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Why does ligand exchange take place?

Why does ligand exchange happen?
Are there specific factors that determine when a ligand can replace another ligand?
When ligand exchange takes place, does that mean that the "previous" ligand is not bonded to the complex ion at all, or both ligands are bonded?

Ligand exchange is in equilibrium, so it is driven by thermodynamics
There are many reasons one ligand would replace another, but it often has to do with Lewis basicity, HSAB (see trans effect), or the thermodynamics of a subsequent reaction.
Ligand Exchange is a broad term for a set of mechanisms that include various types of ligand association and dissociation, and these all occur for different kinetic reasons. I'll try to provide a brief overview, but the answers to all of these questions are varied depending on specific situations.

In the covalent model, both $\ce{1e-}$ "X" ligands and $\ce{2e-}$ "L" ligands are called "ligands", so it's important to point out that we're only talking about L ligands here.
Generally speaking, there are three types of ligand exchange mechanisms: association, interchange, and dissociation, and each mechanism is further broken into the associative and dissociative subclasses, depending on the exact character of the transition state. If the higher energy TS‡ of a ligand association has the geometric character of the dissociation, it is called a "dissociative association" (the Ad mechanism), for instance, but this is somewhat rare. The equilibrium expressions of the Dissociation, Interchange, and Association mechanisms are as follows, respectively:(1)
$$\begin{align}\ce{ML5X &<=>[$k_1$][$k_{-1}$] ML5 + X}\\
\ce{ML5 + Y &->[$k_2$] ML5Y}\\
\ce{ML5X + Y &->[$k_1$]ML5Y + X}\\
\ce{ML5X + Y &<=>[$k_1$][$k_{-1}$] ML5XY}\\
\ce{ML5XY &->[$k_2$] ML5Y + X}\end{align}$$
The exact nature of a ligand exchange mechanism is something that requires computations, and physical chemistry to determine, and you should look at (1) to read more about this. Really, the mechanisms are part of a continuum:
$$\mathrm{D_d \cdots I_d \cdots I \cdots I_a \cdots A_a}$$
Once the ligand is fully displaced, it is no longer part of the inner sphere of coordination, so the "'previous ligand" is no longer bound to the complex, but there exist many ligand exchange mechanisms that involve transient intermediate in which both the leaving and the incoming ligand are bound to the central atom. All associative mechanisms will have such an intermediate, while dissociative and interchange mechanisms will not. Another important example is the formation of a preassociation complex, in which "rapid equilibrium occurs between the incoming ligand and the 6-coordinate reactant to form an ion pair" (1) (or a neutral preassociation complex with strong dipole-dipole interactions, depending on oxidation states). In the case of a preassociation complex, the outgoing ligand is no longer in the inner sphere of coordination, but it is still "bound" to the complex, albeit through weaker noncovalent interactions. Such a reaction has equilibrium like this:
$$\begin{align}\ce{ML5X + Y &<=>[$k_1$][$k_{-1}$] ML5X . Y}\\
\ce{ML5X . Y &->[$k_2$] ML5Y + X}\end{align}$$
Ligand exchange kinetic and thermodynamic arguments also frequently invoke the chelate effect and the trans effect.
Another important concept is that the driving force for individual ligand dissociations and associations involved in catalytic cycles often implicates the driving force of the overall cycle. In catalytic reactions, the rate of ligand dissociation or association is not dependent not on the thermodynamics/kinetics of the individual reaction, but instead on the step in the cycle with the highest activation energy. This step is known as the  turnover limiting step (2). In a catalytic cycle, all equilibria have the same rate. "On a circular track, on average the same number of trains must pass each point per unit time." (2) For example; in this catalytic alkene hydrogenation, the turnover limiting step is the dissociation of the phosphine from Wilkinson's catalyst (3), so this is the step that determines the turnover frequency (TOF). The rate of association of the ethylene ligand later on is driven by the cycle's overall TOF, so such a ligand association would be indirectly driven by a different step in the cycle. If the final step of the cycle is irreversible, the substrate release might also be a thermodynamic driving force.
$\hskip1.4in$(3)
These are just some of the many possible answers to your questions.
(1) G. L. Miessler, P. J. Fischer, D. A. Tarr. Inorganic Chemistry (5th Ed.), Pearson, 2014, p. 442-444
(2) R. H. Crabtree, The Organometallic Chemistry of the Transition Metals (3rd Ed.), John Wiley & Sons, 2001 p. 225
(3) J. Luo, A. G. Oliver, J. S. McIndoe, Dalton Trans. 2013, 42, 11312-11318

Np! I recommend you look at Miessler, or another inorganic book if you want to read more about the specifics of ligand exchange.

Ligand bonds are labile. A ligand isn't welded to the complex. Remember that it is an equilibrium.
To switch gears a bit consider water just having $\ce{^1H2O}$. A drop of $\ce{^2H2O}$ is added. The $\ce{^1H}$ and $\ce{^2H}$ atoms exchange quickly and the solution is mostly $\ce{^1H2O}$, with some $\ce{^1H^2HO}$ and only the faintest trace of $\ce{^2H2O}$.


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Why is the letter J omitted in the spdf... sequence?
Under spherical symmetry, the irreducible representations corresponding to $L = 0, 1, 2, \cdots$ are assigned the letters $\mathrm{S}, \mathrm{P}, \mathrm{D}, \mathrm{F}, \cdots$ after which the letters progress in alphabetical order. (The familiar names of the atomic orbitals are also labelled with this sequence of letters.) Therefore, we have:
$$\mathrm{S}, \mathrm{P}, \mathrm{D}, \mathrm{F}, \mathrm{G}, \mathrm{H}, \mathrm{I}, \mathrm{K}, \cdots$$
but as far as I can tell, $\mathrm{J}$ is conspicuously omitted. I am guessing there is a reason behind this - what is it? Is it to avoid confusion with some other $J$, like the total angular momentum quantum number $J$, or is it perhaps based on some typographical argument (J being easily confused with I)?

Omitting j when alphabetically enumerating things has a long tradition.

First of all, the alphabet did not always exist in the form we know it today.
Quoting Wikipedia:

After [...] the 1st century BC, Latin adopted the Greek letters ⟨Y⟩ and ⟨Z⟩ [...] Thus it was during the classical Latin period that the Latin alphabet contained 23 letters:
[no J, V, W]
[...]
It was not until the Middle Ages that the letter ⟨W⟩ [...] was added
[...]
only after the Renaissance did the convention of treating ⟨I⟩ and ⟨U⟩ as vowels, and ⟨J⟩ and ⟨V⟩ as consonants, become established. Prior to that, the former had been merely allographs of the latter.

In some books, this has consequences till today.

