathrm{petrol,benzene,alcohol}\end{array}

Also, a table of comparison for the other two iron carbonyls for a reference.

\begin{array}{c|c}
\mathbf{Iron~carbonyl} & \mathbf{Solubility} \\\hline
\ce{Fe(CO)5} & \mathrm{soluble~in~all~organic~solvents~like~ether,petroleum.Insoluble~in~water}\\
\ce{Fe3(CO)12} & \mathrm{Insoluble~in~water. Soluble~in~non~polar~organic~solvents.
} \end{array}

We can observe that there is a drastic anomaly in case of solubility of diiron nonacarbonyl in both the tables. Any explanation to this?

Wikipedia (primary reference) suggests a possible reason for $\ce{Fe2(CO)9}$ dissolving preferentially in THF versus nonpolar solvents:  it reacts according to the scheme

$\ce{Fe2(CO)9 + THF <=> Fe(CO)5 + Fe(CO)4 \cdot THF}$

Such a reaction is invoked to account for the dinuclear complex giving mononuclear products with various ligands in THF.  A similar reaction with the corresponding cobalt and manganese species would have to produce radical or ionic products, due to the odd atomic number of the metal, and so would be less favored.

This would not, of course, explain the lack of solubility in nonpolar solvents or nonvolatility of $\ce{Fe2(CO)9}$.  It is possible, however, that the same $\ce{Fe2(CO)9}$ molecules that bind with the oxygen in THF could also bind intermolecularly through the oxygen atoms, in effect delocalizing the covalent bonding and creating more intermolecular cohesion.

We can represent this type of interaction using the SMILES structural notation.  Fr a single $\ce{Fe2(CO)9}$ molecule it would look like this, taken from Wikipedia:

$\ce{O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O}$

For two molecules we would have

$\ce{[O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O][O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O]}$

Then we have a contributing structure where an oxygen atom from the left molecule could combine with the right one, displacing an iron pentacarbonyl molecule:

$\ce{[O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=\color{blue}{O[Fe]}(=C=O)(=C=O)(=C=O)C=O][[Fe](=C=O)(=C=O)(=C=O)(=C=O)C=O]}$

where the blue atoms form an intermolecular bond in this contributing structure.  The displaced iron pentacarbonyl molecule is in the last set of brackets.

The Case of Hydrogen Bonding

The delocalization of covalent bonds, hypothesized above, may be seen in a more familiar context:  hydrogen bonding.  Usually thus is rendered as an electrostatic attraction between oppositely charged atoms.  But it may also be considered as a molecular-orbital interaction:  a nonbonding electron pair overlaps an adjacent molecule's antibonding orbital in exchange for creating an "intermolecular" covalent bond.  With water it would look like this

$\ce{[HOH][HOH]}$

with the hydrogen-bonded, "intermolecular bonded" contribution

$\ce{[H\color{blue}{O}(H)\color{blue}{H}^+][OH^-]}$

where the intermolecular bond is between the blue atoms.  This is also the autoionized structure for water.


How can you tell when a molecule forms a lattice I came across the molecule SiO2 and didn't realize that in fact, it forms a lattice. How could I have found out that, in fact, it forms a lattice?

There is no easy way to guess or predict the answer in general, but experience will give you some clues

The way to tell whether something is a molecule or a covalent solid (Made from a lattice) is to look up the structure. When you first learn chemistry, some things have just have to be learned. But when you learn a few things, some patterns will emerge and you will get better at predicting the properties of other compounds.

The structure will usually tell you the answer. Other physical properties of the compound can hint at the structure. Carbon dioxide is a gas so you should be able to predict it is a small molecule; silicon dioxide is a refractory solid: this is a big clue that it is not a small molecule but is, in fact, a network solid mad from a lattice of Si-O bonds. Nitrogen and Oxygen are gases, so most likely small molecules; graphite and diamond (both forms of pure carbon) are solids so  out likely networks made from large lattices.

There isn't really a simple way to predict this, but once you have learned a few examples, patterns will emerge.


Can you artificially create molecules with a machine? Is it possible to create molecules with machines? For example: you shoot one oxygen-atom $(\ce O)$ on one diatomic hydrogen $(\ce{H2})$ to create one water molecule $(\ce{H2O})$.

This sounds to me like molecular beam chemistry. Within a vacuum chamber, you can create and collimate a focused "beam" of moving atoms or molecules and direct it to hit a stationary target or another beam. You can use spectroscopy to study the species that are formed. The idea is to look for and characterize reactive intermediates or transition states between starting materials and products. But if you were to create a familiar species such as water in this manner, it is not "artificial" water, but the real thing.

Maybe there is a physical chemist out there that can elaborate further on molecular beams.


Do the spectra of compound materials combine linearly with respect to the ratios of the constituent compound spectra? My question stems from reading this article on the FTIR spectrum of oil that has been mixed with various contaminants.

If one has the spectrum for the original uncontaminated oil and the spectrum for water, will the spectrum of a sample that has been contaminated with 10% water (by weight) be equivalent to the addition of original oil sample spectrum and 0.1 of the water spectrum intensities?

(Also, should the unit of contamination be the weight or the volume?)

Yes, you are right. If you have $n$ components in a mixture each with a spectrum $S_1$, $S_2$,...,$S_n$, the observed spectrum $S_{obsvd}$ would be the sum of $S_1$+ $S_2$+...+$S_n$.

In FTIR you either measure the transmittance or the absorbance. The term intensity is more or less used when you deal with molecular or atomic emission of light (fluorescence etc.) As long as there is no chemical reaction among the constituents, the absorbance should be linear with the concentration.


Do companies recover carbon dioxide and water from the purification step of cryogenic air separation, and how? In the 3rd step of cryogenic air separation, filtered air is purified to remove water, $\ce{CO2},$ and hydrocarbons from the $\ce{N2}$-$\ce{O2}$-$\ce{Ar}$ stream. Im interested in knowing what happens with that $\ce{CO2}$ and water. Can both resources be recovered in a useful form? Can the $\ce{CO2}$ be purified? Are Praxair or others currently doing this?

Context: Im doing research on a theoretical industrial process which uses pure $\ce{CO2},$ water, and $\ce{N2}.$ I want to know if all these inputs can be obtained from a standard air separation facility (of course $\ce{N2}$ can). Im particularly interested in $\ce{CO2}$ recovery.

Please note I'm not interested in buying gases. I'm working on a publication about a theoretical system to produce protein from air and solar power. In your answer it would be helpful to provide links to papers or websites which give descriptions of the recovery process (energy costs etc).

It is unlikely that this is an economically feasible $\ce{CO2}$ recovery method, which means commercial gas producers probably don't bother to isolate $\ce{CO2}$ this way (surely not water). The removal of $\ce{CO2}$ and water is performed with a pre-purification unit (PPU) that contains an adsorbing material such as alumina or molecular sieve. When the material nears saturation with impurities, the PPU needs to be taken off-line for regeneration. This is done by flowing warm $\ce{N2}$ over the adsorbent to heat it. It certainly seems possible to attempt capture of the $\ce{CO2}$ at this stage. Water is likely to be released from the adsorbent after the $\ce{CO2}$, allowing in principle a stream containing CO2 and N2 to be isolated. However isolation of $\ce{CO2}$ from this stream is expensive. As explained in a brochure from the CO2 Capture Project:

$\ce{CO2}$ can be separated from other gases by cooling and condensation. Cryogenic separation is widely used commercially for streams that already have high  $\ce{CO2}$ concentrations (typically >90%) but it is not used for more dilute  $\ce{CO2}$ streams.
A major disadvantage of cryogenic separation of  $\ce{CO2}$ is the amount of energy required to provide the refrigeration  necessary  for  the  process,  particularly  for  dilute  gas  streams. Another  disadvantage  is that some components, such as water, have to be removed before the gas stream is cooled, to avoid blockages.
Cryogenic separation has the advantage that it enables direct production of liquid  $\ce{CO2}$, which is needed for certain transport options, such as transport by ship. Cryogenics would normally only be applied to high concentration, high pressure gases, such as in pre-combustion capture processes or oxygen fired combustion.

