text:
Do filled Orbitals also hybridize?
Recently I was watching a video on Valence Bond Theory, on the MIT Open-Course youtube channel. The teacher there said that

For hybridization to occur, electron promotion from fully filled orbitals to empty ones is a must.

But here in a book it is specifically mentioned that

Electron promotion, though common, but is not 'a must' for hybridization.

Also, it is clearly written that sometimes even the filled orbitals take part in hybridization.
I tried finding some examples of such a case on the web, but could not. It would be a great help if someone tells me even one example where the fully filled orbitals also take part in hybridization, with an explanation why it is so.
The link for the 41 min long video is here.

I tried to watch that... $\ce{PH3}$ cannot be described in terms of VSEPR or hybridisation. The bond angle is about $93^\circ$. I could not watch any more after that.

@Martin I never said anything about PH3 and she DID NOT explain PH3 by VSEPR theory. If you had watched the video full you would have got my point. I think since you did not watch the video and neither you bothered even reading my question , so you should not comment anything that is not related to it.

I have carefully read and edited your question, and as I just saw I have to do this again since you changed the tags to the topic that they aren't. I really wanted to answer this question, but now I am more than only annoyed by your suggestive tone.

I can understand Martin's first comment; the lecturer does imply VSEPR and hybridization for $\ce{PH3}$ (a molecule relevant to the question) when she says "the lone pair pushes down the bonds below 109.5°". This is not true; the lone pair is in an orbital which is essentially the same as the phosphorous $3s$ atomic orbital, and hence has quasi-spherical symmetry about the phosphorous nucleus. Such high symmetry cannot induce directional repulsion. Martin's view (shared by many others) is that hybridization doesn't exist in the first place. This is fair as a type of answer.

@NicolauSakerNeto She said that just in a casual way, though she was just trying to explain that the bond angle will be decreased due to the replulsion from lone pair. She is teaching to first year students, she need to specify such detail, that too on open course. But still I dont understand why am I commenting this PH3 thing again and again. I want to know an example where filled orbitals hybridize, nothing more.

@Martin I don't care about the tags. I want an answer to my question, which I don't think you can give ( I am not being rude ). You can go on editing if you like. I am sorry if you found me rude, but I still think there was no need of your comment.

@NicolauSakerNeto how do yo know the lone pair which is "essentially" the same as the phosphorous 3s orbital?

@NicolauSakerNeto I am not opposing hybridisation. Just the concept in which it is usually treated, as a rigid theory that is required to form certain bonds. Since the bonding angle is $93^\circ$ it would be wrong to neglect the influence of the s orbital. || Dear Hardik, after studying five years of chemistry, a doctorate and currently a postdoc in computational/ theoretical chemistry, I think I would be able to answer your question. Maybe my comment was a bit impulsive, and I apologize that I have offended you with it, I just wanted to point out an obvious flaw of the lecture.

@Dissenter Look at ron's answer, explaining Coulson's theorem. You can calculate the hybridisation at the phosphorus centre based on the bonding angle and you will see, that the only place the lone pair can go is the (mostly) s orbital.

@Martin I don't doubt your knowledge at all. You have a large reputation, it indicated you know much more chemistry than me. Just the point was I needed an example, which LDC3 has given.

Whatever you were trying to say, I clearly understood it wrong, I apologize for any inconvenience. I would like to add one last thing, your book mentioned electron promotion, this concept is outdated by at least twenty years now. I highly recommend to switch to a different book, e.g. Chemistry of the Elements, by N. N. Greenwood, A. Earnshaw; Inorganic Chemistry: Principles of Structure and Reactivity, by James E. Huheey et. al. ...

@Martin Yes you are right, this concept is outdated. The book although mentioned it as a separate small note. And the teacher in the video too, thats why i got confused. I will sure try to read the book you recommended.

You mean like ammonia? If the filled orbital did not partake in hybridization, then the molecule would be flat.
Water would be linear if the filled orbitals did not partake in hybridization.

Actually, one of the lone pairs is unhybridized (basically pure $\ce{p}$) and the other lone pair is roughly $\ce{sp}$ hybridized.  The two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized.

@ron If that was the case, then the bond angle would be different fro what you measure.

How so?  The O-H bonds are $\ce{sp^3} hybridized so the expected H-O-H angle would be around 109 degrees, close to the experimentally observed value.

@ron I didn't quite see what you wrote, but it is incorrect since you cannot have $sp^3$ hybridized and $sp$ hybridized and a $p$ orbital on the same atom. $sp^3$ hybridization requires an $s$ orbital and 3 $p$ orbitals, which uses all the orbitals so you cannot also have an $sp$ hybridization and a $p$ orbital on the atom. BTW, water has a bond angle of 104.45°.

Sure you can.  2 sp3 orbitals require half of an s orbital and 1.5 p orbitals, leaving 1 p orbital for one lone pair, and half an s orbital and half a p orbital for the sp orbital the other lone pair to exist in.  This hybridization scheme is consistent with the Photoelectron spectrum data for water.

@ron You said (in your answer above) that it is a mathematical concept. What you are doing here is picking and choosing what you like (and using that) instead of following the rules. If you use a $p$ (or $s$) orbital in one hybridization, you cannot use it any other way. It may help explain data using your scheme, but for others following the rules, it will confuse them entirely.

see http://isites.harvard.edu/fs/docs/icb.topic776365.files/lecture%202.pdf  water is discussed on p. 18.  Let me know what you think.

Let us continue this discussion in chat.

Can't now, should I post a comment here when I have a block of time? Never used chat before.

@LDC3 Thats the answer. In ammonia the 2s and all 3 2p orbitals hybridize and 2s is completely filled. No electron promotion occurs. I now think why I could not get such an easy example. Thankyou so much. Your one line answer was all I wanted.

@ron Your comment was not correct. In water also the 's' orbital which gets hybridized to sp3 is fully filled, and that hybrid orbital holds the lone pair.

Water would not be linear, since the s orbital is filled and the p orbitals are perpendicular to each other. Also there are no "rules" for hybridisation, as hybridisation refers to the contribution of different orbitals of the same atom within one molecular orbital. The source given by @ron shows that quite clearly. It also states the most important thing about hybridisation: It is in response to bonding.

First off, hybridization is a concept chemists developed to help explain reality (their observations).  Just like resonance theory and Huckel MO theory, it is often (but not always) a useful way to explain the world around us. A "rule" on hybridization: hybridization occurs in response to a bonding interaction.  Further, hybridization involves the mixing of filled (or partially filled) atomic orbitals to produce new atomic orbitals that can be used to form "directed" bonds (molecular orbitals) that produce more stable molecular systems than an unhybridized atom might produce.  Finally, energetically speaking, hybridization is a net neutral process.  When low energy AOs are mixed with higher energy AOs, the resultant new AOs have the average energy of the initial AOs.  So if some electrons are "promoted", then some are lowered in energy as well
For example, in the formation of methane, carbon mixes the electrons in the 2s and three 2p atomic orbitals to produce 4 $\ce{sp^3}$ hybridized atomic orbitals.  The energy of these 4 equivalent $\ce{sp^3}$ orbitals is in between that of the starting 2s and 2p orbitals. In the case of methane this use of $\ce{sp^3}$ orbitals to produce a tetrahedrally shaped molecule, produces a molecule that is lower in energy than the alternative molecule formed from unhybridized s and p orbitals on carbon.  In the case of phosphine ($\ce{PH_3}$) the situation is reversed.  The $\ce{P-H}$ bonds in phosphine are directed along the x, y and z-axes.  The central phosphorous atom remains unhybridized.  In the case of phosphine, hybridization does not produce a lower energy molecule, so the central phosphorous atom in phosphine remains unhybridized.
Personally, I like the concept of hybridization and find it quite useful.  But you have to be careful where to apply it.  Hybridization is best when it is used to explain reality after the fact.

Not sure what you're getting at, but I thought my answer might help the OP have a better understanding of the subject.  If you feel the answer isn't helpful, you are free to down-vote and\or constructively criticize it.

It does help, I was just pointing out that you could also include an example that the OP requests. It's easy to veer from the requested answer when there is so much to explain.

I agree with almost everything you say, but not the part where the new resultant AOs have an average energy. It is just not possible to assign energy values to hybrid orbitals, as these are no eigenfunction of the Hamilton operator (They do not produce the correct quantum numbers).

@Martin 2 questions, 1) if I theoretically mix 50% carbon 2s AO and 50% carbon 2p AO to create an sp orbital, why wouldn't its theoretical or estimated energy be 0.5E(s) + 0.5E(p); 2) if I mix two carbon sp AOs to form an sp-sp MO I can experimentally determine its energy (PES spectrum of acetylene) and see if it is halfway between the 2s and 2p energy levels for atomic carbon, right?

Theoretically you could do that, but the value you obtain would be a number without a physical meaning. One has to be very careful in using this number even as an estimate, since it is not an observable state. Hybrid orbitals are directionalized (localised) orbitals, that would violate the spherical symmetry of an atom and also the quantum numbers. The concept of hybridisation is very useful, in describing a general bonding pattern. It will give you wrong results for (excitation) energies, since if you distort the molecule (in any way) bonding and therefore also hybridisation changes.

You will always measure the MO, which would be the linear combination of two sp AO, or in MO theory terms 22s + 22p. I am pretty certain that the contributions of these orbitals are not equal, since the hydrogen cannot polarise the orbital as well as the carbon. And then there is this nasty thing called electron-electron repulsion. This affects filled molecular orbitals, resulting in bonding orbitals less bonding than virtual orbitals are antibonding. So I am pretty sure, that the energy you measure for this MO is nowhere near the energy halfway between 2s and 2p.

@Martin I should have been more clear, I was referring to the sp-sp C-C sigma bond in acetylene, not the C-H bond

Your point was very clear. But both bonds will not involve an idealised sp orbital. The second carbon has a much larger effect on the polarisation on the carbon atomic orbital, than the hydrogen on the other side. In a very idealised theory, the hybrid AO for the CH and CC Bond would be the same. Since these are subject to a different external field (of the nuclei) they cannot be the same any more. Therefore the hybridisation will also change, because as you stated, this is only a response to bonding.

@Martin Yes, of course!  Thank you for reminding me.