The footnotes to the Confession and Catechisms [of the Presbyterian Church], containing the proof texts, are enumerated in the traditional manner, that is, by letters of the alphabet (omitting j and v, as alternative forms for i and u in the Latin alphabet). Source

But even later, long after the letters I and J were considered distinct in terms of proper spelling, their alphabetical order (I preceding J) was not firmly established.
The New General English Dictionary of 1768 had a combined section for I and J, treating both equal with respect to alphabetical order (but not regarding spelling).
Same with Handwörterbuch der allgemeinen Chemie, a German chemistry book printed in 1818.
The latest book, I could find, is the Handwörterbuch der reinen und angewandten Chemie of 1850. This is pretty close to when the letters spdefghik.. must have been defined! (Does anyone know when exactly this was?)
And now, it perfectly makes sense to me. Even if i and j were distinct letters at the time and their order should have been commonly established in the early 20th century, the possibility that some readers could still be confused about what comes first, must have lead to the decision to leave j out.

Update (2017-12)
Concerning the spelling and order of names, I was able to find evidence that is even 100 years younger.
The Berlin telephone book contained a spelling table. The first book of 1890 contained a spelling table that assigned numbers to each letter, omitting the letter J, i.e. I=9 and K=10. That is pretty similar to our orbital labels, isn't it? In the 1903 printing, words were assigned to the letters, but J was still left out. The 1905 printing was the first to include J into the spelling table. (source)
Even the Berlin address book of 1943 did not distinguish between I and J. For instance Jutta is listed before Iwanski. (Interestingly, this book doesn't even use different glyphs for I and J in the Fraktur font.)
Of course, there are many books around that time, and earlier, that sort I before J and that use different glyphs, even if printed in Fraktur (example). Nevertheless, this shows that the convention "I before J", as we know it today, was not firmly established in early 20th century Germany.

http://www.snopes.com/history/american/jstreet.asp

RE: "This is pretty close to when the letters spdefghik.. must have been defined! (Does anyone know when exactly this was?)"  I don't see how the spdfghik.. nomenclature could have been worked out before a decent model of the atom was conceived in the early 20th century. I'm sure line spectra were known before then but why certain lines and why the relative intensities varied would have been a mystery. I have no idea what previous nomenclature system might have been used.

In pretty sure that spectral lines in general go back to the Fraunhofer lines.

Sorry, just one month late. I spent a long time debating over which answer to accept, and whether I should accept one. Ideally, we would be able to find out when the people who named spdf... named it, and ideally they would have left some written record of why they omitted J. In the (assumed) absence of that, this is the next-best. I think this comes closest to what I was looking for, with all the historical sources.  Thanks for sharing!

For the azimuthal quantum number (l) of an atom, there is no "j" because some languages do not distinguish between the letters "i" and "j".
L is the total orbital quantum number in spectroscopic notation and uses capital letters. The nomenclature just follows suit with the suborbital notation and skips J since there is no corresponding j.

Yes, it is because of what @orthocresol mentioned in his post and what you have written. For what it is worth, I looked over the reference given in Wikipedia (Atkins, Paula, Friedman. (2009). 'Quanta, Matter, and Change' (p 106)), and it checks out. see here Would be interesting to know what kind of modern languages do not differentiate between i and j though.

Huh. I always thought it was because i and j look too similar, and j was skipped to avoid interconfusion, as ortho noted.

Mathematicians worldwide do not have the same issue and happily use indexes i and j.

Distinguishing between i and j is, as others have mentioned, a rather recent phenomenon, not unlike the distinction between u and v. Especially in German when typesetting in blackletter, the glyphs for capital I and J would be identical and there are still a few (older) road signs around Germany that use a capital J where the letter is in fact an I going by pronunciation (an example would be Jllertissen, but I can’t remember the actual name of the place I saw it).
Some intelligent OpenType fonts such as UnifrakturMaguntia implement quick switches between the two, so that one may type the ‘correct’ letter but the ‘historically accurate’ glyph is used (one would type Illertissen to get Jllertissen in the example above). The font also includes ‘ready-to-use variant fonts’ differing by century; going by the features activated in the 19th century variant the two capital letters were still considered as one in that age. The English language manual also goes on to say:

In some early fraktur texts, j was used at the beginning of a word
  for both, i and j, while everywhere else i was used for both. [Typesetting adjusted to antiqua rules.]

In der Kajüte ist jemand → Jn der Kaiüte jst jemand.

While the latter was not implemented in historical variants — indicating that according to the designer’s research it was not really common at any point in the history of German blackletter typing — it was nonetheless sufficiently common to be determined worth integrating.
This may serve as further verification that i and j were historically often simply considered variants of each other.

Having i-j problem in German around 1910 would make a lot of sense. In English i-j were sorted out long before atomic theory was developed. // PS - knew I was never going to be an organic chemist because I don't know German and couldn't read Bilstein.

@MaxW Yeah, in English i and j represent completely different sounds as they do in French. In most other continental languages, j is what English uses (consonantal) y for, making i and j much more similar — the full-vowel and half-vowel variants of otherwise practically the same sound.

@MaxW No, the differentiation between i and j was definitely done in German around 1910. While, indeed, I found chemistry books that use a font that does not clearly distinguish I and J (https://books.google.de/books?id=ThpkAAAAcAAJ&dq=chemie&pg=PR1#v=onepage&q&f=false, Inhaber vs. Jod, but i vs. j is clear), there are also plenty of books in roman/antiqua typeface that show chemists had a clear unterstanding of the difference between I and J (https://books.google.de/books?id=YcMPAAAAQAAJ&dq=chemie&pg=PR1#v=onepage&q&f=false from 1831, Jod vs. Interesse).

@mhchem Inhaber is a very clear-cut case much like Jodeln. Iod/Jod not so much since it stems from Greek io-eidés and thus should have an i etymologically but was pronounced much more like j. But the important point in both my answer and my comment was that there was only one glyph to represent both capital I and J in blackletter. So much like in the history of u and v it would only be the surrounding letters that distinguish which sound was to be represented.

When two s-orbitals form bond, we have σ-bond.
When two p(x,y)-orbitals form bond, we have π-bond.
When two d(...)-orbitals form bond, we have δ-bond.
When two f(...)-orbitals form bond, we have φ-bond.
So, similarly, when two g-orbitals form bond, we have γ-bond, h-orbitals form η-bond, i-orbitals form ι-bond, etc...
But what is the Greek character for "j"? There is none. Hence "j" is skipped.

This is interesting, although it would be more useful if you could think of some sources that support your idea.

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How does ferrous iron coordinate nitrogen in a porphyrin ring?
In first semester biochemistry, my professor said that iron forms a coordination complex with nitrogens in the heme group.
I'm failing to understand what this even means since the charge on nitrogen does not change as a result of this bond. My organic chemistry textbook and biochemistry textbook do not even have a basic explanation of what is meant by this, and I can't find anything on the web.
How does iron form a bond with two of the nitrogens in the porphyrin ring, the histidine, and the oxygen?

