Why is the excitation energy different from the energy difference between the orbitals involved in transition? Why is the excitation energy different from the energy difference between the orbital from which a electron is excited to the orbital in which the electron is excited to? Consider that the transition mainly consists of the above specified 2 orbitals only. Do comment if I need to add additional information to the question.

Just look at the Hartree-Fock equations that define orbitals
$$
  \hat{F} \psi_{i}(1)
  =
  \varepsilon_{i} \psi_{i}(1) \, ,
  \quad i = 1, 2, \dotsc \, ,
$$
where $\hat{F}$, the Fock operator, is given by
$$
  \hat{F}
  =
  \hat{H}_{\mathrm{core}}
  +
  \sum\limits_{j=1}^{n}
  \big(
    \hat{J}_{j} - \hat{K}_{j}
  \big) \, .
$$
First, notice that summation in the series goes over the number of electrons, or, in other words, over all occupied orbitals. Secondly, note that $\hat{J}_{i} \psi_{i}(1) = \hat{K}_{i} \psi_{i}(1)$, and thus, the Fock operator for the $i$-th occupied orbital does not include the Coulomb and exchange contribution from the $i$-th occupied orbital itself. The so-called self-interaction is perfectly canceled in the HF model and this is a rather nice feature of this model.

It also means that an electron in each and every occupied orbital "feels" only $n - 1$ electrons in other occupied orbitals, while an electron in a virtual orbital "feels" all $n$ electrons in all occupied orbitals without any exceptions. As a result a virtual orbital is useful only for describing the system in which an electron was added to it but none of the occupied orbitals were depopulated, so that electron in the previously unoccupied orbital still "feels" all $n$ electrons.

Now think about exciting an electron from an occupied orbital $\psi_o$ to a virtual one $\psi_u$. Before excitation an electron in $\psi_u$ would "feel" all $n$ electrons, but after excitation it "feels" only $n - 1$ electrons since $\psi_u$ is not unoccupied anymore. That is clearly a contradiction. So, indeed, $\varepsilon_u - \varepsilon_o$ should not be used to calculate the excitation energy, because exciting an electron from $\psi_o$ to $\psi_u$ leaves behind a hole in $\psi_o$ and $\varepsilon_u$ describes the case when one puts an electron into $\psi_u$ but does not remove one from $\psi_o$.


Calcium detection using XRF I have a mixture of metal oxides in powder form. Mainly Iron oxide, silicon dioxide, and calcium oxide. I know for certain calcium oxide is in the mixture, but my handheld XRF is not detecting it.

I am analyzing my sample through a very thin plastic wrap, polypropylene.

I have even taken a sample of pure lime, and my handheld XRF is not detecting any calcium. Just a bunch of others such as $\ce{Si, Al, P}$ and so on.

How might this be?

Is the plastic having an adverse effect?

My XRF is in 'general metals' mode; does it need calibrating to detect the $\ce{Ca}$?

Ca cannot be detected in General Metal Mode. You have to change to "Mining Mode"
In this mode, you can measure Ca properly.


What is the relation between absolute temperature, and rates of diffusion and effusion of a gas? According to Graham's diffusion law, rate of diffusion is directly proportional to square root of temperature but rate of effusion is inversely proportional to square root of temperature.

Why is this so? I would assume that rate of effusion would also increase with increase in temperature due to increased speed of the gas.

The self diffusion coefficient in a gas is given by $\displaystyle D=\frac{\lambda}{2}\bar c$ and the rate of effusion by $\displaystyle \frac{\rho}{4}\bar c$ where $\bar c$ is the average velocity of the gas molecules of mass $m$, and $\displaystyle \bar c= \sqrt{\frac{8k_BT}{\pi m}}  $, $\lambda$ the mean free path and $\rho$ the gas density. The diffusion constant is $\displaystyle D=\frac{1}{\pi \sigma^2 n}\sqrt{\frac{k_BT}{\pi m}}$ where $\sigma$ is the molecule's diameter, and the rate of effusion $\displaystyle \rho \sqrt{\frac{k_BT}{2\pi m}}$. The diffusion constant can also be written as $\displaystyle D=\frac{m}{\pi \sigma^2 \rho }\sqrt{\frac{k_BT}{\pi m}}$ thus the dependence on temperature is the same for diffusion and effusion if the gas density is held constant.

[ The rate of effusion is calculated by  considering the gas bouncing around in a container and then suddenly opening a tiny trap door of area $dA$ in the wall so that some minute proportion of the gas streams out in time $dt$. The trap door diameter should be no more than 1/10 of the mean free path.   If each molecule travels with average speed $\bar c$ the number leaving would be $n\bar c dA dt$ if there are $n$ molecules /unit volume. The efflux in terms of the mass/unit area/unit time is then $mn\bar c =\rho \bar c$.

However, half the molecules are moving away from the trap door at this instant so the rate is divided by 2. Additionally molecules come out at different angles so it is necessary to integrate over the volume element $\sin(\theta)d\theta d\phi$ and this leads to a further factor of 1/2.]


Are 6-hydroxyhexan-2-one and hexanoic acid isomers? Are the following two compounds isomers?
(A) $\ce{CH_{3}-CO-CH_{2}-CH_{2}-CH_{2}-CH_{2}-OH}$
(B) $\ce{CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-COOH}$

Although they have the same molecular formula ($\ce{C_{6}H_{12}O_{2}})$, I think that this two compounds are not isomers.

Isomers are compounds having same molecular formula, but a different structural formula.

As they satisfy the above condition, they are isomers and they are structural isomers. Structural isomers have a few variety based on connectivity of constituent atoms.

Chain : which has a different arrangement of carbon molecules
Position : different position of same functional group
Functional : different functional group for the same atoms
the others (tautomerism, metamerism and ring-chain isomerism)

They are functional isomers as they have different functional groups. First one has a ketone and an alcoholic group, while the second one has a carboxylic acid functional group.

You will come to see that it is easier to recognise structural isomers compared to the rest (stereoisomerism).

When speaking about isomerism, a picture is worth a thousand words:


What is the maximum distance over which van der Waals forces act? The VDW radius places a lower limit on the distance between atoms mutually attracted to one another. I am looking for the upper limit:  what is the name of the concept or theory that describes the maximum distance over which the various attractive forces (Keesom, Debye, London...) become operable / take effect?

This question is similar to the following Q, which didn't get fully answered:

Range of distance for Van der Waals force

There is no strict upper limit you can place on any type of electromagnetic interaction. That being said, van der Waal's forces are due to the formation of instantaneous dipoles which are a result of perturbations of an otherwise spherically symmetric electron density.

The electric field produced by a dipole decays as $1/r^3$. Therefore, it should not be surprising that the interaction between two instantaneous dipoles decays as $1/r^6$. It's actually not completely obvious this should be the case, but it turns out to be true. At very close distances, the van der Waal's forces actually scale as $1/r^7$ due to an effect from quantum electrodynamics, but chemists don't typically worry about this because the effect is negligible in almost all chemical processes.

The actual distance will vary depending on the system, so it's not like one can give actual numbers for the distance. But, for a system like water, the electrostatic force, which decays as $1/r^2$, can be relevant for distances up to, say, 100 angstroms. So, you can try to back out the scale of the van der Waal's interactions from there.

Another way of saying this is that van der Waal's forces can have an important effect on the first solvation shell of a molecule, but they typically do not have a strong effect beyond that distance.


Why does charcoal absorb smells from the air, and where do those smells 'go' once absorbed? I've read that using blocks of charcoal can remove some odors from the air by placing them in the vicinity of the odors. (Maybe not as efficiently as other forms of carbon, as pointed out in the comments of that link, but something nonetheless.)

What is the chemical process by which charcoal can do that, and once the odors are 'gone', where do they 'go'? In the case of 'toxins in gases', what typically is done to them upon 'deodorizing' by the charcoal block - are they collected, neutralised / transformed upon contact, or something else?

It is a form of carbon that has very large surface. The odors and impurities are held to the surface with Van der Waals forces. The capability of removing notable amounts of these impurities or odors is thanks to the extraordinarily large surface.


IS DMSO-o a σ-donor, a π-donor, or both? DMSO is an ambidentate ligand that can bind in a κO fashion via the oxygen lone pair and in a κS fashion via the sulfur lone pair.

I know that DMSO can behave as a π-acceptor when it bonds a metal at the sulfur, but what is the symmetry of the $\ce{O\bond{->}M}$ MO when DMSO binds at the oxygen? I have some diagrams of DMSO's HOMO and LUMO MOs, and it seems clear that DMSO-κS is a π-acceptor due to the LUMO's potential to overlap with a $\mathrm{t_{2g}}$-type d orbital, but the HOMO looks like it would do a π interaction. This confuses me because I know from IR that the $\ce{S-O}$ bond is weakened when DMSO binds at the oxygen, and there are some lower energy occupied MOs that would have σ symmetry overlapping with $\mathrm{d}_{z^2}$, for instance.

Is DMSO-κO a σ-donor, a π-donor, or both?

(keep in mind that there are some more σ-donor-looking MOs at energy levels below the HOMO)

Also; The reference below (1) says that DMSO's tendency to bind as DMSO-κS is dependent on the metal having a high enough "electron charge density" to have a sufficient back donation contribution to the $\ce{M\bond{->}S}$ π bond. What does this mean? The paper also says that ruthenium and rhodium prefers to bind as DMSO-κS. Based on Pearson's HSAB concept, I would expect that ruthenium would prefer DMSO-κS because it is a soft acid as a result of its larger second-row ionic radius giving it lower charge density (and S is softer than O), but the paper seems to be saying the opposite. What do they mean by "electron charge density"?

(1) Panina, N. S; Calligaris, M. Inorg. Chim. Acta 2002, 334, 165-171.

The first thing to remember is that a σ-symmetric bond is always better than a π-symmetric one. Orbitals that can bond with each other are typically oriented in one direction or another (except for s-orbitals, but those can only participate in σ bonds, anyway). Approaching these orbitals from the direction ‘their lobes point towards’ and in the direction ‘your own lobe is pointing towards’ results in a much better overlap than if you attempted to create a π type interaction. Thus, typically any σ interaction will be preferable to any π interaction between a given set of two bonding partners.

To be perfectly honest, I am not aware of any ligand which does not initially bond as a σ donor.

The question remains what will happen if we add π effects into the picture. π effects can either be of the π-Lewis acidic type (e.g. $\ce{CO}$) or of the π-Lewis basic type (e.g. $\ce{F-}$). In the case of DMSO, the LUMO has much greater contributions on sulfur than on oxygen — in line with the general expectation that the LUMO is centred on electropositive partners while HOMOs centre more around the electronegative partners.

Also in line with what oxygen does in other ligands, I would expect DMSO-κO to be a π base, due to the probability of another populated orbital having significant contributions on oxygen. Contrarily, the picture of the LUMO shows how well sulfur in DMSO-κS can act as a Lewis acid.

For Lewis acidic bonding, i.e. π backbonding (also written as M$\leftrightarrows$L), it is beneficial if there is significant charge density on the central metal, i.e. if the metal has a somewhat Lewis basic character. Often, this will be enhanced by the presence of electrons in d-orbitals of the $\mathrm{t_{2g}}$ type.