With hybridisation, you only hybridise the number of orbitals you need i.e. that are filled. So in methane, with four bonds, you have four electron pairs which need to fit somewhere. So you need four orbitals. Therefore we say that three 2p orbitals hybridise with the 2s orbital, making four sp3 hybrids (the superscripts denoting the number of each pure orbitals involved) each with 1 electron in them. Note that we had to invoke the idea of hybridisation here, since the electron configuration of carbon is 1s22s22p2, and this would only give us two bonds. We know experimentally that this isn't true (in other words, methane exists!) and so have to conjure a trick to work around it.
Alternatively, let us look at the CH3+ cation. This has three bonds, and so three electron pairs. This time we need to invoke the idea of hybridisation to give us three orbitals. So we say that two 2p orbitals hybridise with the 2s orbital, to give three sp2 orbitals. This leaves us with one p-orbital that we didn't hybridise, which would be a non-bonding orbital (in fact, empty), and so this lies perpendicular to the plane of the molecule.
For another example, to clarify further, consider ammonia. Now ammonia has three bonds, but it also has a lone pair. We count the lone pairs in our hybridisation scheme. Therefore we also need to hybridise to obtain four orbitals, like in methane, despite only having three bonds. Thus ammonia is also sp3, mixing three 2p and one 2s orbital. Note however that in contrast to methane, three of these hybrid orbitals are singly occupied, and one is completely filled (the one which contains the lone pair).
Now, we can look at cases by which orbitals do not hybridise.
For heavy atoms such as lead and bismuth (the same groups as carbon and nitrogen respectively) hybridisation does not occur. This is referred to as the inert pair effect. In this case inert pair refers to the unwillingness of the 6s2 orbital to undergo hybridisation. This is because heavy atom self-self bonds i.e. Pb-Pb are incredibly weak. We can imagine that some energy is expended in the hybridisation process because we are mixing s orbitals with p-orbitals of the same principal quantum number, of which the p are always higher in energy (said to be non-degenerate). So the hybrid orbitals are higher in energy than the pure s, and ever so slightly lower than the pure p (but not by much, as the p-character of an sp3 hybrid is 75%). So, the energy gained in forming a heavy atom bond such as Pb-Pb isn't enough to warrant hybridisation, since hybridisation requires energy.
Notice that I've used anthropogenic words to explain this i.e. (we hybridise as opposed to the atoms hybridise). This is deliberate. It's unclear whether hybridisation really occurs or not; indeed more sophisticated theories can be used to explain the phenomena. Hybridisation is more of a caveat used to explain something we couldn't readily explain with the previous level of theory. It's perhaps telling that we only invoke hybridisation when we need to. It's a characteristic of a bad theory when that happens. The idea is invoked readily in chemistry however, so you would do well to familiarize yourself with the terminology.If you're interested more in this topic, look up molecular orbital theory for a more nuanced explanation.

Molecular orbital theory is not a competing interpretation. In MO terms, hybridisation is the mixing of different kinds of atomic orbitals of the same atom into a molecular orbital. It is still invoked as a mathematical concept and only treated as such. There can also by hybridisation in virtual (unoccupied) orbitals. The theory that is employed (by organic chemists mostly) is a very crude approximation. Hybridisation is always a result of bonding and therefore hybrid orbitals cannot be assigned an energy value.

@Martin, I'm not sure I completely agree. In inorganic chemistry, for example, you use the idea of a symmetry adapted linear combination of orbitals (SALCS) to rationalise the bonding in, say, octahedral complexes, instead of saying that they are sp3d2. Nowhere in the MO diagram of such complexes does it mention hybrid orbitals of this kind. Unless of course the process of taking SALCS is considered hybridising, which is something I'm not familiar with?

The symmetry adapted linear combination is somewhat closer to the truth - meaning a less crude approximation - but it is still an approximation. Even everything we know so far about orbitals is an approximation, since knowing the exact wave function is not possible, we can still consider different atomic orbitals contributing to a molecular orbital. Any linear combination of ao towards an mo of the same atom is therefore a hybridisation. You cannot neglect the influence of this when working within the framework of linear combinations.

I see, thanks Martin, I understand your point. I'll edit my contribution accordingly to reflect this.


text:
A 1.00 L gas at 100 °C and 500 Torr contains 70.0 % He and 30.0 % Ne by mass. What is the partial pressure of He?

A $1.00~\mathrm{L}$ gas at $100~^\circ\mathrm{C}$ and $500~\mathrm{Torr}$ contains $70.0\ \%~\ce{He}$ and $30.0\ \%~\ce{Ne}$ by mass. What is the partial pressure of $\ce{He}$?

My answer is $350~\mathrm{Torr}$ whereas the correct answer is $421~\mathrm{Torr}$.
Here is my solution (source; commas , should be points .; I used to write it this way.)
$$\begin{aligned}
  n_\mathrm{total} &= \frac{pV}{RT} = \frac{(1\ \mathrm l)(0.652\ \mathrm{atm})}{(0.082\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}})(373\ \mathrm K)}=0.0215~\mathrm{mol}\\
  70\ \%~\ce{He} &\implies (70\ \%)(0.0215\ \mathrm{mol}) = 0.0151~\mathrm{mol}~\ce{He}\\
  p_\ce{He} &= \frac{nRT}{V}\\
            &= \frac{(0.0151\ \mathrm{mol})(0.082\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}})(373\ \mathrm K)}{1\ \mathrm l} = 0.460~\mathrm{atm}\end{aligned}$$
$$0.460~\mathrm{atm}\cdot\frac{101325~\mathrm{Pa}}{1~\mathrm{atm}}\cdot\frac{1~\mathrm{Torr}}{133.322~\mathrm{Pa}}\approx 350~\mathrm{Torr}$$

You're getting a different answer because partial pressure is proportional to
the fraction of each gas in moles, rather than by weight. Why is that?
Looking at $pV=nRT$, you can see that the ratio between partial
pressures is the same as the ratio between number of moles (since $V,
R, T$ are the same for both gases).
So, if you respect proportions,
your answer will come out the same whether you consider $1.00\ \mathrm{L}$ of the mix
or some unspecified number of liters. This lets you avoid a messy
calculation with $pV/RT$.
You may as well assume that there are 70 g of helium and 30 g of
neon. Find how many moles of each gas that represents. The partial
pressures of each gas are in the same proportion as the number of
moles. With this and the total pressure, you'll get the answer.

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text:
Is "propanol" a correct name in IUPAC nomenclature?
My teacher said that propan-1-ol can also be named propanol in the latest IUPAC recommendations. I doubt this because Wikipedia only mentions propan-1-ol as the IUPAC name (propanol being relegated to the "other names" section).

Is "propanol" an acceptable name?

The last IUPAC nomenclature recommendations were published in 1993. The current draft documents (release date uncertain) are far more lenient in their acceptance of semi-systematic naming of molecules. Please make sure you understand the differences between the Preferred IUPAC Name (PIN), the retained name, and trivial name.
Propanol is not be the preferred IUPAC name, but significantly, neither is it listed in the tables of Trivial and semisystematic names retained for naming organic compounds. It is, however, a widely accepted name for this chemical, and you are far more likely to encounter propanol in industry than propan-1-ol. Nobody uses propan-1-ol in the real world, even if it is the preferred IUPAC systematic name.

If the question in the exam asks: Write the IUPAC name of . I answer the question: Propanol, would the answer be technically correct ?

From IUPAC Nomenclature of Organic Chemistry page : Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a ‘general’ IUPAC name, in the context of ‘general’ IUPAC nomenclature. So, perhaps the question is poorly worded, and should read "what is the preferred IUPAC name".

This is a question that you need to discuss with your teacher. Can it be argued that propanol is unambiguous? Perhaps there is an assumption that substitution is at the one position. If I were marking the exam, and it was testing knowledge of IUPAC nomenclature, I would give you 1 mark for propan-1-ol, and half a mark for answering propanol, n-propanol or propyl alcohol.

You mean there are two types of IUPAC nomenclature: Gen. IUPAC nom. and Preferred IUPAC num.. I didn't know that. Moreover our book only mentions it as IUPAC nomenclature; no general, no preferred. May be the book isn't perfect. Anyways I got your point. Thank you.

I think the distinction is between systematic nomenclature and nomenclature that is accepted because everyone uses it. I'm not a professional chemist, but if people know what you mean when you say propanol then it's okay to use it. If you're teacher is testing you specifically on nomenclature, it might be best to stick to the systematic method unless otherwise directed (as with benzene, for example).

The last IUPAC nomenclature recommendations were in 2013: http://dx.doi.org/10.1039/9781849733069-fp001 "Propanol" is still not an acceptable name for propan-1-ol, because there is ambiguity as to whether "propanol" refers to propan-1-ol or propan-2-ol.

The most recent IUPAC recommendations for Nomenclature in Organic Chemistry were published in 2013 and supersede the previous 1993 recommendations, which are no longer valid.

Based on the 2013 recommendations, the molecule should be referred to as propan-1-ol as described below:

P-15.1.7.2.1: Alcohols are named by attaching the suffix ‘ol’ to the name of the parent hydride, with elision of the final letter ‘e’ in the parent hydride, if present. When alone in the structure, the characteristic group(s) must receive the lowest locant(s) possible, which is (are) cited immediately in front of the suffix

Propan-1-ol is known as the PIN (preferred IUPAC name), that is the name that you should arrive at through proper application of the recommendations.
IUPAC do allow, however, for 'retained' or 'trivial' names for compounds, so long as their use is unambiguous.
Propanol, on the other hand, is neither a PIN nor a retained name (based on the IUPAC guidelines), as it introduces ambiguity. Are you describing propan-1-ol or propan-2-ol?

Although I will admit that there is a grain of truth in the other answer’s statement that propan-2-ol is commonly referred to as isopropanol, hence propanol will often be understood (word carefully chosen) as propan-1-ol.

Also according to the obsolete 1993 recommendations, the correct name has already been propan-1-ol. According to the obsolete 1979 recommendations, however, the correct name was 1-propanol.

@Jan However, the name "isopropanol" is incorrect (and also was incorrect according to previous IUPAC recommendations) since there is no hydrocarbon isopropane to which the suffix "-ol"can be added.

@Loong Yes, I never wanted to claim either name be correct. However, isopropanol is widely used both in and outside of laboratories.

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text:
What is the appropriate way of writing the Iron/thiocyanate equilibrium?

I understand the theory, but am struggling to answer the following discussion questions:
(i)
Why isn't this represented by: $\ce{Fe^{3+} (aq) + 3 SCN- (aq) <=>  Fe(SCN)3 (aq)}$
Why do we dilute the solution with water?
(ii)
I don't understand what the question is asking, can someone please explain
(iii)
I understand this.
(iv)
I am unsure, can someone suggest a reason.
(v)
I understand this.
(vi)
Constant volume of container?
ADDITIONAL QUESTIONS:

I thought ALL equilibrium reactions needed to be in a closed vessel - so if this experiment was conducted in an open test tube, wouldn't this be an error? How would this affect the accuracy of the results - would it be under/over estimated and why?