Nitrogen is coordinating iron; not the other way around. You should look up coordination compounds in general inorganic textbooks. $\ce{[Fe(haem)(His)(O2)]}$ is similar enough to $\ce{[Fe(NH3)6]^2+}$ as a starting point. I’m not sure if this site is the best place to answer the question. By the way, the porphyrinic nitrogens (two of the four) do have a charge associated with them; they are rendered in the image as forming a single bond to iron.

This is something I have asked myself for a long time until I finally realized by myself. I have superficial knowledge on this type of chemistry but I believe this will be enough, so i will explain it in a very simple fashion. Well you have to look at Nitrogen first, nitrogen doesn’t change it’s charge just like you said, which seems weird at first since it forms four bonds. Now think of nitrogen in the lewis formula, what does it look like? Nitrogen has a free pair of electrons available. Think of the d-orbitals in Fe, their character and shape. Think of metals in general, having their electrons in a type of diffuse cloud. Now to the bond, the Iron is capable of grabbing these free electron pairs and use them in a bond, pulling them into it’s direction. This way the charge doesn’t change. The nitrogen “donates” it’s electrons which do not participate in regular bonding like you would see it in ammonia, but in complex bonds. Hope this is enough to clarify :)
Greetings from Germany

This one I think I remember.
There are two nitrogens atoms that act as an unstable free radical and take more two eletrons of iron. The iron atom now has four posiitive charges.
Those four positive charges naturally atracts all negative charges around, the pairs of the three nitrogens and the pair of the oxygen, Iron would stay with four negative charges, if wasn't the fact that the donators atoms also stays with four positive charges, resulting in zero.
Exactly as in the positive ion of ionized water, the total charge is conserved. In water, it is conserved as one positive charge, in heme molecule with iron 2, the total charge is zero. It's not two because of the 2 eletrons of the two nitrogens tha entered the system.

Except porphyrin isn't a radical with unpaired valence electrons.

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How to get (2R,3R)-2,3-dibromobutane from but-2-yne?
What I had in mind was to form but-2-ene with $\ce{H2}$ and Lindlar's catalyst and then simply add bromine ($\ce{Br2}$). But this addition, if I'm not mistaken, is a trans-addition. So I would end up with (2R,3S)-2,3-dibromobutane instead of (2R,3R).
Note that this is (2R,3S)-2,3-dibromobutane:

Have I made a mistake, or is there something wrong in my way of thinking? If my solution is wrong, how could I get (2R,3R)?

I think I figured it out. Lindlar is a cis-addition so one would not obtain a (2R,3S), but rather (2S,3S) and (2R,3R) which happen to be enantiomers.

Yes, pretty much. And if you wanted the other diastereomer (the (2R,3S) i.e. meso diastereomer), you would just need to use Na, NH3 to reduce the alkyne first. That gives the trans alkene which adds Br2 in an anti fashion (via the bromonium ion) to give the meso compound.


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Why is the only H NMR resonance of 1,4-dibromobenzene a singlet?

I am working on a chemistry problem set and the answer key says all H's will show up as singlets. This does not make much sense to me as there are adjacent hydrogens to each carbon which should mean the H's show up as a doublet. What am I thinking incorrectly about this?

All four hydrogens are equivalent and therefore they do not display any coupling to each other. (They do couple, but the coupling cannot be seen in the spectrum.) See: Why don't equivalent hydrogens cause splitting in NMR? for more discussion of this topic.
Just to be extra nitpicky, I wouldn't write "all four hydrogens show up as singlets", since that implies that there are four different singlets. I'd write something like "one singlet is observed in the spectrum, corresponding to all four hydrogens".


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Reduction of potassium ion and hydrogen ion
I am studying a galvanic cell, and the cathode uses potassium hydroxide as its electrolyte. I am getting confused as to what gets reduced when electrons come from the anode.
I believe it would be the hydrogen ions (because the solvent is water) as potassium ion would never want to accept an electron (because it has a configuration of a noble gas as an ion and is therefore very stable). However, the pH of the solution would be very basic because KOH is the electrolyte; thus, the reduction potential value of hydrogen ion should also decrease.
Can someone please tell me what the right answer is?

What are the electrodes made of?

aluminum is the anode and platinum is the cathode electrode

Platinum is generally inert, so I would check half reactions involving aluminum...

no platinum is the electrode that will be in the KOH electrolyte.

Yes, but the platinum is just there to conduct. It doesn't actually do anything chemistry.

ya, so i was asking if the hydrogen ion gets reduced or the potassium ion?

Check the reduction potentials...

based on what I described in the question can you tel me if I am right when I say that it will be hydrogen than potassium even with the higher pH

You're not going to make metallic potassium in water...

but potassium would never want to gain an electron. It has a configuration of noble gas when it is an ion.

You're adding energy to the system. If it weren't in water, you can add an electron. Indeed, the industrial production of potassium is achieved by electrolysis of molten $\ce{KCl}$.

wait I am very confused with your comment above. I am asking if I have a cell that has an anode electrode of aluminum giving 3 electrons to the platinum electrode, i think hydrogen would be more likely to "absorb/accept" these electrons (even at a higher pH when there are not so many H+ ions) than K+. Now, my solvent is water so it could be that OH- could be absorbing the electrons. I just want to know what is most likely: H+ accepting the electrons or OH- or K+.

I'm trying not to outright give you the answer per homework policy since you seem to be moving in the right direction. Adding an electron to potassium ion is highly disfavored (not impossible). Adding an electron to hydroxide isn't great. I think you've already got the correct answer above?

but I am getting a little confused in that because this reaction is so basic, there won't be enough hydrogen ions to be reduced, so what about those extra electrons that exist at the double layer? Will that react with OH-? Sorry if I am being too repetitive

Ah... I see your issue. Even at pH 14, there is still a significant amount of hydronium. And proton transfer in water is lightning fast, so as you consume the hydrogen ions via reduction, you'll get more because the equilibrium is fast. Now, at some point, it will be disfavored, but you'd need to use the Nernst equation to figure out if that's the case.

can you please show me how you can use the Nernst equation to do that?

I'm on my phone, so now is not the best time. Check wikipedia first... That might enough to help you figure it out.


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Spectrochemical series of TMs: why does the field-splitting parameter NOT steadily increase along a period?
Why doesn't the field splitting parameter increase steadily along the period for 3rd-row transition metals, as one would expect due to a better energy match between ligands and the metal as $Z_\mathrm{eff}$ increases?
The series is (in the order of increasing $\Delta$: $\ce{Mn^2+}, \ce{Ni^2+}, \ce{Co^2+}, \ce{Fe^2+}, \ce{V^2+}$ etc. Source: Spectrochemical series – Wikipedia, confirmed by Shriver and Atkins.
I would guess that this is to do with $\mathrm{3d}$ orbitals getting too contracted on the right of the period and, and hence, having a poorer size match, but I couldn't find this explanation mentioned anywhere.