This is complemented by information provided by Universal Industrial Gases Inc.:

Producing carbon dioxide as a commercial product requires that it be recovered and purified from a relatively high-volume, CO2-rich gas stream, generally a stream which is created as an unavoidable byproduct of a large-scale chemical production process or some form of biological process.
In almost all cases, carbon dioxide which is captured and purified for commercial applications would be vented to the atmosphere at the production point if it was not recovered for transport and beneficial use at other locations.
The most common operations from which commercially-produced carbon dioxide is recovered are industrial plants which produce hydrogen or ammonia from natural gas, coal, or other hydrocarbon feedstock, and large-volume fermentation operations in which plant products are made into ethanol for human consumption, automotive fuel, or industrial use.  Breweries producing beer from various grain products are a traditional source.  Corn-to-ethanol plants have been the most rapidly growing source of feed gas for CO2 recovery. CO2-rich natural gas reservoirs found in underground formations found primarily in the western United States and in Canada are another source of recoverable carbon dioxide.


Which out of the two has greater bond angle? Water or diethylether.Why? My theory:

The ethyl groups in diethylether being larger in size when compared to hydrogens in water, should be repelled more by the lone pairs on the central atom oxygen. Hence, the ethyl groups should come closer, thereby decreasing the bond angle.

Moreover, the electronegativity difference between the central atom and the peripheral atoms is more in case of water i.e., $\ce{O-H}$ electronegativity difference is more than that in $\ce{C-O}$ as per Pauling’s scale. Hence, the oxygen atom should pull the bond pair closer, thereby allowing to open up the $\ce{H-O-H}$ bond increasing the bond angle.

Hence bond angle of water should be greater than that of diethylether. Am I right?

PS: I know that in reality dimethylether has greater bond angle than water.

You are mostly correct with your reasoning , and the $\ce{C-O-C}$ bond angle was indeed expected to be lesser than that of $\ce{H-O-H}$ bond angle of water . But it's all theoretical. Experimentally the opposite was observed and was explained on the basis of steric hindrance or steric repulsion of the two bulky $\ce{C2H5}$ groups.

In case of $\ce{(C2H5)2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two ethyl groups. They repel each other and as a result , they do not let the lone pairs of Oxygen compress the bond angle more .

$\ce{H-O-H}$ bond angle is found to be $104.5^\circ$ whereas the $\ce{C-O-C}$ bond angle in diethyl ether is found to be roughly close to $110.5^\circ$. It can be attributed to the steric repulsion that trumps the bond pair-bond pair and lone pair-lone pair repulsion in this case.


Why is the deposition of silver more uniform from [Ag(Cl)₂]⁻ (aq) than from Ag⁺? What determines the uniformity? Just a fun fact I am wondering about, because I saw cyanide solutions used in electroplating.

The uniformity of a deposition depends on more parameters, but in this case the difference is in electromigration effect. $\ce{Ag+}$ ions are atrracted to the electrode surface, but the complex $\ce{[Ag(Cl)2]-}$ is repulsed. Cyanides form  complexes as well: $\ce{[Ag(CN)2]-}$ or $\ce{[Ag(Cl)3]2-}$ and they are also  repulsed.

Silver electrodepostion form $\ce{AgNO3}$ ($\ce{Ag+}$ in solution) is without additives nearly impossible as dendrites are created. Fukunaka, Yamamoto, Kondo, Ionic Mass Transfer Associated with Dendrite Growth of Electrodeposited Silver in AgNO3 Solution, 1989 ECS.

The metal deposition from a complex molecules is described in Wilkinson, P. Understanding Gold Plating. Gold Bull 1986, 19, 75 – 81

Some chemicals, that are usually added to electrolyte, affect the deposition because they create a film as they are adsorbed at  inner Helmholz plane. (E.g. acidic Cu deposition with alcohol addition Popescu et al, Materials Chemistry and Physics 72 (2001) 332–336.)


Dissolution of iron(III) hydroxide Problem
From the theoretical tour of the Mendeleev Olympiad in Chemistry:

a) Calculate the minimal amount of sulfuric acid that should be diluted with water to yield $\pu{100 mL}$ of solution that could completely dissolve $\pu{0.4 g}$ of rust.

Answer

 a) After the dissolution of a $\pu{0.4 g}$ portion of rust, the total concentration of iron(III) in a $\pu{100 mL}$ portion of solution is $\displaystyle\frac{\pu{0.4 g}}{\pu{107 g mol-1}\times\pu{0.1 L}} = \pu{0.037 M}.$ As there are no competing reactions (sulfate ion does not form stable complexes with iron), it is easy to find the minimal solution acidity after the dissolution using the solubility product $K_\mathrm{s} = [\ce{Fe^3+}][\ce{OH-}]^3$:

 $$[\ce{H+}] = K_\mathrm{w}\left(\frac{[\ce{Fe^3+}]}{K_\mathrm{s}}\right)^{1/3}  = \pu{0.0097 M},$$

 that is $\pu{0.49 mmol}$ of sulfuric acid in a $\pu{100 mL}$ solution. In addition, $\pu{0.037 M}\times 3\times\pu{0.1 L}\times 1000/2 = \pu{5.6 mmol}$ of sulfuric acid has reacted with iron hydroxide, therefore, the total amount of sulfuric acid required for the dissolution is $\pu{6.1 mol}$ (2 points).

My thoughts
This task seemed incredibly easy, but the solution overcomplicated some aspects, from my point of view.
The concept of "rust" was simplified in this problem since it was stated from the beginning that here "rust" means, in fact, $\ce{Fe(OH)3}.$ So, I wrote the chemical equation (which, because it is a neutralization reaction, should have an astonishingly large constant and is an equilibrium practically shifted fully towards the products):
$$\ce{2 Fe(OH)3 + 3 H2SO4 -> Fe2(SO4)3 + 3 H2O}$$
$\ce{Fe(OH)3}$ has a molar mass of $\pu{106.87 g/mol}$. Finally, the amount of sulfuric acid is
$$n(\ce{H2SO4}) = \frac{\pu{0.4 g}}{\pu{106.87 g mol-1}}\times\frac{3}{2} ≈ \pu{5.61 mmol}.$$
After checking the solutions, not only I realized I were wrong (even though my final answer was part of theirs), but I also were confused by the explanations given. For some reason, the solubility product was included, even though its constant is so insignificantly small in comparison to the constant associated with the reaction I wrote above.
What I thought was happening in this system was that $\ce{H2SO4}$ would neutralize the rust and that was basically it. While I am familiar with the concept of equilibrium and I understand that $\ce{H2SO4}$ doesn't nullify the concentration of $\ce{Fe(OH)3},$ in the light of the huge constant the neutralization is expected to have, the solubility product, $K_\mathrm{sp},$ should be negligible.
However, it seems that not only it is not negligible, but it also seems to be able to influence the final result significantly. But why is that? Why do we bother calculating the additional acidity from the hydroxide ions resulted in the solubility reaction? Why can't we solve the problem stoichiometrically, the way an inorganic chemist would approach it? Please correct my reasoning wherever you consider it wrong.