The second representation is more accurate because in water there can be no ``naked'' $\ce{Fe^3+}$ ion; as it goes into solution, iron combines with water to form an ion consisting of
a $\ce{Fe}$ atom surrounded by six molecules of water. It is the oxygens that form bonds with the iron atom, in a roughly octahedral arrangement: two oxygen
atoms in each coordinate direction, one on either side. The reaction
consists of the replacement of one of these coordinated water
molecules by the $\ce{SCN-}$ ion; now the iron has bonds to 5 oxygens
and 1 nitrogen atom.
The addition of $\ce{H2O}$ to the formula doesn't mean we are
diluting the solution with water.  It just means we're letting the
water that's  present anyway play a role in the equation, to reflect reality.
It's not clear to me what they are asking in (ii).  The colors change
dramatically when you mix $\ce{[Fe(H2O)6]^3+}$ (pale yellow) with $\ce{SCN-}$
(colorless); the result is deep red. Perhaps some context will help.  Read the discussion of the reaction in the text to see what they say about color intensity.
(iv) Again, what is the context?  If the text is about the spectrospic
measuring of the equilibrium constant, low
concentrations ensure that the reation obeys the ideal law.  In
concentrated solutions the law doesn't work quite right unless you
replace concentrations by activities, something that is only
indirectly measurable (it's whatever makes the formula work).
(vi) Again, we need to know the context.

No, not all equilibrium reactions need to be in a closed vessel. An
aqueous reaction that only takes a short time (so that for example
water won't evaporate appreciably, messing up the concentration) is
just fine in a test tube.

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text:
Equilibrium Graphs

What if I2 and H2 began at the same concentration - then do we start at the same point? ... or is it acceptable to start at different concentrations...

The question probably won't start H2 and I2 at the same C so that it is easier to see the two lines.

What if I2 and H2 began at the same concentration - then do we start at the same point?

Yes (and in textbook problems involving graphs this situation would be avoided for the reason pointed out in brinnb's comment)

is it acceptable to start at different concentrations...

Also yes.  No law against it.
I'm guessing that the graphs are part of a textbook problem. Are you able to figure out which graph can occur?


text:
Atomic radii of Sc, Ti, Fe, and Co
The atomic radius of Sc is $\pu{162pm}$, Ti is $\pu{147pm}$, $\ce{Fe}$ is $\pu{126pm}$, and that of $\ce{Co}$ is $\pu{125pm}$.
The electronic configuration of $\ce{Fe}$ is $\ce{[Ar] 3d^6 4s^2}$, and that of $\ce{Co}$ is $\ce{[Ar] 3d^7 4s^2}$.  The difference in atomic number, and hence the difference in the number of 3d electrons, is 1. So, due to screening, the extra charge is "canceled" and they have nearly the same radii.
The difference in atomic number and 3d electron count between $\ce{Sc}$ and $\ce{Ti}$ is also the same — 1.
So, I want to know why there is a considerable difference between the radii of Sc and Ti but not between the radii of Fe and Co.

There are different notions of atomic radius; the one you're using seems to be the metallic radius, which is half the distance between nearest neighbors in the metal.  This notion is very sensitive to the number of electrons per atom involved in bonding.  Scandium has only 3 valence electrons, while $\ce{Ti}$ has 4.  These all participate, in some extent, in the "electron soup" that holds metals together.  I haven't been able to figure out exactly to what extent, but  It's fair to say though that the 4 valence electrons of Ti bind the nuclei together significantly more tightly than the 3 of $\ce{Sc}$. As a result, the $\ce{Ti}$ atoms get significantly closer together.  (An analogous situation is the covalent radius of $\ce{F2}$, of around $\pu{70 pm}$, versus that of $\ce{O2}$, of around $\pu{60 pm}$; although the covalent radius tends to decrease across a period, it grows from $\ce{O}$ to $\ce{F}$ because $\ce{F2}$ has a single bond while $\ce{O2}$ has a double one.)
As you progress further along the transition metals the delocalization of d electrons in the metal goes down.  That is, although there are more d electrons in $\ce{Co}$ than in $\ce{Fe}$, their effectiveness in binding atoms together isn't really any larger.  As a result, the distance between neighbors (hence the metallic radius) is the same for both.

The series you cited belongs to so known 'metallic' radius, and it depends on crystal structure of the element, which changes through the row. In short, you cited series, that is not suited for consideration of isolated tendencies.
There are, indeed, several types of atomic radii (covalent with different valur for bonds of different order, van-der-waals radii and radius of cutoff that leaves some amount of electronic density inside the atom). When comparing atomic radii in comparable environment, two main trends are observable: growth of atomic size down the column in periodic table because more electronic shells is packed into same atom, and contraction of atoms towards the end of the row. This is a bit trickier to explain. Essentially, completed inner electronic shell isolate outer shells from nucleus, reducing effective charge of the nucleus that the outer shell 'feels'. Given that, at the beginning of the row outer electrons feels effective charge of 1 around already quite big completed shell, while at the end outer electrons feels effective charge of 8 around compacted inner shell. This is further complicated by 'mixed' status of d-electrons, that are isolated from the nucleus by inner shells much more effectively then p- and especially s- electrons, so they are valence-active in transition elements, but valence-inactive in p-elements.

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text:
How to predict the products of a organic chemical reaction
Given a set of reactants, what aspects should I consider when predicting possible products?

You need to build a working knowledge of known reaction mechanisms that include the functional groups of your reactants. Put another way, (1) identify the functional groups present in your reactants and (2) check known reaction mechanisms that proceed by reaction of these functional groups.
Really, the only way to be good at this is to study and memorize organic reactions. There's a number of resources to help with this (e.g. online flashcards, and I really like Chemistry by Design, an app that you can use on a smartphone).

On the topic of flash cards - I think these are an excellent way to learn the patterns of organic chemistry, and with those patterns ingrained, you're much better equipped to deal with the important, applied aspects of the subject. I wasn't happy with the online flash cards available for organic chemistry, so I've been steadily creating high quality sets for students to use on Quizlet - https://quizlet.com/features/OrganicChemExplained

Sharpen up your general chem skills.  One skill is recognizing redox reactions.  Many organic reactions are redox.  In problems where you have the reactants and products and are prompted to give the reagents you can recognize reductions and oxidations.  For the problems where you have the reactant and conditions you can more easily recognize oxidizers and reducers.  You can work on relative acidity and identify the most acidic proton in a molecule.  You can recognize the different types of carbon in a molecule and remember which are more reactive to which conditions.
Also, keep a notebook where you put in 5-10 reactions a day and memorize them.  You probably don't want to start today but you wish you started a month ago and a month from now you'll wish you started today.  Just learning name reactions would be a good start.

Short answer: You can't.
Long answer: for simple substrates and simple reagents there is a big stock of reaction protocols with more-or-less known general mechanisms, that can give a good guess of resulting product. However, in case of complex substrates or tricky reagents (yet unknown) strange transformations may occur. They can get ad-hoc explanations, but.. that's not satisfactory enough. A really good book, considering key intermediates and most important reactions with mechanisms is March's Organic Chemistry (that's probably one of the best books on the subject).

Are March's organic chemistry and March's advanced organic chemistry the same books ?

@user70856 Truthfully, I know one book "organic chemistry", originally written by March and survived several reeditions. No idea, how it is called right now and who edits new editions.

authors: Дмитрий Борисов


text:
How to calculate the partial pressure of oxygen in this mixture of oxygen and argon?

A $\pu{19.3 g}$ mixture of oxygen and argon is found to occupy a volume of $\pu{16.2 L}$ when measured at $\pu{675.9 mmHg}$ and $\pu{43.4 ^\circ{}C}$. What is the partial pressure of oxygen in this mixture?

The answer is $\pu{436 mmHg}$, but I didn't manage to get to it.
I know that it is found by the formula
$$P_i=X_i\cdot P_\text{total},$$
where $P_i$ is the partial pressure and $X_i$ is the mole fraction.
$P_\text{total}$ is given to find $X_i$. I should divide the amount of $\ce{O2}$ by the total amount of substances of the mixture $n_{\ce{O2}}/n_\text{total}$.
$n_\text{total}$ is easy to find with $n=PV/RT$,
but my problem is, that I can't find the amount of substance of $\ce{O2}$. The mass in the problem is given for the mixture, so how to extract only the mass of $\ce{O2}$, or in other words, how to get the amount of substance for $\ce{O2}$?

How is the mixture composed? Find expressions for mass, volume and amount of substance.
\begin{align}
\tag1\label1
  V_\text{total} &= V_{\ce{O2}} + V_{\ce{Ar}}\\
\tag3\label2
  n_\text{total} &= n_{\ce{O2}} + n_{\ce{Ar}}\\
\tag3\label3
  m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}
\end{align}
Transform $V_\text{total}$ with the ideal gas and solve for $n_{\ce{Ar}}$
\begin{align}
  \tag4\label4
  V_\text{total} &= \frac{RT}{p_\text{total}}(n_{\ce{O2}} + n_{\ce{Ar}})\\
  \tag5\label5
  n_{\ce{Ar}}  &= \frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}
\end{align}
Use total mass \eqref{3} and transform it with the with molar mass to amount of substance equation ($m = nM$), then plug in \eqref{5}:
\begin{align}
  \tag3
  m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}\\
  \tag6\label6
  m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + n_{\ce{Ar}}M_{\ce{Ar}}\\
  \tag7\label7
  m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + \left(\frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}\right)M_{\ce{Ar}}
\end{align}
Now there is only one variable left, which you do not know:
\begin{align}
  \tag8\label8
  m_\text{total} &= n_{\ce{O2}}\left(M_{\ce{O2}} - M_{\ce{Ar}}\right) + \frac{p_\text{total}V_\text{total}}{RT}\cdot M_{\ce{Ar}}\\
  \tag9\label9
  n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot  M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}
\end{align}
We need the following data (It would be by far better to only use SI [derived] units):
\begin{align}
M_\ce{Ar} &= \pu{39.9 g//mol}\\
M_\ce{O2} &= \pu{32.0 g//mol}\\
R &= \pu{8.314 L kPa //K mol}\\
\pu{1 mmHg} &= \pu{133.322 Pa}\\
p_\text{total} &\approx \pu{90 kPa}\\
T &= \pu{316.5 K}\\
m_\text{total} &= \pu{19.3 g}\\
V_\text{total} &= \pu{16.2 L}
\end{align}
Plugging this into \eqref{9}:
\begin{align}
  \tag9
  n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot  M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}\\
  n_{\ce{O2}} &= \left(\pu{19.3 g} - \frac{\pu{90 kPa}\cdot\pu{16.2 L}}{\pu{8.314 L kPa //K mol}\cdot\pu{316.5 K}}\pu{39.9 g//mol}\right)\cdot \left(\pu{32.0 g//mol} - \pu{39.9 g//mol}\right)^{-1}\\
  n_{\ce{O2}} &= \pu{0.36 mol}
\end{align}
Using the ideal gas with the data provided yields:
\begin{align}
  n_\text{total} &=\frac{pV}{RT}\\
  n_\text{total} &=\frac{\pu{90 kPa}\times\pu{16.2 L}}{\pu{8.314 L kPa K-1 mol-1}\times\pu{316.5 K}}\\
  n_\text{total} &=\pu{0.55 mol}
\end{align}
Therefore the mole fraction is
$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\text{total}} \approx 0.64.$$
And the partial pressure is therefore $\pu{57.7 kPa}$ (or $\pu{433 mmHg}$ which is close enough to the solution, given that I rounded the medieval witchcraft units).