A greater octahedral field split would be associated with a shorter $\ce{L\bond{->}M}$ bond length. Bond length is a function of:

size of both partners
strength of the bond

In particular, as we are examining metals in $\mathrm{+II}$ oxidation state, the metal ion’s size should increase somewhat steadily along the period while the ligand size should stay the same. This means that we need to be looking at bond lengths to understand the discrepancy.
For reference, here is a typical octahedral $\ce{[ML6]}$ complex ignoring π effects.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
We should immediately be able to see how field split is generated: it is the difference between the $\mathrm{t_{2g}}$ orbitals and the $\mathrm{e_g^*}$ orbitals; the latter of which are antibonding. Thus, having any electrons in $\mathrm{e_g^*}$ will weaken the $\ce{L\bond{->}M}$ bonds. This corresponds nicely with vanadium(II), a $\mathrm{d^3}$ ion, having the largest field split, and manganese(II) ($\mathrm{d^5}$) having the lowest.
However, the final value is dependent on very subtle differences and not easily predicted a priori.


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How does magentic field change spin values of a proton or electron?
I am asking this in context of NMR. Firstly I wanna say that I thought that spin values of a proton or electrons were intrinsic. I didn't know that you can change spin states from one value to another. But now I read that this is not the case. So my question is that I am reading that prior to application of magnetic field, spin values of proton are equal. However when magnetic field is applied, the of the two spin values change? How does this happen? I am not advanced in physics so if anyone can explain this in detail and in layman terms it would be great, Thanks.

The value of the spin is not changed. A proton is a spin-1/2 particle, whether it is in a magnetic field or not.
However, all spin-1/2 particles have so-called "spin up" and "spin down" states. The thing that is changed is the energies of these up and down spin states. You may wish to think of the hydrogen nucleus as a bar magnet in a magnetic field. The case where the magnet is aligned with the field is more stable; when the magnet is opposed to the field it is less stable. For example, a compass pointing North is more stable. If you reach inside and turn the compass needle around 180 degrees so that it points South, you need to input energy to hold the needle in that orientation; that's because that orientation is less stable.
This analogy should be described very thoroughly in various places on the Internet. For example, Chemguide has an entire page on it, replete with diagrams.
(Disclaimer: quantum mechanically a proton does not necessarily have to be either in the spin-up or spin-down state. For more details, consult an NMR textbook. The difference in energy also can be explained by QM and is due to the presence of a magnetic term in the Hamiltonian: $\hat{H} = -\hat{\mu}_z B_0 = -\gamma \hat{I}_{\!z} B_0$ where $\gamma$ is the gyromagnetic ratio, $\hat{I}_{\!z}$ the projection of the nuclear spin along the $z$-axis, and $B_0$ the strength of the external magnetic field. Therefore, the "up" state which has $\hat{I}_{\!z}|\alpha\rangle = +\hbar/2$ has a magnetic energy of $E = \langle\alpha|\hat{H}|\alpha\rangle = -\gamma B_0 \hbar/2$.)

Okay so you are saying that spin values are intrinsic. So a particular proton that is -1/2 will stay -1/2 forever except that within it there is a spin up and spin down. I always thought spin up was +1/2 and spin down was -1/2. I didn't know that within each there is spin up and spin down

A proton has a spin of $1/2$. Positive $1/2$, not $\pm 1/2$! However, this spin can "point" in different directions: so-called up and so-called down. The proper terminology is that the projection of the spin can assume the values $+1/2$ ("up") and $-1/2$ ("down"). However, no matter which way the spin is pointing, its magnitude (i.e. its spin) is always $1/2$. This magnitude is invariant.

Oh okay. That's so weird. I always assumed by -1/2 and +1/2 protons and electrons come in two types. So I just wanted to ask, I read often that spin states are quantized and that u can only have up and down states and not take any values between those up and down. But then other time I read that without magnetic field protons and electrons are spinning at randon and I see arrows depicted not only as up and down but also sideways, making it seem that u can have range of values from up and down.  What are u supposed of make of this?

It may perhaps not be a very good idea to read too much into it, because it can lead to misconceptions. It is a fundamental aspect of QM: before you measure the state of a particle, you cannot know whether it is spin up or spin down. It is best described by a certain percentage of up and down: for example, a proton might be 45% up and 55% down at a certain point of time. However, if you try to measure it, then you can only measure either up and down. You'd have a 45% chance of measuring up, and a 55% chance of measuring down. That's the real meaning of quantisation.

A proton only has an up and down orientation in a magnetic field. In the absence of a magnetic field there is no preferred direction.

That first sentence is found in typical introductory NMR explanations, but does not make any sense from a QM perspective. There is no difference between the states that a proton nucleus is allowed to adopt, with or without a magnetic field. All linear combinations of $|\alpha\rangle$ and $|\beta\rangle$, i.e. $|\psi\rangle=c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$, are allowed states. The only difference is that under a magnetic field, the ensemble average of $|c_\beta|^2$ is slightly different from that of $|c_\alpha|^2$. Whether it is bigger or smaller depends on the sign of $\gamma$.


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Why is the chemical formula for vinegar CH3COOH what it is?
I am quite new to chemistry, but I would like to think that I have mastered the basics. This would include naming and writing out chemical formulas. However, once I had seen the formula for vinegar ($\ce{CH3COOH}$), I was bewildered.
It would have made sense to me that the formula would be more, say, "compacted" into just $\ce{C2H4O2}$. What is the reasoning behind this odd way of writing such formulas and when should I use it (it seems like this could be a huge gap in my knowledge)

When expressing a chemical formula within a text line three different forms are used - Empirical formula, molecular formula and condensed formula. Chemists also use structural formulas (drawings) to show a 2D representation of a 3D molecule. https://en.wikipedia.org/wiki/Chemical_formula

IUPAC, functional groups, and ACID! and no molecule would be worth without its function, chemical properties that is shaped by its structure. How would anyone know if C2H4O2 is a ketone, aldehyde, ester...ether or an acid by just looking at that? As you dive into organic chemistry, there will be thousands of compunds and structures. This naming standard gives wings and freedom to a student, a chemist to do pattern recognition and to know who we are dealing with. In days and years to come you will appreciate this little tables and conventions much =)

Also relevant: http://chemistry.stackexchange.com/q/58910

CH3COOH is an accepted and common form of writing the structure of acetic acid (commonly known as vinegar when diluted in water). You are right in saying the formal notation should be used - which is C2H4O2 - since it gives the reader the types of atoms and their quantities, and these two characteristics define the exact composition of the molecule, and there's nothing wrong with that.
However, the notation CH3COOH takes you one step further and tells you something about the actual structure of the molecule, since it groups together the atoms that comprise a functional group, and a very important one at that. Grouping together COOH is an accepted and convenient form of communicating to the reader: "there is a carboxylic group here", or "this molecule is actually a carboxylic acid", which is something you'd like to know when dealing with a certain substance.
To summarize, C2H4O2 can be the molecular structure for acetic acid or formate (formic ester, as mentioned in another answer). In order to avoid confusion and, more importantly, indicate that there is a carboxylic group in the molecule, it is more convenient and straightforward to note it in the formula than letting readers figure it out themselves.

