Synthesis of Hydrogen Fluoride At almost 300°C, $\ce{CaF2}$ reacts with sulfuric acid and this reaction produces $\ce{HF}$. Is there any way to make a relatively dilute (10-20%) hydrofluoric acid with $\ce{CaF2}$ as the source of fluoride ions without that much amount of heat?

In desperation, for some unknown reason, I need 15% aqueous $\ce{HF}$. What do I really need? A weak solution of $\ce{H+}$, say about pH 2. And a solution of fluoride ions to wreak havoc on some unsuspecting metal or glass(Ooops, I may have given away the secret).

My subconscious suggests that I could get them from two different sources, like $\ce{NaF}$ and $\ce{HCl}$. Yeah, that'll work.

Maybe. Maybe even with $\ce{CaF2}$(The use of $\ce{H2SO4}$ with the calcium salt suggests the reason was to precipitate the $\ce{CaSO4}$ and flash off the $\ce{HF}$, thereby separating, not just generating, the $\ce{HF}$).


Shielding effect of d electrons In transition elements shielding effect is observed . Due to this zinc shows abnormality in atomic size.
But at the same time we also say that gallium has same similar size as  aluminium due to poor shielding effect of d electrons . Why is it that d electrons have good shielding effect in case of zinc but poor shielding effect in case of gallium?

When we are talking about zinc... We have to think about period trends. So in periods along moving from left to right covalent radius decreases. But in case of transition metal..

Two effects works first is $Z_{eff}$. Another effect is electronic repulsion.At the starting the no of electron in $d-$ orbital are less so $Z_{eff}$ is dominant over electronic repulsion but as we are moving alosng d-orbital electronic effect become dominant factor over $Z_{eff}$, So in the middle redius become almost constant and at the end it increases.

When we are talking about gallium and Aluminium radius comparision thats mean we are talking about group trends ,So in group trend abnormal size is due to poor shielding effect and relativistic effect. Normal trend on going down the size increases from aluminium to gallium size decreases because here a poor shielding of 3-d orbital present so $Z_{eff}$ penetarate and valence shell contract so size decreases.


Bond lengths: 1-D particle-in-a-box for 1,3-butadiene When estimating the length for a one-dimensional particle-in-a-box model for a hydrocarbon (such as 1,3-butadiene), do we include the bond lengths between the (terminal) carbon and hydrogen atoms at the ends of the chain?

As porphyrin said in his comment, the $\ce{C-H}$ bond is not part of the delocalized $\pi$ system that makes up the one dimensional box. This is partially because hydrogen does not have a p orbital to contribute to the system. The $\pi$ system that creates the box is the result of overlap between p orbitals found on adjacent carbon atoms, meaning that you would be able to approximate the length of the box by summing up the bond lengths between all adjacent carbons in the delocalized system. While resonance means that there are not actually alternating double and single bonds across the $\pi$ system, all of the $C-C$ bonds will be the same length, which should be close to the average of the bond lengths for the most stable resonance contributor.


Can Isopropyl β-D-1-thiogalactopyranoside be degraded into either 1-deoxy-galactose or just galactose? This may be a silly question, but is it feasible that Isopropyl β-D-1-thiogalactopyranoside (IPTG) when acted upon by a beta-galactosidase could produce 1-deoxy-galactose or just galactose?

The reason I am asking is because I have a crystal of a beta-galactosidase that was exposed to 5mM IPTG for ~6 months at 4 degrees C for inhibitor studies.I have X-ray crystallographic data (Beta-galactosidase + IPTG) to ~1.8 angstroms and can easily place the galactosyl ring into my electron density, however, I cannot place the sulfur atom or the isopropyl group.

This would normally indicate that the sulfur and isopropyl atoms are disordered, however, the possible conformations are sterically restricted to one direction.

As such I am wondering if IPTG could, in fact, have been turned into either 1-deoxy galactose or galactose?

I'm not sure if I understood your question correctly, and my thought is based on this assumed reaction:

Your indication of a resolution of about 1.8 Å reads like the data were collected by Cu K$\alpha$ radiation, rather than by Mo K$\alpha$.  If sulfur was split off from your molecule, than (even more with softer Cu than Mo X-rays) this should have decreased a lot the contribution of anomalous scattering in the X-ray data, as CHO's atomic scattering factors -- in comparsion -- are mainly described by $f'$, and not so much about $f''$. What about looking on value .and. esd of the Flack parameter of your model for your new product, compared to a "typical / reference" model of IPTG?
(By the way, Howard Flack passed away just at the beginning of this year.)


Calculation of fluorescence quantum yield The fluorescence quantum yield of a sample $\Phi$ is defined as the fraction of photons emitted per number of photons absorbed. The quantum yield is often calculated by comparing a sample with a reference substance of known quantum yield $\Phi_\mathrm{r}$ such that
$$
\Phi=\Phi_\mathrm{r}\times\frac{n^2}{n_\mathrm{r}^2}\times\frac{I}{I_\mathrm{r}}\times\frac{A_\mathrm{r}}{A},
$$
where $n$ ($n_\mathrm{r}$) is the refractive index of the sample (reference), $I$ ($I_\mathrm{r}$) is the integrated emission of the sample (reference), and $A$ ($A_\mathrm{r}$) is the absorbance of the sample (reference) at the excitation wavelength (see here or here for details).

It seems odd to use the integrated emission because it is not proportional to the number of photons emitted unless the emission peak is extremely narrow. In particular, the integrated emission is
$$
I=\int_{\lambda_1}^{\lambda_2}\mathrm{d}\lambda\,I_\lambda,
$$
where $I_\lambda$ is the spectral intensity. Because the number of photons for a given energy changes with wavelength ($E=hc/\lambda$) using a quantity that is proportional to the number of photons seems more sensible. In particular,
$$
I_\lambda= n_\lambda \frac{hc}{\lambda},
$$
where $n_\lambda$ is the photon density at wavelength $\lambda$. Consider the quantity
$$
N=\int_{\lambda_1}^{\lambda_2}\mathrm{d}\lambda\, \frac{I_\lambda \lambda}{hc}.
$$
It is directly proportional to the number of photons even if the emission peak is broad. Why is $I$ used instead of $N$?

After some more reading, it turns out that the protocols suggesting to use the integrated intensity are wrong and it is necessary to correct for the variable wavelength across the emission peak. Würth et al. (2013) provide a description of the correct method. First, they state that

  all fluorescence spectra have to be converted to a (relative) number of photons per unit time

Second, they describe the conversion as discussed in the question:

  the radiometric quantity has to be multiplied with $\lambda/(hc_0)$


Does an "excited atom" lose energy gradually? When an electron absorbs energy and is in a higher energy orbit (I guess the atom would be in an unstable state), when the electron releases this energy, would all of this energy be released at the same time, or could it release energy one "level" at a time?

So for example if a Hydrogen atom absorbed a lot of energy and its electron went to some level n = 5, could it slowly make its way back down to level 1?

Yes, it certainly can. In fact, it can do either, traverse gradually across excitation states down to its ground state, or jump from its current state to its ground state, or to any lower state for that matter. So essentially, as long as the target state is equal or lower in energy than the current one the electron can jump to it.

Those jumps have different probabilities though. Determining which process is more likely to occur, and what the probabilities are is a matter of quantum mechanic calculations. Have a look here for example: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/fermi.html


What do the dots mean in HN.CO.CH3 connected to the second carbon of N- Acetylglucosamine? And what do the N and acetyl indicate? What do the dots mean in HN.CO.CH3 connected to the second carbon of N- Acetylglucosamine? And what do the N and acetyl indicate?

These are not dots. They actually are single bonds though I do not know why they are printed that way. As such, they should be represented by a half-long dash (called EN DASH in Unicode).

In the name N-acetylglucosamine, acetyl means that a hydrogen atom has been replaced an acetyl group (i.e. –CO–CH3) and that the replacement occurred on the nitrogen (aka "N") atom.

Overall, it means something like: "consider glucosamine but replace one hydrogen on the nitrogen atom by an acetyl group". This is the standard "substitutive naming" defined by the IUPAC.


Reaction involving opening of epoxides The following question was asked in an exam today:

I chose the option (a), but the correct answer given was (b). Following is the explanation why I chose (b):

Step 1 (Reaction with $\ce{PBr_3}$): $\ce{Br}$ substitutes $\ce{OH}$. Hence, the product is $\ce{CH(CH_3)_2Br}$
Step 2 (Reaction with $\ce{Mg}$ in the presence of dry ether): $\ce{Mg}$ gets inserted between $\ce{CH(CH_3)_2}$ and $\ce{Br}$. Hence, the product is $\ce{CH(CH_3)_2MgBr}$
Step 3 (Opening of epoxide and attacking of grignard reagent): First, the $\ce{CH(CH_3)_2MgBr}$ will break into $\ce{CH(CH_3)_2^-}$ and $\ce{MgBr^+}$. Then, $\ce{MgBr^+}$ will attack $\ce{O}$ of the epoxide. After that, $\ce{O}$ will take its electron away from 2nd $\ce{C}$ (referring to the catalyst used) due to the greater stability of secondary carbocation.
Further, $\ce{CH(CH_3)_2^-}$ will attack the carbocation and hence form the product $\ce{CH(CH_3)_2-CH(CH_3)-CH_2-OMgBr}$
Final step (Hydrolysis): On hydrolysis, $\ce{CH(CH_3)_2-CH(CH_3)-CH_2-OH}$ will be our product. Phew!

Where I am wrong and what should be the correct mechanism?

As you are dealing with Grignard reagents, the carbocation argument is inadequate since the reaction is in basic media (carbocations are brønsted acids, so they would be easily deprotonated to produce an alkene). In fact, this mechanism is concerted (everything happens simultaneously, not in 2 distinct steps such as in the formation of a carbocation followed by a nucleophilic attack).

The Grignard reagent could attack either the methylene $\ce{(CH_2)}$ carbon or the methylidyne $\ce{(CH(CH_3))}$ carbon. The methylene carbon is attacked preferentially because its substituents are smaller than the methylidyne ones (2 hydrogens against one hydrogen and a methyl group). So steric hindrance is the key factor for predicting the product correctly in this question, not carbocation stability.

ps.: I drew a specific enantiomer just for visualization, but in this case, the absolute configuration doesn't matter.


Detectable elements using flame AES I've been looking for extra information concerning flame atomic emission spectroscopy. I need a clear overview of which elements are detectable with this technique, but I can't find anything because flame AES seems to be quite outdated.
If there is no such overview, could there be a list of elements not detectable by flame AES, but detectable by ICP-AES? That would also work for me.

but I can't find anything because flame AES seems to be quite outdated.

Please don't say that you cannot find anything, this is the reason (i) why physical libraries still exist and (ii) right key words are needed for proper literature search and (iii) not everything is available on the web free of cost.
Most of the metals are detectable by flame atomic emission spectroscopy. All you need is the right flame. The instrument, called the atomic absorption spectrophotometer, works perfectly in atomic emission mode. There is no separate instrument for Flame-AES.
The flames are quite "cold" for non-metals like sulfur, carbon, nitrogen, and noble gases. ICP also works poorly for non-metals.