The $K_\mathrm{sp}$ of $\ce{Fe(OH)3}$ is varied from source to source, but a reliable source gives the value of $2.79 \times 10^{-39}$ at $\pu{25 ^\circ C}$. We'll use this value throughout the calculations.
Suppose, $s$ amount of $\ce{Fe(OH)3}$ dissolves some in water according to its $K_\mathrm{sp}$, but assume water is not ionized:
$$\ce{Fe(OH)3(s) <=>[H2O] Fe^3+(aq) + 3OH-(aq)}$$
$$K_\mathrm{sp} = s \times (3s)^3 = 27s^4 \ \Rightarrow \  \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{27}\right)^{\frac 14} = 1.008 \times 10^{-10}$$
This is $[\ce{Fe^3+}]$ in water. Thus, $[\ce{OH-}]$ would be $3s = 3.024 \times 10^{-10}$.  This value is much smaller than autoionization values of water. Therefore we have to consider $[\ce{OH-}] = 1.00 \times 10^{-7}$ for this calculation. Hence, if revise the calculation accordingly:
$$K_\mathrm{sp} = s \times [\ce{OH-}]^3 = (1.00 \times 10^{-7})^3s \ \Rightarrow \  \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{1.00 \times 10^{-21}}\right) = 2.79 \times 10^{-18}$$
Thus, when autoionization of water is considered, $[\ce{Fe^3+}]$ in water at $\mathrm{pH} \ 7$ is much smaller (the actual value). I did these calculations to show the effect of autoionization of water. Similarly, when real $[\ce{Fe^3+}]$ is high in the solution like the case (a) here, $K_\mathrm{sp}$ plays a role.
Let's re do the calculation again considering $[\ce{Fe^3+}] = 0.037 M$. If all $\ce{Fe(OH)3}$ stay dissolved, them we can find the $[\ce{OH-}]$ in the solution using $K_\mathrm{sp}$ calculations:
$$K_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH-}]^3 =  0.037 \times [\ce{OH-}]^3 \ \Rightarrow \  \therefore \ [\ce{OH-}] = \left(\frac{2.79 \times 10^{-39}}{0.037}\right)^{\frac 13} \\ = 4.22 \times 10^{-13}$$
Thus, you need to keep $[\ce{OH-}] = 4.22 \times 10^{-13}$ to keep the solution away from precipitating back to $\ce{Fe(OH)3}$. Thus, $[\ce{H+}]$ should be at:
$$ [\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} = \frac{1.00 \times 10^{-14}}{4.22 \times 10^{-13}} = 0.024$$
Thus, after completely dissolving rust, you need to add additional $\ce{H2SO4}$ to keep the $[\ce{H+}]$ of the $\pu{100 mL}$ solution at $\pu{0.024 M}$. I think you can calculate that amount in $\pu{mmol}$ easily now (Keep in mind that you already added $\pu{5.6 mmol}$ of $\ce{H2SO4}$ to dissolve $\pu{0.4 g}$ of rust).
Note: Any deviation from the given answer should be due to the $K_\mathrm{sp}$ value used here (The given answer didn't give that numerical value).


What is the hybridization state of E (= Sn, C) in ECl₂? What is the hybridization state of tin in tin(II) chloride, $\ce{SnCl2}$, and is the explanation similar for dichlorocarbene, $\ce{CCl2}$?

For $\ce{SnCl2}$, I read in my book that it is sp² but I am not able to explain it. I am not sure about the carbene.

I might have a misunderstanding about hybridisation, I think that the energy required for hybridization has to be compensated by the energy released in forming bond between hybridized orbitals.

Hybridisation is a mathematical concept, you never gain or lose energy due to it. Descriptions of the wave function involving hybrid orbitals or not are equally as valid. The linear combination of orbitals that result in the equilibrium geometry can however be interpreted as hybrid orbitals. Sometimes this may lead to a better understanding of the bonding situation.

From the above, it is obvious that hydbridisation is always a result from the molecular structure, never a cause for it. Hence we can deduce from the known bond angles the following:

Tin(II) chloride has a ∠Cl-Sn-Cl of about 95° (see Wikipedia). We know that p-orbitals are 90° towards each other. Therefore the tin will have more or less unhybrididised orbitals. See also this previous answer of me and this answer of Ben Norris.
Dichlorocarbene on the other hand has a ∠Cl-C-Cl of about 109° from both experimental and calculated data (Phys. Chem. Chem. Phys. 2002, 4 (14), 3282–3288). We know that the angles between sp³ hybrid orbitals are tetrahedral angles of about 109.5°. We can therefore deduce that the carbon employs sp³ hybrid orbitals in first order approximation.

The differences in the molecular structure and the utilisation of the s-orbitals for bonding stem from what is termed the inert pair effect, there is an explanation available in the this Q&A: What is the inert pair effect?


Why do simple molecular substances have weak intermolecular forces and why do giant covalent substances have strong intermolecular forces? Why do simple molecular substances have weak intermolecular forces and why do giant covalent substances have strong intermolecular forces?

I understand that in simple molecular substances the atoms within a molecule are held together by strong covalent bonds, but the intermolecular forces between molecules are weak.

But I don't see what makes giant covalent substances have stronger intermolecular forces? What makes them stronger and have higher melting points?

I thought I understood but I think I must have gone wrong somewhere and now I am very confused. Help would be much appreciated. Also please explain simply, chemistry does not come naturally to me so I might get even more confused.

In simple molecules the intermolecular forces are, in order of increasing strength, dispersion forces, permanent dipole interactions and hydrogen bonding (which has significant covalent character but is generally considered to be an intermolecular force).

By contrast giant covalent repeating structures such as diamond and $\ce{SiO2}$ are not molecular in the same sense as they can theoretically be infinitely large. Therefore they do not really have intermolecular forces but they are simply held together by covalent bonds between the atoms in the structure.


How are hot and cold defined? My teacher asked me this question a few days ago. "How to define hot and cold?" And I said, "The higher the temperature, the hotter you are. The lower the temperature the colder you are." But my teacher said, "That's not 100 percent correct."

So, how do you really define hot and cold?

The concept of temperature arises from the zeroth law of thermodynamics:

If two systems are in thermodynamic equilibrium with a third system, the two original systems are in thermal equilibrium with each other. Basically, if system A is in thermal equilibrium with system C and system B is also in thermal equilibrium with system C, system A and system B are in thermal equilibrium with each other.

From here, many hypothetical scales for temperatures are stablished.

It's said than one body has higher temperature than another when there is a heat exchange from the former to the latter.

But hot and cold are just sensations or feelings defined by people. Something is called hot when it's at higher temperature than our body (~37 °C), so a positive heat exchange will take place on contact. The opposite for cold (heat transference from our body to the cold object).

So, you are right, the higher the temperature of an object, the hotter it will be, but the concept hot is just to define the thermal perception.


Phase transition from gas to solid Question 13 from the ACS chemistry olympiad exam:

  The phase transition from gas to solid is called
  \begin{align}
&\text{(A) condensation.} & \text{(B) evaporation.} \\
&\text{(C) polymerization.} & \text{(D) sublimation.}
\end{align}

I am pretty sure the answer is deposition, which is not listed, but yet the answer is (A) condensation [scroll to the end of the question paper to see the answer key].

Did the chemistry olympiad make a mistake, or is there something obvious that I am missing?

I guess the diagram of phase changes you have in mind looks like this:

Indeed this says deposition is the right name for a gas-to-solid phase transition and I believe that this is the only correct term.

The only thing I can argue why it is obvious here that condensation is the right answer (at least the closest) is that this is the only answer that gives a transition from fluid to solid. B and D are the other way around, and C is not really a phase change (although it could result in one).


Health Risks Associated with Working in Chemistry Labs I've been working in a contaminants food quality lab for about 2 years now and was curious what the chronic health effects would be, should I be worried about cancer?

I would recommend looking at the MSDS information for anything that you work with specifically and/or researching the long term health effects of those specific substances


Quantum Valence Theory Reference At least to me, the contents of chapters 5 & 6 given here are representative of discussions on chemical bonding given in nearly all intro general & organic chem books, the statements in those chapters are mainly memorization (to me anyway). I basically don't find it rigorous enough, leads to more rather than less confusion and doesn't give any in depth discussion (similar to the way the quantum mechanics sections of those books are rote assertion ignoring all of the math).

I'm in the middle of a physics-based quantum mechanics course, so I'm developing the conceptual & mathematical tools to read a serious explanation of lewis bonding, polar-covalent bonding etc... but my book doesn't mention much in detail on this topic either, and apparently not even Pauling's QM book mentions much on the topics.

I'm really hoping somebody would have a quick look at the contents of chapters 5 & 6 of the first link & offer a recommendation of a rigorous reference or four on the contents of those chapters - I really can't face the headache I'll get reading chapters like those again due to all the confusion they lead to by dumbing it down, I need something more rigorous if such a thing exists, thanks a lot :)

Your Essentials of Physical Chemistry seems to be heavy on the "essentials" and light on the "physical chemistry". A textbook used in a physical chemistry course (and not a "physical chemistry for __" course) is what you need. The second portion of the traditional physical chemistry sequence is devoted entirely to the implications of quantum mechanics in chemical bonding. Pauling's book on QM will not cover much modern bonding theory since that book predates most modern bonding theory. The applications of QM to chemical bonding in all but the simplest molecules require computer programs to slog through the maths, since multielectron wave functions suffer from the three(+) body problem and cannot be solved exactly.