Plus one @Don_S simple and conscise.

Carboxylic acid and ester are functional group isomers which means that both of them has the same formula but different functional groups.
Lets take the example of $\ce{C2H4O2}$.
Carboxylic acid : $\ce{CH3COOH}$

Ester : $\ce{HCOOCH3}$

To avoid the confusion formulas like $\ce{CH3COOH}$ are used.

Structral diagrams are at the heart of organic Chemistry. Nice one.

The short answer is that CH3COOH indicates (by convention) that this compound has both a methyl group and a carboxylic acid group. (The group COOH is generally understood to be the C(O)OH group or C(=O)OH where the double bond "=" and the parentheses are implied.) Note that the organic acid in vinegar is acetic acid, but acetic acid is NOT vinegar!!! Vinegar is an aqueous mixture, not a chemical compound, vinegar is biologically derived and generally includes hundreds of other chemical compounds. As far as the basics of chemistry, no you haven't mastered them. Consider the difference between red phosphorus and white phosphorus. White phosphorus ignites at 30°C, red at 300°C. Or consider graphite, buckyballs, diamond, and graphene - all are "just" carbon, but their properties vary enormously. You haven't identified a chemical compound until you know both its atomic formula AND the bonding between its atoms. The common example is dimethyl ether H3C-O-CH3 and ethyl alcohol CH3CH2OH. Their atomic formula is identical, but one is a gas, and the other is a liquid (at STP). One is fermented in our guts and the other is used as a propellant for aerosol cans. To say that vinegar "is" acetic acid is like saying beer "is" ethyl alcohol.


text:
How is the tetrachloridocobaltate(II) anion stable if its crystal field stabilization energy comes to be positive?
I tried to calculate the ligand field stabilisation enthalpy of $\ce{[CoCl4]^2-}$ (as shown in the image) but found it to be positive. What could be the reason behind this? How is it stable with positive crystal field stabilization energy?

You didnt put the right values. Higher orbitals have energy value of $+0.4 \Delta_t$ and the lower orbitals have the energy value of $-0.6\Delta_t$.
This complex has a coordination number of 4, so it can have tetrahedral or square planar geometry. Tetrahedral complexes are favored where attainment of regular complexes is important. For the tetrahedral complexes $\mathrm{d^0, d^2, d^5, d^7}$, and $\mathrm{d^{10}}$ the configurations are regular.
In tetrahedral complexes, mostly $\Delta t < P\ (\Delta_t \approx 4/9 \Delta_o)$ and hence they generally form high spin complexes.

The splitting will be into $3\ \mathrm{t_2}$ (not $\mathrm{t_{2g}}$) higher energy orbitals and $2\ \mathrm e$ (not $\mathrm{e_g}$) lower energy orbitals. The effect of the ligand wont matter here because the value of $\Delta_t$ is small. So the filling will be same for strong field and weak field.
$$\begin{align}\text{CFSE} &= +0.4(3) -0.6(4)\\
&= -1.2\end{align}$$

Thanks for the answer...  It was really a silly mistake..

‘Tetrahedral complexes are favored where attainment of regular complexes is important.’ — This is incorrect. Tetrahedral complexes are formed if the overall complex is better stabilised in a tetrahedral manner than in a square-planar, octahedral (having added ligands), linear (having removed ligands) or any other manner

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


text:
What is the meaning of overbased additive in lubricants?
From the google search that I did, I do not understand what overbased means.
Any comment,link or answer would be useful

In detergents, by overbasing, it is meant that a stoichiometric excess of the metal be present over that required to neutralize the anion of the salt. The excess metal from overbasing has the effect of neutralizing acids which may build up in the lubricant. A second advantage is that the overbased salt increases the dynamic coefficient of friction. Preferably, the excess metal will be present over that which is required to neutralize the acids.

Very likely the same for lube oils. They are to neutralize carbonic acid ,etc , in combustion gases that may leak past the rings. I remember reading of Mg and Ba being used.


text:
how to determine the volatility of an organic compound using the physical properties of organic molecules
How do you tell if a compound has a high volatility?
Also can the boiling point of a compound be taken into consideration when determining if a compound has a high volatility?

Vapor pressure is the physical property that expresses a compound's volatility at a given temperature, and the enthalpy of vaporization describes the temperature dependence of the vapor pressure.
For some compounds, like water, there are tables and formulae readily available for determining the vapor pressure at different temperatures. If you know the compound's vapor pressure at a given temperature and you also know the enthalpy of vaporization for the compound, you can then estimate the vapor pressure of the compound at another temperature.  This method assumes that the enthalpy of vaporization of the compound is constant with temperature, which is only a good approximation over a relatively small temperature range. You would need to know the heat capacity of the compound to correct for the temperature dependence of the enthalpy of vaporization.
Comparison of boiling points is often used to say "compound x is more volatile than compound y" because compound x has the lower boiling point. This statement is only true for the pressure at which the boiling point was given, often 1 atmosphere.  The vapor pressure of a polar compound like a low-molecular weight alcohol has a greater temperature dependence (enthalpy of vaporization) than a non-polar compound like an alkane.  So an alkane and an alcohol with the same boiling point at one atmosphere will not have the same boiling point at another pressure. Correspondingly, an alkane and an alcohol with similar vapor pressures at one temperature can have very different vapor pressures (volatilities) at another temperature.

This answer is in the same spirit that the question is given, which is to just get an idea about volatility of a compound given its chemical formula. I need to make it clear that there are times when you need to quickly compare two chemical structures in your head and decide which one is more volatile. For that, you can use the below criteria, however, for more rigorous and serious calculations, you need to directly evaluate vapor pressure using known experimental data or precise thermodynamic models.
When comparing two molecules:
1- lower molecular weight leads to higher volatility
2- presence of polar groups usually decreases volatility
3- presence of hydrogen bonding sites significantly decreases volatility


text:
What defines basis sets in computational chemistry?
I have a simple question about the Linear Combination of Atomic Orbitals (LCAO):
$$
\psi_{i} = \sum_{r} c_{ir} \phi_{r}
$$
where

$\{\psi\}$: Set of molecular orbitals
$\{c\}$: Set of coefficients
$\{\phi\}$: Set of atomic orbitals

When we talk about basis sets, do we mean here the coefficients or the atomic functions?

Related: http://chemistry.stackexchange.com/questions/59674/

atomic functions.

A basis set is a set of $c_i \phi_i$: $c_1 \phi_1, c_2 \phi_2, c_3 \phi_3$... You sum the functions in a basis set to approximate the wave function.