Another visualization:

Assume $\pu{19.3 g}$ of pure oxygen;  use the ideal gas law to find the volume $V_1$
Assume $\pu{19.3 g}$ of pure argon;  use the ideal gas law to find the volume $V_2$, smaller than $V_1$ since argon is heavier than oxygen
Hopefully, the actual volume will be somewhere in between $V_1$ and $V_2$.
How far along the gap from $V_2$ to $V_1$ is the actual volume?
That's how far you are from zero oxygen to 100% oxygen.
With this fraction and the total pressure, the partial pressure is simple.


text:
Find the color of the solution

The acid-base indicator methyl red has a $K_\mathrm{a}$ = $1 \times 10^{-5}$. Its acidic form is red while its alkaline form is yellow. If methyl red is added in a colorless solution with a $\mathrm{pH} = 7$, the color will be :
a) Pink
b) Red
c) Orange
d) Yellow

Seeing the $K_\mathrm{a}$ = $1 \times 10^{-5}$, I concluded that it is a weak acid. Thus I think that when it is added to a neutral solution ($\mathrm{pH}=7$), the resultant solution must be weakly acidic and thus its color must be red.
However, contrary to my logic, the answer given in my book is option

d) Yellow

Is there a flaw in my argument?

A pH indicator is generally a weak acid and often work in a fashion similar to buffers in that most of the time both the acid and its conjugate base exist in significant concentration. From the Ka of the indicator you should therefore be able to find the "halfway" point of the indicator or the middle of the range of pH values over which the indicator operates.
Think about how indicators work. Indicators work simply by mixing colors. If I have an indicator which is black at pH = 0.0 and white at pH = 2.0, then I know that the protonated form of the indicator must be black-colored, and that the deprotonated form of the indicator must be white colored. This should make sense. As pH increase, hydroxide ion molarity increases, and hydroxide ion deprotonates other molecules.
Thus, at pH = 1.0, we should have a 50/50 mix of black and white molecules. In other words, the indicator shows a gray color at pH = 1.0. At pH = 1.0 we also have a buffer solution - we have the acid and conjugate base - and specifically in a 1:1 ratio.
By application of the Henderson-Hasselbach equation we will realize:
$\ce{pH = pK_{a} + log\frac{base}{conjugate~acid}}$ simplifies to ...
$\ce{pH = pK_{a}}$
I'll let you take the problem from here. Don't forget to apply common-sense considerations.

It's an indicator problem, so trying to balance the pH seems a bit off, since you typically add only a few drops of indicator to determine a solution's pH. Ideally I'd imagine you'd want the indicator to interfere as little as possible with whatever's going on.
Also, different indicators have different ranges. (They don't have to change color at exactly ${pH=7}$.) You're given that the acidic color is red, and the basic color is yellow. The ${K_a}$ helps to determine this switching point:
Regarding simple indicators, if the solution's pH is lower than the indicator's ${pK_a}$ you'll see the acidic color. Vice versa for the basic color.
The ${pK_a}$ that corresponds to ${K_a=1*10^{-5}}$ is simply ${5}$. As ${7 > 5}$, you'd expect the alkaline color to appear (in this case, yellow)


text:
What is the chemistry of this household cleaning recipe?
What would be the product if baking soda and common salt and vinegar are added together?

A lot of the cleaning effect of this mixture is because of the sudden production of lots of CO2 bubbles that loosen and dislodge whatever it is you're trying to clean.

Baking soda is sodium bicarbonate, and vinegar is dilute acetic acid. Common salt (table salt) is sodium chloride. Sodium bicarbonate will react with acetic acid to form sodium acetate (in solution), carbon dioxide gas, and water. Sodium chloride will not participate in the reaction. The reaction equation would be:
$$
\ce{NaHCO3(s) + HCH3COO(aq) -> NaCH3COO(aq) + CO2(g) + H2O(l)}
$$

The sodium chloride probably provides an abrasive.

@Ben It might - but it would need to be at a very high concentration for any to remain undissolved.


text:
How to find the potential of a mixed Iron/ Cer solution?

Given the following half-reactions:
   \begin{align}
    \ce{Ce^{4+} + e^{−} &-> Ce^{3+}} & E° &= \pu{1.72 V}\\
    \ce{Fe^{3+} + e^{−} &-> Fe^{2+}} & E° &= \pu{0.771 V}\\
 \end{align}
   A solution is prepared by mixing $\pu{3.0 mL}$ of $\pu{0.30 M}$ $\ce{Fe^{2+}}$ with $\pu{2.0 mL}$ of $\pu{0.12 M}$ $\ce{Ce^{4+}}$.
What is the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE)? Calculate $\ce{[Ce^{4+}]}$ in the solution.

My attempt at solving the problem (which was incorrect):
First to find the potential (relative to SHE):
\begin{align}
E&=E^\circ-0.05916\cdot\log(\ce{Fe^{2+}/Fe^{3+}})\\
0.771&=E^\circ-0.05916\cdot\log((3\cdot0.3)/(3\cdot0.3 + 2\cdot0.12))\\
E^\circ&=\pu{0.7649 V}\\
\end{align}
I don't think this is correct.
Then:
I am not sure how to find $E^\circ$ cerium or how to find $\ce{[Ce^{4+}]}$.

First it helps to have an idea of what will happen. Since 1.72 V is a
lot more than 0.771 V (a difference of 1V in chemistry goes a
long way), the numbers are telling us that $\ce{Ce^{4+}}$ is a much
better oxidizer than $\ce{Fe^{3+}}$; in other words, in the presence of
enough $\ce{Ce^{4+}}$, almost all $\ce{Fe^{2+}}$ will be oxidized to $\ce{Fe^{3+}}$.
Conversely, in the presence of enough $\ce{Fe^{2+}}$, almost all $\ce{Ce^{4+}}$
will be reduced to $\ce{Ce^{3+}}$.
In equilibrium, therefore, either you'll have almost no $\ce{Fe^{2+}}$
left, or almost no $\ce{Ce^{4+}}$ left: one of the reagents will have been almost entirely consumed. To tell which one, look
at the number of moles of each ion that went into the reaction at the
moment of mixing. Keep in mind that each unit of $\ce{Fe^{2+}}$ reacts
with one unit of $\ce{Ce^{4+}}$.

So now you've figured out which is the limiting reagent (the one that gets
consumed almost entirely). Let's suppose for the sake of argument it's
the $\ce{Fe^{2+}}$. Because you know how many
moles of $\ce{Fe^{2+}}$ went in, and because all of it reacted, you know
how many moles of $\ce{Ce^{4+}}$ got turned into $\ce{Ce^{3+}}$, and you can
find the concentration of $\ce{Ce^{4+}}$ and $\ce{Ce^{3+}}$ in the
equilibrium mix. And you also know the concentration of $\ce{Fe^{3+}}$,
because essentially all the $\ce{Fe^{2+}}$ became $\ce{Fe^{3+}}$. That's
all under the assumption that $\ce{Fe^{2+}}$ is the limiting reagent.
If the limiting reagent is $\ce{Ce^{4+}}$ instead, you can use the same
logic to find the concentration of $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$ and
$\ce{Ce^{3+}}$ in the equilibrium mix.
So now you have 3 concentrations out of 4.  Finding the 4th requires
knowing the equilibrium constant, which you can calculate from the
difference in potentials. (Write the combined full reaction based on the half-reactions.)
You'll find that the 4th concentration is
very small, justifying the claim that one of the reagents
was almost entirely consumed.
If you're with me so far, please update the question with these
results, and if you still need help with finding the potential at that
point, we can take it from there...


text:
How to obtain product ratio from energy differences via Boltzmann statistics
I am currently calculating some reactions and I would like to express the ratio of the products in per cents. I am using the Boltzmann statistic in the following way:

The probability $p_j$ of a particle $j$ to be in a state with the energy $\varepsilon_j$ is given as $\text{(1a,a')}$. The statement $\text{(1b)}$ must therefore hold, whereas $\text{(1c)}$ also follows.
\begin{align}
\text{(a)}  && p_j        &= \frac{N_j}{N}\\
\text{(a')} && \%P_j      &= \frac{N_j}{N}\cdot100\%\\
\text{(b)}  && 1          &= \sum_j p_j\\
\text{(c)}  && N          &= \sum_j N_j\\\tag1
\end{align}
For any system with $k$ particles, that are in thermal equilibrium, the probability to find particle $j$ can be expressed via the Boltzmann statistics $(2)$, where $Z$ is the canonical partition function, the inverse temperature is $\beta = (\mathcal{k}_\mathrm{B}\cdot T)^{-1}$ and the degeneracy coefficient is $g_j$.
$$p_j=g_j\cdot \frac1Z\cdot \exp\left\{-\beta\varepsilon_j\right\}\tag2$$
I can assume that a state with the energy $\varepsilon_0$ is the ground state, therefore no other state can have lower energy. I can therefore assign the ground state as the baseline and switch to relative energies, as demonstrated in in $(3)$.
\begin{align}
\text{def.}&& \varepsilon_0 &= 0.0\\
           && \varepsilon_j &= \varepsilon_0 + \Delta\varepsilon_j' \\
\therefore && \varepsilon_j &= \Delta\varepsilon_j'\\\tag3
\end{align}
As a consequence I can express the canonical partition function in terms of the ground state probability.
\begin{align}
            && p_0 &= g_0\cdot \frac1Z\cdot \exp\left\{-\beta\varepsilon_0\right\}\\
 \therefore && p_0 &= g_0\cdot \frac1Z\\
 \equiv     && Z   &= \frac{g_0}{p_0}\\\tag4
\end{align}
The canonical partition function is also defined in $(5)$, but I am quite sure this one would not lead me to my desired goal.
$$Z = \sum_l^\infty g_l\cdot\exp\left\{-\beta\varepsilon_l\right\}\tag5$$
I can express any probability in terms of the ground state probability following from $(2-4)$.
\begin{align}
   && p_j &= \frac{g_j}{g_0}\cdot p_0\cdot \exp\left\{-\beta\Delta\varepsilon_j'\right\}\\\tag6
\end{align}
As a last step I will switch to macroscopic numbers, as my quantum chemical calculations give me $\Delta{}G$ values.
\begin{align}
 \text{def.} && \Delta{}G_j &= \mathcal{N}_\mathrm{A}\cdot \Delta\varepsilon_j'\\
   && p_j &= \frac{g_j}{g_0}\cdot p_0\cdot
                \exp\left\{-\frac{\Delta{}G_j}{\mathcal{R}\cdot{}T}\right\}\\\tag7
\end{align}