For example, in Physical Chemistry by Peter Atkins and Julio De Paula (W. H. Freeman & Company), Chapters 8 through 20 deal with chemical bonding and phenomena first with quantum mechanics and then with statistical mechanics. There are other physical chemistry texts out there as well. These texts will also cover the computational methods seeking to find 1) exact solutions to approximate multielectron wavefunctions (treating multielectron wavefunctions as linear combinations of single electron wavefunctions) and 2) approximate solutions to exact multielectron wavefunctions.


Why is salt needed when using vinegar to clean pennies? Let's say you have a solution that is vinegar and it has salt (NaCl) dissolved in it. Then you place old dull pennies in it. Then you look at it 5 minutes later and the pennies are clean and looking like new again.

I understand acid can etch the copper compound but why is the salt necessary for this process and what exactly does it do?

The copper surface is oxidized, which slows down the formation of copper acetate. When you add chlorides to the solution, the chloride penetrates the oxide layer and creates a hole in that layer that allows acetate to react with the copper. To find more information on the effects of chlorides on the surface of copper look up copper corrosion and chlorides.


Notation "5% solution" What does 5% aqueous solution of glucose mean? Note that it's just 5% and not 5%w/w. Do we assume it to be w/w?
The question is

If the osmotic pressure of 5% aqueous soultion of glucose is $\pi_1$ and that of 5% aqueous solution of urea is $\pi_2$ then:
A) $\pi_1=\pi_2$
B) $\pi_1<\pi_2$
C) $\pi_1>\pi_2$

Does 5% mean that the solutions are equi-molar/molal?

Generally speaking, percentage aqueous solutions like that are either weight for weight or weight for volume. Either way, in this example, you are comparing how many osmoles there for the same weight of urea ($CO(NH_2)_2$) and glucose ($C_6H_{12}O_6$).

Mole fraction as expressed in a percentage does exist though I have not seen that notation much myself and is usually designated n/n%

From the wording your question I'm guessing that it's a physiology or biochemistry question, in which case they're almost certainly referring to the 5% aqueous solution of glucose as the common IV hydration fluid 5% Dextrose Solution, which is w/w.

Otherwise, it's hard to guess as it's ambiguous!


Chlorination of Propane During monochlorination of propane why is chloropropane the major product formed. As far as I know using free radical mechanism 2-chloropropane is the supposed to be formed as the secondary free radical is more stable than a primary free radical. But the calculations involving reactivity order of chlorination tell otherwise. Why is this so!?

The reason might be one of kinetic versus thermodynamic control.  The actual halogenation step is stage 2 in this scheme:

(reaction scheme from chem.libretexts.org)

The 2-radical might well be more stable, but that doesn't mean that it will react faster than the 1-radical.  If the reaction rate constant for the 1-radical with chlorine is much faster then the 1-chloro- product will predominate.

More likely, though is that it is that the 1-radical is formed in preference over the 2-radical in step 1. The halogen can attack at one of three positions, so statistically speaking you'd expect more 1-radical over 2-radical intermediates.  If this species then reacts immediately in step 2 without rearrangement then you'd get a predominance of the 1-product formed.
(Thanks to Karl for pointing this out).


Atomic Mass of ¹²C before 1961 Question:

  Prior to 1961, the atomic mass unit was defined as 1/16 the
  mass of the atomic mass of oxygen; that is, the atomic mass
  of oxygen was defined as exactly $\pu{16 amu}$. What was the
  mass of a $\ce{^{12}C}$ atom prior to 1961 if the atomic mass of oxygen on today’s scale is $\pu{15.9994 amu}$?

I don't fully understand the logic behind the question. Here's my thinking. Atomic mass of all substances were related to $\pu{16.0000 amu}$ prior to 1961. Therefore, we can make the following expression where $X$ is mass of $\ce{^{12}C}$
$$\dfrac{X}{\pu{16.0000amu}}$$

Today, all atomic mass is related to $\pu{12.0000 amu}$ of $\ce{^{12}C}$. We can model the relationship as $$\dfrac{\pu{12.0000amu}}{\pu{15.9994amu}}$$

What the solution manual does is make the two equations equal to each other and solves for $X$.

$$\dfrac{X}{\pu{16.0000amu}}=\dfrac{\pu{12.0000amu}}{\pu{15.9994amu}}$$

How are we justified in doing this?

Here is another equivalent way of thinking about the problem. The mass of oxygen on the old scale is 16 amu. The mass of oxygen on the new scale is 15.994 amu. When you divide them:

$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} = 1.0000375 \frac{\textrm{old}}{\textrm{new}}$$

Basically, taking a ratio of any value in the old scale to the equivalent value in the new scale (or vice versa) gives you a conversion factor. So for every new amu, there is 1.0000375 old amus. The question is asking you to convert 12 new amus to old amus. Just by looking at the units, it should be obvious what to do:

$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} * 12\ \textrm{new}=12.00045\ \textrm{old}$$

If you explicitly label old and new, it's easy to see that the new units cancel out.

Incidentally, if you keep track of your significant figures, you can see why this redefinition of the amu didn't cause too many problems for chemists—even to five sig figs, the change is negligible.


Units of rate constant (cycles per second vs radians per second) in Eyring equation In the Eyring equation (EE),

$$k = \frac{k_\mathrm B T}{h} \exp\left(\frac{-\Delta G_{\mathrm f}}{RT}\right),$$

the units of $k$ are $\mathrm{s^{-1}}$. However, in general rate constants are usually expressed in $\frac{\mathrm{rad}}{\mathrm{s}}$. For instance, in expressions for damped oscillations of the form $\exp[(\mathrm{i} \omega - k)t]$, where $\omega$ is by definition in $\frac{\mathrm{rad}}{\mathrm{s}}$ and $k$ has to bear units providing consistent dimensions.

By expressing in the EE the energy $k_\mathrm B T$ as $\nu h$, where frequency $\nu$ is in cycles per second $\left(\mathrm{Hz}=\frac{\mathrm{cyc}}{\mathrm{s}} = \mathrm{cps}\right)$, it appears that $k$ is also given in $\mathrm{cps}$ in this equation. The desired units of $\frac{\mathrm{rad}}{\mathrm{s}}$ would be obtained if $h$ were replaced by $\hbar$. However, I have never seen the EE formulated that way. Has anybody an idea in this matter?

I have looked at the original article of Henry Eyring (J. Chem. Phys. 1935, 3, 107). In that article, Eyring explains the derivation of his famous equation. It basically boils down to the following:

You assume an equilibrium between the initial and transition state and consider that once a species crosses the barrier, it goes to the product state:

$$R \leftrightarrows R^{\dagger} \rightarrow P$$

The rate expression is then simply:

$$k = \nu K$$

where $k$ is the reaction rate constant, $\nu$ the crossing frequency and $K$ the equilibrium constant between the IS and TS.

I will not go into detail about how to calculate $K$, but just remember that $K$ can be obtained by calculating the quotient of the partition functions of the system in the transition and initial state, respectively. How $\nu$ is calculated, will hopefully answer your question.

Let us assume that in the transition state, the bonds to be formed or broken are weak. In that case, we can model the reaction coordinate (the direction of the reaction) as a loose vibration. The partition function of a vibration is

$$
f = \frac{1}{1 - \exp \left(\frac{-h \nu}{k_{b}T} \right)}
$$

Herein, $\nu$ is the frequency of the loose vibration. We assumed that this vibration was very weak (loose). In that case we can apply a Taylor expansion to the exponent and cut off the series after the first term. This gives us:

$$
f = \frac{1}{1 - 1 + \frac{h \nu}{k_{b}T}} = \frac{k_{b}T}{h \nu}
$$

Plugging this result back into our original rate expression gives

$$
k = \nu \frac{k_{b}T}{h \nu} K^{\dagger} = \frac{k_{b}T}{h} K^{\dagger}
$$

Here, I have put a $\dagger$ after the equilibrium constant to make clear that I have extracted one partition function (that of the loose vibration) out of the equilibrium constant.

From the dimensional analysis of $\frac{k_{b}T}{h}$, you can see that $k$ should have units of $s^{-1}$ and not something else.


Intramolecular or intermolecular? I am extremely confused about these types of forces.

Some websites state that the forces in the covalent bonds of carbon dioxide are inter. Other websites say they are intra?

Also, some websites say that the electrostatic bond between ionic compounds is intra, others say they are inter.

