text:
How to calculate Ka for hydronium and Kb for hydroxide?
I am wondering how to calculate the $K_a$ for hydronium, $H_3O^+$, and the $K_b$ for hydroxide. Because hydronium undergoes the equilibrium
$\ce{H_3O^+(aq) + H_2O (l) <=> H_2O (l) + H_3O^+ (aq)}$ ,
isn't the constant  exactly 1? (And the same thing for hydroxide?)
However, Wikipedia gives the $pK_a$ for hydronium to be -1.7. Does it have something to do with the "concentration" of water, which is 55.6 M, and upon taking the negative natural log, is -1.74?

To an extent, you are right; the calculation of the $K_\mathrm{a}(\ce{H2O})$ does include $[\ce{H2O}]$.
Consider the following equation:
$$\ce{H2O(l) +H2O(l) ->H3O+(aq) + OH- (aq)}$$
The equilibrium for this equation is:
$$K_\mathrm{a}(\ce{H2O})=\frac{[\ce{H3O+}]\cdot [\ce{OH-}]}{[\ce{H2O}]\cdot [\ce{H2O}]}$$
Since $K_\mathrm{w}=[\ce{H3O+}]\cdot [\ce{OH-}]=1.01 \times 10^{-14}$ at $25^\circ \mathrm{C}^{[1]}$, we can make such a substitution in our equation:
$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]\cdot [\ce{H2O}]}$$
Now here you will have to bear with me. In a usual $K_\mathrm{a}$ calculation, $[\ce{H2O}]$ is disregarded, as it remains mostly constant ($55.5084\ \mathrm{M}$), except when working with extremely concentrated solutions. In this equation, there are two $[\ce{H2O}]$ terms, and only one of them is disregarded (as a solvent). The other one can be considered concentration of the acid. Therefore:
$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]}=\frac{1.01 \times 10^{-14}}{55.5084}=1.819544 \times 10^{-16}$$
Thus we have calculated the $K_\mathrm{a}$ for water, or the $K_\mathrm{b}$ for $\ce{H3O+}$. The relation between these is:
$$K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$$
So:
$$K_\mathrm{a}(\ce{H3O+})=\frac{1.01 \times 10^{-14}}{1.819544 \times 10^{-16}}=55.5084$$
and:
$$-\log(K_\mathrm{a}(\ce{H3O+}))=\mathrm{p}K_\mathrm{a}(\ce{H3O+})=-\log(55.5084)=-1.7404$$
The calculation for the $K_\mathrm{b}(\ce{OH-})$ is exactly the same for the calculation of $K_\mathrm{a}(\ce{H3O+})$ (pretty much just replace all the $\ce{H3O+}$ with $\ce{OH-}$) and yields the same result.

[1] Starkey, R.; Norman, J.; Hintze, M. Who Knows the Ka Values of Water and the Hydronium Ion? J. Chem. Educ. Journal of Chemical Education. 1986, 63, 473.

Including water on both side of the equation is a bit of a mistake here.
For hydronium:
$$\ce{K_a = \frac{[H^+][H_2O]}{[H_3O^+]}}$$
and the source of the proton in question is not likely water (normally a stronger mineral acid).
If you put the water on both sides of the equation you are describing another process, which would be proton transfer equilibrium.  In that case, you are correct the constant would be 1, but all that means is that any water is as good as any other water at moving the proton around.


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What happens when a semipermeable membrane allows only one solute to pass through?
I'm essentially a medical student where we deal a lot with osmosis. But when we are taught, it is done generally with only a single solute in consideration.
What if two different solutes are used on either side? The membrane being permeable only to one, not the other but also free for water to move.
Lets assume a beaker of base 10cm x 20cm, of any height. Add 2 litres of water to it, it would come to a height of 10cm. Dividing the beaker equally into two with the semipermeable membrane of the specifications, we'll be left with 1 litre on each side. Now add 1 osmole glucose on one side and 1 osmole NaCl on the other (you can't add only Na+ ions). Where glucose and water can move but NaCl wont.
I guess it'll be safe to assume that $\ x$ moles of glucose moves to the other side resulting in a change in concentration on both sides. Now the water would also move causing not only a change in concentration due to the change in volume distribution but also a change in hydrostatic pressure.
Let the change in height be $\ y $ cm. Now the only constraint I can think of is that the pressure on either side should be equal at the base. By pressure i mean the sum of both hydrostatic and oncotic pressure. Oncotic pressure is given by $\ P_o = icRT$ and hydrostatic is by $\ P_h = h\rho g$.
Now I write $$\ P_{o1} + P_{h1} = P_{o2} + P_{h2} $$
That'd give $$\ (h+y)\rho g + \frac{n+x}{(h+y)A}RT = (h-y)\rho g + \frac{n-x}{(h-y)A}RT$$
Where $\ h$ is the height 10cm , $\ n$ is the no. of moles which we fixed as 1 mole (or rather, osmole) So rest all are constants and the only variables left here is $\ y $ and  $\ x $ which we need.
So rest all are constants and the only variables left here is $\ y $ and  $\ x $ which we need.
 On solving, the equation comes to a cubic in $\ y$ which is actually not so desirable!! Also we need another equation, since it is an equation with two unknowns. So I showed all this since this is the way I would proceed. But I am now stuck.
P.S This was a post on the biology forum and then the physics forum too, a discussion of which I extended here.

cross-posted http://physics.stackexchange.com/questions/241639/what-happens-when-a-semipermeable-membrane-allows-only-one-solute-to-pass-throug

Why do you think that glucose will contribute to the oncotic difference at all? As in why would it not be equilibriated in concentration on either sides, since there is no other force acting on it?

If you are willing to assume the concentration of the solution in any one partition is the same, then you cannot equate at the bottom. At the bottom, there has to be a net flux from the greqater height-lower height side, which is balanced at the topmost point of the semipermeable-contact by a flux in the opposite direction. You'll have to integrate along the height. (similar to the frank-starling capillary exchange, except that the exchange is balanced)

Now this is getting real complex! Could you post the solved answer @pasani

I do not get your mindset. If the membrane is not permeable to the other species, then there is not need for equating chemical potentials for those species so the presence of them won't have influence at all (except by modify the density of the solution).

How does it go then? Each compartment works separately? So the sodium won't have the slightest effect on the glucose which is trying to equilibrate? Could you please elaborate?

I am afraid you are stuck because you considered glucose as osmotically active.
But you assumed that glucose can freely move across the membrane, which is the opposite of osmotic activity.
So, short-short answer : ignore the glucose, it is an osmolyte, but it is not osmotically active on the membrane you are thinking of; its concentration is going to equilibrate on each side of the membrane :
$$[\mathrm{C_6H_{12}O_6}] = \frac{1 \mathrm{mol}}{2 \mathrm{L}} = 0.5\, \mathrm{mol\, L^{-1}}$$
This solution of glucose is simply the solvent.
Also, you may have inverted terms in the equilibrium equation.
A longer answer :
First we assume that the volume and density of the solution are unchanged by the solutes added (ideal solutions).
We are left with $1 \,\mathrm{osmol\cdot L^{-1}}$ NaCl on one side of the membrane, and no osmolyte on the other side.
The equilibrium between hydrostatic and osmotic pressure ($i c R T $) then writes :
$$ 2 \rho g y = i \frac{n}{(h+y)A} RT  $$
Solving for $y$, we get
$$ 2y = \sqrt{\frac{2inRT}{\rho g A}+h^2}-h$$
Using the following values :  $i=2$, $n=0.5 \,\mathrm{mol}$, $R=8.3\,\mathrm{kg\cdot  m^2\cdot s^{-2}\cdot K^{-1}\cdot mol^{-1}} $, $T=3\cdot 10^2\,\mathrm{K}$, $\rho=10^3\,\mathrm{kg\cdot m^{-3}}$, $g=9.8\,\mathrm{m\cdot s^{-2}}$ ,  $A=10^{-2}\, \mathrm{m^{-2}} $ and $h = 10^{-1} \mathrm{m} $, we obtain :
$$ 2y \simeq 7\, \mathrm{m} $$
This value is actually unreachable given the volume of solvent and the geometry of your beaker. All the glucosed water would be drawn on one side, and equilibrium could not be reached.
The osmotic pressure of a solution containing $1\,\mathrm{osmol\cdot L^{-1}}=10^3 \,\mathrm{osmol\cdot m^{-3}} $ is approximately $\Pi = 2.5\, \mathrm{MPa} = 25 \, \mathrm{bar}$. To equilibrate such a pressure with hydrostatic pressure, you need a column of water of depth $z = \frac{\Pi}{\rho g} \simeq 260\,\mathrm{m}$.
The osmolarity of sea water is about $1\,\mathrm{osmol\cdot L^{-1}}$. In a desalination plant, sea water is purified by reverse osmosis using pressures from 50 to 80 bar. Osmotic equilibrium is of utmost importance in physiology, but this you must know better than me. Unless there is active transport, diffusion of solutes, or extraordinary high applied pressures (which may be relevant for deep-sea creatures), water is always going from the least to the most concentrated side of the membranes, so as to even out the osmolarities. Hope it helped.

Hey! thanks a lot. Made me realize i've misunderstood the concept of a solute being 'osmotically active' all this while! It now makes perfect sense! Yeah sure! it did help. Here's a unrelated but further discussion on the topic https://www.reddit.com/r/askscience/comments/1wnh80/what_determines_whether_a_substance_is/

@SatwikPasani done :)

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What happens at equilibrium for two solutions that produce gasses?
Suppose one has two chemical reaction equations:
$\ce{A_{(aq)} + B_{(aq)} <=>[H2O] C_{(g)} + D_{(g)}}$
and
$\ce{A_{(s)} + B_{(s)} <=> C_{(g)} + D_{(g)}}$
Suppose one puts 0.1 mol of A and 0.1 mol of B into 1 litre of water. The water is contained in a closed rectangular container that can hold up to 2 litres of material.
The produced gasses C and D are lighter than water and so float up out of the water into the top of the container where they can't react . So if some equilibrium is reached for A and B in the water where they do not react any more it has to be of the form $k_{\mathrm{eqc}} = \frac{1}{{c_{\ce{A}}} {c_{\ce{B}}}}$ unless A and B completely and totally react together into gasses or do not react at all to form gasses.
The produced gasses C and D float up out of the water and when they react in the air they produce heavier solids A and B which fall to the water at the bottom and then dissolve into ions.  As there are no A and B products in the air any equilibrium for C and D has to be of the form $k_{\mathrm{eqp}} = {P_{\ce{C}}} {P_{\ce{D}}}$ unless C and D always completely and totally react together to form solids or do not react at all to form solids.
What happens here?
I think I'm making a mistake where I'm not factoring pressure in somehow.

To begin with, solids have vapor pressure as well...

The gist is that thermodynamics still works. Since there are two phases in both reactions, both are really irreversible reactions and will go towards products and not have a reverse reaction. The "equilibrium" in the first case is really based on activities in solution not concentrations. The activity of the gas will be defined as 1 so there is a sort of "equilibrium." In the second reaction the solids would have an activity of 1, and there would be no reverse reaction in any practical sense.