Is a gas leak considered incomplete combustion? This may be a dumb question, but if I leave the stove-top valve open (or I have a gas leak due broken tubing), would that be considered incomplete combustion?

From my knowledge, I figured a reaction is necessary for combustion to occur and that just the leaking of a gas would not qualify.

In climate science we call these leaks fugitive emissions. The chemical composition of the emissions are different. If you just have a leak, the emissions will consist of what ever your fuel is made out of, probably methane. If you burn your fuel and the combustion is not perfect you might sill have a littel bit of the original fuel left but most will have been used up and formed CO (incomplete combustion), $\ce{CO_2}$ (complete combustion) and $\ce{H_2O}$ .


Why do acids favor the oxidizer? And bases the reducer? Simply put, why does adding acid to the cathode up the cell voltage? From what I understand, acids are proton donors. Do these positive protons attract electrons toward the cathode compartment of the electrochemical cell?

The effect that can be observed by adding acids to a solution to increase conductivity is known as the Grotthuss mechanism. Protons have a much higher mobility in solution ($3.62\cdot10^{-3}~\mathrm{cm^2~V^{-1}~s^{-1}}$) than any other cation, basically allowing it to tunnel through a solution. (A similar effect can be noted for hydroxyl anions.)
Basically this allows a much faster charge exchange through hydrogen bond networks, also resulting in a higher cell voltage.

(But then again I am not really sure I understood your question correctly.)


Why is a volumetric flask shaped the way it is? A volumetric flask does not look like an Erlenmeyer flask, but has more of a pear shape. Is it just to distinguish more easily between them while they are on the shelf?

A volumetric flask is designed to store a relatively great volume of liquid with great accuracy. If you want to reach a determinate volume of liquid in a large backer when you add liquid to it so the liquid height increases a little bit the volume increase a lot. What you measure with your eyes for determining the volume is the height of the liquid, so the error in the volume measure is directly linked with the error in measuring the liquid height. The relation between the increase of the height ($\Delta h$) and the volume for a cylinder is very simple:

$$ V= \pi r^{2} \Delta h  $$

So suppose that you make a systematic error $\Delta h$ of 0.5 mm, that is quite a good esteem of average max human accuracy in measuring the liquid height in a cylinder with naked eyes. If you make this error in glassware with different radius you will change the accuracy of your measurements, according to the previous equation:

So if you use ten times greater radius you will have a 100 times greater error!

Why the body has not the conical shape of the Erlenmeyer?

This actually is not so important. Generally, the spherical surface more evenly distributes stress across its surface and is less prone to fractures. This is why we use round-bottom flasks. It doesn't seem there are a lot of differences in the production of both Erlenmeyer and volumetric flask, so it's probably mainly due to a convention. In fact, you can find also volumetric flask with Erlenmeyer style. Like in the image below.


Nitration of N-phenylbenzamide Why during nitration of N-phenylbenzamide the $\ce{NO2}$ group is directed at para-position of the ring attached to the nitrogen atom?
Okay, I understand that this is due to the para directing effect of the $\ce{>NH}$ group. But carbonyl $\ce{>C=O}$ group has a meta directing effect, then why is $\ce{NO2}$ not directed to the meta-position of the benzene ring attached to carbonyl group?

In the following reaction

(1) $\ce{NO2}$ group at meta position w.r.t. Ring 2
(2) $\ce{NO2}$ group at para position w.r.t. Ring 1
(3) $\ce{NO2}$ group at para position w.r.t. Ring 2
(4) $\ce{NO2}$ group at meta position w.r.t. Ring 1

You should not be looking at $\ce{-NH-CO\bond{-}}$ as two different functional groups i.e. as $\ce{-NH-}$ and $\ce{-CO-}$.

Since as in $\ce{-(NH-CO)\bond{-}}$: $\ce{N}$'s lone pair are actually in resonance with the carbonyl group.

The left handed phenyl ring is more activated than the right handed phenyl ring. Hence the electrophile would preferentially attack on the left handed phenyl ring.

The amide group is too weak base to be protonated by the acid and therefore it does not direct the electrophile to meta position.

Ortho position is hindered due to steric hindrance of the another phenyl ring.

And therefore

   the attack takes place at para position of the left handed phenyl ring

so answer should be option B.


What is the Lewis structure for HClO? In many sites, I see that they put $\ce{O}$ in the middle, but our teacher said that we should put $\ce{O}$ near the extremity. Which is the best way?

I completely agree with Gonçalo Justino's answer, but would like to add another shade to it.
In principle there are quite a few possible arrangements thinkable for this molecule - at least in the gas phase. In this thought it simply refers to the fact that they are local minima on the potential energy surface. The most stable of all is the one that has been discussed here before. Hypochlorous acid is also one of the compounds that disrupt the ozone cycle. Because concentrations in the stratosphere are low, we can find most possible arrangements of the three atoms. It should also be noted, that these reactions usually occur via a radical pathway.
Structure of $\ce{HOCl}$
The bonding situation in these "weird" compounds is not obvious at all, classifying it as a purely covalent bound is equally as wrong as calling it ionic.
Interestingly the $\ce{O-H}$ bond is much more covalent than the $\ce{O-Cl}$ bond, when we look at the most stable, i.e. bent, conformation. I am not going into detail about possible linear intermediates, as they are only very short lived species.
The analysis of a DF-BP86/def2-SVP calculation with the quantum theory of atoms in molecules (QTAIM) shows us this somewhat unexpected behaviour.

What we see here is that the bond between the hydrogen and the oxygen is strongly covalent as the electron density is $\rho_\mathrm{BCP}(\ce{O-H})=0.34~\mathrm{e\,a.\!u.^{-3}}$ and the Laplacian has a high negative value, $\nabla^2\rho_\mathrm{BCP}(\ce{O-H})=-2.02$, i.e. much charge accumulation. On the other hand the chlorine oxygen bond has quite some ionic character. The electron density is $\rho_\mathrm{BCP}(\ce{O-Cl})=0.18~\mathrm{e\,a.\!u.^{-3}}$, while the Laplacian is slightly positive, $\nabla^2\rho_\mathrm{BCP}(\ce{O-Cl})=+0.01$, i.e. low charge depletion. This makes this bond about half ionic and half covalent. (An ideal ionic bond would have an electron density close to zero and a much larger positive Laplacian value.)
From this analysis it makes sense, that in the gas phase the molecule more likely dissociates via $$\ce{HOCl ->[h\nu] HO + Cl}.$$
In aqueous solution there are quite a few more equilibria at work and the reaction and stability of this compound is highly pH dependent. In general this compound will decompose whenever it gets the chance.
Structure of $\ce{OClH}$
This structure is about $61.5~\mathrm{kcal\,mol^{-1}}$ higher in energy and therefore quite unstable. But that does not mean it does not exist.
The interesting point here is that both bonds are about the same strength and both can be considered predominantly covalent. The oxygen chlorine bond, $\rho_\mathrm{BCP}(\ce{O-Cl})=0.25~\mathrm{e\,a.\!u.^{-3}}$, has a little higher electron density than the chlorine hydrogen bond, $\rho_\mathrm{BCP}(\ce{Cl-H})=0.21~\mathrm{e\,a.\!u.^{-3}}$. Both have a negative Laplacian, $\nabla^2\rho_\mathrm{BCP}(\ce{O-Cl})=-0.19$, $\nabla^2\rho_\mathrm{BCP}(\ce{Cl-H})=-0.57$, i.e. charge accumulation.

Other possible structures of $\ce{HOCl}$
Any other structure, for example linear or bent $\ce{OHCl}$, are no minima on the potential energy surface. They might exist as excited or transition states, but they are incredibly short lived. One only has to care about these species when dealing with complex gas phase reactions. (There is now way they could be stable in solution.)
Conclusion
While it is theoretically possible to have the oxygen in a terminal position, it is certainly not the ground state. Another aspect to think about is also, how accurately can a Lewis structure represent the molecule. Especially when it comes to small molecules one always has to take it with a grain of salt as most of the theories brake down at this level.
For up to highschool level of chemistry, however, the answer to this question is the one given by so many others before.

If that did not bore you to death, might I suggest additional literature: Kirk A. Peterson, Sergei Skokov and Joel M. Bowman; J. Chem. Phys. 1999, 111, 7446. (mirror)


Salt analysis: Position of lead ion Why is $\ce{Pb^2+}$ present both in Group I and II of the basic radicals?
Is it related to the solubity product of its salts?

Yes, the Ksp for $\ce{PbCl2}$ is ~$10^{-5}$ while for $\ce{PbS}$ is ~$10^{-28}$

So if the concentration of $\ce{Pb^2+}$ is too low to detect by dilute $\ce{HCl}$ precipitation, it is still possible it could be detected as the sulfide precipitate.


Are there contradictions about intermolecular forces in the postulates of kinetic theory? Regarding the postulates of the kinetic theory, in one place I read:

  There are no intermolecular forces except during the collision between molecules.

And yet in another place this appears:

  Collisions between two molecules do not exert unattractive or repulsive forces.

Aren't these statements contradictory? Are there or are there not intermolecular forces during the collision, according to the postulates of kinetic theory?

The kinetic theory of gases is, empirically, a good approximation to how most gases behave most of the time. Its key assumptions are:

there are no attractive forces between the molecules or atoms that make up the gas
the molecules or atoms interact only by elastic collision
the molecules or atoms have negligible size

The two statements you have are (especially the second) are poorly worded versions of these assumptions. For the collisions to be elastic there have to be repulsive forces between the entities that make up the gas.

But don't forget that the assumptions are not strictly true, they are just good approximations under most conditions. Molecules are not infinitely small and attractive forces do exist. But, at least when most gases are far from the point where they liquefy, the approximation is good and it explains the way gases behave. For example, a given quantity of a gas (specifically the number of molecules or moles) always takes up the same volume at the same temperature and external pressure however big or heavy the component molecule or atom is. We observe this to be true (approximately).

The apparent contradiction you worry about is due to a careless way of stating the key assumptions and is not a problem with the (well validated) theory.