So, which is inter and which is intra?

Inter- means in between molecules (like the interstate highway - in between states)and intra- means within one molecule ( like an intrastate highway- within one state). So a covalent bond of CO2 OR an electrostatic ionic bond would be an intramolecular bond. The type of bond that keeps water molecules close to each other ( hydrogen bonding) is an example of an intermolecular attraction.


Cis/trans interconversion of crotyl-boron compounds Why is the following interconversion slowest for crotyl boronates and fastest for crotyl boranes while crotyl borinates are somewhere in-between?

The lack of an octet and the empty p-orbital on boron make trialkyl boron compounds quite reactive.  If we replace one of the alkyl ligands with an alkoxy group (borinate), we can now draw a resonance structure where oxygen shares one of its lone pairs with boron.  In such a resonance structure the boron atom has achieved an octet and the empty p-orbital on boron has been filled.  Consequently borinates are less reactive than the corresponding boranes.  Boronates, with two alkoxy substituents, have an even higher contribution from the resonance structure with the $\ce{B^{-}=O^+}$ double bond and are even less reactive.


Can the enthalpy change of a reaction be changed, and if so, will this affect the value of the equilibrium constant? In the case of an arbitrary equilibrium reaction, is it possible to change the absolute enthalpy of the final products, thus the reaction enthalpy change? For example, say pH affects the charge distribution of the product molecule, changing the attractive force between it and the solvent (feel free to come up with a better example).

Given that this does in fact affect $\Delta H$, would the Gibbs free energy change of the reaction $\Delta G=\Delta H-T\Delta S$ be altered as well, or does the entropy change somehow adjust automatically?

If $\Delta G$, changes, I would assume $K$ changes as well through $\Delta G = - RT$ $\mathrm{ln} K$

Short answer: No and yes.

Enthalpy is a state function. That means that it depends only in the current "state" - the identity of the substances, their concentrations, pressures, etc. That means that the absolute enthalpy of some substance is independent of how you produced it. However, that substance has different enthalpy in different states: gas versus liquid, pure versus in solution. If you change the state of the product(s), you with change their enthalpy. Same goes with any of the other thermodynamic potentials: entropy, internal energy, free energy.

And that of course will alter the equilibrium situation.


Safety concerns - Heating PMMA (acrylic glass) and fluorinated polymers I have some plastic optical fiber, which consists of a thin core of PMMA (polymethyl methacrylate), covered by a fluorinated polymer.

I would like to heat a bundle of these fibers to ~130 degrees, in order to form a simple image guide.

However, I won't start until I know this is safe. Can you help me find out what types of gases I may encounter by heating these materials? Anything known dangerous? What other safety precautions should I take?

One problem is bonding of the bundle.  The contact surface is the fluororesin cladding that allows more nearly perpendicular rays to be totally internally reflected.  If you nick the cladding you get a light leak.  Will the cladding fuse when heated?

If you ligthly bond a bunch of cylinders you get air gaps where they touch.  Those will be a triangular grid off dark spots, 9.3% of the area of the image.

http://en.wikipedia.org/wiki/Circle_packing


Partial oxidation of toluene While reading the preparation of phenol from cumene by reaction with $\ce{H2O2}$ or partial oxidation, I was curious to find out why toluene wasn't used.

I checked the literature and found that benzaldehyde would be formed in the case of toluene [1, 2].

Why does benzaldehyde form in the case of toluene?
Does the peroxide formed after initial oxidation become benzaldehyde due to its instability?
Why doesn't acetophenone form from the peroxide formed in cumene?

References

Sun, H.; Blatter, F.; Frei, H. Selective Oxidation of Toluene to Benzaldehyde by $\ce{O2}$ with Visible Light in Barium(2+)- and Calcium(2+)-Exchanged Zeolite Y. Journal of the American Chemical Society 1994, 116 (17), 7951–7952. https://doi.org/10.1021/ja00096a084.
Wang, X.; Wu, J.; Zhao, M.; Lv, Y.; Li, G.; Hu, C. Partial Oxidation of Toluene in $\ce{CH3COOH}$ by $\ce{H2O2}$ in the Presence of $\ce{VO(Acac)2}$ Catalyst. The Journal of Physical Chemistry C 2009, 113 (32), 14270–14278. https://doi.org/10.1021/jp9028062.

You can justify the formation of Benzaldehyde from Toluene by oxidising with peroxide just by looking at the mechanism of the reaction and think of migratory aptitude of Phenyl group , hydride and methyl group.

Firstly, if we delve into the mechanism of the oxidation of Cumene, we'll find that there is a rearrangement step in the mechanism. The rearrangement occurs when one of the oxygen in hydroperoxide gets protonated and leaves as water. If you now think carefully, there are both the possibilities of migration of Phenyl and methyl group to the oxygen. But Phenyl has certainly more migratory aptitude than methyl, and also the carbocation formed after the rearrangement is stable, as it has two methyl group on the $\alpha$-position to stabilise by electron release and one oxygen atom to stabilise it via donation of lone pair. Thus the migratory aptitude and stability of the carbocation formed both drive the rearrangement to prefer the phenyl migration. Therfore acetophenone is not formed due to unfavoured methyl migration.

On the other hand, the situation with toluene is slightly different. Here you have both the possibilities of hydride as well as phenyl migration. But hydride has almost similar (slightly more) migratory aptitude as Phenyl group, and what is more interesting is that the migration of hydride will lead to more stable benzylic carbocation, which is not possible by phenyl migration. These two effects cooperatively facilitate the hydride migration and therefore leading to the product Benzaldehyde.

Thus the migratory aptitude and stability of the carbocation are the two effects which create the differences in the product formation in case of oxidation of Cumene and Toluene.

(Note : For clarification, see the migratory aptitude order here : https://en.wikipedia.org/wiki/Migratory_aptitude )


Can visible light or infrared radiation excite electrons? So the limit of ionizing radiation seems to be UV light.

I'm wondering if IR light or Visible light can excite the electrons in an atom.

It has to do with the difference in energy level right?
So the closer two energy levels the lower the frequency/energy.

And that is why coordination complexes and/or transition metals are usually colored because once coordinated, the d orbitals lose their degeneracy?

I also just wanted to clarify that this excitation is different from the color we see for conjugated systems. That is because these systems when conjugated can absorb light? And so we see what is not absorbed and reflected back to us?

Different regions of the electromagnetic spectrum correspond to different atomic and molecular processes, each with one or more associated spectroscopies. Here is a general summary, with decreasing frequency/energy going from left to right:

In order to excite a valence electron, the longest wavelength or lowest energy radiation is usually in the visible region, not longer than ~700 nm. I'm not sure if there is a lower limit to this energy, or what the observed lowest energy electronic transition is, but certainly in this region, there is some overlap with the energy required to excite molecular vibrations. For some nice cases you can perform vibrational-electronic spectroscopy (usually called "vibronic" spectroscopy), where you resolve a number of vibrational transitions (peaks) as fine splitting or fine structure for a single electronic transition:

$\hspace{20ex}$

However, we do not say that "infrared radiation can excite electrons", though there is a range of frequencies where the types of excitation couple together. See also rotational-vibrational (rovibrational) spectroscopy; a nice example is hydrogen chloride in the gas phase.

The reason coordination complexes are often colored is because there are valence electronic excitations that correspond to the absorption of electromagnetic radiation in the visible region. The reason many molecules or complexes are white is because their lowest energy electronic transition is in the ultraviolet region. This is more general than losing degeneracy, and holds true for any macroscopic or microscopic system. It is not different from the color we see for conjugated systems. If the lowest energy available transition of a system is in the visible region, the system will appear colored. If it absorbs red-orange light, the opposite side of a color wheel will show you that it should appear green-blue. What you see as the system's color is the remaining color after subtracting the absorbed color. Yes, the kinds of orbitals involved are different (d-d transitions or metal-to-ligand charge transfer versus $\pi\rightarrow\pi^{*}$), but the transition energy matching is the most important thing.

For the "limit of ionizing radiation", ionization refers not to the excitation of an electron, but its removal:

$$
\ce{X + energy -> X+ + e-}
$$

The atom with the smallest first ionization energy is sodium, given as 496 kJ/mol. Converting this to a wavelength gives roughly 0.013 nm or 13 pm, which is a hard X-ray. Normally X-rays only have enough energy to excite core electrons, so this is the minimum wavelength for atomic ionization. Many atoms will require gamma rays, though I also assume that when looking at whole molecules rather than atoms, the ionization potential may be reduced.