So, @MaxW the reaction must be either $\ce{A_{(aq)} + B_{(aq)} -> C_{(g)} + D_{(g)}}$ or $\ce{ C_{(g)} + D_{(g)}} -> A_{(aq)} + B_{(aq)} $ or else I'd have a violation of thermodynamics?  I see how at a macroscopic level this leads to a violation of thermodynamics but at a microscopic level why is that these reactions must be one-way? Also, for particular cases why would the reactions be one way or another?

Both reactions are irreversible because the reactants are in one phase and the products are in a different phase.  // I asked essentially same question. http://chemistry.stackexchange.com/questions/47384/does-an-irreversible-reaction-have-an-equilbrium-between-reactants-and-products

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Resistivity of pure water vs salt water for electrolysis
OK so I have a group project. Were "building" an ammonia plant. Part of the plan is to gather the $\ce{H2}$ we need from the splitting of pure water. We've been told $1000\ \mathrm{kg}$ of $\ce{NH3}$ must be produced an hour.
One of the problems asks for the voltage and current needed to split pure water to make enough $\ce{H2}$. I can do the half reaction, that's easy $1.22\ \mathrm V$.
But I've done this loads of times with kids and a $9\ \mathrm V$. I know that adding a little salt speeds up hydrogen gas production.
Does the $1.22\ \mathrm V$. assume that its pure water and I don't have to worry about the resistivity?

for 1000kg/hr I suggest you do not use electricity.  The amount of wire need to sustain the needed current will be outrageous.

The salt just allows for the current to flow.  The only reason it worked with "pure" water is because of the small amount of salts that were actually present in the water, or dissolved from the sides of the glassware, etc.

I hope you are just designing something and not trying to build it: that is quite a large task for this volume of ammonia!


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Energy of orbitals under a crystal field
It is shown in octahedral crystal field splitting that both the $e_g$ orbitals have equal energy. But how can that be possible ?
The $d_{x^2-y^2}$ is directed along both the axes (x and y) whereas the $d_{z^2}$ is directed along only the z-axis, shouldn't the later have less energy than the former ?

You envisioned the orbitals wrong. For example see: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/crystal.php#8

@MaxW I have read what is given in the link, but couldn't find where I am wrong. Would be helpful if you can point out.

They are equivalent in energy because symmetry says so, and because the calculations give that as an answer. Okay, that wasn’t exactly satisfactory, was it?
The next complicated answer on the road from over-simplified to reality is that the $\mathrm{d_{z^2}}$ extends to both ligands, too: Do not forget the ‘ring’ in the $xy$ plane. By extending towards the ligands on the $z$ axis further than the $\mathrm{d_{x^2 - y^2}}$ orbitals do to their ligands, it is less stabilised than it looks, and the interaction with the equatorial ligands also destabilises.
The really sophisticated and probably most correct answer is that we generally have a wrong view of those two orbitals. They are mathematically equivalent and would actually form a group of three: $\mathrm{d_{x^2 - y^2}}$, $\mathrm{d_{z^2 - x^2}}$ and $\mathrm{d_{y^2 - z^2}}$ — but because we can only have two, wee need to add two together and commonly take the two that give ‘$\mathrm{d_{z^2}}$’. In some images of molecular orbitals you can see that they are much more identical than they initially look.

If they 'actually form a group of three' then the d-orbital manifold would be able to hold 12 electrons.  These images, are they mathematical representations or empirically determined, such as in doi:10.1038/nature03183? I'm pretty sure even the more complicated molecular orbitals people have imaged look surprisingly like we thought they should.

@StevieD No to your first comment, because QC tells us that there is only space for ten. Just because something can mathematically extend in a beautiful way doesn’t mean it does. Partial yes to your second comment. But a professor of mine actually used that explanation and showed us how to look at a specific orbital representation. I have forgotten where to find it so I cannot reproduce it, unfortunately =C.

Yes Jan, I meant that the fact that they only accommodate 10 is evidence that there can only be 5 subshells of d orbitals. Likewise the ml quantum number agrees. The waveform of these orbitals may be represented as the sum of other waveforms, but these don't have a physical representation....I.e. it's just an artifact of the math. When in a ligand field the orbitals split in such a way that the net sum of the orbital energies are 0, that is they maintain the barycenter. If orbitals go up in energy due to repulsion between ligand orbitals and metal orbitals (spacial overlap) then

Others must go down to maintain an overall conservation of energy.


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Why does increasing orbital energy mean an electron can be more easily lost?
Why is the higher the orbital energy, the easier the electron lost?
Also, if the distance between the electron and nucleus is getting father，the   electronegativity is getting weaker, the electron is getting easier to lost?
I feel confused about this two concepts.

Why is the higher the orbital energy, the easier the electron lost?

Orbital energies are measured relative to an electron at infinity (i.e. one which has been ionized), which corresponds to an energy of 0 eV. An electron which is lower in energy than this (i.e. bound to an atom) has a negative energy.

The first six energy levels for hydrogen are shown above. As you can see, the energies converge to 0 as n, the principal quantum number, goes to infinity. In order to ionize an electron it must be given enough energy to promote it to the n = ∞ energy level. Therefore the higher the energy of the electron, the less energy is required to ionize it.

Also, if the distance between the electron and nucleus is getting father，the electronegativity is getting weaker, the electron is getting easier to lost?

The most likely distance of the electron from the nucleus (the maximum in the orbital's radial distribution function) increases with n and so the electron is in a higher energy orbital and is more easily lost.

1) As we go farther from the nucleus, the energy of orbitals increase. Also, the farther an electron is from the nucleus, the more repulsions it has to face because of other electrons(more screening effect) and hence it is easier to remove that electron.
2) As the distance between the nucleus and the electron increases, electronegativity gets weaker. It's true but we use this while comparing electronegativity of various elements and not for a particular element. For any particular element , its electronegativity is always fixed, because it's just the tendency of that atom to attract electrons and nothing else.

For any particular element , its electronegativity is always fixed, because it's just the tendency of that atom to attract electrons and nothing else. No it isn't. It depends on the oxidation state of the atom - or in organic chemistry this is usually viewed in terms of hybridisation rather than oxidation state.

See its related to screening effect which is denoted by $Z^*$ and $Z^*=Z-a$ where z is nuclear charge and a is atomic number as distance increases the effective charge of nucleus decreases resulting in loss of electron to attain stability .Higher energy implying farther
orbitals . and as distance increases electronegativity decreases down the group like $H>Na$ thus na us more reactive than H where H,Na are hydrogen,sodium


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Which substance can be easily adsorbed by charcoal?

Which substance can be easily adsorbed by charcoal?

$\ce{CO2}$
$\ce{SO2}$
$\ce{H2}$
$\ce{He}$

I thought the answer was (3) because the intermolecular forces are weak. So charcoal can easily adsorb hydrogen. But answer given is (2) which has strong intermolecular forces. Is my reasoning wrong?

The adsorption occurs because of the intermolecular forces, not despite them.

Gas molecules bind to activated carbon as a result of London dispersion forces. Therefore, the larger the molecules, as determined by the size and the number of atoms in it, the better they bind to the carbon.
If you were to rank these in order of highest dispersion forces to lowest, that list would read: $$\ce{SO2>CO2>H2>He}$$


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Are highest energy electrons lost first?
In the ground-state, is it always the case that the orbitals with the highest energy donate the electrons when the atom is ionized?

"Donate" is quite an ambiguous term. Think of "donating 20$ to a guy with a knife". You hit an atom with X-ray, it may pretty well lose an electron from any orbital, including the lowermost one, so what?

Which is, of course, the whole basis of Auger spectroscopy- electron from a deep level is kicked out, a higher electron drops in to the empty orbital, releasing enough energy to kick a different electron out at a characteristic energy.

No, sometimes you should consider ions like $\ce{Mn^2+}$. Its electronic configuration is $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^5}$. Although $\ce{3d>4s}$, it loses the electrons in $\ce{4s}$ first.

When 3d electrons exist in the same atom as 4s electrons, the 4s electrons are higher energy.

@DGS, always 4s orbitals have greater orbital energy (in absolute value) than 3d orbitals.

@user1420303 Then why don't they fill first as you go along the periodic table?

@DGS Sorry, I tried to say that they are always <> in absolute value.

@DGS I am unsure about what you mean by energy of orbitals. When I speak about orbital energy I am refering to the autovalue of the Fock equation for this orbital. Orbitals have not energy itself, I mean, the energy as described by physics in mechanics. In my context the filling order it is not difficult explained by this together with the fact that the sum of the orbital energies is not the molecule/atom energy.

Let us continue this discussion in chat.


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Can molten aluminium be used to reduce iron oxide (rust) into iron?
Will surface rust on an iron crucible be converted if used to melt aluminium?
Can one convert bog/lake iron ore into iron using molten aluminium similar to the Thermite reaction?
Or perhaps the real question is what temperature would this occur at?
EDIT: I'm hoping one could use the molten aluminium instead of aluminium dust to achieve the same reaction as occurs in the Thermite process.  I know about Thermite steel rail and copper earth connection welding. My goal is to reduce rust to iron by sprinkling it (with attendant sparks) onto a molten aluminium surface. Will the temperature be too low in practice or will the iron oxidise or blow off before consolidating?  Trials today were inconclusive due to low temperature and small sample size.

Short answer: you can

Related: How can Ellingham diagrams be linked to Gibbs energies of formation?

Short answer: Yes.
Take a look at the Ellingham Diagram.

The line for aluminium oxide is lower than the line for the all the iron oxides. This indicates greater stability of aluminium oxide, or in other words indicates aluminium's greater affinity for oxygen. Thus, based on thermodynamics alone one would expect aluminium to be able to reduce iron oxides.
Based on this figure, one would also expect this reaction to be spontaneous at all temperatures.
Important caveat, the ignition temperature remains critical. Just as a piece of paper will not combust unprovoked, even though the reaction is spontaneous, the thermite reaction needs to reach its ignition temperature.
Additionally, of interest to you would be this paper. Given below is a DSC curve from it

The endothermic peak ca. 660 corresponds to the melting of aluminium, while the exothermic peaks correspond to the thermite reaction.
The activation of the process is reported as 145 kJ/mol, and according to he authors in the study the product is controlled by the diffusion of $\ce{Al}$ in $\ce{FeAl2O4}$

To the extent that this theory suggests the reaction to be spontaneous, it is woefully inadequate.  The reaction is notorious for its high ignition temperature.

@feetwet I agree with you completely. In fact, I plan to update my answer with some additional references :)

If I am reading this correctly it implies that at a temperature somewhere around 850 to 950 degC we can expect the reaction to self start.  If there are issues with diffusion it would also suggest to me that this would work best if the components were mixed before they were expected to react.  Floating the iron oxide on aluminium might be expected to work at the interface though.  I suppose there is hope but it is going to have very hot molten aluminium to get it going.