For Lewis basic ligand bonding, i.e. π forward bonding (also written as L$\rightrightarrows$M), it is beneficial for the metal to be in a Lewis acidic, electron-poor state — often with empty d-orbitals.

In the DMSO-κS context, remember that sulfur carries a partial positive charge and is thus already electron deficient somewhat ($\pm 0$ oxidation state, but it almost has the same electronegativity as carbon, so it is not far away from $\mathrm{+II}$). Therefore, the σ forward bond will be weak per se. Only if sufficient stabilisation through backbonding can be supplied will a DMSO-κS complex be sufficiently stable.


The hydrogen ion in acidified potassium manganate (VII) Recently, I am playing with redox reaction and I come over a compound acidified potassium manganate (VII) ($\ce{KMnO4/H^+}$). I get confused that why they put $\ce{H^+}$ beside $\ce{KMnO4}$. Doesn't the hydrogen ions flow freely in the solution or something else.

And so, when we are writing a chemical reaction with occur $\ce{KMnO4}$, we sometimes exclude the $\ce{H^+}$. Why that so?

I think you have known that the permanganate(VII) ion is a strong oxidizing agent because manganese is in the highest +7 oxidation state.

But in different pH conditions, $\ce{MnO_4^-}$ Will be reduced to different product:

In a acidic solution, it is reduced to the colourless manganese(II) ion, $\ce{Mn^{2+}}$.
$\ce{8H^+ + MnO_4^{−} + 5e^{−} → Mn^{2+} + 4H_2O}$
In a strongly basic solution, it is reduced to the green manganate ion, $\ce{MnO_4^{2−}}$.
$\ce{MnO_4^{−} + e^{−} → MnO_4^{2−}}$
In a neutral solution, it is reduced to the brown manganese dioxide $\ce{MnO_2}$.
$\ce{2H_2O + MnO_4^{−} + 3e^{−} → MnO_2 + 4OH^{−}}$

But the more understandable way to say, I think, is that $\ce{MnO_4^-}$ can be reduced to different oxidation states; however, it is an idea about chemical balance. If there is $\ce{H^+}$, the first reaction dominates (since we put in more reactant). While if there is $\ce{OH^-}$, neither reaction 1 nor 3 happens (1: lack of reactant; 3: too many product), so the third dominates.

Hope this helps.

Addition: in those different reactions, there are also different tendency to take place and make $\ce{MnO_4^-}$ reduced. In fact, the first reaction (happening with acidic solution) have a much larger tendency to take place. Thus, if $\ce{H^+}$ presents, $\ce{MnO_4^-}$ will have a larger tendency to oxidate other compounds. Thus some reactions require $\ce{MnO_4^-}$ with $\ce{H^+}$ instead of only $\ce{MnO_4^-}$.


Does acetoacetic ester give haloform test? According to my class notes, acetoacetic ester doesn't give haloform test, but I feel that it should.
I know that the methyl group on acetoacetic ester is not responsible for the haloform test. First, the active methylene group is attacked by the base and halogen ion is substituted there. After the attack of $\ce{OH-}$ on the carbonyl carbon we will also get an ethanol molecule (from the ester part), which apparently should give positive haloform test.
I am confused. Which is correct?

As @user55119 says it is unlikely the di-iodide is formed under iodoform conditions.
According to this paper1, acetoacetate gives only mono-iodination at the methylene centre under more powerful iodinating conditions than the standard iodoform $\ce{I2/OH-}$. The mono-iodo product in the presence of excess base will exist as the iodo-enolate effectively stopping iodination at the methyl ketone centre. Thus no iodoform product will be formed.
References

Goswami, Papori & Ali, Shahzad & Khan, Dr. M. Musawwer & Khan, Abu. (2007). Selective and effective oxone-catalysed α-iodination of ketones and 1, 3-dicarbonyl compounds in the solid state. General Papers ARKIVOC. 2007. 82-89. (Link)


Calculating the number of neutrons in a sample Which of the following contains one mole of neutrons?
A. $\pu{1 g}$ of $\ce{^1_1H}$
B. $\pu{1 g}$ of $\ce{^{12}_6C}$
C. $\pu{2 g}$ of $\ce{^{24}_{12}Mg}$
D. $\pu{2 g}$ of $\ce{^{23}_{10}Ne}$

Here's a question which I don't know how to answer. I tried using the equation:
moles=mass/mr
but that did not match the answer given, which is (C). I keep doing 2/24 which doesn't appear to equal 1 mole.

The formula

$$\text{amount of substance} = \frac{\text{mass}}{\text{molar mass}}$$

tells you that you have 1/12 moles of magnesium, since you are taking the mass of magnesium (2 g) and dividing by the molecular mass of magnesium (24 g/mol). However, the question is not asking about the number of moles of magnesium.

Each magnesium atom has 24 - 12 = 12 neutrons. Consequently, 1/12 moles of magnesium has (1/12)(12) = 1 mole of neutrons.


Quantitative MO theory- how is the overlap integral and the weighting coefficients of the molecular orbital actually evaluated? Title. In my inorganic chemistry course, we learn about SALCs and qualitative MO treatment, with only a fleeting reference to the S integral and actual molecular wave functions (only hydrogen, $\ce{H2}$, was discussed).

I'm curious about how the S integral is evaluated for MOs in slightly more complex molecules i.e. $\ce{BF3}$, $\ce{CO}$, $\ce{H2O}$, and how the weighting coefficients are found, as they aren't discussed at all at my level.

Can someone explain this? Thanks!

In essence, one picks a computational level, chooses a system to run the calculation on (typically a molecule) and lets a computer and an appropriate program do the heavy numeric and algebraic lifting. No really, that's it.

In more detail: One would want to solve the multi-electron, multi-nucleus Schrödinger equation (that's already an approximation, because we ignore relativistic effects here). That's really difficult. So we make more approximations, such as fixed nuclei, a product ansatz for the wave-function (made up of one-electron wave-functions) and often a finite-size basis set. A basis set are functions that can be linearly combined into molecular orbitals; a large one usually gives better results, but make the computation longer. This choice is part of the computational level. Other, more mathematical and technical choices make up the other part (and this is a science in itself).

For a molecule like $\ce{C2H6}$, the computational time can vary between several days to weeks and less than a second, depending on sophistication of the computational level and its accuracy. In the end, we obtain an approximate wave-function (containing molecular orbitals), from which approximate properties such as energy or NMR shifts can be obtained. The computer is useful in this by calculating integrals (partially analytic, mostly numeric) and keeping track of many very large matrices.

At times, the problems can be worked out using pen and paper, making use of high symmetry and tiny basis sets. This leads to easily interpreted, but not very accurate wave-functions.


Maximum number of atoms in the ground state I was given the following question.

A sample contains $100$ identical and hypothetical $H$-like atoms ( $Z$ may be fractional). Out of the given atoms, some are in the ground state, while others are in a higher energy level ($n=x$). The $P.E$ of the electrons in the ground state is $-192\ \mathrm{eV}$.
The sample is subject to radiations of wavelength $155\ \mathrm{nm}$, and the electrons jump to a higher energy level ($n=x+3$). Upon back-transition, a total of $10$ different spectral lines are observed
If the final state the electrons reach after absorbing photons of $\lambda=155\ \mathrm{nm}$ is $a$ , and the maximum number of atoms which had electrons in the ground state initially is $b$, find $\frac{b}{a}$

By equating ${}^{x+3}C_2$ to $10$, I found $x$ to be $2$. Then, $a$ must be $x+3$, which is $5$.
As to finding the value of $b$, I am stuck. Can we use the fact that $\frac{|P.E|}{2} = |T.E|$? The energy associated with each photon comes out to be $8\ \mathrm{eV}$, so the total energy absorbed is $800\ \mathrm{eV}$ (Since there are $100$ atoms). Maybe we can use this?
Any help would be greatly appreciated.
Edit:- The answers given are $x+3=6$ and $96$ , so I ended up getting the first part wrong as well.

Of the $100$, initially some are in $n=1$, and some in $n=2$ and then some in $n=5$ after absorbing 155 nm then some in lots of other levels after photons emitted. Thus, eventually, none will be left in $n = 5$ but spread  in all in the other levels  including the ground state. However, all other possible transitions occur ($n=2 \to 1$ for example) so all end up in the ground state in total $(100-b)$. However, you do not know the initial amount $b$ so I'm not sure that this question can be answered except that the ratio is $b/(100-b)$.


Using a significantly large metallic sphere, could we draw out the electrons from the negative terminal of a battery? I've pored over many articles on electrochemical cells, and I understand that on the oxidation side of a zinc-copper Daniell cell, the anode is highly negatively charged because zinc participates in oxidation and loses electrons as the ions go into the zinc sulfate.

So since the anode is rich in free electrons, is it possible to take an electrically neutral, large conductor (a metallic sphere for example), touch it to the negative terminal of the battery so that the free electrons will distribute themselves over its surface, discharge the sphere, and repeat this until the negative terminal is neutralized?

Yes, this is possible, you have effectively described using a battery to charge a very weak capacitor, then discharging the capacitor.  The charge stored will be minimal (possibly immeasurable), pico-coulombs or less and the number of cycles to fully discharge then battery may be on the order of billions or trillions, but in theory the process described is possible (more properly: plausible).


Comparing radii in lithium, beryllium, magnesium, aluminium and sodium ions Apparently the of last four, $\ce{Mg^2+}$ is closest in radius to $\ce{Li+}$. Is this true, and if so, why would a whole larger shell ($\ce{Mg^2+}$) be closer in radius to $\ce{Li+}$ than its isoelectronic neighbor ($\ce{Be^2+}$) with just one extra proton?

Many people may think that orbitals, or ‘shells’ essentially work like planet orbits around the sun; i.e. no matter how many electrons populate ‘Mercury’, ‘Venus’ and ‘Earth’ or how many protons are in the ‘Sun’ Venus’ orbit will always have the same dimension.

This is very much not so when studing atomic orbitals. The principle of hydrogen-like orbitals (which are the only orbitals that can be exactly calculated with the Born-Oppenheimer–approximation) is that we are dealing with an electron free to rotate and translate on a potential energy surface which is formed by the electrostatic potential of a proton or a nucleus with a charge of $z+$. Commonly, taking the 1s orbital as an example, there is a distance from said nucleus where the probability of locating the electron in the sphere defined by that radius is larger than a certian cutoff value, e.g. $90~\%$. If we go from a hydrogen-like system to a $\ce{He+}$ cation, we are doubling the charge thus strongly increasing the attractive force between nucleus and electron. As such, the $90~\%$-radius is much closer to the nucleus than before.

This holds true with limitations when no longer observing hydrogen-like atoms but actual multi-electron atoms: the more protons in the nucleus the more contracted the orbitals become. While that effect in itself is rather small, the second one is substantial: the higher an ion’s charge is, the more contracted its orbitals become. Therefore, $\ce{Be^2+}$, which is even considered to be the smallest atom or atomic ion known (yes, smaller than hydrogen) is substantially smaller than lithium.