Finally the questions:

How do I obtain $p_0$ in a way, that I can use it in spreadsheet software like libreOffice calc, so that I satisfy $\text{(1b)}$? If I insert this equation, I would have to solve it iteratively, as I have the variable to solve for on both sides. I could then go ahead and use $Z$ directly.
\begin{align}
  p_0 &= 1 - p_j -\sum_{i\neq0,j} p_i\\
  \color{\red}{p_j}
      &= \frac{g_j}{g_0}\cdot \left(1 - \color{\red}{p_j}
                                 -\sum_{i\neq0,j} p_i\right)\cdot
                \exp\left\{-\frac{\Delta{}G_j}{\mathcal{R}\cdot{}T}\right\}\\
\end{align}
Am I making correct assumptions along the way?
Could the following work? Assign an arbitrary probability to $p_0$, to obtain an arbitrary value for $[p_j]^\ddagger$. What would be reasonable? I picked $0.25$ as an example. Can I then normalize to get the total percentages?
\begin{align}
 \text{def.}  && p_0   &= [p_0]^\ddagger\ =\ 0.25\\
              && [p_j]^\ddagger &= \frac{g_j}{g_0}\cdot \color{\green}{0.25}\cdot
                \exp\left\{-\frac{\Delta{}G_j}{\mathcal{R}\cdot{}T}\right\}\\
 \text{Norm.} && \%P_j &= \frac{[p_j]^\ddagger}{\sum_i [p_i]^\ddagger}\cdot 100\%
\end{align}
This would maybe break down the iterative procedure to a two step process. Is this procedure really equivalent to $\text{(1a')}$?

Closely related: 4 Compounds in Equilibrium with One Another - Determine Their Equilibrium Concentrations

Is the number of states finite or infinite?  And why do you think (5) will not work? Also, are there degeneracies?

It seems to me that (5) -- with the replacement of $l$ by $j$ of course -- is saying the same thing as the procedure you outlined in your last bullet point, but taking out the initial "guess" for $p_0$ (or if you prefer, think of it as taking the initial guess to be 1 instead of $0.25$, then finding the factor by which you must divide it, which is $Z$).

@SilvioLevy In theory the number of states to be considered can be infinite. In praxis, I would not consider any state that is like 25 kJ higher in energy that the reference state, i.e. I would truncate the sum in (5). However, this number can still become quite large, depending on the system size and I am not very good at handling quite large spreadsheets. I do not understand your second comment, how would that change the sum to be infinite. (also there was a small typo)

@SilvioLevy I did not see the edit before it was too late to edit my comment. Thank you for this statement, which I now understand. I am however not sure if it would solve the two step process or the iterative solving of the equations.

My second comment is unrelated to the finite/infinite question.  All I'm saying is that I think your approach in the last bullet point is 100% correct, and that it is also 100% equivalent to computing the partition function in (5) -- or an approximation of it, if the number of states is infinite.  The fraction in your last equation, $[p_j]^\ddagger/\sum_i[p_i]^\ddagger$, is $1/Z$ in the case $j=0$, the ground state.  Unless I'm missing some subtlety. [Likewise I didn't see your 2nd reply before writing this; but since it's a clarification, I'm leaving it.]

@SilvioLevy I would very much welcome it, if you combine your comments in an answer.  It might not solve the problem completely, but it at least clears some of my doubts. Therefore I am of the opinion it deserves at least an upvote. (And it will be conserved for later reference.)

There are two aspects to the problem: theory and practice.  In theory you have a complete answer in the form of the partition function (5).  If you know all the energy levels and their degeneracies, you can at least in principle carry out the sum, and you've solved the problem.  One case where this is practical is where you have a (small) number $n$ of particles with spin $\frac12$ interacting with each other; there are $2^n$ configurations, each with potentially a different energy, but you can still carry out an explicit sum.
That example probably does not help  in your case; in chemistry you usually have infinitely many energy levels.  As you point out in the comments, you can cut off the sum at some point. This may still not give you a practical solution, either because the sum up to the cutoff is still too hard to calculate, or because the cutoff error is unacceptably high (say, if levels are closely spaced and the sum in (5)  converges very slowly).
But assuming there is a reasonable cutoff, the method may very well be practical.
With this in mind here are the answers to your questions at the end:

How do I obtain  $p_0$   in a way, that I can use it in spreadsheet software like libreOffice calc, so that I satisfy (1b) ? If I insert this equation, I would have to solve it iteratively, as I have the variable to solve for on both sides.

The "iterative" aspect is a red herring and the difficulty of having $p_0$ on both sides is illusory, since the $p_i$ are all proportional to $p_0$, and the proportionality factors are known and independent of $p_0$.  You can simply factor out $p_0$, which gives $p_0$ multiplied by the inverse of the partition function; then you equate this expression to 1 and (if you can calculate the partition function) you are done.

Am I making correct assumptions along the way?

Yes, I think you've stated the theory and the assumptions quite accurately.

Could the following work? Assign an arbitrary probability to $p_0$, to obtain an arbitrary value for  $[p_j]^\ddagger$. What would be reasonable? I picked  0.25 as an example. Can I then normalize to get the total percentages?

Yes, that would work; the formulas you wrote under this item are correct.  You can save yourself the trouble of choosing a number like  0.25, as  this number cancels out of the fraction  $\frac{[p_j]^\ddagger}{\sum_i [p_i]^\ddagger}$ -- notice that it appears as a factor in the numerator and also in each summand in the denominator, where it can be factored out.
One last remark: if you have access to symbolic algebra software like Matlab, Maple or Mathematica, you may be able to do the calculation more conveniently than with libreOffice calc. If you're very lucky, you may even find the exact solution to your infinite sum!    (Not all sums can be solved exactly, of course; still it might be worth a try, and you can sometimes guide the software.  It helps to be familiar with generating functions and summation techniques as explained for example in the book Concrete Mathematics, by Graham, Knuth and Patashnik.)
Good luck!


text:
Rationalising the order of reactivity of carbonyl compounds towards nucleophiles
This is immediately following ron's answer from Why is a ketone more nucleophilic than an ester?
One of the most simplest questions you can ask, how can you rationalise the order of reactivity towards nucleophiles, which is given as

acyl halide > acid anhydride > aldehyde > ketone > ester ~ carboxylic acid > amide > carboxylate ion
There might be many concepts out there, that explain this order, the above linked question could only halfway satisfactory answer it in the kind of generality I am looking for.
I am looking here for any model that can be used to explain this reactivity.

I will also later on post an answer to this question using quantum chemical calculations, trying to generalise from example. The key points that will be included are stated in ron's answer linked above.

I like Martin's posts, they make me think - and that is always a good
  thing. Here is where my thinking has taken me:

I believe Martin's approach (basically frontier MO) is fundamentally
  correct. I would be more convinced that his actual numbers are
  meaningful if he were to calculate 1) the carbonyl carbon LUMO
  coefficient and 2) the HOMO-LUMO separation for the following series
  of compounds in order to see if his calculated values correlated with
  this order of reactivity of carbonyl compounds towards nucleophilic
  attack (we could use water as the nucleophile) - but Martin, I think
  such calculations might be a poor use of your time and resources.

While I agree with most of this, I do strongly disagree with the last statement. Which I will hopefully be able to prove soon.
In the meantime I welcome everyone to share their thoughts on the matter.

Do you mean "nucleophilicity" or "electrophilicity"? I associate the power of a compound to act as a nucleophile with the term nucleophilicity whereas you included the "order of reactivity towards nucleophiles" in your question which is a measure of electrophilicity. Or did I misunderstand something?

Here is my "old school" explanation.
Below is a drawing of the reaction coordinate for nucleophilic attack at a carbonyl carbon. The energy well for the starting carbonyl compound is shown on the left.  As the positively-polarized carbonyl reacts with (forms a bond with) the nucleophile we pass over a transition state and fall into a second potential well representing the tetrahedral intermediate.  $\ce{E_{act}}$ is the activation energy required to pass over the transition state.  To whatever extent a substituent might stabilize the starting carbonyl compound, the energy of the carbonyl compound will be lowered, and consequently $\ce{E_{act}}$ will increase.  The central carbon (it was the carbonyl carbon) in the tetrahedral intermediate is fully saturated (just $\ce{sp^3}$ bonds) and without charge, so the substituent X cannot interact through resonance to stabilize the tetrahedral intermediate.  In other words, while substituent X can stabilize or destabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.

Below are the resonance structures we can draw for our carbonyl compound.

If structure III contributes significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structure III contributes little.
Let's start by comparing the ester to the amide.  In both cases structure III involves a 2p-2p pi bond, but the resultant positive charge is placed on oxygen when we start with the ester and on nitrogen when we start with the amide.  Clearly, for electronegativity reasons, the amide case is better with the positive charge on the less electronegative amide nitrogen.  Our prediction: the amide is more stabilized than the ester and therefor should be the slower reacting of the two.
$$\ce{ester > amide}$$
In the case of the carboxylate anion, resonance structure III is equivalent to resonance structure I (note: there is no positive charge in structure III for the carboxylate case).  Since they are equivalent, they carry high weight stabilizing and in describing the carboxylate anion.  The carboxylate anion is so stabilized by resonance structures I and III that it is basically unreactive to nucleophiles
$$\ce{ester > amide >> carboxylate}$$
There's really no difference between the ester and carboxylic acid in terms of resonance structure III contribution so we would expect them to have similar reactivity towards a nucleophile.
$$\ce{ester \sim acid > amide >> carboxylate}$$
Let's compare the ester to the anhydride.  Resonance structure III is equally effective for both compounds, BUT, there are two carbonyls in the anhydride for the oxygen to interact with.  Therefor, the oxygen will only be half as effective at stabilizing any one anhydride carbonyl as it was for the ester carbonyl.
$$\ce{anhydride > ester \sim acid > amide >> carboxylate}$$
In the case of the acid chloride, resonance structure III involves a 2p-3p pi bond, not very effective overlap due to the size difference (also chlorine is electronegative so placing a positive charge on it is not desirable, but the poor pi overlap is what limits the contribution from structure III in the acid chloride case).
$$\ce{acid~ chloride > anhydride > ester \sim acid > amide >> carboxylate}$$
For the ketone, our resonance structure III has a slightly different look, it involves hyperconjugation as shown in the following figure.