Now to touch on the specific example of rhodopsin. It is a protein that contains retinal, which undergoes photoisomerization:

where A is the 11-cis form, and B is the all-trans form:

If we say that the ultraviolet region spans 10 to 400 nm, then retinal absorption is in the UV, not the visible. However, "the absorbance spectrum of the chromophore depends on its interactions with the opsin protein to which it is bound; different opsins produce different absorbance spectra." From that page:

\begin{array}{lcccll}
\hline
\text{Name} & \text{Abbr.} & \text{Photoreceptor} & \lambda_{\text{max}}~(\text{nm}) & \text{Color} & \text{Human variant} \\
\hline
\text{long-wave sensitive}           & \text{LWS}  & \text{cone} & 500-570  & \text{green, yellow, red}  & \text{OPN1LW "red" / OPN1MN "green"} \\
\text{short-wave sensitive 1}        & \text{SWS1} & \text{cone} & 355-445  & \text{ultraviolet, violet} & \text{OPN1SW "blue" (extinct in monotremes)} \\
\text{short-wave sensitive 2}        & \text{SWS2} & \text{cone} & 400-470  & \text{violet, blue}        & \text{(extinct in therian mammals)} \\
\text{rhodopsin-like 2}              & \text{Rh2}  & \text{cone} & 480-530  & \text{green}               & \text{(extinct in mammals)} \\
\text{rhodopsin-like 1 (vertebrate)} & \text{Rh1}  & \text{rod}  & \sim 500 & \text{blue-green}          & \text{OPN2 = Rho = human rhodopsin} \\
\hline
\end{array}

From photopsin:

  The mean absorbance spectra of outer segments of the four classes of human photoreceptors. Curves labelled as follows: '498', mean of eleven rods; '420', mean of three blue-sensitive cones; '534', mean of eleven green-sensitive cones; '564', mean of nineteen red-sensitive cones.

Regarding the mechanism of opsins as triggered by the photoisomerization,

  The photoisomerization induces a conformational change in the opsin protein, causing the activation of the phototransduction cascade. The opsin remains insensitive to light in the trans form. It is regenerated by the replacement of the all-trans retinal by a newly synthesized 11-cis-retinal provided from the retinal epithelial cells. Opsins are functional while bound to either chromophore, with A2-bound opsin λmax being at a longer wavelength than A1-bound opsin.

The reason for the isomerization is that upon electronic excitation, the system can approach a point on the potential energy surface where two or more geometries are degenerate; this point is called a conical intersection. That is, after exciting 11-cis-retinal, certain internal rotation barriers are lowered in energy, the nuclei begin to respond, the system passes through a conical intersection, and further relaxation of the nuclei leads to all-trans-retinal. In general, photochemical processes are very complicated.

References

Kropf, Allen; Hubbard, Ruth. The photoisomerization of retinal. Photochem. Photobiol. 1970, 12, 249-260.
Bowmaker, J. K.; Dartnall, H. J. Visual pigments of rods and cones in a human retina. J. Physiol. 1980, 298, 501-511.
Send, Robert; Sunholm, Dage. Stairway to the Conical Intersection: A Computational Study of the Retinal Isomerization. J. Phys. Chem. A 2007, 111, 8766-8773.


What are some incapabilities that the early spectroscopy have when compared to the more modern method? What are some incapabilities that the earlier spectroscopy (Visible, Ultraviolet and Infra-red) have when compared to the more modern method?

Some of the things that come to mind are:

Resolution  Modern equipment has much higher resolution than older equipment.  Optical component surface smoothness and uniformity, optical component alignment and electrical component stability all permit the user to see more detail and fine structure in the spectrum, which assists in analysis
More stable light sources  Again resulting in more detailed and more reproducible spectra
Improved monochromaticity of the light sources  Further improving spectral resolution
Improved sample cells  Polished glass surfaces provide for better signal to noise making it easier to discern the signal
Signal processing software  Such as Fourier Transform methods which increase the signal to noise ratio making it easier to discern the true peaks in a spectrum
Other computer and software enhancements  That can be applied to sample handling and spectrum processing.  They dramatically reduce the time it takes to record and process a spectrum


Why hybridization occurs in diatomic molecule In diatomic nitrogen
There are two electron configurations I'd like to compare:

1s2 2s2(lone pair) 2p3(the bonding electrons sigma,pi,pi)
1s2 2sp(lone pair) 2sp(sigma bonding) 2p2(pi,pi bonding)

Why is option 2 favored over option 1?
I thought having the lone pair in a 2s orbital would be favored over 2sp.

I thought having the lone pair in a 2s orbital would be favored over
  2sp

That's true, placing the lone pair in a 2s orbital would be put the lone pair electrons at lower energy then if they were in a 2sp orbital.  However you're only looking at a part of the puzzle.  As you note, in this case the sigma bonding electrons would be in a pure p orbital and they would then be higher in energy then if they were in an sp orbital. You need to optimize the total energy of the molecule, so you need to consider all electrons and find the hybridization that gives the lowest possible energy to both the bonding and non-bonding electrons.

Which hybridization scheme a molecule adopts is the result of a delicate balance between electron stabilization and other factors such as bond strengths, steric and electrostatic effects.  The balance of these factors can shift and that is why some molecules hybridize one way and other molecules hybridize differently.


Structure of crystalline boron trioxide I recently came across the statement

  The structure of crystalline $\ce{B2O3}$ consists of $\ce{BO4}$ tetrahedra, two sets of which form two types of interconnected spiral chains, three $\ce{B-O}$ bonds are equivalent, but the fourth one is somewhat longer.

Please explain the meaning of this statement, above is the image of crystalline $\ce{B2O3}$ from Wikipedia.

This statement is relevant for the initially [incorrectly] determined structure of $\ce{B2O3}$ by Berger [1, 2] who used powder x-ray diffraction data.
You can still access the structure and visualize it in 3D (ICSD 24047), but it was shown later by several researchers and summarized in paper by Gurr et al. [3] that this structure has been erroneously determined:

After failing to grow a useful single crystal by this process, Berger (1952, 1953) was forced to use data from powder photographs in an attempt to determine the structure. The basic units of Berger's structure were two different irregular tetrahedra of oxygen atoms each surrounding a boron atom. This structure has been considered unsatisfactory in several respects (e.g. Wells, 1962), particularly as it results in two different $\ce{BO4}$ tetrahedra with six different $\ce{B-O}$ bond lengths ranging from $1.31$ to $\pu{2.14 Å}.$ In this laboratory a least-squares refinement (Hamilton, 1961) on Berger's powder data was attempted from his positional parameters. The attempt was abandoned when the refinement diverged. The tetrahedrally coordinated structure proposed by Berger was also puzzling because in various borate systems there is a regular variation of the fraction of boron atoms in fourfold coordination as a function of increasing alkali content (both in glasses and crystals); yet in each instance there was an obvious trend towards only threefold coordinated boron as the composition approached $\ce{B2O3}.$
[…]
The n.m.r, absorption for $\ce{BO4}$ groups is sharper and stronger than that for $\ce{BO3}$ groups (Bray et al., 1961; Silver & Bray, 1958) and thus the absence of sharp components in the spectrum of $\ce{B2O3}$ I precludes the presence of tetrahedral $\ce{BO4}$ groups in this material.

A new crystal structure was proposed: same space group $P3_1,$ but lacking tetrahedral units (ICSD 16021).
This is the one that is shown on Wikipedia, hence the dissonance.

Figure 1. Comparison between the unit cell content of two $\ce{B2O3}$ structures (boron atoms and boron coordination polyhedra are depicted in yellow color). Left: crystal structure according to Berger (1952). Right: crystal structure according to Gurr et al. (1970).
References

Berger, S. V. The Crystal Structure of $\ce{B2O3}.$ Acta Cryst 1952, 5 (3), 389–389. DOI: 10.1107/S0365110X52001155.
Berger, S. V.; Boeri, E.; Bonnichsen, R. K.; Sörensen, J. S.; Sörensen, N. A. The Crystal Structure of Boron Oxide. Acta Chem. Scand. 1953, 7, 611–622. DOI: 10.3891/acta.chem.scand.07-0611.
Gurr, G. E.; Montgomery, P. W.; Knutson, C. D.; Gorres, B. T. The Crystal Structure of Trigonal Diboron Trioxide. Acta Crystallogr B Struct Crystallogr Cryst Chem 1970, 26 (7), 906–915. DOI: 10.1107/S0567740870003369.