Raman Spectrum of carbon chains: How to identify the vibrational modes So I am using NWChem to calculate the raman spectra of C2H2 and C4H4. For example for the first I obtained the following frequencies:

#  freq [ 1/cm]  S [Ang**4/amu]
------------------------------------------------------------
  1   644.63        0.0087
  2   679.18        0.0076
  3   792.83        0.0001
  4  2146.53        8.0917
  5  3506.05        0.0007
  6  3645.57       37.8189

and for example, when comparing with the expected results from http://www.chem.purdue.edu/gchelp/vibs/c2h2.html , I could say that my mode number four corresponds to C-C stretching.
So now, when I perform the same calculation for C4H4, I get the following:

#  freq [ 1/cm]  S [Ang**4/amu]
------------------------------------------------------------
  1   216.77        2.3873
  2   305.29        0.7775
  3   548.05        0.8095
  4   643.94        1.7042
  5   677.47        0.5218
  6   700.29        6.8906
  7   893.88        0.1890
  8   928.98        0.5379
  9  1033.68        0.1554
  10  1134.09        1.5802
 11  1357.84       16.9774
 12  1504.93       31.2415
 13  1715.33       17.1094
 14  2269.08       38.4659
 15  3293.55       19.5975
 16  3313.05       78.1132
 17  3449.94       44.9757
 18  3578.50       24.6781

My question is this, how could I identify the vibrational mode corresponding to C-C stretching now?
And if I performed the same calculation for C6H6, how could I identify the C-C vibrational mode?
Is there a way of identifying what each mode corresponds to? not only in calculation but also in experiments?

thank you

When you perform a IR or Raman calculation, the primary step is the diagonalization of the mass-weighted Hessian:

$$
\begin{align}
H_{ij,ab}^{\mathrm{mw}} &= \frac{1}{\sqrt{m_{i}m_{j}}} \frac{\partial^{2} E}{\partial a_{i} \partial b_{j}} \\
\mathbf{Hq} &= \lambda \mathbf{q}
\end{align}
$$

where $H_{ij,ab}^{\mathrm{mw}}$ is a $3N\times3N$ Hermitian matrix, $i,j$ are indices running over the atoms, $m_{i}$ is the mass of atom $i$, and $a,b \in x,y,z$ (the $3N$ Cartesian molecular coordinates). The eigenvalue $\lambda_{i}$, multiplied by some conversion factors, is a frequency in the harmonic oscillator approximation, and the corresponding eigenvector $\vec{q}_{i}$ is the normal mode, the collective atomic displacements, for that frequency.

(A calculation of Raman intensities involves extra work, namely the calculation of $\frac{\partial^{3} E}{\partial \varepsilon_{\alpha} \partial \varepsilon_{\beta} \partial a_{i}}$, but the normal modes are identical.)

Aside from performing literature searches for normal mode frequency assignments on related systems, a good alternative is the visualization of these normal modes. Many outputs from frequency calculations can be opened in Avogadro:

Clicking on a frequency in the right-hand side pane (only IR intensities are displayed) can display both the displacement vectors and an animation of the displacement (see the lower right corner box). Avogadro is able to read this portion of the NWChem output file:

          -------------------------------------------------
          NORMAL MODE EIGENVECTORS IN CARTESIAN COORDINATES
          -------------------------------------------------
             (Projected Frequencies expressed in cm-1)
...
                    7           8           9          10          11          12

 P.Frequency      252.66      366.45      606.76      773.89      790.45      808.06

           1     0.06344     0.00000    -0.02127     0.00000     0.00000     0.00096
           2    -0.10753     0.00000    -0.09540     0.00000     0.00000    -0.00247
           3     0.00000    -0.03639     0.00000    -0.01614     0.01014     0.00000
           4    -0.08820     0.00000    -0.09326     0.00000     0.00000     0.00127
           5     0.02291     0.00000    -0.01064     0.00000     0.00000    -0.00238
           6     0.00000     0.00615     0.00000     0.14073    -0.07465     0.00000
           7    -0.13091     0.00000     0.17533     0.00000     0.00000     0.05512
           8     0.01443     0.00000     0.09744     0.00000     0.00000     0.01777
           9     0.00000     0.19419     0.00000    -0.08894     0.11256     0.00000
          10     0.11294     0.00000    -0.04324     0.00000     0.00000    -0.12834
          11     0.07728     0.00000     0.04956     0.00000     0.00000    -0.02839
          12     0.00000    -0.10132     0.00000    -0.04711    -0.12489     0.00000
          13     0.29675     0.00000    -0.23722     0.00000     0.00000     0.83429
          14     0.12532     0.00000     0.00136     0.00000     0.00000     0.22209
          15     0.00000    -0.33078     0.00000     0.53191     0.69392     0.00000
          16     0.24449     0.00000     0.20104     0.00000     0.00000     0.01418
          17    -0.04794     0.00000    -0.02443     0.00000     0.00000     0.00175
          18     0.00000    -0.37449     0.00000    -0.49495     0.26769     0.00000
          19     0.03319     0.00000    -0.05962     0.00000     0.00000    -0.00066
          20    -0.29762     0.00000    -0.33571     0.00000     0.00000    -0.01410
          21     0.00000     0.25757     0.00000     0.27323    -0.15332     0.00000
          22    -0.06568     0.00000    -0.11340     0.00000     0.00000    -0.00250
          23     0.13575     0.00000    -0.12891     0.00000     0.00000    -0.02550
          24     0.00000    -0.29795     0.00000    -0.17367     0.10673     0.00000
...

So, to find the C-C stretching mode (or any vibration of interest), clicking through the list in Avogadro is more convenient than eyeballing the displacements in the output file. With the advent of accurate computational methods, the assignment of peaks in experimental spectra is usually aided by calculations identical to these. It's even possible to plot the computed spectra with some artificial broadening:


Could this be a feasible fuel system? First off, please forgive me, I am a chemistry novice!

Overall Process

Fuel for the system is distilled water
Split water into Hydrogen and Oxygen (electrolysis or similar process)
Exhaust Oxygen
Burn Hydrogen to power a generator/engine
Generator/engine charges battery
Battery powers Step 2 as well as whatever else we want to power

Problems

Energy gained needs to be greater than the sum of lost energy in all steps, as well as the energy required to perform electrolysis
We could compress the exhaust oxygen and use that as an energy source?
Is there a danger of running out of hydrogen? (Forgive my naivety, but I've heard about a helium shortage...)

"Energy gained needs to be greater than the sum of lost energy in all steps"

http://www.lhup.edu/~dsimanek/museum/unwork.htm
 Perpetual motion machines
http://news.sciencemag.org/chemistry/2014/01/rhubarb-battery-could-store-energy-future
 Flow battery, which works.

1) You cannot win (First Law of thermodynamics from Noether's theorems and the isotropy of time)
2) You can only break even on a very cold day (Second Law of thermodynamics from Carnot cycle and the law of large numbers).
3) It never gets that cold (Third Law of thermodynamics.   If there were an entropy difference at absolute zero, T = 0 K could be reached in a finite number of steps. However, at T = 0 K there is no entropy difference, so an infinite number of steps are needed).


Does NaCl reduce the surface tension of water? Does NaCl reduce the surface tension of water? And why?

For example soap does it, and I was wondering if NaCl does the same.

Gibbs adsorption isotherm tells us that (roughly speaking) the decrease in surface tension is proportional to the surface excess concentration of the solute.

Positive surface excess means that the solute tends to concentrate on the surface rather than stay in the solution. The surface concentration has no physical limit and can go arbitrarily high. In soap and other surfactants, it is hundreds of times higher than the concentration in the solution, therefore their effect on surface tension is negative and very strong.

Negative surface excess means that the compound prefers to stay in the solution rather than on the surface. It has a physical limit, though: the concentration can't be lower than 0. Therefore the effect of such compounds on surface tension is positive, but small (not noticeable with naked eye). This is the case with $\ce{NaCl}$ and pretty much all "normal" inorganic salts.


E1 in 6 membered rings According to my lecturer, if the geometry prevents planarization of the C-centre than an E1 is impossible. As chair and boat are the only stable conformations of a 6 membered ring of carbons, does this mean that such a ring can never undergo an E1?

Furthermore, why must the C-centre be planar for E1? I guess it allows for maximal stabilization by hyperconjugation?

Well, the answer to your question, without any complications, is pretty simple

Whatever can be forced into planarity in order to bear the $sp^2$ carbocation will undergo $E1$, this can be easily achieved in cyclohexane, as it can easily have a twisted conformation which is relatively less stable, but yet can bear the carbocation. Same is not the case with bridgehead carbons.


Is there an intramolecular hydrogen bond in guaiacol? Is there a hydrogen bond between the $\ce{H}$ atom of $\ce{-OH}$ group and the $\ce{O}$ atom of the $\ce{-OCH3}$ group in ortho-methoxyphenol? I was told no. But I want to know why.

According to a quick calculation on the DF-BP86/def2-SVP level of theory, there is an intramolecular hydrogen bond with a strength of about $\ce{-26.6 kJ/mol}$ according to a quantum theory of atoms in molecule (QTAIM) analysis.

For an explanation of the evaluation of the bond strength see: E. Espinosa, E. Molins, C. Lecomte. Chem. Phys. Lett. 1998, 285, 170-173.
Analysis done with MultiWFN 3.3.8. Calculations with Gaussian 09 D.01.


Why is the most probable KE for a monoatomic ideal gas exactly 1/3 of its average KE? The most probable value of the kinetic energy for a monoatomic ideal gas can be calculated by setting the first derivative of the probablity density function equal to $0$, this turns out to be $\frac{1}{2}kT$ which is $\frac{1}{3}$ of the average value, $\frac{3}{2}kT$ what is the physical reason behind this?

There is no physical reason. It's bare mathematics. The probability distribution in terms of energy is asymmetric, for it is bounded by 0 from below and not bounded at all from above. Asymmetric distributions are known to have (generally speaking) different values for mean, mode, and median.
Think of the exponential distribution: its average value is some positive number, but the mode (the most probable value) is $\bf 0$.

Moreover, 3 is not even some intrinsic characteristic of the system. If you rewrite the distribution in terms of velocity (the way it is commonly done), you'll have an altogether different situation. All of a sudden, the ratio of mean to mode is... $2\over\sqrt\pi$, how would you like that?


How to calculate the standard Gibbs free energy of formation of bromine vapor at room temperature? The normal boiling point of liquid bromine is $\pu{58.2 ^\circ C}$. At $\pu{9.3 ^\circ C}$ the equilibrium vapor pressure of liquid bromine is $\pu{100 torr}$.
  From this data, calculate the standard state Gibbs energy of formation of bromine vapor at room temperature, $\Delta G ^\circ_\mathrm{f}$, $\pu{298 K}$.

I tried using Clausius Clapeyron equation and get $\Delta H$ but I can't manage to solve this problem.

I just used Clapeyron equation to get around $\pu{32 kJ/mol}$ but I can't really manage to get Gibbs free energy at standard state.

If you calculate the equilibrium vapor pressure at 298K (25 C), you can calculate the change in Gibbs free energy of the vapor in going from this pressure to the hypothetical state of 1 bar and 298K using $dG=VdP=RTd\ln P$.  Neglecting the Poynting correction, the free energy of the liquid at 298 and at the equilibrium vapor pressure is 0.  And this is also the free energy of the vapor at 298 and at the equilibrium vapor pressure.


What is the definition of a "Simple enzyme" exactly? Is it an enzyme that does not require a co-factor? I'm having a hard time making a distinction between simple and complex enzymes. the definitions I've found so far all say this:

  Simple enzymes are enzymes that are only made out of protein, and complex enzymes are ones that, in addition to the protein portion, have a non-protein portion as well, which is called a co-enzyme/prosthetic group (I've seen both terms being used)

Does this mean that any protein that utilizes a co-factor, even if that co-factor is simply a co-enzyme or an activating metal, i.e., not an actual part of the enzyme's structure, is considered a complex enzyme? Or are the two terms restricted to the structure of the enzyme (i.e., whether they have prosthetic groups, be they organic or inorganic, or not)?