Yes, I believe so. Although I would expect the ignition temperature to be > 1000 C...the information I could find showed some variance. However, I have no practical experience whatsoever so it may very well be lower. If you are carrying out the reaction, experiment with the temperature and see what results you get. I'd love to hear them myself :)

Aluminum can indeed reduce iron oxide, but you may not get iron metal. Aluminum and iron constitute one of many pairs of metals that form intermetallic compounds. Reference [1](https://doi.org/10.1007/s11661-003-0215-9) gives a phase diagram:

Thus upon reducing the iron oxide to iron, aluminum can also combine with iron to form several different compounds.
Among the iron-aluminum compounds, $\ce{Fe2Al5}$ plays a significant role in galvanizing steel. Running the steel strip into a pure molten zinc bath forms not a zinc coating, but another series of intermetallic compounds (of iron and zinc this time) which are brittle and easy to powder off. Adding a small amount of aluminum, typically 0.1-0.2% in the molten metallic solution, to the bath forms a very thin layer of $\ce{Fe2Al5}$. This allows the alloying of iron into the zinc to be prevented or moderated, thus obtaining the desired coating properties.
Reference

Han, Qingyou & Viswanathan, Srinath. (2003). "Analysis of the mechanism of die soldering in aluminum die casting". Metallurgical and Materials Transactions A. 34. 139-146. https://doi.org/10.1007/s11661-003-0215-9.

(+1) You always enlighten me. Very cool about how aluminum facilitates galvanizing.

Good find. Thermite is used for welding railway track.  The resulting quality of the steel is adequate for the job (though sometimes an alloy insert in the mould adds hardness to the top) so somehow at least at Thermite temperatures there is not a problematic concentration of these intermetalics but at low temperatures like I would hope for they might become significant.

Just had a look at the boiling point of Aluminium is over 2000C and the melting point of rust is about the same as that of Iron.  It may be that the temperature where we get the Thermite reaction will be hotter than the temperatures that will favour intermetalics.  With enough energy the temperature may boil off aluminium, the burn temperature of Thermite may exceed 2800C so aluminium is likely to be driven off.

@KalleMP that depends on conditions. Moten aluminum can dissolve some iron, and form the intermetallics, without melting the unreacted iron. This would occur at low enough temperature to allow said intermetallics. You must indeed get the termite hot enough to avoid this.

The trick to the thermite reaction of iron oxides with aluminum is the ignition temperature: somewhere over 1600 °C.  Since aluminum melts at 660 °C you have to add (and sustain on contact) a lot of heat before it can begin to steal oxygen from rust.

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Is it valid to use Gibbs Free Energy for gases?
This has confused me for a long time now. Why do we compute Gibbs free energy when we figure out the spontaneity of a reaction involving gases? In my mind, I envision a reaction involving gases to be always in a closed container, which would clearly change the pressure of the gas, but it seems to be that constant pressure is a requirement for using Gibbs free energy.
Now if it is that we are supposed to assume that the reaction involving gases does not occur in a closed container, why do we use molar concentration of gases when computing equilibrium constants? In this situation, it seems to be that volume of gases will change.
I'm suspecting that most of my confusion is on what constant pressure actually means in a reaction, so as a third question, what does constant pressure mean when we are referring to a reaction?

Hello, could you describe a little more specifically which problem you think would arise in this situation, or which kind of calculation you think that is not feasible to do with Gibbs energy? Just to be sure to adress your question with my answer.

Who says that the pressure of the gas has change in a closed container? Certainly, if there is a piston, the volume of the gas can be caused to increase such that the pressure is held constant.

Gibbs is for constant pressure, Helmholtz is for constant volume.

The change in Gibbs Free Energy can be readily determined if pressure is not constant, and the change in Helmholtz Free Energy can be readily determined if volume is not constant.

The Gibbs Free Energy has been found to be the most convenient thermodynamic function to work with when determining the equilibrium constant for a reaction or in establishing the chemical equilibrium state for a chemically reacting system involving multiple reactions.  The condition for equilibrium of these is that, at a given temperature and pressure, the Gibbs Free Energy is minimized with respect to changes in extent of reaction.  This applies to whatever temperature and pressure are currently present in the reactor, irrespective of whether they have been changing. For a mixture of ideal gases, it is possible to show that this Free Energy minimization is equivalent to requiring that the sum of the free energies of the pure reactants at pure gas pressures identical to their partial pressures in the reactor are equal to the sum of the free energies of the pure products at pure gas pressures identical to their partial pressures in the reactor.

Sir can you explain why Gibbs Free Energy is minimized w.r.t changes in extent of the reaction even if pressure and temperature is changing? Everywhere I've seen they derive dG<0 for reactions only at constant pressure and temperature.

Imagine conducting a reaction in a piston that is filled with the gases.  (As an example, think of a syringe with the "needle" port closed off; on the other end of the syringe there is a plunger that is gas tight but can move up and down.)  The atmosphere presses against the other side of the piston/plunger.
If the reaction causes the pressure to rise, even a tiny bit, the piston (e.g. plunger) will move up, expanding the volume of the gas inside the reaction, which reduces the pressure back to atmosphdoing work against the atmosphere.  If the reaction causes the pressure to fall, even infintessimally, the piston (i.e. plunger) will fall, as the atmosphere presses the piston down to maintain the pressure in the chamber to be one atmosphere.
The key is that reactions at constant pressure do not necessarily correspond to reactions at constant volume.  The container the gaseous reactants are held in needs to be able to expand or contract in order to maintain constant pressure, which of course means it changes in volume.   Such containers can still be closed in the sense that no atoms pass from the inside to the outside.

I think it makes no sense if you computes Gibbs free energy in a closed container.As @A.K. and @Chester Miller  said,Gibbs is for constant pressure, the change in Gibbs Free Energy can be readily determined if pressure is not constant.We can computes Helmholtz free energy
The Helmholtz energy is defined as:
A=U—TS
where

A  is the Helmholtz free energy
U  is the internal energy of the system
T  is the absolute temperature  of the surroundings, modelled as a heat bath
S  is the entropy of the system

The Helmholtz energy is the Legendre transform of the internal energy, U, in which temperature replaces entropy as the independent variable.

You've copied and pasted a section directly from Wikipedia here. You might want to elaborate on your answer to the OP's question.


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Which species occupy the equatorial positions in VSEPR's trigonal bipyramidal and octahedral geometry
Which species (electronegative, double and triple bonded atoms, ions, lone pair etc.) occupy the equitorial position in the trigonal bipyramidal and octahedral geometry when we find the shape using VSEPR theory?

Which species are given more priority and why?

This goes some way to explaining it.

In trigonal bipyramidal geometry, roughly speaking, the more bulky substituent goes to the equatorial position. In octahedral geometry, there is no such thing as equatorial position at all.

Less electronegative and double/triple bonded groups tend to occupy the equatorial position, while the more electronegative and single bonded groups tend to occupy the axial position. Ron's explanation accounts for these observations: http://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5

Trigonal planar is a molecular structure that requires five ‘substituents’ around a central atom (counting lone pairs as ‘substituents’). Assuming main-group compounds, that means, we have already left the realm of easily derived molecules and need to resort to 4-electron-3-centre bonds to properly explain the molecule and be in line with the octet rule.
This very good answer by Ron does a good job of explaining what the central atom’s electronic structure should be considered to be and what the 4-electron-3-centre bond is. In a nutshell, assume the central atom to be $\mathrm{sp^2}$ hybridised with an unhybridised $\mathrm{p}$ orbital. This additional $\mathrm{p}$ orbital then interacts with two electronegative substituents forming a 4-electron-3-centre bonds; this bond can be rationalised in a Lewis formalism by looking at these two resonance structures:
$$\ce{X-\overset{+}{A}\bond{...}\overset{-}{X} <-> \overset{-}{X}\bond{...}\overset{+}{A}-X}$$
The bond order of a 4-electron-3-centre bond is approximately ½, in good agreement with a single bond and a non-bond forming equivalent resonance structures for each $\ce{X\bond{...}A}$ fragment. Each outer atom has a partial negative charge while the central atom has a partial positive one.
This already means a few things:

a lone pair can never be in axial position as this position is formed by a $\mathrm{p}$ orbital. If a lone pair were in axial position, no second ligand can be there and the structure is actually better described as trigonal planar.
the more electronegative a substituent is, the more likely it is to occupy the axial positions as these positions require stabilising a partial negative charge. Hence, fluorine is most frequently found here.

This is also somewhat of an example for Bent’s rule — the more electropositive a substituent is, the higher the $\mathrm{s}$ orbital participation for the orbital forming the bond to said substituent. Conversely, the more electronegative the substitutent is, the higher the $\mathrm{p}$ content. Naturally, a pure $\mathrm{p}$ orbital is on one end of that spectrum, clearly preferring electronegative substituents.


text:
dsp2 and sp3 hybrid orbital shapes in coordination chemistry
Why is $\mathrm{dsp}^2$ square planar but $\mathrm{sp}^3$ tetrahedral in shape? I am unable to find a reason anywhere. I know how to figure which hybridisation a compound is, but not why.

How do you think you find the hybridisation of a molecule?

@bon I find the number of electrons, then check if the ligand is strong field or weak field. If it is strong I pair the electrons. The left over orbitals are used by the ligand, so they all get hybridised.

‘Stop using hybridisation to explain complexes’ is the answer to your question.

Square planar usually happens if there is a reason to keep two electrons (sometimes even one) in a $\mathrm{d_{z^2}}$ orbital for stabilisation reasons.


text:
Nomenclature of benzene derivatives with two functional groups
In several places, I have seen that while naming benzene derivatives, no priority for substituents is followed. For example, I've seen both 2-hydroxyaniline and 2-aminophenol being used.
Is there an order? Or is the order different from aliphatic compounds?

Aniline is technically a trivial name. Phenol is IUPAC-approved, but it's still kind of trivial. The naming will depend on the person using it and the situation it's used in.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the retained names “aniline” and “phenol” are preferred IUPAC names (PINs). Parent structures having retained names are called functional parent compounds.

For both retained names, substitution is permitted at any position. However, substitution of a functional parent compound is limited to substituent groups cited as prefixes in accordance with the seniority of the functional group that is explicitly expressed or implied in the functional parent compound name. (In systematic names, the presence of this characteristic group would normally be expressed as a suffix.)

P-15.1.8.2.1 Substitution rules for Type 2a retained names
Type 2a retained names include functional parent compounds whose name expresses or implies a characteristic group expressed as suffix in systematic names.
P-15.1.8.2.1.1 Substitution by substituent groups, expressed as prefixes, having a lower seniority than that denoted by the suffix is allowed.
(…)
P-15.1.8.2.1.2 The senior characteristic group or functional class name must be
  expressed in the name.

Therefore, the correct name for the compound given in the question is “2-aminophenol” (not “2-hydroxyaniline”) since hydroxy compounds are senior to amines.

Thanks, and sorry for the late reply. In my textbook they have named a compound as 2-aminotoluene. Since NH2 has higher priority, this is wrong, right? Should it be 2-methylaniline?

@user93868 Yes, see http://chemistry.stackexchange.com/q/47562/7951


text:
Does acetic acid give a positive result with the iodoform test?
Acetic acid contains a $\ce{-COMe}$ group, so theoretically, it should give a positive result in the iodoform test (i.e. a yellow precipitate of $\ce{CHI3}$ being formed).
But apparently the iodoform test is used to distinguish between acetic acid and other compounds with a $\ce{-COMe}$ group. So, logically, acetic acid must not give a positive result. Why is this so?