Separate toluene from acetone and isopropyl alcohol I found a cellulose thinner that is made of toluene 60-80%, acetone 10-20%, and 2-propanol 5-10%. As acetone and 2-propanol are water soluble, but toluene isn't, would it be possible to just add a lot of water and then remove it, and then end up with just toluene?

Multiple washes with smaller quantities of water will be more efficient than "just adding a lot of water" in a single wash. The toluene will be "purified" further with each successive wash. Some toluene will be lost with each wash also to the degree of it's solubility in water at the temperature of the extraction (about 0.03% soluble in water at room temperature). As stated in a comment however, distillation would be a superior method for recovering the pure toluene.


Acid catalysed mechanism of the addition of alcohols to alkenes I believe that this mechanism is false.
The protonated ether would in my opinion give a proton to methanol rather than $\ce{H2SO4}$ because the protonated $\ce{CH3OH}$ is more stable than $\ce{H2SO4}$.
Can someone confirm my thought?

The short version: You are correct, the schematic drawing is not the correct mechanism and it does also not provide the most likely path. For a training exercise it is close enough though. It shows the most important steps and it conserves stoichiometry.

The depicted scheme does not show a reaction mechanism in the sense of the definition given in the goldbook. A reaction mechanism always also has to include transition states, that lead to the respective products. It is rather a reaction sequence, that shows the most important steps. In this sense, the stoichiometry has to be conserved and in this sense it is as correct as it could be.

Noteworthy is, that the catalytically active species is not sulfuric acid, but instead a mixture of hydronium and protonated methanol, $\ce{H3+O/H2+OCH3}$. Sulfuric acid is such a strong acid, that it is capable of protonating any solvent that has a lone pair, even itself via autoprotolysis (in concentrated $98\%$ solution). $$\ce{2H2SO4 <=> [H3SO4]+ + [HSO4]- <=> H3+O + [HS2O7]- <=> H2O + H2S2O7}$$
The use of concentrated sulfuric in organic synthesis is usually not advisable, since its dehydration property can quickly decompose many organic compounds that contain hydroxyl groups. The use of diluted sulfuric acid (either in water or alcohols) is therefore much more common.

It is only logical, that sulfuric acid will protonate methanol, which can be considered a preceding equilibrium. $$\ce{H2SO4 + HOCH3 <=> H2+OCH3 + [HSO4]- }$$
Since methanol is also the solvent and the acid is only available in catalytic amounts, it is feasible to assume, that the equilibrium lies entirely on the right.

It is also important, that the last step must be also in equilibrium, since it is only the protonation/deprotonation of an ether. Here again we have much more methanol than product, so that the equilibrium will most likely be entirely on the product side.

A more descriptive mechanism would be the following:

For simplicity and sufficiently accurate bond formation and breaking is indicated by arrows, omitting the transition states. Resulting a slightly simplified picture. However, the preceding equilibrium - as you noted - should not be omitted. This formulation of the mechanism is particularly important for many simple organic reactions, as it focusses entirely on the reactants and products (and to a minor extent on the reactive intermediates).

When it comes to more complex reactions, e.g. with asymmetric catalysts it is often much better to include the transition states.

From this it can also be seen, that the choice of sulfuric acid is somewhat arbitrary, as most acids would do the same. However, sulfuric acid is often used, since bisulfate and sulfate are very weak nucleophiles.

Additional note
In all schemes only the most likely path is shown. Any possible side reactions, like concerted [2+2] cycloaddition, have been omitted.


Chemoselectively dihydroxylate a terminal alkene in the presence of an internal alkene? The goal is to dihydroxylate the exo alkene, and then convert to a ketone.

Any suggestions would be apprecitated to either chemoselectively convert the terminal alkene, or directly convert the terminal alkene to a ketone.

Thank you!

The Lemieux-Johnson oxidation (details here) using catalytic $\ce{OsO4}$ or $\ce{RuO4}$ turned over with $\ce{NaIO4}$ will do the transformation you want in one step. These conditions should be selective for the exo double bond.


Does aniline breakdown in groundwater produce benzene Can the benzene ring attached to aniline be detached in an aqueous solution (specifically groundwater) under certain conditions? Would the two compounds be detected as separate entities in a laboratory analysis?

Aniline is not easily converted to benzene in aqueous conditions. A google search turns up several papers describing the breakdown products of aniline in groundwater but benzene is not one of them. Methanogenic and denitrifying bacteria seem to play a key part in aniline breakdown, releasing methane and carbon dioxide as byproducts. The other main products seem to be polymerisation products resulting from condensation and oxidation reactions.

In answer to your question about detection; no you would not expect to see benzene as a breakdown product in a sample of groundwater that has been contaminated by aniline.

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1519315/


Will oxidation/reduction of copper still occur in an electrochemical cell if it is not in an aqueous environment? Notation of the reduction and oxidation of copper typically includes the aqueous state. Illustrated below:
$$\ce{Cu(s) -> Cu^2+(aq) + 2e-}$$
This event usually occurs at 0.34 V. What if the copper was in an organic environment instead? I ask because I am doing cyclic voltammetry of a capacitor with copper as the current collector and need to measure the non-faradaic capacitance. Our current setup is using $\ce{KOH}$ as the electrolyte and we are basically just seeing a non-reversible/quasi-reversible CV like below:

It is certainly possible for metals, including copper, to corrode in non-aqueous environments.  The details depend on the precise nature of the solvent, electrolytes, and other potentially complexing species present.  For example, in polar protic solvents (like water) that don't have any complexing ligands for copper, the equilibrium potential for copper oxidation will likely be the same as in water.  But oxidation potentials will likely shift in solvents that are non-polar, in ways that depend strongly on the composition of the electrolyte.

However, if you are using potassium hydroxide as the electrolyte, your conditions might be more aqueous than you think.  You can think of "hydroxide" as a type of water that likes to exist at high pHs.  And hydroxide itself generates high pHs.

Elecrochemical water oxidation occurs via this reaction:

$$\ce{2 H2O -> O2 + 4 H+ + 4e-}$$

The pH-dependent equilibrium potential is $1.23~\rm{V} - 0.059\frac{\rm V}{pH}\times pH $, which at a pH of 14 comes out to 0.4 volts.  So you might also possibly be irreversibly oxidizing hydroxide to oxygen during your electrolysis.


Does increasing one of the electrolytes in a galvanic cell increase the voltage? I would like to know how you could increase the voltage in zinc and copper galvanic cell:
$$\ce{Zn(s) + Cu^2+ → Zn^2+ + Cu(s)}$$
I know that temperature and surface area is a factor that affects the voltage, but I would like to specifically know about how concentration changes the voltage.
If I want to increase the voltage, which electrolyte do I change? The anode or cathode? And what's the reason for that?

As suggested in the comments, the key is Nernst equation. The fundation is the Gibbs free energy given by:
$\Delta G = \Delta G^0 + RT \ln Q_r$
where $\Delta G^0$ is the standard free energy for the process, while $Q_r$ is the reaction ratio. Remember that the reaction ratio is the analogous quantity of the equilibrium constant, but with the non equilibrium concentration. Remember also that the Gibbs free energy is proportional to the voltage. Then you need to increase the reaction ratio.


Significant figures in a calculation of atomic mass from isotopic abundance I had this question:

Naturally occurring europium consists of two isotopes with a mass of 151 and 153. Europium-151 has an abundance of 48.030 and, and europium-153 has a natural abundance of 51.970. What is the atomic mass of europium? Do not include units. Be sure to round to the correct number of significant figures.