What is the definition of a “Simple enzyme” exactly?

This is not a term that is commonly used. There are commonly used terms for an enzyme lacking its cofactor (apoenzyme) and having its cofactor (holoenzyme). Also, the term "enzyme complex" is used for the enzyme bound (non-covalently) to other molecules.

  Does this mean that any protein that utilizes a co-factor, even if that co-factor is simply a co-enzyme or an activating metal, i.e., not an actual part of the enzyme's structure, is considered a complex enzyme? Or are the two terms restricted to the structure of the enzyme (i.e., whether they have prosthetic groups, be they organic or inorganic, or not)?

Yes, just apply the definition you were given. If the metal/co-enzyme/prosthetic factor is absent, the enzyme does not work.


How does a double displacement reaction work? When a double displacement reaction occurs with compound AB and CD, what determines that B and C are going to bond instead of A and C?

Double displacement reactions involve ionic compounds, i.e. salts formed from positive cations and negative anions. Consider the following setup,
$$\ce{\color{blue}{A^{+}}\color{red}{B^{-}} + \color{blue}{C^{+}}\color{red}{D^{-}} -> \color{blue}{A^{+}}\color{red}{D^{-}} + \color{blue}{C^{+}}\color{red}{B^{-}}}$$
where the cations are indicated in blue and anions in red (and all charges are $\pm 1$ to reflect your question). All salts are comprised of cations and anions, hence $\ce{\color{blue}{AC}}$ wouldn't form since both are cations. $\ce{\color{blue}{C}\color{red}{B}}$ would form instead.


Are metals hydrogen bond acceptors or donors? In a carbonyl or $\ce{-OH}$ group, it's clear the oxygen acts as a hydrogen bond acceptor.

Do metals act as hydrogen bond acceptors or donors and if so, how can I tell as what it's acting?

Hydrogen bonds are three centre 4 electron interactions that lower the energy of the molecular orbitals involved. We can hypothetically imagine two scenarios,

Metal donor (M$^{\delta ^{-}}-H^{\delta ^{+}}...C$)
Metal acceptor (M$^{\delta ^{-}}...H^{\delta ^{+}}-C$)

(where C is just something else). Metal-Hydrogen bonds where the metal is a donor are scarce since the metal hydride bond is the opposite polarity of the metal hydrogen bond donor, however they do exist.

Metal acceptors are likely to be low oxidation state late row metals with filled $d$ orbitals to donate via $\sigma$ symmetry interaction.

You get these types of interactions all the time in crystal chemistry where we find bridging hydrogens! So in short, it depends where the metal is in the period, if its $dz^2$ is available to donate etc.

If you had a specific example I would be happy to try and work it out, but it looks like you are aware of any literature I would consult!! :)


Why does sunlight cause plastics to become a faded blue color? Due to the UV absorption, I knew that the plastics can be easily affected, especially PP. But is it normal that the change is always into faded blue from white color? and Why?

Because I made an experiments, but the results made me surprised. One sample of opaque white PP material was put on sunlight, one sample of same material having same conditions was put in lightproof black color bags. Then measurements was conducted every hour till 8 hours by using spectrophotometry which works in CIE Lab color space.

According to the results, b-value of sample exposed to sunlight had changed into the minus value, describing blue color in L a b color space, while L- and a-values are relatively stable. The b-value of another sample in lightproof bags had changed into the plus value, describing the yellow color in L a b color space, while L- and a-values are again relatively stable. So we can see that the effect of sunlight is only on b- value, which describes the values of blueness and yellowness of the sample.

My question is that what is the phenomena behind this?

What do you mean by reflected and transmitted lights? Even looked with naked eyes, easily seen that it becomes slightly faded blue color. The object is opaque white color PP material.

I don't think its actually blue, I think the blue is a result of Rayleigh scattering.

Over time the dyes are destroyed by light removing any color. simultaneously the plastic is degraded and becomes rough on the surface and acquires an inhomogeneous refractive index within the structure due to the destruction of bonds and oxidation and the end of chains. These two structural changes cause scattering, and since blue is scattered more than the other colors due to the relation of scattering with wavelength to the fourth power, the material appears blue when looking at the reflected light.  If you look at transmitted light going through the plastic it should appear yellow.


Best ingredients for a homemade desinfectant I am trying to determine the best recipe/ingredients for a homemade desinfectant/germ killer. I've been researching these as of now:

Isopropanol (or even Ethanol)
White vinegar
Borax
H2O2
Specific Essential oils (that are supposed to kill germs)

What I want is something liquid, that you can either spray on your hands or on a toilet seat for example, and kills the most germs possible while being easy to make/mix. It should not be corrosive/toxic etc.

Which of these products can be mixed together? Is there any other products I didn't research? Will mixing any of these cancel the effects of the other?

Thank you!

H2O2 is caustic and reactive, but of the list above, it would be the most complete germicide. H2O2 is an oxidizing agent. H2O2 compares well with bleach (bleach is arguably, the king of disinfectants). However, typically the usage level of H2O2 is higher (3%) than other disinfectants. You cannot mix bleach with H2O2 (H2O2 reacts rapidly with bleach). Also, you cannot mix bleach with acid, although many people will mix vinegar and bleach in tiny amounts to give it an extra kick (which it will)... this is DANGEROUS. H2O2 can generally be mixed with (most) acids- especially organic acids like citric acid or vinegar.

Essential oils (especially phenolics like clove, cinnamon, and/or thyme oil) are the most germicidal e-oils, but e-oils are kind of like gasoline (plastic dissolving-staining etc.). Other, terpenic e-oils (like peppermint, pine, citronella, lemongrass, etc.) are somewhat germicidal. Limonene and pine oil are often used as a germicide, although it is comparitively weak to most essential oils. However, they is not as caustic or toxic as clove (for example)- although they do have some toxicity. Pine oil, thyme oil, peppermint, and limonene are popular choices for germicidal use. I have seen thymol and limonene mixtures before. Peppermint is common for school mop-products (becuase in schools, its common to clean up vomit, and peppermint helps quell nausea). Cinnamon, thyme and peppermint are common in mouthwash. Usage levels for e-oils is usually much less than 1%... they are not as effective (but still helpful) as other disinfectants at such levels, but the lower usage is balanced between the noxious-toxicity of the oil and the efficacy of the germicidal effect. Since antibiotic e-oil products would need to contain a noxious and cost prohibitive amount of e-oil to be effective (as a stand-alone antimicrobial), most antibiotic e-oil products (pine-sol for example) contain ammonia (high pH), and simply list pine as a scent (it would help kill bacteria- but to claim antimicrobial on a bottle, the product must be EPA registered and there are lots of problems with e-oils... again, cost, toxicity, and efficacy to name a few). You can mix e-oils with soap and alcohol. Other chemicals (acids, bases, and oxidizers) can/will break e-oils down, but not very rapidly. And the break down/byproducts would probably still have some degree of antimicrobial activity.

Vinegar is used by many DIY people, but it is not a popular commercial product because it smells kind of bad, it's not very effective, and if it is mixed with bleach (WARNING) it can cause a serious health hazard. Also, 5% acetic acid is not extremely germicidal (it will work to kill bacteria in food after a few days); many spores will survive for hours in acetic acid. Vinegar is most effective against non-spore forming (gram negative) bacteria- which are often the more nasty kinds of bacteria (Salmonella, E. coli, Serratia, Pseudomona, etc.)... but it's less effective against spore forming (gram positive) bacteria (Candida, Bacillus, Streptococcus, Micrococcus, etc.). Again, it is an effective preservative in food... as is 5% ethanol (ie., beer). 50% acetic acid is much more effective and it works faster... as does 50% ethanol. pH is a factor, so vinegar is a bit more effective than ethanol, but in general, a high pH is much more effective than a low pH. So, ammonia or lye solutions are more effective than organic acids.

70% isopropanol and 60% ethanol are a common hand and wound disinfectants. They are fairly antimicrobial and antiviral, but again, spores (and other gram positive bacteria) can survive in dilute alcohols (for a little while).

Borax is not germicidal (I don't care what the popular opinion on the internet is). The insecticidal efficacy of borates is also relatively poor, but since it doesn't leave or break down, it is effective over time (weeks, years, etc.)... persistence is why it is used as an effecitive termiticide for wood-framed houses. And although borax is claimed to be effective for roaches... mixed with bait and such- I have tested it thouroughly and it's not very good. But more at the subject here- I would not consider borates to be antimicrobial... unless you consider pH to be the antimicrobial agent. In which case, ammonia or lye are extremely antimicrobial (they really are). But you can't mix lye and ammonia for cleaning (DANGER). Also, bleach has a relatively high pH (about 11).

Notably, but not mentioned, SOAP is the originial antimicrobial and insecticide. It works great; pound for pound- it works better than vinegar. It's been used for centuries, and it has very low toxicity. But, since it has very low usage levels (generally less than 1% in water) and it is more for removing bacteria- so it is not purely germicidal. Generally speaking, soap more strongly inhibits gram positive, spore forming bacteria (at say 0.1% concentration) and inhibits most gram negative bacteria at around 4% concentration. Also, soap manufacturers don't need to pay hefty tolls for registering all their products as antimicrobial because people buy soap anyway. But adding a strong antibiotic (usually triclosan, which tends to work better on gram negative bacteria) they can easily skate through EPA with an affordable antimicrobial product (affordable especially compared with the cost of the claim which costs MILLIONS of dollars).

To claim "antibiotic" with EPA, a product must kill most bacteria (I think it's something like 99.9%, minimum, within 5 minutes against several standard/representative microbial species). You really can't do that without something like 60% ethanol, or 0.15% bleach (ie., 3.33% of a 5% bleach solution), or 3% H2O2, or 2% clove oil (ooh that stings!) or 0.1% triclosan in soap (which gets diluted 100x to about 0.001% in common use) or some antibiotic molecule in high enough concentrations to kill.

Some antimicrobial combinations are synergistic, some are antagonistic, some are additive, and some are deleterious. Most of the reliable combinations of over-the-counter antibiotic materials have already been tried and the best products (soap, hydrogen peroxide, alcohol, bleach, ammonia, and yes, even fragrance) have stood the test of capitalism and time.

Here is some good reading about this: Guideline for Disinfection and Sterilization in Healthcare Facilities, 2008


Directionality of ring current-induced magnetic fields in purines What would be the directions of ring current-induced magnetic fields in purine-like heterocycles in a strong magnetic field?
Would be the antiparallel state be a default state for these substances in the absence of strong magnetic field at warm temperatures due to heat motion?

This is an interesting problem, I think it is best for a moment before I get stuck in that we considered a slightly different compound. I will consider it with an external magnetic field applied to it. Napthalene, now napthalene has two aromatic hydrogen environments in its NMR spectrum. These are very similar in chloroform to benzene in terms of chemical shift.

This makes me think that the ring currents go in the same direction in this compound.

It may be better to consider the ring current as flowing along the perimeter of the aromatic system rather than the loops formed by the six membered rings.