Since there are 3 hydrogens, there are three such hyperconjugated structures,  They are not super significant resonance structures, but there are 3 of them, they do place the positive charge on a proton, so they do count for some stabilization.  To be honest, I would have guessed that it creates more stabilization than resonance structure III in the acid chloride case, but I wouldn't have known where it fits between the acid chloride and the ester.  Wherever I put it, I'd put the aldehyde to its left.  There is no resonance structure III for the aldehyde.
I wind up with
$$\ce{acid~ chloride > anhydride > ester \sim acid > amide >> carboxylate}$$
and $\ce{aldehyde > ketone}$ fitting in somewhere between the acid chloride and the ester

+1, a very good analysis, I think. For what it's worth, Fleming's book Molecular Orbitals and Organic Chemical Reactions basically agrees with your relative ordering. W/r/t anhydrides and aldehydes, Fleming says "[...] there is a more delicate balance with acid anhydrides, which are similar in electrophilicty to aldehydes," so perhaps anhydride ~ aldehyde > ketone is reasonable.

Thanks Greg, good info about aldehydes in your comment.  It'll be interesting to see how Martin's calculations handle it.

I finally got around to write up my answer, I would appreciate any comment. And while you are at it you could suggest a solvent which could be used for calculations. (I was thinking of DEE or THF maybe.)

Why is acid above amide?

@Abcd Because resonance structure III (as shown at the top of my answer) contributes much more in the amide case, than in the acid chloride case.  This makes the amide carbonyl more stable \ less reactive as compared to the acid chloride carbonyl.

@Ron I was referring to carboxylic acid being above amide not acid chloride.

@Abcd  Oh sorry Abcd, my bad.  But the answer remains the same, resonance structure III carries more weight for the amide than the acid due to the difference in electronegativities of oxygen and nitrogen.

Are Hyperconjugative effects of ketone less stabilising  than anhydride's cross conjugation? And if so then ketone must be more reactive with nucleophiles.

For this approach I am basically employing Frontier Molecular Orbital Theory (FMO) to predict the reactivity of carbonyl compounds towards nucleophiles.
For the purpose of this explanation I have chosen water as nucleophile. In principle we are looking at the addition of an electron rich particle to an electron poor system. In this case, water will attack the positively polarised carbonyl carbon to form the tetrahedral reactive intermediate. The activation barrier of following step, i.e. dissociation of the leaving group (or proton for aldehydes and ketones), must be lower in energy, because otherwise the reaction will not proceed. It is therefore sufficient to look at only the initial step of the reaction, and more rigorous we can even only look at the ground state properties of the reactants.
In FMO theory there are basically three main observations:

Occupied molecular orbitals of the reactants repel each other
Opposite (partial) charges in the reactants attract each other
Occupied orbitals of one reactant interact with unoccupied orbitals of the other reactant (and vice versa) causing attractive interaction.

The first point is fairly obvious and will not cause much change in the interaction of the two molecules.
The second point is also straight forward. In all carbonyl compounds the carbon will be carrying a (partial) positive charge, while the nucleophile will be having a (partial) negative charge.
The third point is not so obvious. It can be expected, the LUMO of the carbonyl compound (electron poor carbon) and the HOMO of the nucleophile (electron rich) will interact to form the new bond. This will also be the strongest interaction. The opposite combination is also thinkable, but the energy difference between these orbitals will be much larger, as we will see. In other words, the closer the energy difference between these orbitals will be, the more attractive the interaction will be. Since we keep the nucleophile constant, we can predict a order of reactivity towards that one specific nucleophile based on the energy of the LUMO of the carbonyl compound. (This comparison will hold, even if we change the nucleophile.)
The following table assembles the energies of the frontier orbitals in $\mathrm{eV}$ as well as the most important partial natural atomic charges. As a model system, the ethyl substituted compounds $\ce{Et-CO-R}$ have been chosen. The calculations were performed with Gaussian09 Rev. D1 employing Becke's exchange functional in conjunction with Perdew's correlation functional. The density fitted def2-SVP basis set was used. For charges the Natural Bond Orbital (NBO) analysis was used.
For convenience I have not included any solvent, so these calculations assume gas phase, vacuum and absolute zero.
\begin{array}{lrrrr}
\ce{R =}       &\ce{HOMO}&\ce{LUMO}&\Delta E &\mathbf{q}(\ce{C})&\mathbf{q}(\ce{O})\\\hline
\ce{-Cl}       & -7.00  & -2.11  & 4.89  & 0.55  & -0.46  \\\hline
\ce{-O-OC-Et}  & -6.35  & -1.85  & 4.50  & 0.82  & -0.52  \\
               & & \text{(L+1)} -0.85 &  &  &  \\\hline
\ce{-H}        & -5.79  & -1.79  & 4.00  & 0.41  & -0.51  \\\hline
\ce{-Et}       & -5.55  & -1.48  & 4.07  & 0.57  & -0.54  \\\hline
\ce{-OH}       & -6.33  & -0.88  & 5.45  & 0.81  & -0.58  \\
               &  &  &  &  &(\ce{OH})  -0.68 \\\hline
\ce{-OEt}      & -6.15  & -0.74  & 5.40  & 0.81  & -0.58  \\
               &  &  &  &  &(\ce{OEt}) -0.56 \\\hline
\ce{-NH2}      & -5.53  & -0.24  & 5.29  & 0.67  & -0.60  \\
               &  &  &  &  &(\ce{N}) -0.83 \\\hline
\end{array}
These kind of calculations cannot handle anions very well, so I have also omitted the carboxylate ion. For obvious reasons, I would say its tendency to react as an electrophile is negligible.
The HOMO of water is one of the lone pairs at oxygen, at $-6.35~\mathrm{eV}$. Correlating the LUMO energies from the table, the expected order of reactivity is
$$\ce{acid~chloride > anhydride\approx aldehyde > ketone > acid\approx ester > amide}$$

With NBO we can look at localised orbitals, which all look very similar and also give very similar numerical results. The $\pi^*$ antibonding LUMO is composed of $67\pm2\%$ carbon $\ce{p}$ AO and $33\pm2\%$ oxygen $\ce{p}$ AO for all compounds. Here are the FMO shown for the anhydride.

Unfortunately I was currently not able to find any reaction barriers for this approach. My educated guess is, that the chosen nucleophile is too weak to coordinate to this compound in the gas phase. I believe I will have to include a solvent for transition state stabilisation, but I am not sure if I can afford the calculations at the moment.

This is very nice work, very informative and it demonstrates how powerful your approach is (wish I could do it on my home computer - maybe someday). The one suggestion I have is that it would be nice to have a second table where you show ΔE  for the water-carbonyl HOMO-LUMO and LUMO-HOMO interactions.

Am I correct that the HOMO and LUMO numbers in your table reflect the carbonyl HOMO-LUMO?  What was the HOMO and LUMO for the nucleophile?  As you noted, "The opposite combination is also thinkable, but the energy difference between these orbitals will be much larger, as we will see."  Where is the delta E for the nucleophile(LUMO)-carbonyl(HOMO) interaction?

Oh, thanks for the comment, I forgot about that while writing. I will try to add that when I get back to the office, perhaps on monday.

There are several different things going on here in this rather long series. I will go through them in turn.
Acid chlorides and acid anhydrides
These are going to be reactive because they have a good leaving group. Chloride ion is so stable that it may break away from the molecule and give $\ce{RCO+}$ and $\ce{Cl-}$ spontaneously, not to a great extent, but sufficient for this to have no limiting effect on the reaction rate.
Similarly an acid anhydride may break down to $\ce{RCO+}$ and $\ce{RCOO-}$. We know that the anion is again quite stable, but not as stable as $\ce{Cl-}$. This is apparent from the fact that $\ce{RCOOH}$ is an acid, but not as strong an acid as $\ce{HCl}$.
$\ce{H}$ and $\ce{R}$ are not stable anions, so aldehydes and ketones are much less reactive than acid anhydrides and chlorides. Of course, the $\ce{H}$ and $\ce{R}$ groups remain attached to the molecule, which is not the case with the chlorides and anhydrides. Being an addition reaction instead of a substitution reaction, there is a decrease in entropy.
An interesting case here is the iodoform reaction where $\ce{R-CO-CH3}$ reacts with iodine to form $\ce{R-CO-CI3}$. $\ce{CH3-}$ is a poor leaving group, but the $\ce{CI3-}$ ion is a good leaving group and breaks away from the molecule, leaving $\ce{RCO+}$, which then reacts with water to form $\ce{RCOOH}$, while the $\ce{CI3-}$ reacts with water to form $\ce{HCI3}$.
The thing to note here is that there is significant thermodynamic drive here for the good leaving group to be substituted by a poorer one.
Aldehydes and ketones
An important factor that makes aldehydes more reactive than ketones is that there is less bulk blocking the approach of a nucleophile. There is less steric hindrance.
Esters and carboxylic acids
We are now back to substitution reactions. But $\ce{RO-}$ or $\ce{OH-}$ are much poorer leaving groups than the $\ce{Cl-}$ or $\ce{RCOO-}$ that we saw with acid chlorides and anhydrides.
There is another factor at play here. The $\ce{C=O}$ double bond is conjugated with the $\pi$ electrons on the other oxygen, which significantly stabilises the molecule. This causes a resistance to both $\ce{S_{N}1}$ reactions (where the $\ce{RO-}$ or $\ce{ROH}$ spontaneously detaches from the molecule leaving $\ce{RCO+}$) and $\ce{S_{N}2}$ reactions (where the nucleophile attacks the molecule, giving a tetrahedral intermediate before the $\ce{ROH}$ or $\ce{H2O}$ is kicked out.) Either way that $\pi$ system has to be broken down, giving a kinetic barrier to reaction (high activation energy.)
Amides and carboxylate ions
In certain reaction mechanisms OH- may spontaneously leave a carboxylic acid. $\ce{NH2-}$ is more basic and so is very unlikely to leave of its own accord. $\ce{O^{2-}}$ is even more basic (unheard of in solution.)
Other reaction mechanisms require that the molecule is protonated first. The next step may be nucleophilic attack ($\ce{S_{N}2}$ type mechanism) or spontaneous cleavage of the molecule into $\ce{RCO+}$ and a leaving group ($\ce{S_{N}2}$ type mechanism).
If $\ce{COOH}$ is protonated, the leaving group will be $\ce{H2O}$, a neutral compound.
If $\ce{CONH2}$ is protonated, the leaving group will be $\ce{NH3}$, a basic compound.
If $\ce{COO-}$ is protonated (to an acid) the leaving group will be $\ce{{}^{-}OH}$.
Generally the more acidic/less basic a group is, the more stable it will be when separated from the $\ce{RCO+}$ carbocation. Here we have a neutral molecule water (a good leaving group), a weak base ammonia (poor leaving group) and a strong base (very poor leaving group) so the reactivity order is as would be expected.

authors: Šhūbhåm Ťhäkûř


text:
The relationships between n , T , P , V for a gas?
The symbols :
T : temperature //
V : volume //
P : pressure //
n : number of mole
what it is clearly known about the relationship between one another is :
P is inversely proportional to V ( from Boyle's law : PV=constant )
V is directly proportional to T ( from Charles's law : V/T=constant )

but the other ones I am not sure about them, So i might say :
P is inversely proportional to T
T is inversely proportional to n
p is directly proportional to n
What do you think?