So I do the math for the calculation:
$(151 \times 0.48030)+(153 \times 0.51970) = 152.0394$
I submit 152.0 because, as I understand it, the products of the math in the parentheses should be considered to have 3 sig figs each - ending in the tenths place. So when I add those two sums up, it should end in the tenths place too given addition/subtraction sig figs rules. This is marked as incorrect, and I worry I am misunderstanding something.
I emailed the professor, and he said it was a word question and I needed only to select the fewest sig figs of the ones provided (151 and 153 have 3), so my answer should have been 152 without the 0 in the tenths place. Is this usually the case? What confuses me is I don't think I was allowed to do this for other word problems, but I could be wrong. Is there a difference between word problems and other problems, and is 3 really the amount of sig figs I should have had?

First, for reasons I'll point out later this is a poorly made-up problem which was only designed to test significant figures, not to teach anything real about chemistry.
Second there are no special rules for "word problems" using significant figures.
The gist is that significant figures are a simple method to do error propagation in calculations. It is essentially based on the notion that the error is +/- 1/2 of the last significant digit, as if the value had been rounded to that digit. So 151 means that the value should be between 150.5 and 151.5.
The whole point of significant figures is to prevent you from dividing 151 by 3.00 and getting some ridiculous precision like 50.333333333333333333333333333333.
Now on to the problem. 151 and 153 are mass numbers for the europium  isotopes. Mass numbers are poor values to use for the isotope  masses. If fact the known masses are 150.9198502(26)  and 152.9212303(26). The numbers in parenthesis give the standard deviation of the measurement. So for 150.9198502 the standard deviation is 0.0000026.
Likewise according to the Wikipedia page, $\ce{^{151}Eu}$ has an abundance of 0.4781(6) and $\ce{^{153}Eu}$ has an abundance of 0.5219(6). The point is that the isotopic variation in samples taken from all over the earth limits the precision of the atomic mass for any element, not the mass of the isotopes which are know with much greater precision.
The same Wikipedia page linked earlier gives the atomic mass of europium as 151.964(1) which is inconsistent with the given percentages. I do not know the how/why there are so many more significant figures in the atomic mass (and I am too lazy to turn this answer into a research project to figure that part out...).


Reaction between barium peroxide and hypochlorous acid When $\ce{BaO2}$ reacts with $\ce{HClO},$ the question tells us that that oxygen, barium chloride and one other product is formed.
The equation I came up with is
$$\ce{BaO2 + 2 HClO -> BaCl2 + 2 O2 + 2 H2}\tag{R1}$$
The equation given, however, is
$$\ce{BaO2 + 2 HClO -> BaCl2 + 1 1/2 O2 + H2O}\tag{R2}$$
Why is $\ce{H2O}$ produced and not $\ce{H2}?$ More importantly, what clues/rules are there that would have helped me to predict these products?

You may have confused barium peroxide with barium metal.  The metal can indeed displace hydrogen from hypochlorous acid, not to mention (for a metal as reactive as barium) from the water in which the acid is dissolved; but metal peroxides will produce oxygen and water from the acid.  So your missing product is water, not hydrogen.
Tricky half-reactions
If we want to balance this with half-reactions we need to take some care, because we need to simultaneously cancel protons and electrons with a weak acid like hypochlorous acid.  Given the reduction half-reaction based on chlorine being reduced from the acid:
$\ce{HClO + H^+ + 2 e^- -> Cl^- + H2O}$  Eq. 1
We must set up the peroxide oxidation to oxygen so that the electron/hydrogen ion ratio is also $2:1$.  Thus a proposed reaction
$\ce{BaO2 -> Ba^{2+} + O2 + 2 e ^-}$  Eq. 2
would be balanced but you don't have the proper ratio of electrons to protons matching the weak hypochlorous acid reduction.  Add water to get the necessary hydrogen ions, and note that this means you change the oxygen production:
$\ce{BaO2 + H2O -> Ba^{2+} + 2 H^+ + (3/2) O2 + 4 e ^-}$  Eq.3
In other words, water is also oxidized to generate the hydrogen ions necessary to effect the reduction of the weak hypochlorous acid, thus producing that extra half-mole of oxygen per mole of barium peroxide.
Four electrons to two protons in Eq. 3 matches the ratio of two electrons to one proton in Eq. 1.  So now we can double Eq. 1 and add to Eq. 3 getting the reaction that avoids both unbalanced protons and unbalanced electrons.  As we saw above, one mole of barium peroxide gives 1.5 moles of oxygen in the overall reaction:
$\ce{BaO2 + 2 HClO -> Ba^{2+} + 2Cl^- + (3/2) O2 + H2O}$.


pH value of a buffer solution of HClO and NaClO So,
I have 200mL of HClO 0,64M.
I mix it with 0,1mol of NaClO.
How should I calculate the pH?

My question is about this: should I keep attention about changes made to the solution volume after adding NaClO?

So far this is what I've done.

Considering:

$\ce{NaClO + H2O -> Na+ + ClO-}$

With n(NaClO) = n(ClO-) = 0.1mol,

I calculated the molarity of the conjugate base:

[ClO-] = 0.1mol/0.2L = 0.5M.

Then I applied the Henderson-Hesselbalch equation:

pH = pKa + log([ClO-]/[HClO]) = 7.53 + log(0.781M) = 7.422

In this case I didn't consider the variation to the solution volume due to the addition of NaClO. So, is this correct?

You're close.  Everything is correct, except that when you take the ratio of concentrations in the H-H equation that ratio is not in moles.   It's just a number, because you divide moles by moles .  Rule of thumb: logarithms and exponential should never involve anything with units.  So don't include the molar unit under the logarithm and you're good.


Why is Kc not affected by change in pressure? Why is $K_c$ not affected by change in pressure? I know the mathematical explanation, but I don't really understand the reason when only looking at $K_c$. The explanation I know is with the reference to $K_p$, but is it possible to explain without $K_p$?

Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is turn is determined by concentration (seen in the rate laws), temperature, activation energy, and how often molecules collide with the correct orientation (together these last three determine the rate constant, $k$, as shown in the Arrhenius equation).

The equilibrium constant, $K_c$ is the ratio of the rate constants, so only variables that affect the rate constants can affect $K_c$. Pressure doesn't show in any of these relationships.

$$K_c=\frac{k_\text{forward}}{k_\text{reverse}}=\frac{[\text{products}]^y}{[\text{reactants}]^x}$$

$$\text{rate}_\text{forward}=k_\text{forward}[\text{reactants}]^x$$

$$\text{rate}_\text{reverse}=k_\text{reverse}[\text{products}]^y$$

$$k=A\mathrm e^{-\frac{E_\mathrm a}{RT}}$$

It may also be useful to think about different ways pressure can be changed.

The volume of the reaction can be changed. This will affect concentrations of reactants and products, and the reaction will likely have to shift left or right to reestablish equilibrium. If volume were decreased, the rates of the forward and reverse reaction will each increase due to higher concentration. If the exponents in the rate laws are different, the amount of increase for each will be different. Although the rates change, the rate constants do not, so $K_c$ doesn't change either.
The temperature can be changed. This will affect the rate constants of the forward and reverse reactions according to the Arrhenius equation. An increase in temperature will increase $k_\text{forward}$ and $k_\text{reverse}$, but by different amounts because their activation energies are rarely equal. This changes $K_c$, which is the ratio of the two rate constants.
More gas can be added without changing the volume of the container.

a. If the gas is one of the reactants or products, this would affect one of the concentrations, and the reaction will have to shift to reestablish equilibrium. Again, this changes one of the rates, but does not affect the rate constants, so $K_c$ is unaffected.

b. If the gas is inert and doesn't react with anything involved in the equilibrium, the concentration of each reactant is unchanged. Neither the rates nor the rate constants are affected, so the system should still be at equilibrium and $K_c$ is unaffected.


What is the mechanism of APTES mono-layer formation on glass substrates? The reagent APTES is a fairly common "ink" for microcontact printing, a technique that forms covalent bonds between the silanols found on the surface of the glass and the silane in the the APTES. It's also been demonstrated that the mono-layer then polymerizes somewhat, forming bonds between neighboring silanols.

What is the mechanism by which the printing and polymerization occur?
What is the reason for using APTES instead of a more reactive trichlorosilane?

Here's two links (1, 2) that show the simple scheme on forming monolayers. The essential figure is also shown here.

For the second part of your question... I can only point out this article that describes which properties change, if you make the molecules more bulky or add chloride, etc. I would guess that APTES has well known properties and they are "neutral" regarding common applications.