When directly comparing two atom sizes of elements that are above and below each other in the periodic table, naturally under otherwise identical conditions the lower one will be larger; even though it has at least 2 (lithium) more protons it has also started populating an additional shell which is further removed from the nucleus. Similarly, if you compare $\ce{Li+}$ and $\ce{Mg+}$, magnesium will still be substantially larger. (In this case, the additional remaining electron in magnesium’s 3s orbital is the deal breaker but note that the same argument applies to $\ce{C+}$ and $\ce{P+}$ where no shell is completely empty.

As noted before, however, if we substract another electron, i.e. if we compare the isolobal $\ce{Li+}$ and $\ce{Mg^2+}$, the latter’s orbitals shrink substantially. By empirical observations scientists have discovered what is known as the diagonal relationship: two ions that are top-left bottom-right diagonally placed in the periodic table, whereby the bottom-right one’s charge is greater by one, have approximately the same radius.


Wurtz reaction with two different alkyl halide $$\ce{C5H11Cl + C4H9I + 2Na ->}$$
Is this possible in a Wurtz reaction? I want to add 2 different alkyl halides with two different halogens. I'm curious if two different types of halogen on the alkyl reactants is possible.

The alkyl groups mentioned could be primary, secondary or tertiary
The last one's will not be going by Wurtz reaction (these will rather undergo elimination to give alkenes); only primary & secondary can.
Even here, we can have a mixture of products:


Regioselectivity in nitration of a 1,2-disubstituted aromatic compound For 1,1,4,4-tetramethyl-1,2,3,4-tetrahydronaphthalene:

What would be the products following a standard nitration with $\ce{NO2+}$?

My reasoning is thus:

Since the substituent group is alkyl then it is ortho- or para- directing. But surely there are steric hindrance factors to be considered due to the bulk of the ring which would mean para- is favoured instead of ortho. However, when I try to draw out the structure it seems the arrows are meta- instead!

The molecule is symmetrical so the two positions on the far side of the alkyl groups are exactly the same. Similarly, the two positions closest to the alkyl groups are the same. For the far positions, each one is para with respect to one alkyl group and meta with respect to the other. This is not a contradiction because ortho, meta and para are purely relative terms.

I suspect that you are right that substitution will mostly occur at the far sites. The directing effect of the alkyl groups is essentially the same for every site because they are all ortho or para to one group and meta to another. Therefore, the steric effect of the bulky methyl groups adjacent to the near positions will disfavour substitution there.


Iodine test for Starch The starch and iodide test is to detect amylose, a polysaccharide. We add an iodide solution to a sample and observe a color change with the starch; it turns dark blue.

While researching information online, I understood that $\ce{I_{2}}$ is not soluble in water, hence it needs to react with iodide ions present to at least create $\ce{I_{3}}$, which is soluble in water, and also forms polyiodide complex, as research shows. I also understood that this linear molecule of iodide migrates to the center of the helicoid polysaccharide structure. Then, with some electronic interactions, there is a modification in the electron configuration which produces the dark blue color.

My questions are as follows. What are the electron's movements? Which atoms are implicated in this transfer of electrons? What is the electron configuration change that results in the color change? In other words, I don't understand the electron's interactions between iodide and amylose that are creates the color change.

The amylose present in starch is responsible for the deep blue colouration.  The structure of amylose consists of long polymer chains of glucose units connected by an alpha acetal linkages. All of the monomer units are $\alpha$ -D-glucose, and all the alpha acetal links connect $C1$ of one glucose and to $C4$ of the next glucose.

As a result of the bond angles in the acetal linkage, amylose actually forms a spiral structure.

X-ray diffraction studies$^{[1]}$, demonstrate that the amylose is organized as left-handed helices, which have outer diameters of 13 Å and a pitch of 8 Å, with each turn of the helix corresponding to six 1,4-anhydroglucose units. The iodine components are linearly arranged in the 5 Å wide inner cavity of the helices with an $\ce{I–I}$ distance of approximately 3.1 Å.

Starch-iodine complexes show dichroism of flow$^{[2]}$, i.e light with its electric vector parallel to the flow lines is more strongly absorbed than light with its electric vector normal to the flow lines. The dichroism of flow is shown to require that the long axes of the iodine molecules in the
complex be parallel to the long axis of the starch iodine complex.

The exact structure of the polyiodide chain remains contentious, however, the following species are commonly implicated as key elements of the substructure:

$$\ce{I2 + I^- -> I3^-}$$

So, ignoring the structural specifics of the polyiodide chain, one notes that if an iodine molecule enters the internal channel of the amylose
helix, it may be surrounded by about six glucose residues. The following image is from reference (3):

Bear this image in mind, and note that the hydroxyl oxygen is pojected into the helix cavity.

Also, when iodine is dissolved in ether or alcohol, a new strong absorption band appears in the near ultraviolet region. In relation to the study of the benzene-iodine complex, Mulliken postulated the complex formation
between iodine and ether or alcohol, and explained this new band by the concept of intermolecular charge transfer spectra$^{[3]}$.

The "simple" idea behind a charge transfer complex is given below:

$$\ce{I2 + D -> I^{\delta -}...D^{\delta +}}$$ Reference 4 gives more details. (It also likens this bonding to a "hydrogen bond" if that helps, you understand). In this case, D is the donor atom (electronegative atoms like nitrogen, oxygen etc.)

The iodine-donor distance is usually in 2~3 Å range. Thus, based on all of this one can conclude that possible charge transfer interactions between oxygen atoms that are projecting in towards the helical cavity and the iodine molecules in the polyiiodide chain are responsible for the colouration. You can view how the oxygen molecules are pointed towards the chain of iodine here. (I recommend rotating the diagram so that the axis of the cavity is towards you, and you are essentially looking down the tube)

References

1.https://www.researchgate.net/publication/244335699_The_complex_of_amylose_and_iodine

http://pubs.acs.org/doi/abs/10.1021/ja01244a017
http://scitation.aip.org/content/aip/journal/jcp/22/3/10.1063/1.1740076 (I recommend reading this if you are not satisfied with the brief, qualitative discussion I have provided here)
Inorganic Chemistry (Nils Wiberg) (google books, relevant portion is part of the preview)


Szabo/Ostlund exercise on spin orbital orthonormality I have a question about exercise 2.1 in Szabo and Ostlund's Modern Quantum Chemistry, which concerns the inner product between different spin orbitals: $\require{begingroup}\begingroup\newcommand{\br}{\mathbf{r}}\newcommand{\bx}{\mathbf{x}}$

Exercise 2.1. Given a set of $K$ orthonormal spatial functions, $\{\psi_i^\alpha(\br)\}$, and another set of $K$ orthonormal functions, $\{\psi_j^\beta(\br)\}$, such that the first set is not orthogonal to the second set, i.e.,
$$\int d\br \, \psi_i^{\alpha*}(\br)\psi_j^\beta(\br) = S_{ij}$$
where $\mathbf{S}$ is an overlap matrix, show that the set $\{\chi_i\}$ of $2K$ spin orbitals, formed by multiplying $\psi_i^\alpha(\br)$ by the $\alpha$ spin function and $\psi_j^\beta(\br)$ by the $\beta$ spin function, i.e.,
\begin{align}
\chi_{2i-1}(\bx) &= \psi_i^\alpha(\br)\alpha(\omega) \\
\chi_{2i}(\bx) &= \psi_j^\beta(\br)\beta(\omega)
\end{align}
with $i = 1, 2, \ldots, k$ is an orthonormal set.

I feel quite confused about this question. Consider the spin-orbital integral
$$\langle \psi_i^{\alpha} \alpha (\omega) | \psi_j^{\beta } \alpha (\omega) \rangle  = \langle \psi_i^{\alpha}  |  \psi_j^{\beta } \rangle  $$
which is not necessarily orthogonal.
The result can only be orthogonal if both ket and bra vectors are in the $i$-indexed set. If so, why does the exercise mention a $j$-indexed set in the beginning?

Consider the spin-orbital integral
$$ \langle \psi_i^{\alpha} \alpha
(\omega) | \psi_j^{\beta } \alpha (\omega) \rangle  = \langle\psi_i^{\alpha}  |  \psi_j^{\beta } \rangle
$$
which is not necessarily orthogonal.

Yes, it is not necessarily orthogonal, but the thing is that you don't need to consider such integrals in the exercise. Read more carefully: spatial orbitals from the first "alpha" set can be multiplied only by the $\alpha$ spin function while spatial orbitals from the second "beta" set can be multiplied only by the $\beta$ spin function. Thus, there are only three types of integrals you need to consider:
$$
\langle \psi_i^{\alpha} \alpha | \psi_j^{\alpha } \alpha \rangle \, ,
\quad
\langle \psi_i^{\alpha} \alpha | \psi_j^{\beta } \beta \rangle \, ,
\quad
\langle \psi_i^{\beta} \beta | \psi_j^{\beta } \beta \rangle \, .
$$


Dyes that can stain oil and water Is there such a thing as a dye that can stain both oil and water?  In other words, does anyone know of an amphiphilic dye?  Maybe a surfactant that is also a pretty shade of blue?  Or maybe turn one color in oil, but another in water, but stain both well?
Thanks

The diazo dye (1), known as Solvent Yellow 124 or Somalia Yellow is soluble in aliphatic hydrocarbons. In the European Community, it is used to mark low-taxed diesel fuel (for heating purposes). In order to prevent its use in cars, samples are acidified. This results in acetal cleavage and protonation of the diazo-fragment, yielding the red, water-soluble iminoquinone species (2)

Since this is probably not exactly what you wanted, what about dyes that exist in a neutral and a zwitterionic form, depending on the polarity of the solvent?

Brooker's merocyanine is one of the most prominent examples and show strong solvatochromism, i.e. the colour changes with the polarity of the solvent


Do molecules with axial chirality have "stereogenic units"? This question is a follow-up to a recent question on discodermolide stereochemistry that I answered here.  In the course of reading IUPAC's definition of "stereogenic center", I wondered if axially chiral molecules such as BINOL have a "stereogenic unit" according to IUPAC.

The part of IUPAC's definition that may (or may not) apply to BINOL is:

  A chain of four non-coplanar atoms (or rigid groups) in a stable conformation, such that an imaginary or real (restricted) rotation (with a change of sign of the torsion angle) about the central bond leads to a stereoisomer.

Here is the structure of (S)-BINOL:

Does the four-atom chain consisting of the O-bond carbons plus the transannular C-C-bound carbons qualify?  I can't quite work out what is meant by "with a change of sign of the torsion angle" and "stable conformation".  If it does qualify, suppose temperature is increased to the point where spontaneous racemization happens (and suppose that any side reactions or decomposition can be ignored).  Is that chain no longer a stereogenic unit?

Yes, axially chiral compounds have a stereogenic unit. Biphenyls (such as BINOL) and allenes are common types of compounds that exhibit axial chirality. As you describe, for BINOL the stereogenic unit is the set of atoms including the carbons joining the aryl groups and one carbon further from the biaryl junction. IUPAC's description is simply meant to define a torsional angle, or thought of another way, define two planes around an axis.