All statements like "P is inversely proportional to V" should be qualified with "other variables being equal". With this qualification you can read off all the proportionalities from $pV=NRT$.  Any two variables on the same side are inversely proportional; any two on opposite sides are directly proportional.  So $p$ is directly proportional to $T$.  Your other assertions are correct.

Just remember the general gas law
$$\mathrm{PV=nRT}$$
or rearranging terms we see that
$$\mathrm{\frac{[PV]}{[nT]}=R}$$
where R is a constant.  Therefore,
$$\mathrm{\frac{[P_1V_1]}{[n_1T_1]}=\frac{[P_2V_2]}{[n_2T_2]}}$$
this very general set of ratios relates all of the variables in state (1) to the same variables in state (2).  If volume and number of moles don't change, then the equation reduces to
$$\mathrm{\frac{[P_1]}{[T_1]}=\frac{[P_2]}{[T_2]}}$$
and so on and so forth.  All of the proportionalities are clearly layed out in one general equation.


text:
Evaporation of water
What are the conditions for the evaporation of water?
I know that for the boiling of water, the vapor pressure of water must equal the partial pressure of water vapor.
But what about for evaporation? Is it the same condition above in addition to the requirement that the relative humidity cannot be 100%?
Can we have boiling when the relative humidity is 100%?

Evaporation is a surface phenomenon, boiling is a bulk phenomenon.  Evaporation will always be occurring for any liquid with an exposed surface.  Water molecules in a liquid sample have a distribution of kinetic energies.  It takes less work for a surface water molecule to break free from its neighbors, then for a water molecule in the bulk to escape, since the former has fewer neighbors (no molecules on top).  Those molecules with enough kinetic energy to break away from their neighbors will escape.
For water to boil its vapor pressure must equal the atmospheric pressure.
The only requirement for evaporation to occur is an exposed liquid surface.  I don't think relative humidity affects the boiling point, but I suspect it would affect the rate at which a liquid will evaporate from an open container.  Evaporation is always occurring, but to notice a change in the amount of material in an open system, more material must leave the surface then returns to the surface and the rate of return will be affected by the composition (water content) of the air above the open sample.

So is delta G for evaporation always negative?

@Dissenter No, $\Delta G$ is not negative for metals, or if it is, then it must be very close to zero.

$\Delta G$ is negative if the process occur spontaneously, which is to
say if the  partial pressure of the substance's vapor happens to be
less than the vapor pressure of the substance at the given temperature.
$\Delta H$ for evaporation is always positive, of course; but because
the entropy of a gas is greater than the entropy of the same substance
in the liquid phase, the value of $\Delta G = \Delta H - T\Delta S$
for evaporation can be positive, negative or zero. If negative, net
evaporation occurs; if positive, net condensation; if zero,
equilibrium.

Water evaporates (in the sense that more water leaves a body of liquid or solid water than comes back to it) whenever the partial pressure of water vapor in contact with the liquid or solid water is less than the vapor pressure of water at the given tempearture.  This is the same as saying that the relative humidity is less than 100%, by the definition of relative humidity.

Can we have boiling when the relative humidity is 100%?

You can.  Suppose you have a flask of water with a narrow neck and surrounded by thermal insulation.  If you heat the water to the boiling point you will soon displace all the air, so both the liquid and gas phases are 100% water, and the water will continue to boil as long as you keep supplying heat.

Would there be any net evaporation at a relative humidity of 100%?

No net evaporation when the relative humidity is 100%; the rate at which molecules leave the liquid matches the rate at which (other) vapor molecules enter the liquid.


text:
Are there any full worked examples of DFT calculations?
I just started learning DFT and now I am totally confused.
Assuming I want to use B3LYP:
\begin{align}
v_s\left(\textbf{r}\right) &= v_\text{ext}\left(\textbf{r}\right) + \int d^3r\frac{n\left( \textbf{r}\right) }{\left|\textbf{r} - \textbf{r}^\prime\right|} + v_{\text{XC}}\left[ n \right]\left(\textbf{r}\right)\\
v_{\text{XC}}\left( r \right) &= \frac{\delta E_\text{XC}}{\delta n \left(\textbf{r}\right) }\\
\end{align}
I know that we use B3LYP for approximation of $v_{\text{XC}}$ - but I have no idea about how - and then we use some basis sets to run an SCF calculation to minimize the energy.
I downloaded a bunch of articles and bought some books and read them, but all of them give very detailed discussion about all the basics (e.g. what a functional is, or Kohn and Sham theorem, etc.), but no one talks about how to use them.
I am looking for a reading material or video that explains all the calculation, step by step, for example, some material that shows all the calculations for a CO molecule, from the beginning to the end of the SCF calculation.

So, I repeat it one more time: you are not even supposed to understand how DFT calculations are done step by step. There are scientists who basically spent all their scientific carrier to understand it. And they share their knowledge with everyone else not only by means of scientific papers, but also in a form of ready to use programs. Thus, for everyone else, the question is how to properly and efficiently use a program, and not how it works under the hood.

Do not get me wrong. You can (and even should) understand what is going in under the hood, at least at some basic level: you need to know what this whole business about SCF is, how DFT works in principle, what are the main differences between functionals and basis sets. But understanding how DFT calculations are done step by step is far away from these fundamentals, and it will require a lot of time.

@Wildcat, First of all I appreciate your help.As you may remember from my previous post,I have a polymer that absorbs 470nm light and breaks apart. What I need to do is to simulate this process and later add a radical group to it. Actually I think I need to add this Electromagnetic energy as a constant to the Kohn-Sham equation and then minimize it to find ground state energy and optimized wave function. I downloaded GAMESS but I think these softwares are not that flexible to change their equations ( I have not worked with them but just watched some videos about GAMESS and Gaussian ).

It will be awesome if there is any software that is capable of solving modified equations. Is there any ?

Oh yes, that is true that both GAMESS and Gaussian are not so suitable for using user-defined functionals. But now my question is why do you even want to define and use your own custom functional instead of "standard" functionals already available in programs? What is so wrong with "standard" functionals that you think you need to use a "non-standard" one?

Here is a related question http://chemistry.stackexchange.com/questions/8140/defining-and-testing-custom-made-dft-functionals. Pay attention to the last sentence in @Aesin answer, which basically says that "you should really make sure you have a scientific justification for doing this".

@Wildcat ,I checked that related question. The list of programs is very interesting but which one fits my work? Here is the exact problem: We have developed a new polymer for dental composites ( that white filling material ). This polymer has very good properties except its light absorption property ( at 470 nm) that gives it a blue color under fluorescent lamps !!! We want to simulate this light absorption phenomenon to find best way to solve it. Any of those softwares have standard functionals for this purpose? If not, can I define this energy in any of them?

@Wildcat: Regardless of the original intent of the question, I think an answer would be valuable -- theoretical explanations are too often sloppily notated and incomplete.

Did you understand Hartree-Fock, the principles of Ground state DFT and the computational adaption of Kohn and Sham. These are very important prerequisites for explaining how a functional is designed, where are it's flaws and strengths. B3LYP is one of the worst (hybrid) functionals, which was originally implemented to prove a certain point (hybrid functionals are possible). It was not optimized for any specific test set and it performed reasonably well. It is one of those functionals, that are right for the wrong reason.

When you are talking about light absorption, then you are basically talking about excited states calculations. Not a good job for DFT in the first place. So you will have to switch to the time dependent Schrödinger equation. There are some handy codes available, like TDDFT and CIS and the most powerful and time consuming CASPT2. It is still a very active field of research and only applying some of these methods might sometimes need a lot of time. (I have studied computational chemistry now for more than five years and my comprehension of CASPT2 is quite limited.)

@Martin T-DFT was a very good suggestion. Would you suggest a good reading resource for that please ( book / video/ articles)

I do not have enough experience on the subject, so I would start with the wikipedia page first to get an overview. If I really wanted to continue, then I'd get a book on the matter. But these are expensive so I might just have a look at the key papers (refs at wiki). If you have fundamental knowledge on quantum chemistry, you could go to a workshop to learn the insights and maybe how to use a program and interpret the output. Or start with TDHF.

I'm erring on the side of what @Wildcat said.  Unless you're someone who does well jumping into the thick of things and working your way out, I'd suggest beginning with the fundamentals and work your way toward the center.  This means building a solid foundation on QM theory, the Hartree-Fock approximation and the SCF procedure.  Once you've mastered this, then moving onto the Kohn-Sham SCF equations would be the next logical step.  (See: http://journals.aps.org/archive/abstract/10.1103/PhysRev.140.A1133)  Also, most everything you need to know about SCF is in Szabo & Ostlund.

@Aug  Take a look at a book called "A Chemist's Guide to Density Functional Theory."  I've read bits of it in the past before and it was very informative.  Glancing at the table of contents, it looks like it may cover a lot of the things you are seeking.  Google around for a PDF of this book.  It may be available (or it may not... I have no idea).

@LordStryker You learned something from Szabo & Ostlund? He uses very complicated notation. I prefer Levine, Jensen or Cramer. Jensen's book also covers the fundamental approaches to DFT. Koch & Holthausen's does not cover the TD ansatz, but it dives substancially deeper into DFT and the performance of the models.

@Martin I used Szabo much more than I did Levine.  A large part of my dissertation was aided my the former but I did use Levine in some of the derivations.  I find that Levine overcomplicates much of the presentation because (to me) it seemed overly wordy and dense.  Levine is definitely good and anyone who is serious about learning this type of material should be drawing their knowledge from both references.