It is important when considering aromatic rings to escape being trapped by six membered rings. A lot of aromatic molecules have six membered rings but there are plenty of five and seven membered aromatic rings. It is more important to understand that a pi system which is cyclic and has 4n+2 pi electrons is the defining thing which makes a ring system aromatic.

If napthalene had ring currents going in different directions I would expect that if the interconversion of the sites (by ring current reversal or flipping of the molecule) was slow then I would expect to see four different chemical shifts in the proton spectrum.

If this exchange of sites was fast then I would expect to see two sites which are each the average of the two sites. I would also expect the chemical shifts of the napthelene to be very different to that of benzene.

If we allow ourselves a little diversion for a moment, the cyclopropanes have a rather rare proton chemical shift, this is becuase the C-C bonding in cyclopropane has some pi character and the hydrogen nuclei are above the ring in such a place that the ring current has the opposite effect on the NMR shift to the magnetic field due to the ring current in benzene.

Now if we move onto purine, in case anyone is not sure what the shape of it is then here it is.

If we look at the literature (S2-75) in the supporting literature of Thomas H. Graham, Wensheng Liu, and Dong-Ming Shen, Org. Lett., 2011, 13 (23), pp 6232–6235
DOI: 10.1021/ol2026813. This compound in DMSO has three very similar proton NMR shifts. This suggests that the ring current goes the same way around in both rings. These NMR shifts are all in the aromatic area and are further left (higher ppm) than benzene.


When and how in anaerobic digestion is water dissociated? Anaerobic digestion and methanisation of organic feedstocks is a complex series of processes that result in biogas. One way to look at this is a very top-down, stochiometric approach as in the Buswell equation

$$\ce{C_aH_bO_c + \left($a - \frac{b}{4} - \frac{c}{2}$\right)H2O -> \left($\frac{a}{2} + \frac{b}{8} - \frac{c}{4}$\right)CH4 + \left($\frac{a}{2} - \frac{b}{8} + \frac{c}{4}$\right)CO2}$$

There's also a Buswell-Boyle equation that takes nitrogen and sulfur into account, but right now I just want to understand one aspect.

During the process, at some point, water is dissociated. When does this happen (hydrolysis, acidification, ...), what role does the water play and what are sample reactions that illustrate this?

This is not a trivial question to answer, as many diagrams are often simplified and do not state the full chemical reactions that occur, such as

from Aerobic and Anaerobic Digestion and
Types of Decomposition education webpage from the Mountain Empire Community College.

However, according to this image (below) from the Ecocorp page - Metabolic Chain of the Anaerobic Process for the Utilization of Organic Wastes, water is a reactant in the acetogenesis
 stage and is removed as a product in the methanogenesis stage.

(Sample reactions are included in the image)

(Just in case the image becomes a dead link), the chemical processes involving water from the diagram are:

acetogenesis

$$\ce{CH3CH2COOH + 2H2O -> CH3COOH + CO2 + 3H2}$$

methanogensis

$$\ce{4H2 + CO2 -> CH4 + 2H2O}$$


Can the direction of electron flow in a voltaic cell be reversed? I am wondering if the direction of electron flow in a voltaic cell be reversed such that the cathode becomes the anode and vice versa. I understand this is not possible for two standard half-cells, but what about non-standard half-cells at STP ?

Can going to non-standard conditions reverse the cell potential of a voltaic cell?

I have already answered that question, in the affirmative, for a simple tin and lead galvanic cell, here: https://chemistry.stackexchange.com/a/116734/79678.

What follows is an elaboration with some specific values and illustrative figures. In fig. 1, the tin and lead voltaic cell is shown:

In fig. 1, the tin and lead ion concentrations are each exactly $\pu{1 M}$ and the cell potential is $\pu{+0.014 V}$, with tin as the anode. All necessary equations are shown in the figure and the usual simplifying assumptions are made: unity activity coefficients, zero cell resistance, etc. The electron flow is from $\ce{Sn}$ to $\ce{Pb}$.

In the next figure, the lead ion concentration is $\pu{0.3363 M}$ while the tin ion concentration remains at exactly $\pu{1 M}$. This results in zero cell potential, as shown in fig. 2:

Since the cell potential is zero, it is a moot point to call tin the anode: the cell is not doing (net) electrochemistry. There is no net electron flow.

In the final figure, the lead ion concentration is exactly $\pu{0.1 M}$ while the tin ion concentration remains at exactly $\pu{1 M}$. This results in a cell potential of $\pu{+0.016 V}$, as shown in the figure:

But now lead is the anode and tin is the cathode. The electron flow is from $\ce{Pb}$ to $\ce{Sn}$.

So changing the ion concentration to non-standard values can reverse the electron flow direction. But this was an especially easy case because the two metal electrodes had standard potentials that were relatively close. For the Daniell cell, with $\ce{Zn}$ and $\ce{Cu}$, it would not be feasible.


Hybridization in Light of Resonance According to this video:

http://www.youtube.com/watch?v=ns4Py96Bkn8&NR=1

1) Is this true? (You might recall I asked a similar question about $\ce{sp^2}$ hybridization implying $\ce{sp^2}$ hybridization for all directly linked atoms a while back, but I couldn't remember my prof's exact wording). I think this YouTube video is saying the same thing my prof said.

2) Does this rule imply that both the oxygens in the below diagram are $\ce{sp^2}$ hybridized? Each is connected to an $\ce{sp^2}$ carbon atom. Each has lone pair(s).

3) Even if we didn't know that the rule existed, we should still know that both oxygens are $\ce{sp^2}$ hybridized because of the following resonance possibility, correct?

ETA:

If the above is true then this doesn't make sense:

1) Is this true?

It's more or less true in the cases were hybridization methods \ techniques work.  For example, when the hybridization of water is discussed it is often said that the central oxygen is more or less $\ce{sp^3}$ hybridized and therefor the two lone pairs on oxygen are in (roughly) $\ce{sp^3}$ orbitals.  Current thinking based on photoelectron spectral data suggests that, instead of this traditional view, one lone pair appears to be in an orbital that is predominantly $\ce{P}$ and the other in an orbital that is predominantly $\ce{S}$.  So hybridization models don't always give the correct answer.  I'm not sure what the hybridization of an atom with a lone pair that's attached to a double bond really is.  But it is probably OK to say what the teacher said in an introductory class in order to keep things simple for a while.

  2) Does this rule imply that both the oxygens in the below diagram are
  sp2 hybridized?

Assuming that the traditional hybridization approach is applicable here (e.g. is the ester oxygen behaving like water or is it "traditionally" $\ce{sp^3}$ hybridized?) then yes, both oxygens are roughly $\ce{sp^2}$ hybridized.  Remember, hybridization is not a step function.  There are hybridization indices between 2 and 3.

  3) Even if we didn't know that the rule existed, we should still know
  that both oxygens are sp2 hybridized because of the following
  resonance possibility, correct?

Some data on esters suggests that the $\ce{C-O-C}$ angle is around 115 degrees.  Somewhat between 109 degrees ($\ce{sp^3}$) and 120 degrees ($\ce{sp^2}$).  And maybe an intermediate hybridization is what we should expect based on the resonance structures for an ester.


Contradiction of general rules of bond character The rules the ionic character of metal halides of halogens is given in order $\ce{MF > MCl > MBr > MI}$.

While judging polarisability we say that a compound that has large cation and small anion will have more covalent nature. So here for same cation, anion with small size should have more covalent character but the trend is totally reverse. Is there any explanation to this?

Although atomic bond in a compound such as $\ce{M+X-}$ (e.g., $\ce{Na+Cl-}$) is considered to be 100% ionic, in reality, it also has some covalent character. An explanation for the partial covalent character of an ionic bond has been given by Kazimierz Fajans in 1923 (Ref.1). According to Fajans, if two oppositely charged ions (say $\ce{M+}$ and $\ce{X-}$) are brought together, the nature of the bond between them depends upon the effect of one ion on the other. In inorganic chemistry, these Fajans' rules are used to predict whether a chemical bond will be covalent or ionic, and depend on the charge on the cation and the relative sizes of the cation and anion, summarized in the following table:
$$
\begin{array}{c|c}
\hline
\bf{\text{Ionic}} & \bf{\text{Covalent}} \\
\hline
\text{Low positive charge }(\text{e.g., } \ce{Na+}) & \text{High positive charge }(\text{e.g., } \ce{Al^3+})  \\
\text{Large cation }(\text{e.g., } \ce{K+}) & \text{Small cation }(\text{e.g., } \ce{Li+})  \\
\text{Small anion }(\text{e.g., } \ce{F-}) & \text{Large anion }(\text{e.g., } \ce{I-})\\
\hline
\end{array}
$$
When the two oppositely charged ions, $\ce{M+}$ and $\ce{X-}$, approach each other, the positive ion ($\ce{M+}$) attracts electrons on the outermost shell of the anion ($\ce{X-}$) and repels its ($\ce{X-}$) positively charged nucleus. This results in the distortion, deformation, or polarization of the anion. If the polarization is quite small, the forming bond between them would be more ionic, while that would be more covalent, if the degree of polarization is large.
The power of an ion (cation) to distort the other ion is known as its polarization power and the tendency of an ion (anion) to get polarized by the other ion is known as its polarisability. Based on the (cation’s) polarizing power or (anion’s) polarisability, the formation of covalent bond is favored by the following factors:

Small positive ion (Cation): The smaller the size of cation, the greater is its polarising power. Due to greater concentration of positive charge on a small area, the smaller cation has high polarising power. Thus, the greater will be the covalent nature of its bond with an anion. This explains why $\ce{LiCl}$ is more covalent than $\ce{KCl}$.
Large negative ion (Anion): The larger the size of anion, the greater is its polarizability, i.e. susceptibility to get polarized. It is due to the fact that the outer electrons of a large anion are loosely held and hence can be more easily pulled out by the cation. Thus the greater will be the covalent nature of the bond with a cation. This explains why iodides (e.g., $\ce{Kl}$), among halides (e.g., $\ce{KF}$ and $\ce{KCl}$), are most covalent in nature.
Large charge on either of the two ions: As the charge on an ion increases, the electrostatic attraction of the cation for the outer electrons of the anion also increases. As a result, the cation’s ability for forming the covalent bond increases (see the similarity with second point). For example, the covalency increases in the chlorides of $\ce{Na+, Mg^2+,}$ and $\ce{Al^3+}$ in the order of $\ce{NaCl < MgCl2 < AlCl3}$.
Electronic configuration of the cation:  For the two ions of the same size and charge, one with a pseudo-noble gas configuration (i.e. 18 electrons in outer-most shell) than a cation with noble gas configuration (i.e. 8 electrons in outermost shell) will be more polarizing. Thus copper(I) chloride is more covalent than sodium chloride although $\ce{Cu+}$ ion ($\pu{1.04 \mathring A}$) and $\ce{Na+}$ ion ($\pu{1.05 \mathring A}$) have similar size and charge.