  If it does qualify, suppose temperature is increased to the point where spontaneous racemization happens (and suppose that any side reactions or decomposition can be ignored). Is that chain no longer a stereogenic unit?

This is a great question that forces us to think about what we mean by 'stable' and under what conditions is something stable.

Reference to 'stable conformation' means that the two conformations cannot interconvert at ambient temperature. In BINOL, there is restricted rotation around the biaryl bond, which means that the two stereoisomers cannot interconvert under ambient conditions.

Let's think though about a much simpler compound with four carbons in a row that can be used to define a torsional angle: butane. The gauche conformation actually meets the definition of chiral: it's mirror image is not superimposable on itself. However we don't consider butane to have a stereogenic unit (or consider butane to be a chiral compound) because those two conformations can interconvert readily at room temperature.

So those are two ends of the spectrum if we consider room temperature: BINOL has a stereogenic element but butane does not.

Let's say we heat up (S)-BINOL to a temperature that allows for rotation around the biaryl bond. As you say, the sample will racemize. At that temperature, the compound no longer has a stereogenic unit because rotation is no longer restricted. Allow the sample to cool, and the stereogenic unit returns, although the sample will be racemic.


Do Chemists "discover," "invent," or "create" things? I am an ESL teacher, and am designing an activity to teach my engineering students the differences between the words "invention," "discovery," "innovation," and "creation," words that are often confusing for non-native speakers (and as I'm finding out, native speakers.)

I have looked on the internet for usage and it seems to be all over the map: Nobel either invented dynamite, created it, or discovered it, and Teflon was called a discovery.

In my view, physical laws are discovered, as are molecules, fundamental particles, and forces because they already exist naturally.  Teflon cannot be said to be discovered because it does not exist naturally; neither does DDT and thousands of other molecules and compounds.  Is there a plant that produces nitrous oxide?

Based on the usage of "creation" I feel that this word more aptly describes the "process" of chemistry.  Teflon was created, not discovered or invented.

As chemists what usage do you prefer?  Do you prefer the use of inventor, creator, or discoverer?  If there are multiple answers can you help me see why you choose that word?

discover
invent
create
innovate

Create is totally weird with chemistry, you can create item, but substance?

Innovate is more about improving something existing.

This leaves us with discover and invent, which in case of chemistry seems to me almost equivalent. For "invent" I would imagine targeted effort towards a new substance with desired properties, whereas "discover" evokes more trial-and-error approach.


How does metallic bonding in alkaline earth metals work? Metallic bonds are said to be delocalised.

Lithium crystals have each lithium atom in direct contact with eight other lithium atoms. Each atom could therefore be in a hybrid of eight chemical bonds with each other.

But a similar explanation does not work for atoms like beryllium.

Some possible arrangements of electrons that beryllium could be a hybrid of configurations of might be:

$$\ce{Be^{+} + Be^{-}}$$

$$\ce{Be^{2+}_2 + Be^{2-}_2}$$

$$\ce{Be^+ + e^-}$$

$$\ce{Be^{+}_2 + e^-_}$$

$$\ce{Be^{2+}_2 + 2e^{2-}_}$$

$$\ce{Be^{2+}_2 + e^{2-}_2}$$

$$\ce{Be^{2+}_2 + e^{2-}_2}$$

But beryllium is diamagnetic and all these suggest paramagnetic properties or a $e^{2-}_2$ pair which doesn't seem right to me although I'm not fully sure how the free electron hybrids would work.

Another possibility is that the electrons are dragged up into the p orbitals.

For example, the atoms would have valence electron configurations of:

$$\underset{sp}{[\uparrow \vert \uparrow]} \underset{p}{[\; \vert \; ]}$$

which would result in hybrids of $\ce{Be2}$ or $\ce{Be_4}$ I think.

However, the atoms would have to be dragged up really high so that they aren't paramagnetic.

Let me list out the molecular orbital theory bonding and antibonding orbitals for diberyllium.

$$ \underset{\sigma}{[\uparrow \downarrow]} \underset{\sigma^*}{[\uparrow \downarrow]} \underset{\pi}{[\; \vert \;]} \underset{\sigma}{[\; ]} \underset{\pi^*}{[\; \vert \;]} \underset{\sigma^*}{[\;]}$$

If I wanted a bonding material that was not paramagnetic I'd have to lift up the energy to something like:

$$ \underset{\sigma}{[\uparrow \downarrow]} \underset{\sigma^*}{[\;]} \underset{\pi}{[\; \vert \;]} \underset{\sigma}{[ \uparrow \downarrow]} \underset{\pi^*}{[\; \vert \;]} \underset{\sigma^*}{[\;]}$$

which is probably too high energy to matter for any purposes.

Metallic bonding is essentially delocalisation of covalent bonds to the logical extreme and cannot be considered as "two-centre, two-electron". The metallic lattice is a network of cations immersed in a "sea" of electrons. As the highest energy orbital is the spherical s, there is no directional influence and hence why beryllium has the efficient hexagonal close packed structure.


Are there any fluorophores with lifetimes in the micro to milli second range? I'm looking for a fluorophore with exceptionally long lifetime to use in an oxygen optode (the longer the lifetime the less timing precision needed in the electronics).

So far the longest I have encountered in my searches is 600nS for Ru(bpy)3[PF6]2 found in this table but that resource appears less than exhaustive; Is there a more comprehensive data sources available where I could compare lifetimes and relevant wavelengths for all known fluorophors?

.. also just to clarrify the fluorophore needs to undergo quenching by oxygen ...

According to the article Indicators for optical oxygen sensors; Michela Quaranta, Sergey M. Borisov, Ingo Klimant

  Among all the indicators that have been used for optical oxygen sensors, Pt(II) and Pd(II) porphyrins are the mostpopular luminophores since they possess strong phosphorescence at room temperature, moderate to high molar absorption coefficients and large Stokes’shifts. Additionally, phosphorescence lifetimes are rather long (microsecond to millisecond) and can be tuned by varying the nature of the central atom.

They then go on to include a table (Table 3 on page 131) of 36 dyes based on that chemistry, 35 with lifetimes in the microseconds and 1 (PdTFPP) with a lifetime of 1.65 milliseconds.

In relation to my quest for a data source to compare fluorophors I managed to stumble upon the website fluorophores.org that currently includes 952 substances and can be filtered by excitation, emission, quantum yield, and lifetime parameters.


Does the explosion of TNT include oxidation? This comment below the question Rocket explosion compared to kT of TNT; has one ever knocked something over at a distance? suggests that

  ...TNT includes it's own oxidizer...

Explosion isn't the same as combustion and you don't need to add a separate oxidizer to an explosive for it to do its thing.

But technically speaking does the explosion of TNT include any steps that could be considered as oxidative? Can some intermediate products be considered to be acting as oxidizers?

3 nitrogroups of TNT are rich of oxygen, but the aromatic ring has only carbon and hydrogen atoms.

Therefore oxygen of nitrogroups oxidizes carbon and hydrogen atoms.

There is no principle difference between oxidation and reduction agens being separate molecules, or being different parts of the same molecule.

Note that TNT, in contrary to e.g nitroglycerine, is oxygen deficient, so the amount of available oxygen is not enough for complete oxidation to water and carbon dioxide.

TNT:

$$\ce{2 C7H5N3O6 -> 3 N2 + 5 H2O + 7 CO + 7 C}$$
or
$$\ce{2 C7H5N3O6 -> 3 N2 + 5 H2 + 12 CO + 2 C}$$

Nitroglycerine

$$\ce{4 C3H5N3O9 -> 6 N2 + 10 H2O + 12 CO2 +  O2 }$$


Electrode for methanol Could you advice an electrode for electrophoretic deposition in methanol which won't spoil methanol and also wont be spoiled by methanol. I googled a bit and found that
1 copper, platinum and gold are catalysts of methanol oxidation
2 aluminum reacts with methanol
3 silicon has poor conductivity...

What can you advice?

Glassy carbon may be a good choice. Looking at this, electrooxidation of methanol doesn't seem to be possible at reasonable potentials on glassy carbon, and it's a common working electrode material, relatively inexpensive, durable, and available in many shapes and sizes. Graphite could also work, though it may need to be replaced periodically as graphite electrodes tend to be less robust.


Comparing relative negative inductive effect of substituents (dialkyl amine, amine, alkoxide, fluoride) Which of the following is correct with respect to the negative inductive effect of the substituents?
  1. $\ce{-NR2 < -OR < -F}$
  2. $\ce{-NH2 < -OR < -F}$
  3. $\ce{-NR2 > -OR > -F}$
  4. $\ce{-NH2 > -OR > -F}$

Since it would be directly proportional to electronegativity, $\ce{-F}$ will be highest. That cancels out (3) and (4).

Now, to me both (1) and (2) seem correct. But the official answer (since this is from an entrance exam) is (2).

Are both (1) and (2) correct? Or is there something wrong with (1)?

The inductive effect is the polarization of σ (sigma) bonds within a chemical species caused usually by differences in electronegativity between the neighboring atoms. The higher the polarization it causes, the higher the inductive effect. In this case, the decreasing order of electronegativity is:

$\ce{-F > -OR > -NH2}$

So, for practical purposes, the correct answer should be option 2.

However, substituents far from the atom you are considering can also affect the inductive effect and an amino ($\ce{-NR2}$) group can be more withdrawing than an alkoxide ($\ce{-OR}$) group.

For example: $\ce{-N(-CF3)2 > -OCH3}$

This amino substituent is probably more electron-withdrawing than the alkoxide in the right, because the first is bound to two highly withdrawing groups ($\ce{-CF3}$) while the second is bound to an electron-donating group ($\ce{-CH3}$)


Can acid concentration be determined from Ka values in a titration? Say I have a sample of vinegar and I want to find the concentration of ethanoic acid in it. So I perform a titration using a $\mathrm{pH}$ meter, and using that data I can draw a $\mathrm{pH}$ curve and find quantities such as the initial $\mathrm{pH}$, the half-equivalence point, the equivalence point, and the $\mathrm{pH}$ at the equivalence point.

I feel like my teacher hasn't explained this very clearly, but given the $K_{\mathrm a}$ value for ethanoic acid and the data collected from the titration, is it possible to determine the initial concentration? If so, how do I proceed?

(I'm aware of the simple stoichiometric method, just curious whether this approach will work.)

The Henderson-Hasselbalch equation will help:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm a} + \log {[\ce{A-}]\over [\ce{HA}]}$$

Once the $\mathrm{p}K_{\mathrm a}$ is known, the equation will tell you the ratio of acetate to acetic acid (ethanoic acid). The ratio changes at various $\mathrm{pH}$ values.

The $\mathrm{p}K_{\mathrm a}$ is 4.76, so for a $\mathrm{pH}$ below 3.75 the dominant species in solution will be acetic acid; at pH 5.76, the dominant species in solution is the acetate ion. If you want to make a solution at a particular $\mathrm{pH}$ value, use the equation to give you the ratio of acetate to acetic acid needed.


MOPAC: can we run THERMO immediately after the geometry optimzation? With MOPAC, I am trying to set up single run with geometry optimization followed by calculation of THERMOdynamic parameters.

Is there viable way for that?