In LCAO, it is the set of atomic orbitals (AOs) that is the basis, and the coefficients are the basis expansion coefficients. For example, take the hydrogen molecule, with 1 atomic orbital on each atom:

The lower energy MO, the bonding one, will be (excluding normalization):
$$
\psi_{\mathrm{1s}} = \phi_{\text{left}} + \phi_{\text{right}}
$$
and the higher energy MO, the antibonding one, will be (excluding normalization):
$$
\psi_{\mathrm{1s}^{*}} = \phi_{\text{left}} - \phi_{\text{right}}
$$
This means that for the bonding MO, $c_{\text{left}} = 1$ and $c_{\text{right}} = 1$, while for the antibonding MO, $c_{\text{left}} = 1$ and $c_{\text{right}} = -1$. If your basis was different (say, the atomic orbitals were of different radial extents for the two atoms), then the coefficients would be different to accommodate that. They aren't part of the basis set.
Since you tagged this with computational-chemistry, I assume you also want to know about basis sets in computational chemistry. Conceptually, they are identical to LCAO-MO bases, but what may be confusing is that each atomic orbital itself may be composed of multiple functions (called primitive functions), rather than just the single function for each AO as seen above. This leads to another set of coefficients, called contraction coefficients, describing how the primitive functions are linearly combined to form a contracted function.
An "implementation" detail is that the functional form of AOs is usually Gaussian functions, which have a parameter in the exponent (here, $\alpha$):
$$
\phi_{r}(x) = e^{-\alpha x^2}
$$
This means that to fully define an atomic orbital basis composed of contracted Gaussian-type orbitals (CGTOs), one needs both the contraction coefficients and exponents. They usually look like the following. For each angular momentum type, there may be one or more contracted functions, each composed of one or more primitives. For example, in the hydrogen definition below, there are 2 functions to describe s orbitals, one of which has three primitives and the other is not contracted. The contraction coefficients are in the first column and the exponents in the second column.
****
H     0
S   3   1.00
     13.0107010              0.19682158E-01
      1.9622572              0.13796524
      0.44453796             0.47831935
S   1   1.00
      0.12194962             1.0000000
****
C     0
S   5   1.00
   1238.4016938              0.54568832082E-02
    186.29004992             0.40638409211E-01
     42.251176346            0.18025593888
     11.676557932            0.46315121755
      3.5930506482           0.44087173314
S   1   1.00
      0.40245147363          1.0000000
S   1   1.00
      0.13090182668          1.0000000
P   3   1.00
      9.4680970621           0.38387871728E-01
      2.0103545142           0.21117025112
      0.54771004707          0.51328172114
P   1   1.00
      0.15268613795          1.0000000
D   1   1.00
      0.8000000              1.0000000
****

The last note is that when performing a Hartree-Fock or similar calculation, the contraction coefficients and the exponents don't change. That would mean the basis set changes. The basis set is fixed, it is the MO coefficients that change, the same coefficients that appear in the LCAO-MO equations.
Image taken from here. Basis set is Def2-SV(P).

Isn't the first column the exponents and the second the contraction coefficients?

Short answer: The set of atomic orbitals $\lbrace\Phi\rbrace$ is the basis set.
Long answer: One can get confused because several primitive functions can be contracted into an atomic orbital, which is done by - wait for it - linear combination. The reason for contraction when applied to Gaussian functions is that Gaussians do not exhibit a cusp and thus describe the near-nucleus region badly. This also shows in the total energy. Therefore, several primitive Gaussians can be combined to form one atomic orbital, which also saves computational time because degrees of freedom are given up. (They are sometimes reintroduced by uncontracting the basis set, which can be sensible for the calculation of nucleus-related properties such as NMR shifts.) However, since the atomic orbitals are then characterized by several exponents and contraction coefficients, coefficients are also part of a basis set.


text:
Boron and diols
How does the Boron react with diol? Is this the right way it reacts? What kind of interaction is this?
Why does such a reaction occur?

Your starting material (1) is 2,1-Benzoxaborol-1(3H)-ol. This can be considered as the intramolecular half-ester of 2-hydroxymethyl-phenylboronic acid.

Under suitable conditions, esterification of 1 with the diol will initially yield the ester 2.
Transesterification, that is closing one five-membered ring while opening the other in this case, may lead to 3.

That said, I do think that even very mildly basic conditions (e.g. aqueous bicarbonate) would suffice to quantitatively convert the mixture into the boronate anion shown in the question.

@NicolauSakerNeto You're probably right and I didn't think it though.

@KlausWarzecha In the absence of base 3 will be the preferred product? In the presence of base, would the tetrahedral product be as shown in my question or would it be 3 with an additional OH attached to B? I am a little confused about how the products are/will form.

Boron is a Levis acid while he alcohol is considered a Levis base. It is able to have a reaction between them. The problem is that the structure of the product of reaction do not show a -OH group.

Why should the product have an OH group in it? Does the initial OH group attach leave as water?


text:
What is the Iso electric point of the given amino acid?

I am not sure about three but for two acidic parts we take the mean of the two..  But what about three?

Thank you for editing! I have removed the comment, and upvoted your question. If you receive a useful answer, consider accepting it. A recent overview of isoelectric point calculation techniques has been published along with a database. It is available here; L. P. Kozlowski. 'Proteome-pI: proteome isoelectric point database'. Nucleic Acids Research, ($2017$), 45, D1, D1112$-$D1116. Hopefully someone more competent will provide a summary. Good luck, and welcome!

Though in the current case it's more straightforward; apply the definition of isoelectric point, consider raising the pH from $0$. Everything is protonated, and there is a positive charge. When you raise pH, the right-hand side group is first to deprotonate. Thus $$pI = average(2; 4).$$

Thank you very much@Linear Christmas. I am using a smartphone and am unable to find the "accept" option.. It would be very kind of you if you could tell me where to find it..

Sure you would have found it if there were one in this case ;). I only provided comments (which are not really intended as answers on SE) because I thought it would be a good idea to provide an answer that also deals with isoelectric points of peptides and longer chains. Might post an answer later on but I would have to go through the literature which will take time. Until then, stick around the site and you are sure to find lots of interesting content. As always, glad I could help!

Chatting requires a minimum of $20$ reputation  on any SE site (see here for an overview). Don't be discouraged, I am sure you'll reach $20$ in no time. You can still listen in on existing chatrooms; the main English chemistry chatroom can be found here.


text:
What results from the reaction of 4-phenylbutanoyl chloride in presence of AlCl3?
Please note that this is 4-phenylbutanoyl chloride:

My guess was that another molecule of 4-phenylbutanoyl chloride would react with the phenyl, thus forming a structure like the one below:

But this molecule seems somewhat awkward, and I am unable to name it. Is my guess correct? If yes, what would be its name?