References:

Kazimierz Fajans, “Struktur und Deformation der Elektronenhüllen in ihrer Bedeutung für die chemischen und optischen Eigenschaften anorganischer Verbindungen,” Naturwissenschaften 1923, 11, 165–172 (https://doi.org/10.1007/BF01552365).
Noam Agmon, “Isoelectronic Theory for Cationic Radii,” J. Am. Chem. Soc. 2017, 139(42), 15068–15073 (https://doi.org/10.1021/jacs.7b07882).


A function for getting the nuclide's nuclear binding energy? Is there a function $f(A)$ or $f(Z, A)$ that gets the energy of nuclear binding of one nuclide in atom? Or other way to get it? My chemistry book says only to get it from the chart which is very impractical.

Nuclear physics is very complicated and full of unsolved problems. For example, we seem to understand rather poorly the effects of three-body forces between nucleons, and describing them quantitatively is probably out of the question. Therefore, I expect that nuclei binding energies for most isotopes are solely determined empirically and have no good theoretical curve to originate from. Maybe there are statistically fitted curves to the data we have, though they may well not have a simple form and likely are only good approximations for relatively small intervals.

Edit: As it turns out, equations indeed exist, though as expected they are quite unwieldy and are semi-empirical.


Free Expansion - Ideal Gas vs Real Gas By free expansion, I am referring to gas kept in a piston-cylinder arrangement freely allowed to expand against vacuum.
It is clear to me that free expansion is an irreversible process because if it were not then it could get compressed as well at equilibrium, but we know it doesn't as it would be violating the 2nd law of thermodynamics.
I know that all the parameters work 'W', Change in Internal Energy '∆U', and Heat transfer '∆Q' would all be zero in the case of an ideal gas.
But, what I want to know is how will all the parameters vary if gas is real?
Also, I wanted to ask whether the answer will change if the surface between the piston and cylinder is not frictionless?

For a real gas, the internal energy varies with both temperature and molar volume.  The variation is described by the equation:
$$\mathrm dU=C_V\,\mathrm dT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]\mathrm dV,$$ where $U$ is the internal energy per mole and $V$ is the molar volume.  This means that, unless the equation of state is such that pressure is directly proportional to temperature at constant molar volume (like an ideal gas), the internal energy will depend on molar volume.


Is there a way to fine tune a laser to excite the electrons of a particular molecular bond into dissociating? I don't know if this makes any sense, but I came up with this in my head: if each molecular bond in any molecule had a specific *signature wavelength of dissociation", this would be the exact wavelength of the laser beam necessary to excite the electrons involved in that one particular bond and no other bonds in the molecule into dissociating homolytically. Is this theoretically sound in any way?

Obviously if you have a diatomic then yes, you can put energy in and the bond will break. There are three cases; (a) as long as the photon has enough energy it can excite a dissociative excited state, (grey in sketch) that is a state that is only repulsive. The absorption is continuous here so its energy has to be above a certain value. (b) The second case has a curve crossing in the excited state. This means that two excited states cross , a bound state and a dissociative one. Then if you hit the energy at the crossing point, dissociation will occur, but not at higher or lower energy. This is called pre-dissociation. (The read & grey line crossing show pre-dissociation). It occurs in many molecules not just diatomics, e.g. benzene. (c) The third case is when a bound state is excited above its dissociation limit, e.g. above approx 10000 in the blue curve, 5500 on the black curve . However, reaching these energies may be hard because the Franck-Condon factors may be vanishingly small. (roughly speaking the FC factors ensure vertical transitions)

Now the case with polyatomic molecules is tricky. Many experiments have been tried with varying success, the area is sometimes called 'mode selective chemistry' and/or 'coherent control'.

In a molecules there are many vibrations, called normal modes. Although we think generally of one bond vibrating, in actuality many bonds vibrate in unison (i.e. in a fixed phase relationship) in a normal mode. These modes are governed by the symmetry species of the molecule. Each normal mode is supposed to be orthogonal, so that vibration of one does not effect vibration of another. But in fact vibrational anharmonicities allow vibrational quanta to flow from one mode to another. Thus exciting one normal mode soon involves energy spreading out into all other modes and thus even if there is enough energy to break a bond it soon becomes diluted.

In coherent control experiments one uses one or more very short light pulse ( < 100 femtoseconds) to excite the molecules but control the phase of the light (hence pulse shape) to control how long the molecule takes absorb the photon. This is done in the hope that once the energy starts to spread out (as a vibrational wavepacket) it can be controlled to end up in just the bond you want to break, i.e. to try and use the vibrational spreading out tendency to advantage. The pulse shaping is done using an electro-optic or acousto-optic device in a feedback loop using say, a genetic algorithm linked to a detector that observed product formation.


The nuclear Overhauser effect I am having some trouble with understanding the NOE. The carbon NMR spectra can't be integrated, because the protons attached to the carbon atoms will make primary carbon atom relaxation faster than tertiary, so their peaks will have bigger intensity. With a bigger acquisition time, the peaks will be more proportional. In proton NMR, nearby protons will cause the target proton relaxation to be faster. When NOE is applied, those nearby protons will be irradiated and the target proton peak will have a slightly bigger intensity. Why will it have a bigger intensity?

Consider the simplest possible NOE experiment, with two dipole-coupled protons, the
target I and nearby proton S.

Before the experiment there are four populated spin states: $\alpha_{I}\alpha_{S}$, $\beta_{I}\alpha_{S}$, $\alpha_{I}\beta_{S}$ and $\beta_{I}\beta_{S}$, where $\alpha$ indicates $m_{i} = \frac{1}{2}$ and $\beta$ indicates $m_{i} = -\frac{1}{2}$. The relative populations of the states at thermal equilibrium (from Boltzmann distribution) are:

$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}$,  $\alpha_{I}\beta_{S}: 1$
$\beta_{I}\beta_{S}: 1 - 2\Delta $

S is irradiated so that the population difference between the $m_{S} = \pm \frac{1}{2}$ states goes to zero. This results immediately in the following populations:

$\alpha_{I}\alpha_{S}, \beta_{I}\alpha_{S}: 1 + \Delta $
$\beta_{I}\beta_{S}, \alpha_{I}\beta_{S}: 1 - \Delta $

What happens next typically for small molecules is fast "flip-flip" cross-relaxation, restoring the $\alpha_{I}\alpha_{S}, \beta_{I}\beta_{S} $ states to their equilibrium populations. So we have the populations:

$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}: 1 + \Delta$
$\alpha_{I}\beta_{S}: 1 - \Delta $
$\beta_{I}\beta_{S}: 1 - 2\Delta $

The population difference between the pairs $\alpha_{I}\beta_{S}, \alpha_{I}\alpha_{S}$ and  $\beta_{I}\beta_{S}, \beta_{I}\alpha_{S}$ has now increased by 50%, from $2\Delta$ to $3\Delta$. This corresponds to a 50% increase in intensity of the peak. Transition between these pairs obviously corresponds to spin-lattice relaxation.


Does Gay Lusaac's law work both ways 1) With $n,V$ constant, increasing $T$ results in an increase in $P$ :
$$P = k_1T$$

2) With $n,V$ constant, increasing $P$ results in an increase in $T$ :
$$T = k_2P$$

We can do the first process by simply heating the confined gas.

However I'm not able to think of a way to do the second process.
How can I make  "increasing $P$" the cause of "increasing $T$" ? (with out changing volume or adding more matter)

Basically there are only 3 ways of increasing the pressure.

1. By decreasing the volume of the gas (i.e. decreasing V)

2. By adding more amount of gas (i.e. increasing number of moles "n")

3. By increasing the temperature of the gas (i.e. increasing T)

Now your question being, would increase in pressure "P" keeping "V" and "n" constant result in the rise in temperature "T". So at the first place you need to increase the pressure.

By keeping V and n constant point 1 and point 2 discussed above is ruled out.

Now you want to observe T from changing P that means T, i.e., temperature over here cannot be used to increase P because you want to see how temperature varies with pressure, i.e. temperature here is to be observed.

So temperature cannot be used here to rise pressure. How would you increase pressure now? Out of the three factors which could have increased the pressure, two of the factors are constant and the last factor, i.e., temperature factor is to be observed and hence cannot be used to rise the pressure. So how will you now increase the pressure? It's not possible to increase the pressure without using any of the above factors.

As you are not able to increase the pressure in the first place, then how are you expecting to see the variation in temperature?


If CO2 is nonpolar how come much more dissolves in water than O2? Why is it that $\ce{CO2}$ is considerably more soluble in water than $\ce{O2}$ is?

$\ce{CO2}$ is nonpolar but dissolves in water which like $\ce{CO2}$ being nonpolar, doesn't make any sense.

Is it the bond polarity and that the bond polarity is more important in solubility?

If not than why is $\ce{CO2}$ more soluble in water than $\ce{O2}$.

Taken from my answer to your original question

There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$.  Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds.  Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility.  Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid.
$$\ce{CO2(aq) + H2O <=> H2CO3(aq)}$$
This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.


Energy profile of cyclohexene conformations The energy profile for cyclohexane flipping looks like this:

How does the corresponding diagram for cyclohexene look like?

I found the answer in the book: "Practical Aspects of Computational Chemistry"
. In chapter 19: Unusual Properties of Usual Molecules. Conformational Analysis of Cyclohexene, Its Derivatives and Heterocyclic Analogues.

The value for ring inversion of cyclohexene is smaller than the one of cyclohexane (which is near 10 kcal/mol):

  Value of ring inversion barrier is within 4.2–
  7.9 kcal/mol for molecular mechanics and 5.2–6.6 kcal/mol for quantum chemical
  methods.

The inversion proceeds as follows:

  The ring inversion process in cyclohexene includes three stage. The first
  stage involves the transition from an equilibrium half-chair conformation to a twist-boat conformation with a significant increase in energy. The second stage
  may be described as a very easy transformation of one twist-boat conformation to
  another via boat almost without any changes in energy. The third stage includes the
  transition from a twist-boat conformation to a half-chair conformation accompanied
  by a significant decrease in energy. The difference in energy between the twist-boat and the boat
  conformation is less than 0.01 kcal/mol. The boat conformation
  is just central point of this plateau and it is only the most convenient point for a description of the saddle point for the ring inversion
  .

The interconversion diagram should be something like this:


Why do PES graphs list 4s after 3d? Coming straight off of some electron configuration worksheets, our teacher started to teach us how to read photoelectron spectroscopy (PES) graphs and I'm having a bit of trouble with the two.
During the previous lecture on electron configuration, we were told that electrons occupied the most stable energy level available. That came in this order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, etc.

But then, during the photoelectron spectroscopy lecture, we were told that the 3d electrons have higher ionization energy than the 4s electrons and, in fact, that this pattern continues, just counting up the orbitals: 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, etc.

Why do electrons fill orbitals in the former pattern, but leave orbitals in the latter pattern?
Why would the 4s orbital fill before the 3d if it is less tightly held by the nucleus, and thus less stable?
I assume I have been taught some simplified versions and my teacher has my interests at heart, but I'm a bit confused.