OK, in between I found it, it is the GEO_DAT keyword, for example, reading in previously converged PM7 geometry of methane:

 AUX LARGE CHARGE=0 SINGLET  GEO_DAT="metan_geomopt_pm7.arc" THERMO(298,298)  PM7


What is the explanation of the energy gap law in radiationless transitions? The rate of radiationless transitions such as internal conversion and intersystem crossing decrease with the size of the energy gap between the states, according to

$$k\propto \exp\left(\frac{-\gamma\Delta E}{\hbar\omega_M}\right)$$

The papers by Jortner[1] that I have found which derive this expression involve taking steepest descent integrals of the flux autocorrelation function, but this is not a very intuitive explanation. (And in particular is not suitable for undergraduate chemists.)

How can this expression be understood qualitatively from the perspective of the sum over Franck-Condon factors:

$$k_{sl} = \sum_j \frac{2\pi}{\hbar} |V_{s0,lj}|^2\delta(E_{s0}-E_{lj})$$

References:

Jortner, J. Radiationless transitions. Pure Appl. Chem. 1971, 27 (3), 389–420. Section IX, weak coupling. DOI: 10.1351/pac197127030389. Mirror: pdf via publications.iupac.org, pdf via the Internet Archive

These are a few 'rough and ready notes' and some data on radiationless transitions.

Intersystem crossing $S-T$ and Internal Conversion $S_1-S_0$ are described by the Fermi Golden Rule:

The rate constant decreases exponentially with energy gap between S and T in solution. In contrast in isolated molecules the gas phase rate increases with energy gap, see figures below.

Molecules in condensed phase

Molecules isolated in the vapour phase

The pure singlet and triplet states cannot interconvert because of the strict spin selection rule. An interaction, which is spin-orbit coupling, mixes the states together to make new states that contain a bit of each initial state. The singlet now contains a little triplet and triplet contains a little singlet. As the S1 and S0 states are orthogonal to one another without an interaction; $\int S_1 S_0 d\tau = 0$ so no transition. States must have different symmetries if coupled by a photon similarly  a vibration with the correct symmetry can also couple S1 and S0 to cause internal conversion.

Transitions occur because an interaction mixes basis states together (singlet and triplet, say) to form new states containing part of each. These new eigenstates are shown schematically below, the thick lines on the lower pair show the amount of each base state in the mixed state.

One state can couple to many others, and when there are very many final states (below) we think of them as a quasi-continuum of levels. The radiationless transition rate is the rate from one state (optically excited) to these many final states.

Equation to solve is $\displaystyle (H_0+\Omega)\psi=i\hbar \frac{\partial}{\partial t}\psi$ to get rate constant $W_{kn}$ between levels $k,n$

Robinson & Frosch simplified the calculation of Fermi Golden Rule with approximations;
(a)Assume that the interaction energy is the same to all $n$ levels ,
(b)The (vibrational) Franck-Condon factors take an average value.
(c) The density of states $\rho$ is made to fit data! ( $\rho$ = number of states / wavenumber)

The approximate formula is: $\displaystyle W_E =\frac{2\pi}{\hbar}|V|^2F_c\rho(E)$  where $W$ is rate constant at energy $E$, $V$ is interaction matrix element between $k$ and $n$ , $F_c$ is the average Franck Condon factor and $|V|^2F_C=\int\psi_1\Omega\psi_2d\tau$. The $\Omega$ is spin-orbit operator for intersystem (singlet to triplet) crossing, kinetic energy operator for internal conversion.  The term $\rho(E)$ is the effective density of states at energy $E$.

This formula is not good for accurate calculation but indicates what is important.

The full and detailed calculation shows that :

(a) In the initial state, say singlet, only certain vibrational modes are effective in causing a radiationless transition.

(b) These modes are separated out and called promoting modes, at least one is present, and can only be identified by experiment.

(c) These promoting modes are the only modes which do not have zero electronic coupling terms.

(d) promoting modes have zero displacement between the two electronic states and are non-totally symmetric

(e) In the final state accepting modes receive the vibrational energy, these are modes of largest frequency so that as few quanta as possible of each are excited. Usually these are CH or CC vibrations.

There are two cases molecules in condensed phase and isolated molecules in gas phase.

(a) Condensed phase

The energy gap law  $\displaystyle K_{ST}=Ae^{-\beta\Delta E}$ is observed experimentally and is obtained theoretically from $W_{kn}$ by taking a Boltzmann distribution over initial levels and then the maximum value of resulting expression. It turns out that this term is greater than all the others combined. (btw, this energy gap law has nothing to do with Arrhenius equation.)

The full expression is

$$W=|V|^2\sum_m \left[\prod_n\frac{X_n^{m_n}}{m_n!}\right] \delta (\Delta E -\hbar \omega_k-\hbar\sum\omega_n m_n)$$

where $\displaystyle X_n=\frac{\mu_n\omega_n}{2\hbar}(Q_n-Q_k)^2$ is the displacement between vibration $n$ between energy surfaces $Q_n\to Q_k$. The term in the square bracket is the 'Franck Condon Factor density of states' assuming harmonic oscillators.  $\omega_k$ is the promoting mode and has no displacement between the two states. The $\delta$ term ensures energy conservation, i.e. $\delta(0)=1$, else is zero.

This equation is very difficult to use, but to a good approximation the complicated $\Sigma\Pi$ term can be replaced by its maximum value and a significant simplification is obtained.

We now take $X$ as being the displacement of the vibration of maximum frequency, $\omega$, usually CH or CC vibration. $\Delta E$ is singlet -triplet energy gap. The expression for the energy gap law is

$$W=|V|^2e^{-X}e^{-\gamma\Delta E/\hbar\omega}, \qquad \gamma= \ln\left(\frac{\Delta E}{\hbar\omega X}\right)-1$$

and $\gamma$ has values in the range 3 to 5 typically and is effectively a constant.

Its success is that it describes fall off with energy gap and also D/H difference in rate constants due to difference in frequency, see first figure above.  The difference to the gas phase is the averaging over the Boltzmann distribution.

(b) Gas phase

In the gas phase with a large molecule, say benzene, we are in the statistical limit with a huge number of final states, effectively a continuum, then the calculation is from one initially excited level $k$ to many levels $n$ in the final state.

$$W_{kn} \sim |V|^2\sum_n \big|\int\psi_n\Omega \psi_k d\tau\big|^2 \delta (\Delta E_n-\Delta E_k )$$

the equations to use in this case can be found in Beddard et al Chem. Phys. Letts. 1973, v 18 p 481, if you cannot get to this paper I can post again.


Why do charges result on atoms when coordinate bonds are formed? I have not been able to get my head around it yet.

For instance in the reaction between ammonia and borontrifloride why is it that a positive and a negative charge results? Is it that ammonia donates is lone pair completely? If not so, if sharing of lone pair donated by ammonia occurs why then should charges result?

The first thing to note about the charges that form during the formation of a coordinate bond: They are a formality.

The second thing to note about the charges that form during the formation of a coordinate bond: They are a mere formality.

There is little correlation, if any at all, between the formal charges assigned to two atoms that have formed a coordinative bond and the actual charges observed in the compound. Take $\ce{H3N-BH3}$: Nitrogen is so much more electronegative than boron, that the electrons are still closer to the nitrogen. Thus, nitrogen has $\delta ^-$ while boron has $\delta ^+$.

So where does the formalism come from? When drawing Lewis structures, after having filled in all bonds and lone pairs, check every single atom and mentally cleave all its bonds homolytically. Count the electrons, subtract the amount it should have as a neutral atom and the difference is its formal charge. Boron would have four electrons if all four bonds be cleaved homolytically, but it should only have three, so it gets a $1-$ charge.

This is only a formalism so that you can basically derive all structures the same way. It has no physical significance.

There are enough people who were unhappy about the ‘wrong’ charge distributions that they proposed different methods to draw these compounds: some used arrows from donor to acceptor, others just drew a simple bond, knowing that it would be different from a covalent bond. Each method has its advantages and disadvantages and all have survived in at least one niche.


Van der Waals Radius clarifications I just read about van der Waals radius and had some questions about it.

Would it be correct to say that in general, van der Waals radius decreases as we move from left to right in the periodic table?
The van der Waals radius of the noble gas in a period is larger than the radius of halogen that precedes it. Why is it so?
What is the difference between van der Waals radius for monoatomic (say neon) and non-monoatomic (say fluorine) gases?

Let me start by providing a reference to a nice collection of van der Waals' data (actually the site provides a lot of other interesting data about the elements, you might want to poke around). As to your questions,

Yes, that is a general trend, but as you can see there are many exceptions.  The trend exists for the same reason that covalent radii generally shrink as we move from left to right across the periodic table, every time we add another proton and electron to an atom, the electrons do not effectively screen the outermost electrons from the positive nucleus so they become drawn in closer to the nucleus.
& 3. There are a lot of different ways to measure the van der
Waals radius of an atom, but in recent times the preferred method is
to examine the crystal structure of a molecule.  Using $\ce{F2}$ as
an example, the crystal would be composed of fluorine molecules.  if
we measure the crystallographic distance between two adjacent, but
non-bonded fluorine atoms and divide by two, we would have a single
value that could be used to estimate the van der Waals radius for
fluorine (we would probably want to measure a number of similar
values for fluorine from other molecules, maybe using $\ce{CH3F,
NF3, ClF,}$ etc., for example, and average them together). Our
crystal might look something like the drawing below.

Because diatomic and polyatomic molecules are polarizable, the bonded fluorine atoms will likely be slightly polarized when packed in the crystal in order to stabilize (lower the energy) the overall system.  This polarization will cause non-bonded fluorines to draw a bit closer to each other then they would if there were no polarization.  A crystal of a noble gas (let's say neon) will just be a collection of monatomic neon atoms, no bonds, no polarization.  Therefor there is no "attractive" force to draw adjacent neon atoms closer together like there was in the case of fluorine.  So even before we make our measurements we might expect the van der Waals radius for neon to be larger than that for fluorine, which turns out to be the case.


How does potassium nitrate work as an oxidiser? Potassium nitrate is very popular compound in pyrotechnics, used as an oxidiser. It can be mixed with sugar to make a "rocket candy", with coal and sulphur to form black powder. You can stick it on rope to create functional fuse.

The saltpeter releases oxygen in all these reactions. But what are the results of reaction? I learned that $\ce{KNO2}$ is also oxidiser. So does the nitrate turn tu nitrite and then reacts again? Or does it form something else?

I have noticed that some strange bubbles come from my homemade fuse (photo will follow). Since potassium nitrate has melting point at 334°C (according to Google) it's likely the saltpeter melting.

And what about fumes? Does potassium nitrate or nitrite evaporate to turn the air poisonous or irritating?

If we assume potassium carbonate ($\ce{K2CO3}$) as the only solid reaction product, the overall reaction can be described as:
$$\ce{2 KNO3 + C12H22O11 + 11 O2 -> N2 + 11 CO2 + 11 H2O + K2CO3}$$

when using sucrose for candy. With sorbitol, an equation like

$$\ce{2 KNO3 + C6H14O6 + O2 -> N2 + 5 CO2 + 7 H2O + K2CO3}$$

seems fitting. In both cases, it is assumed that:

the nitrate is completely reduced to nitrogen
additional oxygen is needed

The latter is plausible, the rocket candy is supposed to burn layer-wise at the surface in contact with air. This allows for propulsion rather than one instant explosion.