The carboxylic acid is deprotonated. You can't do a nucleophilic acyl substitution on it any more

The iodoform reaction proceeds by the mechanism shown below. (The final proton transfer need not occur between the two partners, any other hydroxide could abstract the acid’s proton and any water molecule can protonate iodoform.)

This can fail for carboxylic acids or esters in one of two ways:

The first step. The $\mathrm{p}K_\mathrm{a}$ value of the proton being abstracted is $26.5$ for acetone and $29.5$ for ethyl acetate according to the Bordwell $\mathrm{p}K_\mathrm{a}$ table. This can be attributed to the second oxygen of an ester donating electron density into the carbonyl group and thus lowering the tendency to form the enol(ate). But we can still argue that enough ester will be deprotonated for the iodination to occur. (Note that this fails entirely for a free acid, since the molecule is already charged. A second deprotonation to produce the endioldiate is very unfavourable.)

The final fragmentation step. In esters, $\ce{R}$ also contains an oxygen atom, so there are a total of three pathways by which the tetrahedral intermediate can fragment (one being redissociation of the hydroxide). In the ketone’s case, only one is productive, namely the formation of iodoform which precipitates and drives the equilibrium towards its formation. The ester can fragment under displacement of the alcoholate (productive, gives an acid) which is much more favourable than displacing a $\ce{CI3-}$ group due to the $\mathrm{p}K_\mathrm{a}$ value of both’s conjugate acids.
The resulting acid is merely deprotonated. The attack of a nucleophilic (hydroxide) species to the resulting anion is strongly disfavoured.

It is also to be noted that an ester can always be hydrolysed to the corresponding acid under basic aquaeous conditions. Thus, at any intermediate an acetate anion could generate inhibiting any further reactions.

Regarding (1) I don't think it has to be deprotonated to form the enolate - it could probably occur via an enol as well. In any case the carboxylate still has a lower tendency to enolise, so it doesn't change the analysis.

@orthocresol true that.

beautiful answer as always.


text:
What are these stains on my stainless steel pans?
I've had enough of the homemaker advice of baking soda or barkeeper's friend products - their advice isn't working.  I need a chemist!
Please see the two photos below.  This pan isn't quite as clean as I could possibly get them if I scrubbed and scrubbed with a Brillo pad, but it's close - and as clean as I can get it would only last for about one use before it needed more intense scrubbing.
The brand of this pan is Cuisinart which is relevant only for the particular variant of stainless steel they might be known to use, but I don't think my experience with this brand is much different than I've had with, say, Calphalon stainless steel pans.
I refer to these blackened areas as stains because they have no significant dimension to them, as opposed to "baked-on-crud" which has a dimension and is more expected/accepted by me than the stains.  The stains are rougher than the shiny parts though, impacting use.

1) Why am I getting these stains?  Am I doing something wrong (heating technique or kinds of food chemicals that adhere to SS)?  Are these pans inferior and really I need to buy very high-end/expensive pans (for a different grade of stainless steel) for them to be indestructible?
2) With knowledge of the chemistry involved that leads to this visual effect on the stainless steel, is there an industrial chemical (like those in carburetor cleaner or something) that could clean this stuff off without damaging the pan and which I could completely remove (to avoid toxicity in my food)?

These might be products of the Maillard reaction. Can steel wool remove it?

Steel wool cannot get all of it off

What were you cooking in the pan?

This is over time.  Bacon, eggs, other meats in some kind of oil or butter or cooking spray.  And each is single-use.  I can't use the pans twice without scrubbing the gunk off the bottom every time.  Food never just slides out and I can't flip the eggs over like I used to be able to do.  Only time it doesn't do that is if I'm, like, simmering a water-based broth.  Makes me sound like a terrible cook, but really I don't think I'm that bad :-)

Are you preheating the pan with the oil/spray over a high burner? It looks to me as if the gunk is burnt on the pan.

Yeah, I also think it's burned onto the pan - or at least burned into the pan.  But I don't think I'm using the heat so wrong to cause this.  How do I release the burned-on-whatever chemically?  Because mechanical force and abrasion isn't working.

First I think there is something wrong with your cooking technique. You shouldn't be getting burnt gunk on the bottom of the pan after every use. // Since the pan is stainless steel it doesn't have a coating like a non-stick pan. So you can be fairly aggressive to get it clean. In order I'd try (1) Some oven cleaner that you can brush on. Don't get it on handle or outside of pan which probably has a chrome finish. (2) Some abrasive cleaner like Ajax with a rag. (3) Some very fine wet grade sandpaper

Does the stain form on the bottom as well? If so is this less hard or just more convenient to scrub off.

What is ratio (Diameter of burner) / (Diameter of pan) ? It looks to me that you have a small pan on a large burner. The bottom of the pan should have a thick Al layer to distribute heat. But the sidewalls of the pan will be the thin stainless steel. So it seems that then stainless steel pan sides are getting too hot and burning whatever splatters up on them.

It could be that metals in your tap water are plating out;  i.e. the more active iron in stainless steel is being replaced by a less active metal such as copper. Manganese also tends to precipitate, leaving stain, and these, in particular, are dark colored, but can be removed.
Have your tap water tested, either at a nearby school or with a commercial kit(s) that include copper, manganese and lead tests. If there is a high concentration of anions, you might want to install an ion-exchange water softener.

Interesting.  This is Washington DC tap water, in case that makes it unlikely.  Also, my sink disposal collar looks like stainless steel but it doesn't have any discoloration (though it never gets the kind of heat that a pan does).

Washington, DC may have manganese in the water, particularly where there are older pipes:  https://www.dcwater.com/waterquality/faqs.cfm, https://www.dcwater.com/waterquality/household_water_quality.pdf

(I re-piped the whole house with copper - used to be galvanized.  Water quality seems much better since then.)

Your "but can be removed" link contains a suggestion of "A mixture of white vinegar and baking soda" which I find suspect - is this a special case?

Also, these stains seems to me to appear during cooking, not during cleaning.


text:
Carbon Dioxide Solubility in Water
I did a lab where I tested the solubility of carbon dioxide in water at different temperatures. I took 50 mL of carbonated water at different temperatures and then I titrated that with sodium hydroxide solution. The indicator I used was phenolphthalein with an endpoint of 8.3 pH. I'm unsure how to calculate the amount of carbon dioxide that was dissolved. I'm stuck about deciding which equation to use.
This equation would mean that there is a 1:1 ratio.
$$\ce{CO2(aq) + NaOH(aq) → NaHCO3(aq)}$$
And this equation would mean that there is a 1:2 ratio.
$$\ce{CO2(aq) + H2O(l)->H2CO3(aq)}$$
$$\ce{H2CO3(aq) + 2NaOH(aq)->2H2O(l) + Na2CO3(aq)}$$
Which one would be the correct equation to use?
Also, there are literature values for the solubility of carbon dioxide in water at different temperatures. Do those solubility values apply to my lab? I ask this because I measured the amount of carbon dioxide present in a given amount of carbonated water, I didn't add carbon dioxide to pure water to measure how much can actually dissolve.

What was the endpoint pH of your titration?  How were you monitoring it?  Colorimetrically?  If so, with what indicator?  Or did you monitor with a pH probe?

I used phenolphthalein, the end point was around 8.3.

Great -- could you edit the question to include that detail?  It's very important in determining the right answer, as you might infer from A.K.'s answer.

Sodium carbonate ($\ce{NaCO3}$) is basic and gives solutions with a pH of around 10.  Sodium bicarbonate ($\ce{NaHCO3}$) creates solutions with a pH of approximately 7.3.  This is in fact what the human body uses to regulate pH.  You want to used the sodium bicarbonate since it will yield a nearly neutral solution.

Compare your endpoint $\ce{OH}$ with the relevant $pK_a$ values.  From Wikpedia:
$$\begin{align}\\
\ce{CO2 + H2O = HCO3- + H+}  & & {pK_\mathrm{a}=6.3}\ \ \\
\ce{HCO3- = CO3^2- + H+}  & & {pK_\mathrm{a}=10.3}\\
\end{align}$$
The first of these equilibria is for $\ce{CO2}$ as the solute.  Remember that a base will deprotonate a weak acid only as the pH surpasses the $pK_\mathrm{a}$ for that acid.  So an endpoint pH of $8.3$ allows only the bicarbonate to form.


text:
What are 1A_g and 1B_u excitons?
I was going through a paper on organic LED chemistry [1]. I saw that the authors referred to "$1A_g$" and "$1B_u$" excitons (and also "$2A_g$" and "$2B_u$" excitons), terms which I don't really understand because I am from an engineering background. The terms are used on the 2nd-to-last paragraph of page 4 of the paper.
Can anyone help me understand this terminology?
[1] W. Barford, R. J. Bursill, and R. W. Smith, Phys. Rev. B 2002, 66, 115205

What is the paper?  Please consider editing your question to include at least a link to it.  Even better would be to both link and then excerpt a paragraph that uses the terms you are confused by.

Also, it appears you cross-posted your question to physics.se: http://physics.stackexchange.com/questions/241852/what-are-1ag-and-1bu-excitons  Please DO NOT do that.  Ask only in one place.

You're looking at symbols that describe the symmetry of the excitons. Have a look here for so-called character tables that decipher the symbols. All of this falls under the broader category of group theory.


text:
Vinegar and baking soda: Cleaning agent or neutralized sludge?
I've seen it suggested countless times that combining vinegar and baking soda is a great home cleaning solution, but this always seemed like nonsense to me.
I recognize that baking soda, a base, can eat away at certain kinds of grime and that vinegar, an acid, can operate similarly on different kinds of grime.
Mixed together, of course, the two create a neat foaming reaction and I think this is where the alleged myth would originate - people see the violent foaming and assume cleaning is occurring.
Isn't it true that the two are primarily reacting with each other and yielding only a relatively inert and neutralized sludge?  That the mix reacts with any grime less than they would if both were applied in isolation against the grime?  I assume that the only plausible benefit from this mix would be in any pockets of non-neutralized components floating on the foam and physically reaching new areas of grime that the individual volumes would not cover.
But I guess there could be more to cleaning effectiveness besides pH!
Am I right or are they?

I also think that by combining vinegar and baking soda you would get a less effective cleaning agent than both of them were before.

Sodium acetate? Might as well use sodium chloride. The only difference is that sodium acetate would be slightly alkaline. Pretty much useless, though.

Hmm, apparently acetic acid and bicarbonate can survive together in solution in very small amounts, at least according to these guys. I doubt it's enough to really do much.

@Jason All in all, you are right and the "folk chemistry" is wrong.

There is some physical abrasion related to the foaming, but I doubt it is highly effective at 'cleaning'

Keep in mind that you are not going to get a perfectly neutral result, you are cleaning with a salt and some of the remaining ingredient that was in excess. If you use balsamic vinegar it will smell nicer :-)

As far as cleaning agents, mixing baking soda and hydrogen peroxide will give you a pretty powerful one due to all the extra oxygen that's released. Doesn't store well, though, trust me.