For a given atomic kernel, the orbitals energies are not fixed like a set of shelves. Their energy depends on filling themselves and also on filling the other orbitals by electrons, because of kernel shielding and mutual repulsion.
The mutual energy is $4\mathrm{s} \lt 3\mathrm{d}$ when empty. But as kernel charge gets increased and $\mathrm{4s}$ and then $\mathrm{3d}$ are being filled, their energy order gets switched.
The typical case is the 3d energy drop $\ce{K -> Ca -> Sc}$, where the order is:
$$\mathrm{K: 4s \lt 4p \lt 3d}$$
$$\mathrm{Ca: 4s \lt 3d \lt 4p}$$
$$\mathrm{Sc: 3d \lt 4s \lt 4p}$$
well described on  charts in SE Chemistry Q what-is-d-orbital-collapse, which are related to the effective kernel charge acting upon electrons on various orbitals.


DNA-Sequencing (detection of the change of pH-value) after emulsion PCR I have troubles with understanding the exact order of the steps for DNA-sequencing after the Emulsion PCR (EmPCR). The EmPCR fills each bead with the same DNA strand (copies it $n$ times, till the bead is entirely filled up). But how is then the sequencing done with these beads?

The sequencing is described as follows (in easy words): Each nucleotide is provided to the DNA strand. If a nucleotide is accepted, the $\ce{H+}$ concentration rises and can be detected. Hence, the complementary nucleotide was present at the DNA strand.

But to be able to attach a nucleotide to the DNA, it has to be broken up. But the EmPCR already filled up every copied DNA strand. Where is it then broken up again?

In molecular biology, it is a very important that DNA should be replicated in order to facilitate sequencing. Basically, this is a very important concept that DNA should be replicated, simply in order to amplify signals. No matter the method of sequencing, without a proper amount of amplification, it's near impossible to detect each base call (signal). All Next-Generation Sequencing (NGS) techniques replicate DNA before sequencing is done.

Emulsion PCR is a PCR variation that some NGS technologies (e.g., NGS flatforms: Roche/454’s GS FLX, Titanium; Life/APG’s SOLiD 3; and Polonator G.007) use to replicate DNA sequences. It is conducted on a bead surface within tiny water bubbles floating on an oil solution.

Procedure (Source: https://binf.snipcademy.com/lessons/ngs-techniques/emulsion-pcr):

Fragmentation of DNA: The library is first fragmented either by sonication (high sound energy) or nebulization (forces DNA through a small hole) to fragments ranging from 300 to 800 base pairs.
Adapters are ligated: Adapters are then ligated onto the DNA fragments. These allow the strands to bind to the emulsion beads.
Denature to single strands: The double stranded DNA's with adapters are then denatured by heating the DNA up to $\pu{95 °C}$. Denaturing DNA simply means to go from double stranded DNA (dsDNA) to two single strands (ssDNA) - the hydrogen bonds keeping the two together are broken.
Formation of clonal bead populations: Each bead coated with streptavidin (e.g., Streptavidin-coupled Dynabeads), which is resistant to organic solvents, denaturants, detergents, proteolytic enzymes and extremes of temperature and $\mathrm{pH}$. Over a billion beads are used with a primer that matches the adapters attached earlier. The ssDNA is then attached to these beads. Each bead is emulsified in a water-in-oil droplet with PCR reagents (DNA polymerase, primers, buffers, dNTPs):

Emulsion PCR (emPCR) amplifies DNA strands on beads: Within these droplets, PCR is conducted. This involves the steps Denaturation, Annealing, Elongation. Firstly, the strand is elongated with DNA polymerase and dNTPs. Then the double-strand is denatured, allowing for the strand to ligate to another site on the surface of the bead. Eventually, 1 million copies of the target is amplified on the surface of each bead. The water-in-oil droplet is approximately 1-um. Follow the figure to see how each bead is able to replicate DNA on its surface.

Emulsion Breaking: After the DNA strands are amplified, the emulsion from the preceding step is broken using isopropanol and detergent buffer. The solution is then vortexed, centrifuged, and magnetically separated. The resulting solution is a suspension of empty, clonal and non-clonal beads, which will be filtered in the next step.
Bead enrichment: After PCR is conducted, you are left with a mixture of some beads that have amplified DNA attached on its surface, and some that do not. We may take out the enriched beads by attaching streptavidin coated magnetic enrichment bead. With a magnet, we can then pull out the beads with amplified DNA. There are other methods of bead enrichment that include using larger beads that are able to bind to beads with amplified DNA. After centrifugation, the beads with amplified DNA and without can then be separated.

Bead Capping: Attach a capping oligonucleotide to the 3' end of both unextended forward EmPCR primers and the RDV segment of template DNA. This helps in cover-slip arraying, which is used to polony sequencing, and prevents fluorescent probes from ligating to the ends.
Result: The beads with amplified sequences are then placed on a slide and are sequenced. Due to their high density of the same DNA molecule, the signal is amplified, allowing computers to read the sequencing data.

For example, the method used by the Roche/454 sequencer to amplify single-stranded DNA copies from a fragment library on agarose beads. A mixture of DNA fragments with agarose beads containing complementary oligonucleotides to the adapters at the fragment ends are mixed in an approximately 1:1 ratio. The mixture is encapsulated by vigorous vortexing into aqueous micelles that contain PCR reactants surrounded by oil, and pipetted into a 96-well microtiter plate for PCR amplification. The resulting beads are decorated with approximately 1 million copies of the original single-stranded fragment, which provides sufficient signal strength during the pyrosequencing reaction that follows to detect and record nucleotide incorporation events. sstDNA, single-stranded template DNA (Ref.1):

Keep in mind that in pyrosequencing method, each incorporation of a nucleotide by DNA polymerase results in the release of pyrophosphate, which initiates a series of downstream reactions that ultimately produce light by the firefly enzyme luciferase. The amount of light produced is proportional to the number of nucleotides incorporated (up to the point of detector saturation). Above diagram shows an example of four-color mapping sequence. For additional info, should read all of Ref.1-3.

References:

Elaine R. Mardis, “Next-Generation DNA Sequencing Methods,” Annu. Rev. Genomics Hum. Genet. 2008, 9, 387–402 (doi: 10.1146/annurev.genom.9.081307.164359).
R. Kanagal-Shamanna, “Emulsion PCR: Techniques and Applications,” Methods in Molecular Biology, Volume 1392: Clinical Applications of PCR;  R. Luthra, R. Singh, K. Patel, Eds.; Humana Press: New York, NY, 2016, pp. 33–42 (https://doi.org/10.1007/978-1-4939-3360-0_4).
Michael L. Metzker, “Sequencing technologies —the next generation,” Nature Reviews Genetics 2010, 11, 31–46 (doi: 10.1146/annurev.genom.9.081307.164359).


Conversion of 1,2-dione to a 7 membered oxepane? Has anyone heard of forming a seven membered ring from the rearrangement of a 1,2-dione? What I am talking about specifically is benzil rearrangement:

Could anyone give me guidance for how on earth something like this would form?

By this late date Coherent has probably resolved the structure of his product (9). From the limited information provided, it appears that a photolysis of benzil (1) was conducted in hydroxylic solvent (methanol?). Let's assume for the sake of argument that the product has the structure that Coherent has proposed. Coherent asks, how can it (9) be formed? First of all, hydroxyl radicals need not be invoked. Structure 2 represents an excited state of benzil that may cyclize to diradical 3 whose resonance structure is diradical 4. Rearomatization of 4 provides enol 5, tautomerization of which affords 2-phenylbenzofuran-3-one (6). Now the reorganization 4 --> 5 need not be intramolecular as indicated (red hydrogen) but intermolecular or, less likely, via the solvent acting as a source of hydrogen atoms (methanol C-H bond DHo = 96 kcal/mol. Probably too strong a C-H bond to deliver a hydrogen atom to stabilized radicals).
Benzofuranone 6 is a known compound. If benzofuranone 6 is truly formed in the reaction sequence,  it should independently afford the final product 9 upon irradiation conditions. Cleavage of benzofuranone 6 by the Norrish type I mechanism leads to diradical 7, which may recyclize to 8 and subsequently rearomatize to 9. [I am not overly excited about going from a 5-membered to a 7-membered ring, but then again who knows.]
Finally, compound 9 should be able to be synthesized from O-benzylsalicylic acid by intramolecular acylation.


Increase in Potential Energy When Bonds Are Broken Why is it that the potential energy of a system increases when the bonds present in it are broken? (i.e., when a solid turns into a liquid.) I thought the potential energy of the system was mostly contained in these bonds? If these bonds are broken, then why does the system’s potential energy increase?

First, changing a solid into a liquid does not involve breaking bonds in the normal sense of the word, i.e. covalent bonds. Mostly intermolecular interactions are broken as heat is put in to melt something. These may be hydrogen bonds e.g. water, $\pi-\pi$ interaction e.g. benzene or dispersion as in cyclohexane and similar non polar but (slightly) polarisable molecules.

Starting with free atoms, when these are brought together and a molecule is formed then energy is released to the surroundings and a stable molecule is formed, say, H$_2$, or HCl etc. The figure shows the potential energy plotted vs atom separation. The energy is zero when the atoms are separated at infinity; in reality about 10 bond lengths is equivalent to infinity.

To dissociate the molecules, potential energy has to be added as shown by the dissociation energy. Any energy above this is released as kinetic energy of the atoms.

The heats of formation of different types of molecules have a wide range of values. Ignoring vibrational energy, which is small compared to bond enthalpy, this is in effect moving from the total minima in the potential wells of reactants (of the whole molecule not just a bond) to the minima in the products. This may involve absorption or release of heat to the surroundings and so may involve gain or release of potential energy overall. It just depends what the reaction is. (In calculating the Free energy changes entropy changes have to be considered also.)


How many carbons in this skeletal structure? How many carbons are in this skeletal structure? I was told to draw a circle at the start of the structure and at each point the line turns off to get the total number of carbons but doing so would result in 6 carbons which I know is wrong but why is it wrong? And where on the structure are the carbons located?

Thanks for clearing this up for me!

Your intuition is not wrong.

There are indeed 6 carbon atoms in the molecule (3-methyl-pentane).

With bondline/skeletal notation, there's a carbon atom at all the vertices; at the ends, and every time the line changes direction.

Counting the carbon atoms then becomes as simple as counting the vertices.

This becomes more apparent if you explicitly write all the atoms like illustrated below:


Classifying acid-base reaction between aluminium hydride and lithium hydride So I was dealing with a reaction the other day, which is also briefly written in the title:

  $$\ce{AlH3(s) + LiH(aq) -> Li+(aq) + AlH4-(aq)}$$

The strange thing is:

Arrhenius definition cannot explain the reaction, since there is no $\ce{OH-}$ in the reaction formula.
Also Brønsted–Lowry definition cannot explain the reaction, because the stuff which $\ce{LiH}$ gives to $\ce{AlH3}$ is not a proton (hydrogenium, $\ce{H+}$) but a protide (hydride, $\ce{H-}$).
And as far as I know, this is the exact definition of Lewis acid and base written in my chemistry textbook:

  Lewis acid: A substance which can receive unshared electron pair
  Lewis base: A substance which can give unshared electron pair

leading to a conclusion that the above reaction also cannot be explained through the Lewis definition. Below are my questions:

Is there anything wrong in my thought?
Is that really an acid-base reaction? If it is, what kind of acid-base reaction is it?