Whether the nitrite is formed as an intermediate (or which organic intermediates are formed) is beyond my knowledge. However, once this partial reduction (nitrate to nitrite) has happened, I rather doubt that the reaction goes back again (to regenerate the nitrate).

As far as the bubbles are concerned, is it conceivable that the fuse wasn't completely dried and thus water was driven out?


Why is the pressure correction in the van der Waals equation proportional to (n/V)^2? The van der Waals equation for real gases is stated as follows:

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

For the coefficient $b$, we can reason out that more the number of molecules, the more volume will be occupied by the molecules (in turn reducing the free space available for the motion of the molecules). It must be linearly proportional to $n$ because $b$ represents the sum of the excluded volume of each molecule in a mole. Therefore, we must multiply it by $n$ to obtain the total excluded volume.

The coefficient $a$ is said to represent the strength of the intermolecular attractive forces. Intuitively, we can say that it must be proportional to the number of molecules per unit volume. The more the number of molecules around a molecule, the more attractive forces it shall feel.

Therefore, the correction term for pressure must be proportional to $\frac{n}{V}$. But we know that in reality, it is proportional to the square of $\frac{n}{V}$.

Why does it vary squarely?

Summing up the argument from the Wikipedia derivation, the reason it is proportional to $(\frac{1}{V_\mathrm m})^2$ where $V_\mathrm m=\frac Vn$ (the molar volume) comes from how the molecules experience the attractive force. The only molecules that experience a net attractive force are those close to the edge of the container because as we move toward the center of the container, the attractive forces on either side of a molecule cancel out. The force toward the center experienced by a molecule at the edge should be proportional to $(\frac{1}{V_\mathrm m})$. But we can also say that the number of molecules at the edge of the container should be proportional to $(\frac{1}{V_\mathrm m})$. So, by combining the proportionality constants into a single constant $a$, we obtain that the correction to the pressure should be $(\frac{a}{V_\mathrm m^2})=\frac {an^2}{V^2}$ .


Distinguish between two possible configurations for angular momentum in carbon atom Consider the possible values of $S$ and $L$ for carbon configuration $1s^2 2s^2 2p^2$ and the corresponding rapresentations with arrows indicating the spins (consider only $S=0,L=0$ and $S=0,L=2$, as for $S=1,L=1$ some are missing).

I can't understand which one of the two representations indicated in red belong to $S=0,L=0$ state and which one to the $S=0,L=2$ state (I took a guess making the picture).

The spin wavefunction is in all cases antysymmetric ($S=0$) and equal to
$$\chi =\left(\, |\!\uparrow_1 \downarrow_2\rangle-|\!\downarrow_1 \uparrow_2\rangle\,\right)\div\sqrt 2$$

Therefore the spatial wavefunction $(L)$ must be symmetric for Pauli principle and in the second red configuration is

$$\Psi =p_0 (1) p_0 (2)$$
in the first one  is

$$\Psi= \left[p_- (1) p_+ (2)+p_+ (1) p_- (2)\right]\div\sqrt2$$

The wavefunctions are different so there must be a difference between the two red configurations, but which of them should belong to $S=0,L=2$ and the other to $S=0,L=0$ and why?

You are separating the spatial and spin part of your two wavefunctions. I don't think that is a good idea here, because it does only work for one of the two configurations.

The wavefunction of a configuration is given by a Slater determinant build up from spin orbitals. In the following I will neglected the normalization constant of $\frac{1}{\sqrt2}$ for simplicity. The wavefunction for the first configuration ($\uparrow - \downarrow$) is

\begin{align}
   \Psi_1 &= \begin{vmatrix}
     p_-(1)\alpha(1) & p_+(1)\beta(1)\\
     p_-(2)\alpha(2) & p_+(2)\beta(2)\\
     \end{vmatrix}\\
    &= p_-(1)\alpha(1)p_+(2)\beta(2) - p_+(1)\beta(1)p_-(2)\alpha(2)
\end{align}

Note that this differs from your suggested wavefunction by the spin part of the orbitals, and there is no way to separate them out.

And for the other configuration ($-\uparrow\downarrow -$) we have
\begin{align}
   \Psi_2 &= \begin{vmatrix}
     p_0(1)\alpha(1) & p_0(1)\beta(1)\\
     p_0(2)\alpha(2) & p_0(2)\beta(2)\\
     \end{vmatrix}\\
    &= p_0(1)\alpha(1)p_0(2)\beta(2) - p_0(1)\beta(1)p_0(2)\alpha(2)
\end{align}

Both, $\Psi_1$ and $\Psi_2$ look quite similar now. However, since the spatial part of both orbitals in $\Psi_2$ is the same here, we can factor them out

\begin{equation}
  \Psi_2 = p_0(1)p_0(2)\underbrace{[\alpha(1)\beta(2)-\beta(1)\alpha(2)]}_{\chi}
\end{equation}

And we get what you have suggested for this configuration. But this does only work for $\Psi_1$!

Now, why is $\Psi_1$ assigned to $L=2$, while $\Psi_2$ is $L=0$?

I assume your confusion is because if you add up the orbital quantum numbers $m_l$ you get $0$ in both cases ($1-1=0$ and $0+0=0$). However, those sums give you $M_L$, not $L$. The many-electron quantum number $L$ has $2L+1$ components with quantum numbers $M_L=-L,-L+1,\dots 0, \dots L-1,L$. So for $L=2$ we have $M_L=-2,-1,0,1,2$, much like there are three orbitals $p_-$, $p0$ and $p+$ for $l=1$

So $\Psi_1$ would be the $M_L=0$ component of $L=2$, while $\Psi_2$ is the $M_L=0$ component of $L=0$.

But there is one issue with such assignments: You cannot really do it, as the configurations get mixed together. The point is, that the atom as a whole needs to be spherical symmetric (unless there is some external electric or magnetic field). But the individual $p$ orbitals are not. For example $p_0$ is (in the usual convention) aligned along the $z$-axis and zero in the $xy$ plane.

Mathematically speaking the configurations form a basis (Hilbert space) in which the actual electronic states are expanded. This is known as the configuration interaction method.

So for example the electronic state with $L=0$ would be
\begin{equation}
\Psi_2 = \frac{1}{\sqrt3}(|\uparrow\downarrow - -\rangle + |-\uparrow\downarrow - \rangle + |--\uparrow\downarrow \rangle)
\end{equation}

where each $|\dots\rangle$ denotes a Slater determinant.


Why can't antiaromatic compounds just escape planarity and become non-aromatic to avoid destabilization? Why can't just all anti-aromatic systems lose planarity (and hence conjugation) and be non-aromatic like cyclooctatetraene ?

Would they not be more stable that way? Because anti-aromatic systems tend to be highly destabilized.

What can be a good working rule to predict whether an anti-aromatic system will remain anti-aromatic or whether it will somehow manage to lose planarity and become non-aromatic?

They do try to escape planarity if possible. As a simple example, consider cyclooctatetraene. It has 8 electrons, which is a non-Huckel number, considering it's a multiple of four, and would correspond to an antiaromatic system, if the ring was planar, and allowed delocalization. However, cyclooctatetraene isn't planar. It's rather "tub-shaped", as seen in this structure here:

Click the image for interactive 3-D model

The electrons aren't delocalised either. Cyclooctatetraene is known to add bromine, and generally behave like a cyclic polyalkene in many aspects.

The only cases where molecules don't escape planarity is when they cannot. This happens in the case of cyclobutadiene. Such a molecule is so small, it cannot be distorted to such extents. As a result, it remains planar, antiaromatic, and immensely unstable:

Click the image for interactive 3-D model


Order of molecular mass of solvent; given boiling point Given three solvents X, Y, Z. Order of boiling point of pure solvent X < Y < Z and same order of molal elevation constant ($K_\mathrm{b}$). Find the order of molecular mass.

So I got this question in my tuition test and the solution is unclear. This question is troubling me a lot.

I can see how this question is troubling.

It don't think there is a solution.

$K_b = \frac{RT_b^2M_w}{1000\Delta H_v}$

Where $T_b$ is boiling point, $\Delta H_v$ is the molar heat of vaporization, and $M_w$ is molecular weight of the solvent.

(see Colligative Properties by W. R. Salzman)

So you can see that knowing $K_b$ and $T_b$ is insufficient to find $M_w$, knowledge of $\Delta H_v$ is also required.

Experimentally, from Table of cryoscopic and ebullioscopic constants you can also see there is no solution.


Why do we call O2 oxygen? I have been taught that oxygen is a chemical element, in other words a certain type of atom that has 8 protons in its nucleus. So why is O2 called oxygen? It is not a type of atom but rather a molecule.

I think what you may find most helpful is to know a bit of the history of element discovery and atomic theory.

The first pure substance containing only the element oxygen to be isolated was dioxygen ($\ce{O2}$), in 1774, though it was called "dephlogisticated air" until 1777 when Lavoisier used the term "oxygen" for the first time. This was some 30 years before John Dalton even proposed the first empirical atomic theory. Back then, we only barely had an understanding of stoichiometry, such that Dalton famously claimed the molecular formula for water was $\ce{HO}$. The fact that dioxygen is a substance made of molecules containing two atoms of oxygen probably wasn't widespread knowledge until at least 1811, with the gas stoichiometry experiments of Amadeo Avogadro.

Basically, for a point in time, we knew that there was a substance composed of a single type of atom, which could not be broken down into anything simpler. This fit the then-prevalent definition of an element; "a pure substance that could not be decomposed into any simpler substance". We knew that Lavoisier's "oxygen" had to be $\ce{O_n}$, for some n, but we had no reason to assume $n \neq 1$ for decades. By the time we figured out $n = 2$, the name "oxygen" was already widely used to refer to dioxygen. The fact that $n = 3$ also forms a stable compound in ambient conditions (ozone) would also not be known until 1867. A similar story happened with (di)nitrogen (octa)sulfur, (tetra)phosphorus, and so on. The only elements which form stable monoatomic substances in reasonable conditions are the noble gasses.

There is an interesting aspect to consider behind all this. There are some (such as Eric Scerri) who claim we are doing Chemistry a disservice in muddling together the properties of elements and the pure substances which they make. Nowadays our definition of an element is solely dependent on the number of protons inside an atomic nucleus, with no reference to reactivity or in what form the pure substance can be found. In this sense, the elements do not have "reactivities", "melting points", etc.; these are all properties of the pure substances. The only true properties of the elements are things such as electronic distribution, ionisation energies, and so on. However, it is common to see periodic tables stating the melting and boiling points the pure substances of each chemical element, and even Wikipedia bundles the physical properties of dioxygen with the atomic properties of elemental oxygen. For better or for worse, we're stuck with this subtle ambiguity in nomenclature.


Why do polar aprotic solvents favour SN2 over E2? I was reading about Substitution and Elimination Reactions and I came across the following on MasterOrganicChemistry:

  Polar protic solvents tend to favor elimination (E2) over substitution (SN2). Polar aprotic solvents tend to favor substitution (SN2) relative to elimination (E2).