My guess was that an other molecule of 4-phenylbutanoyl chloride would react with the phenyl [...]

Your guess is not that bad and it seems that you have identified the type of reaction.

 Apparently, it is a special type of electrophilic aromatic substutution, known as the Friedel-Crafts acylation. In your guess, one molecule of the starting material would serve as the acylating agent (through the acyl chloride), while another is acylated at the phenyl.

What if you only had one molecule of the staring material? Can you imagine a similar reaction?

 The alkyl chain of $\ce{CH2}$ that links the acyl chloride unit to the phenyl is flexible! There's no reason why the Friedel-Crafts acylation could not happen as an intramolecular reaction, where acylating agent and the attacked phenyl group are part of the same molecule.

What would the the outcome?

 The intramolecular reaction would furnish tetralone.


text:
Aluminum powder reaction with air
I've made aluminum powder in a ball mill a few times, but my last batch began to smoke and caught on fire soon after I poured it from the tumbler. Nothing unusual was afoot. It was poured into a paper plate. The ball mill aggregate had steel bearings and a few rounded rocks, same as I had used before.
The only difference was I didn't place any charcoal along with the aluminum during the milling? So why would there be spontaneous combustion?

And the aluminum was from foil, same roll I had used before.

I suspect that when you use carbon, it is effectively diluting the freshly activated aluminum so that it cannot reach autoignition temerature.

I read the charcoal coats the powder to prevent the reaction, but wasn't sure exactly why the Al combusts with air contact. I imagine it is reacting with oxygen or moisture.

It would depend on temperature, but I suspect that with carbon the reactions $$\ce{Al2O3 + 3C -> 2Al + 3CO,\ Al2O3 + 3CO -> 2Al + 3CO2,\ 4Al + 3C -> Al4C3}$$
will also have an effect ($\ce{C ->[O2] CO + CO2}$ as well). (reposted comment due to silly mistake)

It rapidly oxidized. If the ball mill is sealed for long periods of time the oxygen inside the ball mill is consumed coverting aluminum to aluminum oxide. When the lid is opened the aluminum is rapidly oxidized with a exothermic reaction. If the temperature gets hot enough it can spontaneously ignite.
The reason for the carbon is it coats the aluminum particles to prevent them from oxidizing.


text:
What decides -1/2 and +1/2 spin values of subatomic particles?
This question is in context of nmr. I know that prior to application of magnetic field and RF energy, protons are found in -1/2 and 1/2 spin values. What decides whethere a particular proton has positive or negative values. I know these values are not intrinsic to a proton since applying energy can cause them to change their spins so what decides whether proton has one value over other in first place.

Contrary to what Einstein may say, God throws a die.

lol okay... One question. Does this have to do with pauli exclusion principle as well? I always think this would be dealing with electrons only.

@Jan ALL particles of matter are subject to the Pauli exclusion principle, whereby two identical particles can never have the same exact quantum state including spin.  Electrons are just the best known case because forcing them into different quantum states directly affects chemical properties; but the principle applies to protons and neutrons (and the quarks that make them up) too.  What doesn't follow the Pauli exclusion principle?  An entirely different class of particles that carry forces instead of making up matter, like pi mesons.

@OscarLanzi Okay, I should stop posting comments about things I know too little about. (But there was something that differentiated electrons and the like from protons and the likes, wasn’t there?)

Two identical fermions, not two identical particles, surely? (Granted, electrons, protons, and neutrons are all fermions.)

Spin is an intrinsic property of any nucleus, and can be 0, 1/2,1, 3/2, 2........ A nucleus of spin I can adopt 2n+1 stable states or orientations. For 1H, I=1/2, and this means it can adopt two possible stable spin states; +1/2 or -1/2. The energy levels of these two spin states are different (and proportional to the strength of the magnetic field); one is slightly higher than the other, and so there is a population difference between the two states. However, the energy difference is incredibly small, and difference in populations (called the Bolztzmann distribution) is only about one in a million. The individual nuclei can hop back and forth between the two states quite readily, provided there is sufficient energy transfer (into or out of) the system. Indeed, this is the basis of the NMR experiment - we can manipulate spins between the two spin states with relative ease.
For a nucleus with a spin greater than 1/2, there are more than 2 stable spin states that the nucleus can exist in. Deuterium (2H) has I=1, and can adopt one of 3 stable spin states. The distribution of spins amongst these states is approximately equal (less the BD of approx one in a million). Any individual nucleus will switch between these states. Lutetium176 has a spin of 7, and so can adopt 15 different stable spin states.


text:
Why n-butanol is not used as a safer alternative to ethanol in the alcoholic beverages?
The LD50 of n-butanol adjusted for its intoxicating potency is about 1.5 times that of ethanol. This would mean to me that n-butanol is therefore 50 % safer than ethanol. Why is it not used to replace ethanol in the alcoholic drinks?

Economics. Fermentation from natural fruits is easy (and therefore cheap) and produces ethanolic beverages. Anything else is hard and expensive. And people who want to drink enough to damage themselves (therefore risk getting close to the lethal dose) care about the cost more than they care about anything else.

LD50 is a poor metric for toxicity. The harms from ethanol occur at much lower levels and there is little to suggest that alternative alcohols would be better (or have the same positive effects either).

Alcoholic beverages historically all stem from some kind of natural fermentation. The first step for all of them is to take something containing carbohydrates (grapes, wheat, …), mix it with water and yeast and wait for alcoholic fermentation to kick in. The yeast will then reduce the sugars to alcohol in a process than can be represented by the following simplified equation:
$$\ce{C_nH_{2n}O_n -> n/3 C2H5OH + n/3 CO2 + $E$}\tag{1}$$
Wherein $E$ represents the energy liberated and used to co-synthesise $\ce{ATP}$ for the yeast to use in growing.
$$\ce{ADP^3- + H2PO4- -> H2O + ATP^4-}\tag{2}$$
Thus, ethanol naturally happens to be part of alcoholic beverages — taking grapes and wheat would give wine and beer, respectively.
Some alcoholic beverages are then further enriched by distillation, reducing the water content or increasing the ethanol content. This adds nothing to the original solution, it only serves to remove some particles; ethanol being one of the prominent ones not to be removed.
To replace ethanol with, for example, n-butanol would require a complicated setup in which you attempt to extract the ethanol once your alcoholic beverage is completed and then re-add butanol. While some distilleries may be able to market these drinks as exceptionally high-class (because the process will be <insert profanity here> expensive), for most this will not even be considered an option.
Note, by the way that alcohol-free beer is typically made by extracting the alcohol from conventionally brewed beer (which is why up to $0.5~\%$ alcohol iirc are permitted to remain; everything else would drive up the process costs exorbitantly). Many people complain about alcohol-free beer not tasting like ‘proper beer’ because aroma compounds have also been extracted as ‘collateral damage’ of the process.