One other possibility: the carbon dioxide bubbles create what is basically a scrubbing action. Chemical elbow grease!

You are absolutely right. Your conclusion is a perfect description of the phenomena and of the myth. I repeat it here, just for the fun : Mixed together, of course, the two create a neat foaming reaction and I think this is where the alleged myth would originate - people see the violent foaming and assume cleaning is occurring.

I agree that using vinegar and baking soda mixed together is a waste of time and chemicals and will not work well to remove soils. I use 9% (Cleaning Vinegar) to remove stains fairly effectively especially hard water born stains. I also use baking soda as a mild abrasive to scour glass, marble and similar polished surfaces. Bob, Alpine Ca.

Well, cleaning is a very broad subject. This is a question which could make a great Science Fair Project. Because vinegar is just a dilute solution of acetic acid (ignoring trace compounds present), and because both baking soda and sodium acetate are pretty soluble, it wouldn't probably be practical to try to make a saturated acidic solution of BS. That is, any precipitates will be made from alkaline solution, under normal circumstances.
There's two questions which would need to be answered:

Does fresh V+BS slurry act the same way as old V+BS slurry?
Are either better than water + BS?

Well, hmm. To be honest, IDK, but I'd guess "maybe" for both (which is why it would be interesting to see what happens). Let me explain:
Acetic acid dissolves soap scum (mostly K, & Ca salts of fatty acids) and dissolves water hardness (mostly Ca & Mg carbonates) So starting off getting rid of some or all of that junk could definitely be a good thing (and acetic acid is only going to be around for a very short time if BS is present in excess). But we know that acids attack metals, so leaving vinegar on metal isn't a good idea...and guess what? Once the BS fully reacts, the acid will be gone! I can't think of any other plausible reason that V+BS would be better than W+BS.
One other potential reason would be if the vinegar changed the abrasiveness of the BS, making it more effective (which is another thing you could test for), but I doubt it would. (possible but not quite plausible, if you know what I mean). Anyway, what you've got left is W+BS after the reaction, so what's the worst it can do?
I do know for sure that most cleaners (with the exception of many metal cleaners) are alkaline because higher pH tends to dissolve (and/or lift-off (which means form a colloid of or weaken the adhesion to the surface-to-be-cleaned of)) organics better than acidic solutions. Cleanser, which is usually just mostly silica, certainly is more abrasive than BS, but that also means it will tend to scratch (or fog) more surfaces. I don't know, but I'll bet most cleansers are alkaline for the reason mentioned above. (alkaline solutions are NOT more slippery, but they form soaps out of your skin, and those soaps are more slippery. So, the slipperiness you feel when you touch an alkaline (basic) solution is just your skin dissolving :)

I'll also note that water hardness is the sum of the concentrations of calcium and magnesium in their aqueous (ionic) states (carbonates are usually the source, but the definition of hardness is incorrect); and that bases (not acids) are used to precipitate the hydroxide salts of calcium and magnesium. The acid will dissolve the solid carbonate species, but you have to get rid of the calcium/magnesium.

One application of baking soda and vinegar is in cleaning a glass stovetop. You start with baking soda and water. The dissolved baking soda acts as a base, and the undissolved baking soda acts as mild abrasive.
Once you are done cleaning, you need to get rid of the baking soda, and might want to adjust the pH of the cleaned surface. So at this point, you can add vinegar, which reacts with the baking soda to yield sodium acetate, which is soluble in water.
$$\ce{NaHCO3(s) + CH3COOH(aq) -> Na+(aq) + CH3COO-(aq) + CO2(g) + H2O(l)}$$
This only works if you use baking soda by itself first, and then add the vinegar once the dirt is already off the stovetop.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Analysis of multiple spectra to determine an unknown structure
I've been trying to solve the structure of a molecule from a $^{13}\ce{C}$ and proton NMR as well as a MS spectrum, but I've been stumped, I've been staring at it for a while with no improvement.
I've attached the images below. From the MS, I understand the Mr of the molecule to be 122, the $^{13}\ce{C}$-NMR suggests six different environments.
The part that confuses me the most is on the NMR. There is a 3 hydrogen doublet and a 4 hydrogen singlet.  I went on to assume that the structure might be some sort of aromatic and symmetrical, and maybe 2 carbons in the same environment so that the formula might be $\ce{C7H10N2}$, this accounts for the Mr and the fact there is an exchangeable proton peak.
The spectra I've been provided are poor quality, so I can't even work out the coupling constants, or properly view the peaks.
I still can't figure out what it might be. Can you point me in the right direction please. I have a feeling I've been confusing myself over something that's simple.

Some points to consider to help you along a bit. Your 4H singlet is not a singlet; it is an overlapping set of peaks. Also, what you have labelled as a d at 4.8 is definitely not a d; looks a lot like a quartet to me. Your expected molecular formula is not correct. Look at your 1H spectra an summarise what groups you have; how many aromatic protons, how many alkyl protons, and which groups can be joined together. For instance, your 3H doublet indicates a particular group, and  what is is next to. The chemical shift and splitting of the 1H at 4.8 likewise tells you what and where it is.

The spectra provided are adequate to determine the structure, and there is no need to work out coupling constants. There are only two sets of peaks that show splitting anyway, and this provides enough information for your purposes.

CPE Lyon student? :P

@long oh it makes more sense now that you pointed out the 4.8ppm peak is a quartet. I thought the shoulders were just due to the low resolution. This clears things up thanks.

A very quick solution to this problem before my train stop.
Mass Spec gives MW of 122.
1H spectrum gives 5 aromatic, a 1H quartet at 4.8ppm, a 1H exchangeable proton at 2.4ppm and 3H doublet at 1.5ppm. The 4.8/1.5 splitting indicates a substituted ethyl chain with the C1 group having two substituents (CHRR'-CH3). The chemical shift of this 4.8ppm proton indicates some substantial substitution, such as a heteroatom. We also have a labile 1H, which could come from an alcohol or an amine.
13C spectrum shows monosubstituted benzene ring (4 carbon environments with ipso carbon downfield shifted). Also shows two alkyl carbons corresponding to the ethyl chain we have worked out above.
To summarise data to date:

1 monosubstituted benzene (C6H5)
1 ethyl chain, doubly substituted on C1 (CH-CH3)

That gives us a molecular formula of C8H9 which adds up to 105. That leaves 17...... which includes one labile proton, leaving a mass of 16.
As this is clearly a homework question, I'll leave you to join the pieces.

thank you, it seems that my confusion stemmed from thinking the 4.8ppm peak was a doublet instead of a quartet - I thought the shoulders were due to poor resolution

The quartet coupling is quite broad, and this comes about because there is also some coupling through to the -OH (but not visible due to exchange broadening) and also through to the aromatic ring.


text:
Why does bond length decrease with increased basis set quality for SCF methods?
From Reviews in Computational Chemistry Vol. 5, for $\ce{H2O}$ using HF, MP2, CCSD, and CCSD(T), all lengths in angstroms:
$$
\small
\begin{array}{rcccc}
\text{Basis}  & \text{HF} & \text{MP2} & \text{CCSD} & \text{CCSD(T)} \\
\hline
\text{STO-3G} & 0.989 & 1.013 & 1.028 & 1.028   \\
\text{3-21G}  & 0.967 & 0.989 & 0.993 & 0.994   \\
\text{DZ}     & 0.951 & 0.979 & 0.979 & 0.980   \\
\text{6-31G*} & 0.947 & 0.969 & 0.969 & 0.971   \\
\text{DZP}    & 0.944 & 0.963 & 0.961 & 0.962   \\
\text{TZ2P}   & 0.941 & 0.958 & 0.956 & 0.959   \\
\text{Experimental} & 0.957
\end{array}
$$
In case one is skeptical due to effects from the lone pairs and the changing bond angle, here are results for $\ce{BH}$ using Hartree-Fock:
$$
\small
\begin{array}{rc}
\text{Basis}  & r_{e} \\
\hline
\text{cc-pVDZ} & 1.235950 \\
\text{cc-pVTZ} & 1.222062 \\
\text{cc-pVQZ} & 1.220306 \\
\text{cc-pV5Z} & 1.220010 \\
\text{cc-pV6Z} & 1.219877
\end{array}
$$
and for $\ce{LiH}$ using Hartree-Fock:
$$
\small
\begin{array}{rc}
\text{Basis}  & r_{e} \\
\hline
\text{cc-pVDZ} & 1.619017 \\
\text{cc-pVTZ} & 1.607865 \\
\text{cc-pVQZ} & 1.605530 \\
\text{cc-pV5Z} & 1.605519
\end{array}
$$
Going on the NIST CCCBDB and looking at calculated geometries, I cannot find a counterexample, at least when considering only HF, as adding dynamic correlation introduces an effect to counterbalance this bond shortening, since electrons can then properly repel each other to levels beyond the mean-field approximation, and DFT offers no insight beyond HF here.
One potential reason that doesn't receive much attention is intramolecular BSSE, where one part of a molecule borrows the basis functions from another. However, in the case of my calculations, I would argue that only cc-pVDZ could be considered a "deficient" basis, and the trend is still seen when leaving it out. I also believe that this is more plausible for molecules with distinct "domains" that can interact with each other, like one amino acid in a peptide forming a hydrogen bond with another AA, etc.
The current rationale I have is that as the basis set becomes more complete, and functions with higher angular momentum (polarization functions) are included, the overlap between the atomic wave functions is increased. This increased overlap results in higher electron density in the bonding region, which strengthens it, causing it to become shorter, like going from ethane to ethene to ethylene.
Specifically, I'm looking for a literature reference where an explanation is given for the phenomenon, not just examples with or without comparison to experiment.

I think that your rationale is right. I would think it in terms of energy instead to make a little more precise. Thinking about a diatomic molecule, the nuclei-nuclei repulsion function of distance does not change with the basis set used. So the difference is strictly in terms of electronic part. At large distance the atoms behave more similar to an isolated atom, so few spherical harmonics are enough (for a relative description). As atoms get closer it is needed more flexibility to represent the very different probability density that there is in the space region located between atoms. With a smaller basis set this wave function is approximated poorer. If so, it is expected that the energy rises in such region if one consider the variational principle.
Here I calculated for the oxygen molecule in the region close to equilibrium distance (about 1.2 A) the difference in electronic energy between HF/STO-3G and HF/6-31* basis set:

The behavior close to 1.4A should be just an artifact in convergence (I did not take care about that).
Sadly I never found a discussion about that in the literature. Up to my understanding there is not to much that can be said about that.


text:
Non-Linear Solubility Trend
I am doing a lab on carbon dioxide solubility in water and I have a non-linear trend which seems to agree with the accepted trend. I am examining the change in solubility as a function of temperature. In the experiment I titrated carbonated water with sodium hydroxide and used molar ratios to determine the carbon dioxide content. The following graph shows the accepted trend:

(Graph from: http://www.rocketscientistsjournal.com/2006/10/co2_acquittal.html)
I've been thinking but, can't seem to figure out why the trend is non-linear. Is there any intuition behind this trend?