I understand the first part. But I have trouble understanding the part with the polar aprotic solvents.

Consider the following example (from the same website):

Why is E2 not favoured? I can understand why polar aprotic solvents favour SN2:

  A polar aprotic solvent like acetone or dimethylformamide preferentially solvates cations, leaving an almost "bare" nucleophile. This increases its nucleophilicity

But I could find no explanation as to why E2 does not happen because of the solvent. What keeps the nucleophile (cyanide anion here) from abstracting a beta hydrogen?

Competition between substitution and elimination is one of the many points of confusion for introductory organic students.
In general, they are taught that there are four factors that will determine the outcome of a reaction where there may be competition between substitution or elimination.
They are:

Type alkyl halide (methyl, $1^o$, $2^o$ or $3^o$)
The strength/basicity of the nucleophile
The solvent
The temperature

I think that many students over-emphasize the impact of any one of these factors.  It is more useful to look at the overall impact of the factors and evaluate the reaction as a whole.

Let's examine the case in question:

Primary alkyl halide - generally favours substitution
The nucleophile is $\ce{^-CN}$ which is a good nucleophile but a weak base.
Polar aprotic solvent
Low temp.

The big factor here is not the solvent.  All of the factors point toward substitution, especially the alkyl halide and the nucleophile.  Remember that elimination reactions are essentially acid-base in nature.  If the nucleophile is not a strong base, elimination reactions are less likely.


Can an ether be selectively hydrolysed in the presence of an ester? What reaction, or protection/deprotection can be used to transform the ether group of a molecule  that also has an ester compound into an alcohol while keeping the  ester unchanged? For example:

The usual way of removing ethers, by acid hydrolysis, would also hydrolyse the ester. Could that preserve the ester under certain conditions?

There are certain ethers which are somewhat easier to remove than others. If you stipulate that the ether must have a CH2 group immediately adjacent to oxygen, then a couple of possibilities, off the top of my head, would be:

Benzyl ethers (R' = Ph), or substituted derivatives thereof (e.g. PMB, R' = 4-methoxyphenyl). These are typically removed with catalytic hydrogenation, or under reductive Birch-type conditions. A related choice is the benzyloxymethyl group (R' = OBn), which also comes off with hydrogenation. Electron-rich benzyl ethers can be removed by oxidation (e.g. DDQ, or CAN).
Allyl ethers (R' = CH=CH2) can be deprotected with Pd(0) or Ni(0) in Tsuji–Trost style chemistry.

If you don't need a CH2 group next to oxygen, then there are many more choices:

Silyl ethers can be removed with fluoride ion.
THP ethers can be cleaved under very mildly acidic conditions (a weak acid such as PPTS is sufficient). Esters usually need slightly more forcing conditions to be hydrolysed.

However, in general, a random ether (e.g. R = Et) would likely prove difficult to deprotect in a selective fashion. You could try using BBr3, but it might end up being very messy. Otherwise, ordinary ethers generally require very forcing conditions to be removed. There is a review on methods of effecting ether cleavage here: Tetrahedron 2005, 61 (33), 7833–7863 and a number of ways to cleave simple alkyl ethers are listed, but many of these are very harsh and are probably not very general.


Which salt is a surfactant and lowers the surface tension of the water? I am doing my experiment on the surface tension of the water. How to show the decrease of surface tension of water with the increase in the concentration of salt? Is there any salt that lowers the surface tension of the water?

Cetyltrimethylammonium bromide (CTAB) and its cheaper chloro homologue are just two examples of cationic detergents commercially available and frequently used e.g., for hair shampoos.  They are called cationic because beside lipophilic alkyl groups their head group is positively charged:

(source)
Note, there equally are anionic and nonionc detergents, too.
There are multiple approaches to demonstrate their effect to lower the surface tension of water.  Just a qualitative illustration is placing carefully a paper clip on water which will sink after the addition of a detergent:

(source)
If you are up to quantify the effect, the measurement of the contact angle between the liquid and the solid is a frequently used approach.  Again, the Journal of Chemical Education (2017JChemEduc230) published a lab class showcasing the relevance of surface tension e.g., when spraying leaves of lettuce.  A second nice demonstration I recall from this journal was to record the rising meniscus in pasteur pipettes (2016JChemEduc1647) in variation of the concentration of sodium dodecyl sulfate (a popular anionic surfactant).


How to use this salt analysis chart to determine the cations present in a given solution? With the only possibilities being $\ce{Ba^2+, Ca^2+, Cu^2+, Fe^2+, Fe^3+, Pb^2+}$, what cations are present in a solution which:
a) Forms a white precipitate with sulfate and with fluoride, no precipitate with chloride, and a blue precipitate with hydroxide which dissolves in ammonia solution

I have the answer with me, but I am not too sure how to use the above chart for determining cations in a solution, to answer the question. I can sort of make it out, but I am not completely sure.

For this question rather than reading the question and trying to understand what is going on all in one go, I suggest just work your way through the flow chart and refer to the question as you go along so that you don't overload your brain with excess information and confuse yourself.
So starting from the top, it asks if a precipitate will form if $\ce{HCl}$ is added. Looking from the information in the question, there is no precipitate formed. Therefore we can rule out $\ce{Pb^2+}$ being in the solution. Now following the steps in the flow chart, it asks if a precipitate forms if sulfuric acid is added to it. Looking from the data in the question, it says that a white precipitate is formed with sulfate. Therefore, we follow the next step which asks if a precipitate is formed when $\ce{NaF}$ is added. In the question it states that a white precipitate forms with fluoride. Therefore we can conclude that $\ce{Ca^2+}$ ions are in the solution.
However, if we look again into the question, it states that a blue precipitate also forms when mixed with hydroxide ions. Therefore, if we look back to the flow chart regarding the section on testing with $\ce{NaOH}$, it can be seen that traces of $\ce{Cu^2+}$ ions also exist in the solution. Therefore, both $\ce{Ca^2+}$ and $\ce{Cu^2+}$ cations are in the solution.
As you can see, once you know the general method, it is really quite simple and self-explanatory.


Is there a way to "electroplate" wood with copper? Maybe wood can be coated with copper (II) sulphate first, or roasted in order to form a conductive charcoal (carbon) layer.

The most common way to apply metals by electrodeposition to nonconductive materials is to apply a "strike" of underlying metal, usually nickel or copper, via a method like electroless plating. In electroless, the electrons for the reduction of the metal ions to the zero-valent state are supplied by a reducing agent in solution:

$$
\begin{align}
\ce{Red &-> Ox + 2 e^-} \\ \\
\ce{Ni^{2+} + 2 e^- &-> Ni^0}
\end{align}
$$

This "strike" coating is then used to provide the electrical connection necessary to apply the metal of interest—here, copper—by typical galvanic electrodeposition methods.


Cis and Trans Isomerism in Resonance-Stabilized Compounds Would it be valid to describe $N_2O_5$ isomers as cis or trans? As a matter of fact, does this molecule even have isomers, given that all its outer N-O bonds are resonance stabilized and therefore equal in length and strength?

Also for non-resonance stabilized compounds, would it be okay to describe say the N to O double bonds as cis or trans with respect to each other? I've just never heard of cis/trans being used this way.

Thanks for the picture, now I understand your question. All of the atoms in $\ce{N2O5}$ don't need to lie in a plane as is shown in the picture in your link.  At room temperature there would be rapid rotation about the 2 bonds connecting the central oxygen to the two nitrogen atoms, so there is no "fixed" geometry from which you could use the cis and trans references.  For example, imagine the conformation with one $\ce{NO2}$ group in the plane of the screen and the other perpendicular to the screen, even if each $\ce{NO2}$ group had one $\ce{N=O}$ double bond and one $\ce{N-O}$ single bond, the terms cis and trans would still have no meaning in this conformation. Further, as you noticed, all 4 bonds between nitrogen and the exterior oxygens are equivalent due to resonance, so again terms like cis and trans would have no meaning.  Finally, I can't imagine a situation where an $\ce{N=O}$ double bond (e.g. where the oxygen is unsubstituted) could have cis-trans isomers because the 2 lone pairs on the oxygen are equivalent.


When disqualifying trans fat, are we qualifying cis-fat as healthy? I have a question about fats. This may sound cliché! Because we are in the era of "trans-fat" free, "unsaturated" food. Not to mention it's the period where people get sicker and die faster (oh may be, the statistics weren't obvious in the past...).

Fats are important part of our diet and major source of energy (the mighty omega-3 fatty acids). But, The trans-fat, it's a linear molecule, it tends to stick and stacks up, builds up bad cholesterol, so that's not good for our cardiovascular system...

Does it mean, the opposite configuration, cis-fat is healthy for our body?
If cis is what's naturally found, by default it's unsaturated fat isn't it?

PS:

It's that time of the year again, for a community project we (the people who are new to biology, organic, biochemistry education) need to explain that to kids of age 7 and 10. They are going to go for shopping with us and we have to pop our eye balls to make sure they don't buy the wrong stuff. Because everything they adore seems to contain trans-fat from breakfast cereals, syrups, chips to doughnuts...list goes on. Saying something isn't bad not going to convince them an atom! So we gotta explain why they should go for trans fat free, unsaturated food or in the worst case just choose a saturated food instead of a trans-fatty thing and prevent any "kids-melo-drama".
I am struggling with this explaination and I doubt it's going to be any easier for the kids.

The breakfast cereals reference.

If cis is what's naturally found, by default it's unsaturated fat isn't it?

Both cis and trans are unsaturated.  "Unsaturated" means having at least one C-C bond that is not a single bond (but instead a double or triple bond).  The terms cis and trans refer to the geometry of the molecule about the double bond.

Trans fat also occurs naturally in milk, for example fats of vaccenic acid

Also, since a fatty acid can have more than one double bond, it can be both cis and trans, for example fats of Rumenic acid found in milk.

  When disqualifying trans fat, are we qualifying cis-fat as healthy?

Cis-fat can also have negative health consequences.  For example, according to the review article Dietary fat, insulin sensitivity and the metabolic syndrome Clinical Nutrition vol. 23 pages 447-456.

  monounsaturated and ω-6 polyunsaturated fatty acids have less negative effect on insulin sensitivity than saturated fat

In other words, even consumption of too much unsaturated, naturally occuring, cis fat can have negative health consequences, such as insulin resistance.

Your question says:

  Because everything they adore seems to contain trans-fat from breakfast cereals, syrups, chips...

but is this really true?

From what I've seen, cereals have little or no fat of any kind, let alone trans-fat.

Similarly, maple syrup has no fat.

Even potato chips don't necessarily contain any trans fat:

  So we gotta explain why they should go for trans fat free, unsaturated food or in the worst case just choose a saturated food instead of a trans-fatty thing

The amounts of partially hydrogenated trans fat in foods has been greatly reduced in the last 10 years.  Milk and meats naturally contain trans fat.

I wouldn't want you telling my kids they can't drink milk or eat a hamburger.  Sure it's bad to eat too many hamburgers.  The focus should be a balanced diet.