The XC potential for DFT actually consists of two terms: $V_x$ and $V_c$. Depending on which XC functional you choose, the exchange part is either computed exactly (using Hartree-Fock) or using fitted parameters or in a combination of both. In the case of B3LYP the exchange is 20% Hartree + 8% Slater + 72% Becke88. The correlation functional consists of three (3) different terms (LYP) in a different form. If you take a different functional, say M06-2X, it has different functional type (54% Hartree exchange, 200+ parameters!).
Coming to the next part: You use a basis set only to compute the kinetic energy and exchange terms in the system. Let me walk you through a HF routine first:
The Hamiltonian consists of four terms(nuclear kinetic energy is zero because of Born-Oppenheimer Approximation) : $T_{elec} + V_{nn} + V_{en} + V_{ee}$. The nuclear-nuclear potential energy is constant (because they are 'clamped'). The first and third terms are fairly trivial, and are computed with relative ease. The electron-electron potential term consists of Pauli's exchange term, and the coulomb repulsion term. They are the computationally expensive part of the HF calculation, and the eigenvalues (energies) are solved self-consistently.
So, when you obtain the HF energy of a system, it is obtained variationally with respect to all the energy components in an interacting system of electrons and nuclei. Coming to DFT:
The first term you mentioned $V_{ext}$ is the external potential, which is usually the nuclear potential for an unperturbed system. The second term is equivalent to Coulomb repulsion. The third term is evaluated using the XC potential of your choice(every DFT method differs only at this part). You evaluate the energies for each and every orbital using the effective potential $V_s$ which you mentioned. Note: You evaluate energies for every orbital separately.
This is because of the KS theorem that equates the density created by non-interacting particles to be the same as that of interacting particles. So, we solve Kohn-Sham equation instead of Schrodinger equation, under a constraint that the sum of square of all the orbitals gives the density:
$$\rho(\mathbf r)=\sum_i^N |\phi_{i}(\mathbf r)|^2$$
This means that the energy computed using DFT is minimum with respect to a non-interacting system of electrons. This is the biggest assumption of DFT, which is quite valid too. As a result, DFT provides a good account of correlation energy at a very minimal computational expense (HF accounts for ZERO correlation)!!!
I took a computational chemistry course, where DFT was a small part, so I am not able to explain more in detail. I used Prof. Kieron Burke's ABC of DFT book, which he has uploaded online in his site for free use. It is a good source to start. There is also a video lecture of Prof. David Sherrill on Youtube: https://www.youtube.com/watch?v=5orzn-XA29M .
I missed the last question: For any molecule:

Compute the kinetic energy (using the KE functional)
Compute the external potential ($V_{ext}$)
Compute the Coulomb repulsion using an initial guess density ($n(r)$)
Compute the XC potential from the XC functional
Use Kohn Sham equations to find out orbital energies (under the constraint that the density ($n(r)$) is equal to the density of interacting particles (electrons). In this case, the density must be equal to the total number of electrons.
Using the newly obtained density, recompute the effective potential ($V_s$)
Perform until self-consistency is achieved. Then total energy is the sum of occupied orbital energies times the occupancy number.

Hope this helps!

(HF accounts for ZERO correlation) This is wrong. HF accounts for Exchange Correlation, exactly. It therefore accounts for almost all correlation energy. This is also a reason why it is used in Hybrid functionals. But otherwise, this is probably as close as you can get for this kind of question. (I am out of votes, but I will come back and give you the thumbs up. Please change the statement I referred to, though.) [It would also be nice if you could add the link to Burkes webpage.]

With all due respect, I absolutely disagree. I still stand by my statement that HF accounts for zero correlation. Please look at the fock operator in HF method: It accounts only for exchange and coulomb repulsion. It accounts for NO correlation energy. The definition of correlation energy itself is: E(Correlation) = E(Exact) - E(HF) Sherrill's notes
Please look at this Wikipedia section as well.

Also, hybrid functionals use HF routine only to (partially or fully) evaluate the Exchange energy. The coulomb energy and correlation energy has no relation with the HF routine whatsoever. This is mainly because HF will not give any correlation contribution, and is used as a zero correlation reference for almost all correlated methods (including DFT). Source - Google books

@user3786990 The equation you give is sloppy at best even if it is touted by nearly everyone.  Martin is correct that HF accounts for exchange correlation (between electrons of parallel spin). This business that HF accounts for ZERO correlation is completely false but CONVENIENT to say.

@LordStryker : I agree with your statement. I should have said ZERO correlation between electrons of opposite spin. But for all practical purposes, we worry about the correlation that is not accounted for, and we call this correlation energy as the correlation energy, as we try to reach the exact limit, isn't it?
I should also point that Martin said "HF accounts for Exchange Correlation, exactly", and hence I gave my comment. Thanks for pointing my mistake. My bad.

There is a confusion here due to the crap nomenclature: correlation energy is per definition the energy difference between the HF limit and non-relativistic Schrodinger solution. This is how it was defined by Lowdin! Therefore per definition HF contains zero of it. On the other hand, DFT theory loves to talk about exchange-correlation which is exchange AND correlation. In hybrid functionals, we use the exchange from HF.

@Greg This is untrue. Löwdin referred to correlation effects as for all electrons independent of spin. He continued "The exchange energy will therefore take care of a rather large part of the original correlation energy, referring to particles with parallel spins." http://journals.aps.org/pr/pdf/10.1103/PhysRev.97.1509 However, the IUPAC rules in your favour with a rather sloppy definition of the correlation energy.

@Greg You are welcome. It is one of those well laid traps in exams, where students forget about exchange correlation. (We must not forget, that we tried the Hartree-Product first.) When the chemists talk they usually refer to coulomb correlation energy and usually more specific to basis set coulomb correlation energy, because who is working on the exact Hartree-Fock solution anyways.

Let me be the elephant here and answering your question:
To actually perform calculations technically speaking you don't really need all this stuff. You need them to able to read papers, design scientifically sound research, understand what the input parameters mean and discuss your results, but not for the manual part of the labor. There are several + 1 commercial and free software are available which does these things for you as those are mentioned in the very useful comments. If you find an unscratched itch to add some energy term that is not in the software and you don't know how, it is kind of a sure sign you don't know what you are doing. It may sound harsh, but the truth is that you need to understand much more about the calculations before you can just add terms here and there, and at that level you will be able to use the developer version of those programs anyway.
It is a funny side effect of education: generally QChem is taught by people who develop code, who talk all day about HF theory and numerical details, but mostly used by students who does not need to develop code, but use already existing programs. You come out from the lecture, learnt all those equations, you think you need to use somewhere, but the cold truth is that you have to open a window an click-click here and there or open a textfile and put some cryptic switches. At manual level, QChem calculations are much more about know-how and experience than writing equations.

This doesn't really answer the question, unfortunately.

Could you add a bit more to it regarding the actual problem the OP is facing?

Speaking of QChem, the OP should browse the QChem manual which is packed full of incredibly helpful theoretical background on nearly every method the program implements.  Find it here: http://hogwarts.usc.edu/~krylov/manual.pdf  Density functional theory starts on page 75.

OK, I reiterate it: the OPs question seems to based on a misunderstanding. This misunderstanding is that an actual calculation is a very involved process where someone actually have to touch those equations. It is not. For calculations like optimizing CO, he shouldn't read more papers or textbooks, he need to read the manual or a tutorial of the qchem software of his choice. E.g. if he choose to use Gaussian, "Exploring Chemistry with Electronic Structure Methods" is a good tutorial which will tell him to write "#p B3LYP/6-311G opt" or similar to a file and done.

It sounds like you are a lot like me and have to do the work of solving every part of a problem before you feel like you understand it. Actually putting numbers in the equations instead of just looking at them.
Unfortunately, I have not found any fully worked examples on Quantum Chemical DFT. However, I would highly recommend the book by Szabo and Ostlund "Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory". They go through full calculations using self consistent Hartree Fock theory. They not only provide the equations, but get this, they actually put numbers in the matrices so you can figure out if you are doing the linear algebra and integrals correctly!!! If you really want to understand QC calculations this is a great book to have. I built a QC engine for two hydrogen atoms using this book. If you would like my notes I would be more than happy to share them with you, or anyone else for that matter.
Once you have Hartree Fock and the self consistent field approach down, I do not believe there is a huge leap to DFT. Though I have to admit I have not yet built a DFT engine.


text:
Hosoya Z Index and Correlation with Boiling Point
I have recently read about the Z topological index found by Hosoya. The papers/sources The Z index is the sum of the coefficients of the matching polynomials (no. of ways to choose an independent edge from the set of edges in the graph - if we let each carbon atom be a vertex and each bond between a pair of carbon atoms is an edge.). The paper states that in particular for alkanes or saturated hydrocarbons, there is a correlation found between the Z index (which measures the extent of branching - in what way??) and the boiling point (bp) of these alkanes.
I know that the bp of alkanes increases with its size (due to strong force of attractions). However, I am confused about how the Z index measures the extent of branching, and any underlying reasons that the Z index will correlate with the bp and other physiochemical properties.
The papers I have read so far fail to enlighten me on these questions above. Please help.
Thank you.

Some ideas I have so far:
  Since Z index measures branching (or at least correlates to branching since more compacted graph will have a smaller sum of matching coefficient?), and branching correlates to the bp of alkanes (due to less surface area sometimes if you have more branching), perhaps thats why the Z index correlates to bp...

Refer to Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers?

Your putative answer of less branching means more surface area contact, and hence larger London forces (used to be called van der Waals forces) which leads to higher bp's is the most useful view of this issue.

For a given hydrocarbon C$_n$H$_{n+2}$ more branched hydrocarbons have lower boiling point. This is because they have less surface area to have Van der Waals interactions to entangle with each other (ropes have more chances to entangle with each other than balls). For example C(CH3)4 (n-pentane) is a gas, but all other pentanes are liquids.
Their method is qualitative only. For instance, o-p-nitrophenol can be steam distilled and has much higher vapor pressure at a given temperature than p-nitrophenol, but the table would predict that they have the same properties. The difference is due to intramolecular h-bond present in o-, but ansent in p- isomer.


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Are there any hazardous products that would come from mixing glass cleaner with lubricant?
Are there any known safety issues if you mix:

Mr.Muscle glass cleaner
GT85 spray lubricant

Nothing compromising safety will happen. Now, these are two completely different cleaning agents based on alcohol/water and organic-fuel respectively. Most likely will turn white  because these substances are immiscible. Therefore, it will not be a good idea to apply a mix for you cleaning tasks.  This will be  specially so  over wood or any rough porous surface likely as they will absorb the substances resulting in unappealing white or smear white spots.  Application over polished metal surfaces will be ok.