Nonlinear with respect to what? Amount of water? Amount of CO2? Temperature? Is it a closed system? Please be specific about what you are asking, as well as the experimental setup...

Sorry, it is updated now.

The solubility can be thought of as an equilibrium. The Van 't Hoff Equation describes the relationship between any equilibrium constant and temperature.
$lnK_2 - lnK_1 = \frac{\Delta H }{R} (\frac{1}{T_1}-\frac{1}{T_2})$ . Thus, the graph is not linear.

I just fixed the equation --there was an extra negative.

Solubility is linearly proportional, however, to pressure.


text:
Gibbs Free Energy and Entropy
I understand Enthalpy, the exchange of energy between products and reactants. But what is Gibbs Free Energy and Entropy? I know that Gibbs free Energy is the difference between the change in Enthalpy - [(the change in entropy)*the Temperature]. However I am not able to understand the physical meaning. What does this really mean? On the molecular scale where is it stored? If you can make it less abstract than it is right now, that would help a lot.
I am assuming once I understand Entropy, I should be able to understand Gibbs Free Energy. So, my question on Entropy is, where is entropy stored? I know it is the total disorder in the ENTIRE system, not just one molecule, the units for Entropy are J/K, and by disorder I mean the preference of the system to go to a chaotic/disordered state.
So, when solving for Gibbs Free Energy, what are we finally getting by subtracting the energy from the system (dS) from the reactions exchange in energy (dH).

Like you said Gibbs free energy is the enthalpy minus TdS so basically it's the energy available (free energy) after you take the thermal energy out of the system. The TdS term is the energy due to random thermal fluctuations so sometimes it's more useful to remove that from the enthalpy and deal with the free energy especially when trying to determine spontaneity of reaction.

Entropy is a measure of randomness in the molecules in a compound. When randomness increases, entropy increases.
This randomness consumes some energy of the compound because energy is required to maintain its randomness. This is represented by TΔS. This energy cannot be used for useful purposes. Thus the total energy which can be obtained from a compound is the total energy( ΔH) minus the TΔS term. This maximum useful energy that can be obtained from a compound is called Gibbs Free Energy (ΔG = ΔH-TΔS).

Entropy: it's complicated
There is no quick answer to what entropy means. It makes sense to take a course in thermodynamics to get a better sense; rather than trying to understand entropy from a single definition, you have to apply it to multiple scenarios and consolidate the results with your own experience of the world.
Gibbs energy of reaction: it is related to equilibrium
It is fairly easy, though, to state what the Gibbs energy of reaction $\Delta_\mathrm{r} G$ is useful for. $\Delta_\mathrm{r} G$ tells us in which direction the reaction would have to proceed in order to reach equilibrium (negative: proceed forward to equilibrium; positive: proceed in reverse toward equilibrium). In an isolated system at constant pressure in the absence of non-expansion work (i.e. no applied electrical power source, no photo-chemistry etc.), a reaction does not proceed in a direction away from equilibrium (a reaction with positive $\Delta_\mathrm{r} G$ would break the second law of thermodynamics under these constraints).
There is a relationship between $\Delta_\mathrm{r} G$ and the (non-expansion) work a reaction a reaction is able to do (or needs to be done on it to make it go away from equilibrium):
$$\Delta_\mathrm{r} G = w_\mathrm{max} $$
If $\Delta_\mathrm{r} G$ of a reaction is negative, it can do work ($w$ is negative as well), and if it is positive, you have to do work on it ($w$ is positive as well). How much work (or how little in case we have to do work) is given for non-attainable ideal conditions by the equality above.
$\Delta_\mathrm{r} G$ is concentration-dependent (depends on the current value of the reaction quotient $Q$):
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + R T \ln(Q)$$
If all species are at standard state (or more generally if $Q = 1$), we talk about the standard Gibbs energy of reaction, $\Delta_\mathrm{r} G^\circ$. The equilibrium constant and $\Delta_\mathrm{r} G^\circ$ are directly related, via:
$$\Delta_\mathrm{r} G^\circ = - R T \ln(K)$$

We got the state function Entropy from 2nd law of Thermodynamics as we searched for some mathematical way to deal with spontaneity and equlibrium criteria. But we got a problem with Entropy function that we will have to check change of Entropy of the system (not isolated one) and surrounding combined for a process to comment on spontaneity of the process for the system. It's not fair to have changes of surrounding for a process in our calculation. So we then tried to figure out some other ways to find spontaneity and equilibrium criteria for a system by only checking the changes of the system. To that end, we found the state functions G,A which provide us criteria for spontaneity and equilibrium for a system under specific constraints by their changes for the system only and not the surrounding. Treat Gibbs free energy as a state function just to understand the spontaneity and equilibrium criteria, it would be easier to think. Don't think it in the way you think about U(Internal energy) or H(Enthalpy), it may be difficult.

Entropy is a measurement of disorder in a sense mostly related to probability and statistics. A system with greater entropy contains a greater amount of possible microstates. For instance, two gas chambers with the same amount of gas molecules will have different entropy values if one of them contains different types of gases. The molecules in the chamber with more gas types can be displayed in many more ways.
In fact, by understanding entropy one can truly understand the statistical nature of all chemical processes and understand why Gibbs free energy can be seen simply as a measurement of the likeliness of a state existing or not and ΔG can be seen as the likeliness of a process happening or not. This can be shown by an analysis of the famous equation:
ΔG = ΔH – TΔS
We can begin by remembering that ΔH represents the heat exchange involved in a reaction at constant pressure, if it’s negative, the system releases energy and, if it’s positive, the system absorbs energy. Now, enthalpy is “stored” in the form of chemical bonds or interactions that are present in the system and, with the progress of a reaction or other processes like solvation, interactions are formed or broken and this energy is released or absorbed.
And what does –TΔS represents? As an endothermic reaction progresses, heat is transferred to the system from the surroundings, but why does that happen? Why is that spontaneous and not, say, the opposite reaction, which would be exothermic and release heat to the surroundings (supposing the reverse reaction is possible)?
Well, chemists and physicists know that the entropy of the universe always increases (second law of thermodynamics), that is just a way of saying mathematically that all that is most likely to happen will happen. So –TΔS is the term that accounts for the effect that the exchanged heat will have in the system or surroundings when it is transferred.
In the case of an endothermic reversible reaction, we have  ΔH > 0, so the reaction can only be spontaneous if TΔS is big enough. That means that the reaction can only be spontaneous if it leads to an increase of disorder in the system (increase in the number of possible microstates of the system) that is big enough to compensate the decrease of disorder in the surroundings, associated to the heat loss by the surroundings, which is accounted for by ΔH.
ΔG will only ever be 0 if the system is in equilibrium and if ΔSu (entropy variation of the universe) is also 0, meaning that the variations of entropy both in the system and in the surroundings caused by the two opposite reactions occurring at the same time are exactly compensated given the constant reaction rates achieved in equilibrium. The system is not only in equilibrium with itself, but also with the surroundings.
The key is to realize that you don’t need to measure the ΔS of the surroundings and yet you obtain information that ultimately depends on it. This is one of the great merits of this equation.
So ΔG can be seen as just a convenient way for chemists to evaluate the balance of factors that “decide” if a process is spontaneous or not, mathematically equivalent to simply checking which processes lead to an increase in the entropy of the universe.


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Why does plastic tear easier when you pull it quickly?
If you've ever taken the plastic from a plastic bag and pulled on it, you might know what I mean. If you pull quickly, the plastic tears relatively easily, but if you pull it slowly, it gets stretched out into an indestructible plastic string. Once stretched, pulling the plastic as hard as you did to tear it before yields no results.
Similarly, adhesive bandages are pulled off the skin (almost) painlessly if pulled quickly, but seem to stick even under the same level of force if it is applied more gradually.
What explains this phenomenon?

Polymers have a special property called viscoelasticity, which means they may strain elastically (elasticity) then flow (visco). The models below are mechanical representations of this phenomenon using springs and dashpots (dampers).

$\qquad\qquad\qquad$ Maxwell Model   $\qquad\qquad\qquad\qquad\qquad\qquad$ Kelvin–Voigt Model

$\qquad\qquad\qquad$Time Stress Deformation Plot
This property of viscoelasticity allows plastics, when pulled slowly (low strain rate), time for the polymer chains to rearrange and the plastic to flow which means it is less rigid but more elastic (think stretching silly putty). When the plastic is pulled quickly the plastic is more rigid and breaks (like breaking silly putty). AS I have alluded you can observe this with silly putty which flows when strain is low and will stretch to long lengths while snap into two pieces when pulled rapidly.
So far I have only explained why plastics will stretch or snap, but not why they are hard to tear after stretching. This is caused because as the plastic is stretched the polymer chains align as shown below. As it turns out polymers are hardest to tear perpendicular to the direction of the molecules because you have to break the bonds of the chain rather than some bond breaking and some chains separating. As a result, the plastic becomes hard to tear across the stretched part, but it is now easier to tear along the stretch (parallel) since in is mostly chain separating to do so, and indeed a tear in a stretched bag will follow the direction of a stretch.

But what's the typical length of a polymer chain in a plastic bag?  If it's short on the macroscopic scale, then even in the last case the main thing that could be happening when you tear a bag isn't that you are breaking carbon-carbon bonds, but rather that you are forcing the chains to separate from each other.  In that case, the reason the bag is harder to tear would instead be that the untangled chain arrangement means you have greater inter-chain attraction (from VDW forces?) to overcome.  Not disagreeing w/ you, just wondering what's the evidence that supports what you describe.

Evidence is that recycling plastics involving macerating chopping or grinding processed plastic significantly lowers the molecular weights of the new polymer. Chains are shortened. Eventually the polymer properties are too low.


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What does it mean to define entropy as a measure of a systems internal energy per unit temperature which is unavailable to do work?
Entropy was defined in one of my findings as "the measure of the internal energy of a system per unit temperature which is unavailable to do work". Can someone help me understand this definition better cause I have always defined entropy as the degree of disorder in a system.

The answer to this question goes back to the Gibbs equations according to which $$\mathrm{d}U=T\,\mathrm{d}S-p\,\mathrm{d}V$$ and for a reversible process $$đq = T\,\mathrm{d}S$$ So, if you have a change in internal energy $\mathrm{d}U$ and it does work $p\,\mathrm{d}V$ then it has also absorbed $T\,\mathrm{d}S$. The verbiage you are quoting is very awkward and is not very useful, nevertheless for your future studies you will be better off if you will never think of entropy as "disorder" in any sense. Entropy is not disorder and such concept will never help you solve any real problem. Also, never think of a body containing "heat"; a body has energy depending on its state but independent of external constraints; and it has entropy that depends both on its state and its external constraints. For example, an ideal gas after it has expanded into a larger volume has the same energy as before it has done no work and absorbed no heat but its entropy is larger than it was in a smaller available volume. The entropy is the same in all possible molecular configuration consistent with the volume, so it is the same whether the molecules occupy uniformly the available volume or because of a large fluctuation spontaneously collapse into a smaller volume form the larger one. So it is the constraint (the available) volume that determines the entropy not how the molecules are distributed in it in any given instant of time. No disorder.