Is there a thermodynamic driving force for racemisation? Why does racemisation occur?

My doubts are:

Is it due to entropy?
If it is due to entropy, how is the reduction of optical activity by racemisation a consequence of increased thermodynamic stability?
Will the entropy increase or decrease? I have learnt in thermodynamics that it is increase in entropy that favours thermodynamic stability.

Thermodynamics is not usually helpful in understanding racemisation: think mechanisms and kinetics instead

The thing about enantiomers is that, from a thermodynamic point of view, they are the same so any process is not being driven by differences in the energy between the molecules.

What matters, if racemisation is to occur, is that there is some accessible pathway that allows interconversion. For example, in acidic or basic ethanol solution, (3R)-3-phenyl-2-butanone will racemise via the (achiral) enol form of the molecule. There is a small amount of the enol present which interconverts to the (chiral) molecule but without remembering which chiral molecule the enol was formed from. This will, ultimately, give a racemic mixture. If no such mechanism existed the molecule would not interconvert. What matters is that some such pathway exists.

The notorious medicine thalidomide is an interesting example. One enantiomer is a useful therapeutic, the other a dangerous teratogen. But making a pure enantiomer doesn't help as the two forms interconvert in the body (by a complicated mechanism).

These conversions are not driven by the different stabilities of the end products but by the inability of non-chiral intermediate molecules in an equilibrium with the chiral starting materials to remember the chirality of their progenitor. This, in effect, increases entropy. But that entropic force is only possible if there is an available reaction mechanism that passes through a non-chiral form allowing the increased disorder that results from a chiral starting material turning into a racemic mixture.


Chromium from electrode not oxidizing in electrolysis I performed an electrolysis of an aqueous sodium carbonate solution with a platinum electrode and a chromium-vanadium-steel-alloy electrode. The solution turned yellow at the chromium-vanadium-alloy electrode (I suspect that it is sodium metavanadate (NaVO3))

Then I also performed an electrolysis in an aqueous sodium carbonate solution with a platinum electrode and an inox electrode (the inox electrode is composed of iron, chromium,...) . A green compound (insoluble) was formed and when I later added acetic acid, a brown compound (I think Iron(III) acetate ) was formed.

But in both cases there were no chromium compounds formed (at least of what I know) So why didn't the chromium react (why didn't the chromium get oxidized)?

More information:
Voltage was around 13 Volts, the inox electrode was stainless steel (see http://en.wikipedia.org/wiki/Stainless_steel) and has according to wikipedia a minimum of 10.5% chromium content by mass. I read online that a regular chromium-vanadium-steel-alloy has a vanadium  content  of approximately  0.181%  and a chromium  content  approximately  1.00%. (however, I'm not sure about these last two figures) What is true is that there wasn't a lot of yellow sodium metavanadate after an electrolysis of around 20 minutes. ) The electrolysis solution contained sodium carbonate(Na2CO3) and water(H2O) the electrolysis was performed at normal conditions (1atm - 25°C)

Some of the ways you aren't at standard condition.
Metal solution so the activity of the metals are at their concentration in the alloys
The solutions have far below 1mol/L of their respective metal ions
The oxygen that you are reducing in basic conditions is probably about 258 micromol per litre-
whereas it should be 1 mol/L. The potential might be too high- its hard to tell the solution resistance is an effect.

If you were at standard condition with low voltage then the first experiment makes perfect sense-
From wikipedia full list of electropotentials
V2+ + 2 e−
    V(s)     −1.13

Cr3+ + 3 e−
    Cr(s)    −0.74
Vanadium is preferred.


Why is there an orbital angular momentum if the electron isn't properly revolving around the nucleus? I'm a tenth grader who's just been introduced to this subject of atomic structure, so please help.

I recently read that an electron doesn't have circular motion around a nucleus, its motion is kind of bizzare, buzzing and basically all over the place. (As long as it's around the nucleus) But we use the azimuthal quantum number to talk about an electrons orbital angular momentum.

Now, to be sure I didn't know what that meant, so I looked it up. I can't figure out why is there a question of orbital angular momentum if the electron isn't revolving around the nucleus.

Also, is the azimuthal number somehow related to the energy an electron has? I can't figure out if it's the radius that matters or the energy of the electron, so can someone please explain that?

In quantum mechanics, the classical concept of trajectory of a particle is abandoned. The quantum state of a particle i.e. the electron is characterized by a wave function $$\psi(\pmb{r}, t)$$ that contains all information about the particle and that extends over space. The simplest case where angular momentum appears in the context of quantum mechanics is the one of a particle in a “ring”. Now you have a wave function which extends over a circular region of space which implies that your particle has angular momentum. When you apply  the same idea to the Hydrogen atom you will find the orbital angular momentum.

The energy levels of the Hydrogen atom is given by
$E_{n} = E_{0}/n^{2} $,
where $E_{0} = 13.6 eV$ and n = 1, 2, 3, ... is the principal quantum number and describes the size of the orbital. Therefore, we see that $E_{n}$ is independent of the orbital angular momentum. If the atom is in the presence of a magnetic field then things change.


Does conductivity change in a solution due to rust? I plan on conducting an experiment in which I place an iron nail in water and analyze its rate of corrosion into rust with different concentrations of NaCl.

My original plan was to cover the water in a beaker with oil then use a dissolved oxygen probe to measure the drop in oxygen levels as the reaction went on. However, I was thinking that maybe conductivity would change as the reaction went on?

In short, in a solution containing iron, salt, oxygen, and water, will conductivity change as the iron rusts?

This is an interesting question. As iron rusts, you will have a couple of reactions. First, as you expect is the oxidation of iron:

$\ce{Fe ->Fe^2+ + 2e^-}$ or $\ce{Fe ->Fe^3+ + 3e^-}$

Then, either one of these may precipitate as their oxide, or hydroxide or their chloride.

On the flip side, somewhere else on the same piece of nail, there has to be a reduction. This may be one of:

$\ce {2H+ +2e- -> H2} $ or $\ce{2O2 + 4e- +4H+ ->2H2O}$

The overall reaction will probably be a combination of the above depending on the various rates of reaction. A priori, I would say it is very hard to predict which mechanism would be operative and therefore it is hard to predict which way conductivity would go.

Another point is even if the overall mechanism was such that there was some imbalance that would affect the conductivity, that would be very hard to measure. The changes would be a very small fraction of the total conductivity which would make it a hard measurement.

The typical way this is measured is "weight loss", which involves cleaning the corrosion products off the metal specimen and measuring the weight after given time periods. You can find lots of cleaning procedures and details of weight loss measurements over the net.


Structure of the bifluoride anion I came across this structure of $\ce{HF2-}$ and realized that the bond shown between H and F is a hydrogen bond.

$$\Large\ce{[F-H-F]-}$$

HF hydrogen bond are the strongest hydrogen bonds we know, but still they are not real bonds.
Writing the molecular formula as $\ce{HF2-}$ will give anyone an idea that the bonds between H and F are only covalent bonds.

Why do we write the molecular formula of $\ce{HF2-}$ as this and can you throw more light on its structure?
Have I correctly understood the structure?

Quoting from a comment of your’s:

  Hydrogen cannot make two covalent bonds since it has only one orbital in which an electron is present.

This sentence is correct for simple two-electron-two-centre bonds; i.e, if each of the two bonds going to hydrogen were to represent a 2e2c bond, hydrogen would have a double negative charge.

However, the $\ce{HF2-}$ anion does not contain two 2e2c bonds. Rather, it should be understood as a 4-electron-3-centre bond, similar to the situation in $\ce{ClF3}$. This can be understood by the following graphic:

Figure 1: Scheme of the 4-electron-3-centre bond in $\ce{HF2-}$. Image taken from Professor Klüfers’ web scriptum of his general and inorganic chemistry course.

The lowest molecular orbital, $\sigma_1$, is a fully bonding $\ce{F-H-F}$ orbital with two electrons. The second, $\sigma_2$ is nonbonding; no (accessable) hydrogen orbital exists that could complete it. The final one, $\sigma_3$, is fully antibonding. This is similar to the orbital scheme of extended linear π systems, where the number of nodal planes increases by 1 with every higher orbital. And it should be understood as such: the nodal plane of $\sigma_2$ traverses the hydrogen atom. Thus, while formally nonbonding, $\sigma_2$ is actually weakly bonding for both $\ce{H-F}$ bonds. If we calculate the overall bond order, we still count the electrons in $\sigma_2$ as nonbonding, giving a bond order of $0.5$ for each $\ce{H-F}$ bond.

The molecule can be understood in 2e2c Lewis terminology by using the following two resonance structures:

$$\ce{F-H\bond{...}F- <-> ^-F\bond{...}H-F}$$

‘Multiple bonds’ going away from a single hydrogen is not restricted to the 4-electron-3-centre bond known in $\ce{HF2-}$; they are also known and well-studied in boranes, although they are typically 2-electron-3-centre bonds in those electron-deficient compounds.


Dimers and polymers of electron deficient compounds Why does $\ce{BeH2}$ polymerize into $\ce{(BeH2)_{n}}$ in solid state whereas $\ce{AlCl3}$ dimerises into $\ce{Al2Cl6}$ in the vapour state ?

First of all, you have it all backwards.

You sound like the $\ce{AlCl3}$ says, "Oh, I am in solid state now! Let me turn to crystal (polymer) structure! [a while later] Oh, I am in vapour state now! Let me turn to dimer structure!"

Where in reality it says, "I don't have enough kinetic energy to break the bonds faster than they form, so I can only be in the crystal structure, in which I cannot move around, so I am in solid state now. [a while later] I have enough kinetic energy now! I can now break bonds faster than they form, but still not enough to break the bonds between the dimer, so let me exist as dimer. Also, I can move around now, since the intermolecular forces are not that strong. I am now a gas."

Almost everything polymerizes when the kinetic energy is low, forming a solid. The difference is only whether the bond responsible for the polymerization is van der Waals' forces or covalent bond or ionic bond.

For $\ce{BeH2}$, the bond is dative covalent as that makes the most sense.

For $\ce{AlCl3}$, the bond is ionic as that makes the most sense.

When the energy is high enough, $\ce{BeH2}$ decomposes, so all discussions henceforth are meaningless.

For $\ce{AlCl3}$, dimer has less energy than monomer, so dimer is formed. Remember that things like to be in states of low energy. Dimer enables it to move around, hence it is a gas.

When the energy is even higher, even the dimer bonds break and you have $\ce{AlCl3}$ monomers.


How to read the used volume from a burette? I conducted an acid-base titration with vinegar and sodium hydroxide today, where sodium hydroxide was used as the titrant (solution in the burette). I collected the data for the initial and final burette reading to work out the volume of sodium hydroxide used. Is the volume used just the final burette reading, as the measurements are counted backwards?

It depends ;-)

The scale on a burette typically is ascending from top to bottom. With other words: zero is close to where you fill your $\ce{NaOH}$ solution in, the largest value is close to the valve. The total amount added (to your vinegar) is the difference between the final and the initial reading.

But if your burette was initially filled up to the zero mark, this would just mean to subtract zero from the final reading (which is just the final reading).