This is indeed a Lewis acid-base reaction.

The hydride ion has a lone pair which is, in general, not bound to anything. The aluminium atom has at least one unpopulated orbital (typically taken to be a $\mathrm p_z$ orbital in a planar $\ce{AlX3}$ environment). This makes hydride a Lewis base and aluminium a Lewis acid. The association of hydride to aluminium is thus a Lewis acid-base reaction.


Why is DBU the most popular base with an amidino group? Does the basic strength have something to do with the ring size being 7 for one of the rings? Structure of DBU

DBU is used as a very strong base(which is due to resonance stabilization of amidinium cation) with steric bulk. Is any other compound used instead? If not, why not?

DBU is a popular amidine base because it can be deemed to be of lower nucleophilicity than cheaper amidines e.g. formamidine.  It has been sometimes erroneously labelled as "non-nucleophilic" but seeing as it reacts with carbon dioxide and sulfonyl halides and anhydrides,  this cannot be supported. As has been pointed out, the price of bulk chemicals such as caprolactam (made from cyclohexanone which in turn comes from phenol) and acrylonitrile makes DBU accessible.
The 7-member ring structure isn't special per se but assists with steric bulk to reduce nucleophilicity of the free base and increase the nucleophilicity of the counter anion when acting as a (large) amidinium ion. This occurs through a reduced association of the ion pair.
For cost weighed up against basicity,  tetramethylguanidine is better value as an organic base if the extra strength is required.


Why color of alkali metal peroxides, superoxides and ozonides deepens down the group? This is basically a continuation of the question-"Why is potassium monoxide (K2O) coloured?"

I knew that color of alkali metal oxides deepens down the group:

Lithium oxide ($\ce{Li2O}$) is the lightest alkali metal oxide and a white
solid. It melts at 1570 °C.
Sodium oxide ($\ce{Na2O}$) is a white solid that melts at 1132 °C and
decomposes at 1950 °C.
Potassium oxide ($\ce{K2O}$) is a pale yellow solid that decomposes at
350°C.
Rubidium oxide ($\ce{Rb2O}$) is a yellow solid that melts at 500 °C.
Caesium oxide ($\ce{Cs2O}$) is a yellow-orange solid that melts at 490°C.

The reason is due to what @Oscar said in the answer of previous question:

  If we go further down the alkali metal oxides the band gap gets
  smaller, and more visible light with longer blue/green wavelengths may
  be absorbed. Then the color becomes stronger and more reddish.

But, it is observed that not only color of normal oxides deepens down the group but the peroxides and superoxides also follow the trend:

Lithium peroxide ($\ce{Li2O2}$) is a white solid that melts at 195 °C.
Sodium peroxide ($\ce{Na2O2}$) is a pale yellow solid that melts at 460
°Cand boils at 657 °C.
Potassium peroxide ($\ce{K2O2}$) is a yellow solid that melts at 490 °C.

Sodium superoxide ($\ce{NaO2}$) is a yellow-orange solid that melts at
551.7°C.
Potassium superoxide ($\ce{KO2}$) is a yellow solid that decomposes at
560°C.

Lithium ozonide ($\ce{LiO3}$) is a red solid which is produced from caesium
ozonide via an ion-exchange process.
Sodium ozonide ($\ce{NaO3}$) is a red solid which is produced from caesium
ozonide via an ion-exchange process.
Potassium ozonide ($\ce{KO3}$) is a dark red solid which is produced when
potassium is burned in ozone or exposed to air for years.
Rubidium ozonide ($\ce{RbO3}$) is a dark red solid which is produced when
rubidium is burned in ozone.
Caesium ozonide ($\ce{CsO3}$) is a dark red solid which is produced when
caesium is burned in ozone.

What is the reason behind the deepening of the color? Is it due to the same reason as @Oscar pointed out? Or is it any other reason? Although superoxides and ozonides are paramagnetic and are colored, the peroxides are diamagnetic but still they are colored. Why?

Most likely, we are seeing the same band structure effects.

There is an additional trend, whereby the compounds become darker and more reddish as one adds more oxygen.  In oxygen-chain species such as $\ce {O2^{2-}} $, antibonding molecular orbital are occupied.  So The valence band containg these electrons is raised.  That lowers the band gap enabling absorption of more visible light.

Incidentally, the visual effects described here are limited to alkali metal (when oxygen is the electronegative element).  When we move to the corresponding alkaline earth oxide the monoxides are white and the peroxides ($\ce {Mg}$ and heavier) are white too.  Less strongly ionic bonding and a lower cation/anions ratio lead to less negative charge on the oxygen species, thus lower valence band energies and gaps  remaining large enough to remain uv-absorptive only.


Why is it wrong to use the concept of hybridization for transition metal complexes? I have asked a lot of questions on coordination chemistry here before and I have gone through a lot others here as well. Students, including me, attempt to answer those questions using the concept of hybridization because that's what we are taught in class and of course it's easier and more intuitive than crystal field theory/molecular orbital theory. But almost all of the times that I attempted to use the concept of hybridization to explain bonding, somebody comes up and tells that it's wrong.

How do you determine the hybridisation state of a coordinate complex?

This is a link to one such question and the first thing that the person who answered it says: "Again, I feel a bit like a broken record. You should not use hybridization to describe transition metal complexes."

I need to know:

Why is it wrong? Is it wrong because it's oversimplified?
Why does it work well while explaining bonding in other compounds?
What goes wrong in the case of transition metals?

Tetrahedral complexes
Let's consider, for example, a tetrahedral $\ce{Ni(II)}$ complex ($\mathrm{d^8}$), like $\ce{[NiCl4]^2-}$. According to hybridisation theory, the central nickel ion has $\mathrm{sp^3}$ hybridisation, the four $\mathrm{sp^3}$-type orbitals are filled by electrons from the chloride ligands, and the $\mathrm{3d}$ orbitals are not involved in bonding.

Already there are several problems with this interpretation. The most obvious is that the $\mathrm{3d}$ orbitals are very much involved in (covalent) bonding: a cursory glance at a MO diagram will show that this is the case. If they were not involved in bonding at all, they should remain degenerate, which is obviously untrue; and even if you bring in crystal field theory (CFT) to say that there is an ionic interaction, it is still not sufficient.
If accuracy is desired, the complex can only really be described by a full MO diagram. One might ask why we should believe the MO diagram over the hybridisation picture. The answer is that there is a wealth of experimental evidence, especially electronic spectroscopy ($\mathrm{d-d^*}$ transitions being the most obvious example), and magnetic properties, that is in accordance with the MO picture and not the hybridisation one. It is simply impossible to explain many of these phenomena using this $\mathrm{sp^3}$ model.
Lastly, hybridisation alone cannot explain whether a complex should be tetrahedral ($\ce{[NiCl4]^2-}$) or square planar ($\ce{[Ni(CN)4]^2-}$, or $\ce{[PtCl4]^2-}$). Generally the effect of the ligand, for example, is explained using the spectrochemical series. However, hybridisation cannot account for the position of ligands in the spectrochemical series! To do so you would need to bring in MO theory.

Octahedral complexes
Moving on to $\ce{Ni(II)}$ octahedral complexes, like $\ce{[Ni(H2O)6]^2+}$, the typical explanation is that there is $\mathrm{sp^3d^2}$ hybridisation. But all the $\mathrm{3d}$ orbitals are already populated, so where do the two $\mathrm{d}$ orbitals come from? The $\mathrm{4d}$ set, I suppose.

The points raised above for tetrahedral case above still apply here. However, here we have something even more criminal: the involvement of $\mathrm{4d}$ orbitals in bonding. This is simply not plausible, as these orbitals are energetically inaccessible. On top of that, it is unrealistic to expect that electrons will be donated into the $\mathrm{4d}$ orbitals when there are vacant holes in the $\mathrm{3d}$ orbitals.
For octahedral complexes where there is the possibility for high- and low-spin forms (e.g., $\mathrm{d^5}$ $\ce{Fe^3+}$ complexes), hybridisation theory becomes even more misleading:

Hybridisation theory implies that there is a fundamental difference in the orbitals involved in metal-ligand bonding for the high- and low-spin complexes. However, this is simply not true (again, an MO diagram will illustrate this point). And the notion of $\mathrm{4d}$ orbitals being involved in bonding is no more realistic than it was in the last case, which is to say, utterly unrealistic. In this situation, one also has the added issue that hybridisation theory provides no way of predicting whether a complex is high- or low-spin, as this again depends on the spectrochemical series.

Summary
Hybridisation theory, when applied to transition metals, is both incorrect and inadequate.
It is incorrect in the sense that it uses completely implausible ideas ($\mathrm{3d}$ metals using $\mathrm{4d}$ orbitals in bonding) as a basis for describing the metal complexes. That alone should cast doubt on the entire idea of using hybridisation for the $\mathrm{3d}$ transition metals.
However, it is also inadequate in that it does not explain the rich chemistry of the transition metals and their complexes, be it their geometries, spectra, reactivities, or magnetic properties. This prevents it from being useful even as a predictive model.

What about other chemical species?
You mentioned that hybridisation works well for "other compounds." That is really not always the case, though. For simple compounds like water, etc. there are already issues associated with the standard VSEPR/hybridisation theory. Superficially, the $\mathrm{sp^3}$ hybridisation of oxygen is consistent with the observed bent structure, but that's just about all that can be explained. The photoelectron spectrum of water shows very clearly that the two lone pairs on oxygen are inequivalent, and the MO diagram of water backs this up. Apart from that, hybridisation has absolutely no way of explaining the structures of boranes; Wade's rules do a much better job with the delocalised bonding.
And these are just Period 2 elements - when you go into the chemistry of the heavier elements, hybridisation generally becomes less and less useful a concept. For example, hypervalency is a huge problem: $\ce{SF6}$ is claimed to be $\mathrm{sp^3d^2}$ hybridised, but in fact $\mathrm{d}$-orbital involvement in bonding is negligible. On the other hand, non-hypervalent compounds, such as $\ce{H2S}$, are probably best described as unhybridised - what happened to the theory that worked so well for $\ce{H2O}$? It just isn't applicable here, for reasons beyond the scope of this post.
There is probably one scenario in which it is really useful, and that is when describing organic compounds. The reason for this is because tetravalent carbon tends to conform to the simple categories of $\mathrm{sp}^n$ $(n \in \{1, 2, 3\})$; we don't have the same teething issues with $\mathrm{d}$-orbitals that have been discussed above. But there are caveats. For example, it is important to recognise that it is not atoms that are hybridised, but rather orbitals: for example, each carbon in cyclopropane uses $\mathrm{sp^5}$ orbitals for the $\ce{C-C}$ bonds and $\mathrm{sp^2}$ orbitals for the $\ce{C-H}$ bonds.
The bottom line is that every model that we use in chemistry has a range of validity, and we should be careful not to use a model in a context where it is not valid. Hybridisation theory is not valid in the context of transition metal complexes, and should not be used as a means of explaining their structure, bonding, and properties.