I was always in dilemma with the thermal energy associated with entropy as being unavailable to do work..... I mean then how could Carnot engine work? Thermal energy is absorbed in the engine from the  hot reservoir it's associated with entropy change. Is this thermal energy unavailable to do work? No; A part of this heat/thermal energy gained due to the entropy change is used for work. So, I couldn't believe that phrase unavailable to do work.


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electrolysis of aqueous zinc nitrate
Can anyone explain why in the electrolysis of aqueous solutions such as zinc nitrate, the oxidation  half equation shows hydroxide ions forming oxygen and water -   can this half equation be balanced producing oxygen and hydrogen ions - if not why not?

Indeed it can. It might even be more appropriate that way.
Also, welcome to Chem.SE.

You mean, why we write $\ce{4OH- -> O2 + 2H2O + 4e-}$ and not $\ce{4H2O -> O2 + 4H+ + 4e-}$?
Indeed, you may find both these reactions in the table of standard electrode potentials (at $E^0=+0.401$ and $+1.229$, correspondingly; also, note the reversed direction). In some sense, the second equation may even be considered more appropriate, since $\ce{H+}$ is a more abundant species in the solution than $\ce{OH-}$ (zinc salts would produce slightly acidic solutions due to hydrolysis).
But the first reaction was chosen not without reason. See, when we electrolyze something, it is anions that rush to the anode to get oxidized. Water molecules are neutral; electric field would not drag them in either direction. So this might be the reason why the equation is written this way, even if the amount of $\ce{OH-}$ is minuscule and the overall reaction looks more like the second one.

Thanks for that, but what about   2OH−⟶O2+2H+ 2e−

That's out of question. There is no such thing as H, to begin with. If you mean $\ce{H2}$, then it is still impossible. You just can't have both oxidation and reduction in one half-reaction.

I really appreciate your help.  The question that is bothering me is a “selective discharge” question .  The example has used aqueous copper sulfate.  They have given the reduction of Cu2+  as one half equation and the oxidation of OH-    as the other.  However the text has written the oxidation half equation as 4OH- →O2  + 2H2O + 4e    My query is can the oxidation equation be written as
2OH-  →O2   +   2H+    +   4e

Now you propose a mix between the two. Well, no, nobody writes it this way; it would look kinda unnatural.


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Why can't lone pairs remain in unhybridised orbitals?
I have seen that lone pairs are almost invariably counted for determining the hybridisation of an atom.
My question is why can't they remain in unhybridised orbitals.

Possible duplicate of Why does ammonia have sp hybridization?. Though it is not an exact duplicate, that would be the answer to your question in any other case.

@ringo Related yes, duplicate no. Also I find the answers given there highly unsatisfactory.

There are many examples of lone pairs in orbitals that are unhybridized.  See this earlier answer for several examples.  Another example, phosphine is discussed here.


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Galvanic cell chemistry
I made a galvanic cell with the anode electrode being aluminum and the electrolyte being aluminum sulfate. The salt bride was magnesium chloride. I am a little confused on what my oxidation equation would be for this set up.
So I did some research and I found out that aluminum sulfate in water will go to aluminum hydroxide and sulfuric acid, so I proposed this half cell equation:
$$\ce{2Al_{(s)} + Al2(SO4)3_{(aq)} + 6Cl- + 6H2O_{(l)} -> 3H2SO4_{(aq)} + 2Al(OH)3_{(s)} + 2AlCl3 + 6e-}$$
So this would then become:
$$\ce{2Al_{(s)} + 2Al^3+ + 3(SO4)^2- + 6Cl- + 6H+ + 6OH- ->
6H+ + 3(SO4)^2- + Al(OH)3_{(s)} + 2Al^3+ + 6Cl- + 6e-}$$
The spectator ions cancel off to give:
$$\ce{2Al_{(s)} + 6OH- -> 2Al(OH)3_{(s)} + 6e-}$$
This equation makes sense but aluminum hydroxide should form a precipitate. However, when I ran my cell, there was no visible precipitate, so it makes me question whether my anode half cell reaction is correct.
What am I doing wrong? Is the process I used to describe the half reaction right?

I think $\ce{OH-}$ is too little

What is your reduction reaction? You didn't mention that anywhere.

I think that what @user6006786 suspects is right. Maybe you can measure the pH and use the Ksp to test the hypotesis.

The aluminum may not hydrolyze all the way to a neutral hydroxide.  It could form hydroxo complexes that remain in solution, especially if the solution is made acidic by the same hydrolysis.
Consider a half-reaction in which the Al is combined with only one or two hydroxide ions.
Some other tips:
1)  Since you have an acidic solution don't put hydroxide ions on the left.  Put water and balance with hydrogen ions on the right.
2)  Sulfate ion is a weak base.  It will combine with hydrogen ions to make bisulfate.  So add sulfate on the left and incorporate the hydrogen ions noted above into bisulfate ions.

Your equations are right, but look at it this way:
$\ce{2Al_{(s)} + 6H_2O_{(l)} -> 2 Al^{3+}_{(aq)} + 6 H^+_{(aq)} + 6 OH^{-}_{(aq)} + 6e^-  <-> 2Al^{3+}_{(aq)} + H_2O_{(l)} + 6e^-}$.
The reaction produces equal amounts of hydrogen and hydroxide ions, and they reconvert to water. Thus, there is not as much hydroxide in the solution to precipitate.


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How to calculate transition dipole moment from two known wavefunctions
I am interested in calculating the transition dipole moment (TDM) from the information from two wavefunctions of different states. This is somewhat similar to calculating the molecular dipole moment which was previously answered here:
How to calculate molecular dipole moment from a known wavefunction?
The information I have is the molecular orbital coefficients of alpha and beta from states 1 and 2 and the multipole matrix.
In considering just one Cartesian direction X, calculating the dipole moment along X would be something like
$$\mathbf{P_a} = \mathbf{C_a} \cdot \mathbf{C_a^T}$$
$$\mathbf{P_b} = \mathbf{C_b} \cdot \mathbf{C_b^T}$$
$$\mathbf{P} = \mathbf{P_a} + \mathbf{P_b}$$
$$\mathbf{\mu_{matrix}} = \mathbf{P} \cdot \mathbf{\text{multipoleMatrixX}}$$
$$\mu_{electronic} = trace(\mathbf{\mu_{matrix}})$$
$$\mu_{nuclear} = \sum_{\text{all atoms}}(Z_{nucleus} * \overrightarrow{r_x})$$
$$\mu_{X} = - \mu_{electronic} + \mu_{nuclear}$$
where $\mathbf{C_a}$ and $\mathbf{C_b}$ are the occupied alpha and beta molecular orbital coefficients, $\mathbf{P_a}$ and $\mathbf{P_b}$ are the alpha and beta density, $\mathbf{P}$ is the total density, and $\mathbf{\text{multipoleMatrixX}}$ is the multipole matrix for the X direction.
I might think that to calculate the transition dipole moment I might just change the densities to the transition densities doing something like I have show below. The transition denisty is based on page 10 of the article here:
$$\mathbf{S_a} = \mathbf{C_{a1}} \cdot \mathbf{C_{a2}^T}$$
$$\mathbf{S_b} = \mathbf{C_{b1}} \cdot \mathbf{C_{b2}^T}$$
$$\mathbf{P_a} = \mathbf{C_{a1}} \cdot \mathbf{C_{a2}^T} \cdot \mathbf{S_a^{-1}}$$
$$\mathbf{P_b} = \mathbf{C_{b1}} \cdot \mathbf{C_{b2}^T} \cdot \mathbf{S_b^{-1}}$$
$$\mathbf{P} = \mathbf{P_a} + \mathbf{P_b}$$
$$\mathbf{\mu_{matrix}} = \mathbf{P} \cdot \mathbf{\text{multipoleMatrixX}}$$
$$\mu_{electronic} = trace(\mathbf{\mu_{matrix}})$$
$$\mu_{nuclear} = \sum_{\text{all atoms}}(Z_{nucleus} * \overrightarrow{r_x})$$
$$\mu_{X,TDM} = - \mu_{electronic} + \mu_{nuclear}$$
where $\mathbf{C_{a1}}$ and $\mathbf{C_{a2}}$ are the occupied alpha molecular orbital coefficients from state 1 and 2, $\mathbf{C_{b1}}$ and $\mathbf{C_{b2}}$ are the occupied beta molecular orbital coefficients from state 1 and 2, $\mathbf{S_a}$ and $\mathbf{S_b}$ are the alpha and beta overlap matricies, $\mathbf{S_a^{-1}}$ and $\mathbf{S_b^{-1}}$ are the inverse of the overlap matrices.
I am unsure how to calculate the TDM since it is in disagreement with known calculated TDM values; some of these transitions should have a TDM of zero due to symmetry considerations. I have been unable to find any discussion on calculating the TDM from two wavefunctions, and most discussions I have seen are in the context of some perturbation on the ground state (i. e. TD or CI). I would be grateful if anyone could provide some suggestions or references to look at.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. You might want to take a shot of expressing the equations you use in these terms. It is a bit hard to follow as of now.

Can I assume that $\mathbf{C}_{a}$ is a vector and hence $P_a$ should be a scalar, i.e. $P_a = \mathbf{C}a\cdot\mathbf{C}a^{\mathbf{T}}$. Or are $\mathbf{C}{a/b}$ matrices, hence $\mathbf{P}{a/b}$ are those, too. I'd assume the latter. However I don't know what you mean with $\mathbf{multipoleMatrixX}$. And with $\text{X coordinate}$ do you mean $\mathbf{r}_x$, $\vec{r}_x$ if you prefer arrows?

$\mathbf{P}$ and $\mathbf{C}$  are both matrices. I tried to represent matrices as bold and have scalars be not bold. $\mathbf{MultipoleMatrixX}$ is the multipole matrix along the X direction. The $\overrightarrow{r_x}$ is meant to be the X coordinate

Right now, your use of MathJax makes it even harder to understand, that's why I was asking. It looks like computer pseudo code and not like mathematics. So $$\mathbf{P_a} = dot[\mathbf{C_a},transpose(\mathbf{C_a})]$$ is $$\mathbf{P}_a = \mathbf{C}_a\cdot\mathbf{C}_a^{\mathbf{T}}?$$ And I have no idea what a multipole matrix along the X direction is, can you please include the definition of such a thing.

That is the correct interpretation, I have updated the formatting. The multipole matrix is the dipole integrals, a further explanation is in Modern Quantum chemistry by Szabo and Ostlund on p. 151.

Why not just subtract dipole moment vectors calculated for different states independently?

Unfortunately, it is not that simple. The of diagonal elements are needed. I believe this is to account for the phase,  i. e. if the relevant orbitals are perfectly out of phase the coupling is zero, regardless of what the state  dipoles are.

I hate to admit that I have been working on this on and off ever since you posted the question, since it is IMO a great question and right up my alley, and now I have motivation to finish my answer. Anyway, although it is not necessary, could you (in a comment) go into more detail about what disagreement you're seeing? A link to some code? It isn't necessary to answer the question, just curious.

@pentavalentcarbon So, it is basically for using constrained DFT (CDFT) or the maximum overlap method (MOM) to optimize an excited state, in the trial cases I have focused on simple things, namely formamide. In formamide the pi-->pi* transition should be symmetry forbidden the Z direction. The formamide and method reference I am using is from the paper  Gilbert, A. T. B.; et. al. J. Phys. Chem. A 2008, 112 (50), 13164–13171 DOI: 10.1021/jp801738f.
The script I am making is for Q-chem, but could probably work with other programs (NWChem, GAMESS).

You may find this video very helpful, especially around 5:30 : https://www.youtube.com/watch?v=YG9eXUzlHRk

Are your states orthogonal or not? If not, you need a different algorithm to calculate the expectation value (http://journals.aps.org/pr/abstract/10.1103/PhysRev.97.1474).


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Nucleophilicty and Basicity of OH and F

Why is $\ce{OH-}$ more basic than $\ce{F-}$?

From what I understand, $\ce{O}$ has larger radius so it should be more stable (meaning less basic) than $\ce{F}$. When they form anions, since $\ce{O^{2-}}$ is larger, it can distribute its electrons over larger volume than $\ce{F-}$, meaning $\ce{OH-}$ should be less basic. Is this because of partial negtaive charge on oxygen?

In a polar aprotic solution, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?

I learned that basicity (not nucleophilicity) is about stability. So can this be explained by electron density? Is explaining it with polarizability right?

In a lecture, my teacher said $\ce{RO-}$ is relatively stable because of the high electronegativity of $\ce{O}$.

I think it should be unstable because of high electron density caused by high electronegativity. What's right?

When people say something is stable, are they referring to its acid/base characteristics, or can stability also be referring to nucleophilicity/electrophilicity?

Why is $\ce{OH-}$ more basic than $\ce{F-}$?

To answer this question, let us consider the $\mathrm{p}K_\mathrm{a}$'s of the conjugate acids of these anions: $\ce{H2O}$ and $\ce{HF}$. $$\mathrm{p}K_\mathrm{a}(\ce{H2O})=15.7\ \\  \mathrm{p}K_\mathrm{a}(\ce{HF})=3.17$$
You didn't need me to tell you that $\ce{HF}$ is a stronger acid than water though, but it is -- $10^{12.53}$ times stronger to be exact. This means that this is also how many times stronger $\ce{OH-}$ is than $\ce{F-}$

In a polar aprotic solution, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?

Small, electron-dense ions are always great nucleophiles, except when the solvent crowds the nucleophile with hydrogen bonds, as in the case of $\ce{F-}$. It is so heavily surrounded by solvent molecules in polar protic solvents that it can't make an attack very well. $\ce{I-}$ is not as heavily solvated in these solvents, and it is very easily polarizable, making it a better nucleophile in these solvents.
In polar aprotic solvents, the small $\ce{F-}$ ion can easily slip between solvent molecules and make attacks. Now in comparison, $\ce{I-}$ is a much poorer nucleophile.

In a lecture, my teacher said $\ce{RO-}$ is relatively stable because of the high electronegativity of $\ce{O}$.

Again, we can answer this by looking at $\mathrm{p}K_\mathrm{a}$'s. The $\mathrm{p}K_\mathrm{a}$ of ethanol, for example, is $15.9$! Only $10^{0.2}=1.58$ times weaker an acid than water! That means that ethoxide will deprotonate everything $\ce{OH-}$ will and then a little more, so it's not really that much more unstable.

When people say something is stable, are they referring to its acid/base characteristics, or can stability also be referring to nucleophilicity/electrophilicity?

The two are pretty well related. You have exceptions like how $\ce{F-}$ is a poor nucleophile in water despite being a good base, but as a rule of thumb, I would say people mean acid/base reactivity when they talk about "stability."

Thank you. it makes sense when you explain it using pKa. but can't you explain another way using movement of e- like I wrote in my question?

Yes, but both $\ce{F-}$ and $\ce{OH- /OR-}$ have unpaired electrons. What is more important is to understand how reactive these electron pairs are.

and you said "Small, electron-dense ions are always great nucleophiles", but I don't understand. high polarizability means good Nu, thus small, elctron-dense ion should be poor Nu. where am I confused?

Charge density actually makes for a better nucleophile, and $\ce{F-}$ is much smaller than $\ce{I-}$.  If you think about it, which will be better at fitting into a molecules structure to force a reaction, something small or something big?

I actually thought bigger atom would be better Nu. because bigger ones pulls their e- less. If you are saying it's wrong, then where does the polarizability come from?

I guess you're talking about easiness of small atom moving around solvent molecules. but I hardly heard about this. In these so many variable, I'm still confused.

You are right in saying that $\ce{I-}$ has less control over its outermost electrons, thus making it more polarizable. Polarizability only determines how good a nucleophile something is in a polar protic solvent, otherwise it is basicity that determines nucleophilicity. Try reading here.

I actually read about this but i'll try this one more time. and I have no Idea why polarizability only applys in polar protic solvent. actually I thought polarizability means much more less in the polar protic solvent because of strong influence by hydrogen bonding. I'm not even sure why other factors matter other than hydrogen bonding to solvent molecules.

@ringo you are saying F- is a good nucleophile than I- but F- has less tendency to lose electrons for a reaction due to strong electronegativity .


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Why are some addition reactions syn and some anti?
Is it experimentally observed or do we know proper reasons for why some addition reactions are syn and some anti?
Like for halogenation to a double bond, I believe that the ring formed will cause hindrance and so only anti can be formed.
But for hydrohalogenation, either can happen. Is this because there is no hindrance in any case?
Similarly for hydrogenation, it is always syn? Why is that? What is the mechanism behind hydrogenation?

As you have correctly surmised, the stereochemistry is controlled by the reaction mechanism. In the case of halogenation, addition to the halonium intermediate is by an $S_N2$ type reaction so the resultant stereochemistry must be anti. In the case of hydrogenation, the reaction happens at a metal surface, with both hydrogens added from the same side of the double bond.

I see, what about HX (hydrohalogenation)?

I want to say it's because the intermediate is a planar carbocation so there is no selectivity of attack but I'm not 100% sure.

There can be ion pair effects in the addition of HX which favour syn addition but attack at both sides can (and will) occur. The ratio will depend on the exact substrate, solvent, reaction conditions etc. For catalytic hydrogenation, the addition is always syn and the mechanism is easily available online, look it up yourself.


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Wurtz reaction on acidic compounds
Why can Na not be used for the Wurtz reaction of compounds having an acidic group like alcohol or carboxylic acid?

Well why do you think? What happens when you react an acid with sodium?

But Wurtz reaction happens in solvents like ether right? Is Na basic there too?

With alkanols, sodium will form alkoxides (and hydrogen gas).
$$\ce{2CH3OH + 2Na -> 2 CH3ONa +H2 ^}$$
With alkanoic acids, sodium will form alkanoates (and hydrogen gas).
$$\ce{2CH3COOH + 2Na -> 2 CH3COONa +H2 ^}$$
If your starting material contains both an $\ce{OH}$ group and a bromo substituent, the alkoxide formed upon the reaction with sodium may attack the carbon bearing the $\ce{Br}$ atom in a substitution reaction to yield a cyclic ether.


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Is there an algorithm that counts Joback groups in a structure?
So I was wondering, there are a lot of cheminformatics tools (eg Babel) out there, to analyze and convert representations. I also saw some codes which counts certain groups, but ignores the carbon chain, which is obviously not suited for a Joback script. My google research only turned me a "homemade" matlab code found at Reddit, but it is no way general, as the first step of it is to exclude most of the rings.
So is there any open cheminformatic tool for this problem?

It might be helpful to add some details to your question about what a 'Joback group' is - it seems like https://en.wikipedia.org/wiki/Joback_method is relevant? In any case, you are probably looking for a /descriptor/ which is simply a value calculated for a structure.

Yes, by Joback grousps I mean the groups used in Joback's method, but any other group contribution method would do. As to your comment, I am not looking for a calculated descriptor, but for a list of the groups that the molecule contains.

Right, great - maybe your question could be "is there an algorithm for calculating group contributions" or something. In any case, I would expect the first step for any Jobeck descriptor would be to find the groups. Looks like some kind of simple substructure search would do it.

So it looks like 'Joback groups' are some set of functional groups that you want to find in molecules.
Sounds a lot like substructure search, for which there are many cheminformatics tools (like SMSD, for example) however, since you mention OpenBabel:
http://open-babel.readthedocs.org/en/latest/WritePlugins/AddNewDescriptor.html
talks specifically about 'group contribution methods' and gives a simple example of how to get the counts for a set of functional groups.


text:
Melting temperature and electronegativity trends in transition metals
So I know that generally the melting temperature increases across the row (let's look at just the first row for now) but what are the deviations? And why do they occur? I've looked everywhere online and can't find a clear answer to this.

I read that one, they say that Cr and Mn have ½ full 3d sub-shells and are therefore more stable but I can't understand how that would lead them to have a lower melting temperature


text:
Why is the 5d orbital instead of the 4f orbital filled in lanthanum?
For orbitals with the same value of $n + l$ (e.g. the $\mathrm{3d}$ and $\mathrm{4p}$ orbitals), we usually expect the orbital with the lower value of $n$ to be filled earlier. Hence, for example, the $\mathrm{3d}$ orbitals are filled before the $\mathrm{4p}$ orbitals in the transition metals.
However, the electronic configuration of lanthanum is $[\ce{Xe}](\mathrm{5d})^1(\mathrm{6s})^2$, and not $[\ce{Xe}](\mathrm{4f})^1(\mathrm{6s})^2$.
I know that the $\mathrm{5d}$ and $\mathrm{4f}$ orbitals are of similar energies, but is there any better explanation for this?

http://chemistry.stackexchange.com/q/8357/27435 there is a similar question

According to Ground-state configurations of ionic species I through XVI for   Z = 57-74 and the interpretation of 4d-4f emission resonances in laser-produced plasmas Phys. Rev. A 25, 275 (where "I" means the neutral atom) :

We begin therefore by showing in the left-hand
  columns of Table I the configurations given by
  Martin et al. for lanthanum through lutetium.
  Here it is seen that at the fourth spectrum the 4f
  level has in all cases fallen below 5d and 6s. This
  is due to the well-known phenomenon of 4f wavefunction
  collapse [reference 5]. The effective potential in
  which the 4f electron moves is made up of an attractive
  Coulomb term and a repulsive centrifugal
  term, which, in general, combine to give a double
  well potential, the height of the intervening barrier
  depending on the nuclear charge Z. For neutral
  atoms the field in the region of the outer potential
  is essentially hydrogenic. The occurrence of the
  neutral lanthanides coincides with the appearance
  of bound 4f levels in the inner well, which, in turn,
  is due to the deepening of the well and the lowering
  of the potential barrier as a result of the increasing nuclear charge.

Here, "reference 5" is Spectral Distribution of Atomic Oscillator Strengths Rev. Mod. Phys. 40, 441 which goes into quantitative equations for the attractive Coulomb term and a repulsive centrifugal term.

Upon reading Martin et al, I immediately thought Martin … ^^'