We should not  be concerning ourselves with natural fermentation. I thought this sort of hypothetical question containing a giveaway hint "alternative" should be clear to inquire about ways to create beverages other than by natural fermentation.

@aiag If you were asking about artificially created beverages, you should have stated so in your question. However, that would mean that your question would be very likely to get closed as too broad.

"Alternative" is  a giveaway.

@aiag We’re not playing a mistery game where every word is in some way significant. Write a clear and unambiguous question to get an answer. The question you have written as is is clear — but obviously not in the way you intended. Two answers ‘mis’-interpreting it should lead you to start thinking why the misinterpretation could have happened.

Butanol is about $60/L , given that you need about ten times less, it's just as affordable as the cheapest spirits you can buy. So I'd say it's more tradition and convenience than economics.

@WolfgangBrehm You would still have to extract the ethanol from the alcoholic beverage before adding the butanol, driving up the cost of butanol spirits to I don't even know where exactly. So I don't know how that would become an argument.

@Jan Obviously you would just add it to something, like juice or soda. You'd use it like Vodka, no one imitates beer with Vodka, you make tasty cocktails and get drunk.

@WolfgangBrehm Interesting idea that didn't cross my mind when reading the question. I nevertheless consider it a tangent because 1. the question used 'replace' and 2. the vast majority of alcoholic beverages sold are produced directly via fermentation and facultative distillation. Pre-mixed drinks represent but a minute fraction.

First: While people certainly drink alcoholic beverages because they like the taste a lot of people actually want to get intoxicated. So using an alcohol which works less is contra-productive.
Second: We don't usually prepare those beverages by mixing pure ethanol with other things. Most alcoholic beverages are produced by fermentation and mostly ethanol is produced in this process. It wouldn't be easy to use butanol instead, even if we want to.

You probably missed the part where I calculated toxicity, adjusted for intoxicating potency. Adjusted means taking it into account. Once you take into account that n-Butanol has 6x stronger inxicating potency and adjust its toxicity by that, you would realize that using 1/6th of the amount of ethanol in the same drink would be ~50% less toxic, but at the same time having exact same intoxicating potency.

And 2nd, asking this question on Chemistry.SE, I do not really care for why people drink and what they think of their drinks, or how those drinks are produced.

@aiag well, if they are easily produced with ethanol then there's your answer to the question in the title... I don't know what you want to hear. There's also no real problem with acute alcohol intoxication.

It is blatantly obvious that the existing alcoholic drinks are produced "easily". The question is, why a much less toxic alternative is not considered. There are numerous examples of how our society went great lengths to eradicate substances less lethal than ethanol.

@aiag Because there's no real need for it. I guess you looked at the acute toxiticity? There are hardly any people dying from acute ethanol poisoning (and most of them are alcoholics). So you would need to look at the long time toxiticity. And you are also talking about getting rid of something that is a very big deal in several cultures and replace it with something else. Try taking away beer from germans or wodka from russians. You think that's something you could do?

This question could be valid under alcohol tax. After all, Russians allegedly drinks body lotion to get themselves intoxicated, so it's natural to speculate as to why a safer alternative is not used. Of cause, the answer provided below is all just speculation...

Potency

Looking into the article cited by Wikipedia on potency of 1-butanol, the 6x potency is actually a molar potency. Molar mass of 1-butanol is approximately double that of ethanol, which means per weight you'll get ~3x potency instead. Still good but probably less impressive than previously thought.
Interestingly, the paper cited also reach a conclusion that the structure of alcohol doesn't matter as long as it pass the biological membranes. Thus the question could be extended to other non-toxic alcohols as well.

Toxicity

With great reluctance the LD50 of ethanol (~7000mg/kg) and 1-butanol (790–4,360 mg/kg) was compared directly. It looks like 1-butanol is more toxic than ethanol. After adjustment for potency as in (1), 1-butanol doesn't look like it offer any safety benefit compare to ethanol
It's hard to compare toxicities of different compounds, unless they have same biological target. Using LD50 could be a way to get a very rough estimate when the LD50 of two compound differ much, say Botulinum toxin vs Alcohol, but it's not really useful otherwise. After all, LD50 consider only lethal effect, not other disturbances to the body. Also because of this LD50 is usually (always?) an estimate from other animal models (hence the range given).
A quick example, table salt (NaCl) has LD50 of ~3000mg/kg. This doesn't convince people that ethanol is less toxic than salt. In fact it's much rarer to have salt intoxication due to differences in consumption pattern, sane people will not keep eating more and more salt until intoxicated.

Enjoyment factor

Despite similar molar potency, not all alcohols are made the same. Especially flavor / smell, as the nose is one of the best chemical detector available. I bet someone has drunk rubbing alcohol as a kid, which taste like, well, rubbing alcohol. Methanol should smell like ethanol but it could have some plastic like taste. 1-butanol is reported to have a banana odor. Messing with flavor of the beverages is probably not the way to make customers happy, or make fake alcohol.
Another thing with 1-butanol is it's solubility with water. Ethanol is miscible with water, meaning you can make any concentration from 0.000...1% to 100% drink with no problem. 1-butanol is not. With 1-butanol the maximum concentration is ~ 7% (weight per volume), which is again not very delighting for alcoholics.

Other

As others have mentioned, 1-butanol is not naturally present in drinks. It could be made by fermentation, but this is still an active area of research (mainly for biofuel using Clostridium species of bacteria instead of yeast, if Wikipedia tells truth). More likely, it is synthesized from petroleum and doped into drinks that are not psychoactive in hope to provide some psychoactive effect (is this even legal?). It is not used to substitute ethanol from beverages as removing ethanol from products already with itis going to be expensive and remove some flavors already present.

Well it doesn't sound that good to use 1-butanol for alcoholics. Maybe a better way to do things would be to use alcohol dehydrogenase inhibitors like Fomepizole to reduce metabolism of ethanol to acetaldehyde, which cause most of the toxic effect of ethanol, specifically a Group 1 carcinogen that contributes to hangover. This also allows you to "get more with less", have greater effect with less drinks. Wait this thing cost U$D1000 per dose? Oh well.

The production of bio-butanol remains a tricky technical challenge. The absolute highest n-butanol titres produced by fermentation are below 3% by volume (compared to the 10-15% easily achieved for ethanol), and those titres were achieved in lab conditions using hyper-specialized microbes, which tend to be outcompeted or to evolve away from butanol production in scale processes. They also produce acetone as by-product, which is rather problematic and requires separations more specialized and selective than basic distillation. In short, the overall industrial intensity (both OPEX and CAPEX) of producing butanol for drinking would be a multiple of that for producing ethanol for drinking, possibly an order of magnitude higher. This would carry over into products in the form of enormously increased price.
Pursuing it, therefore, would require extremely strong motivations to do so, and it seems unlikely that butanol-containing drinks would be competitive in the market.

To answer your question directly, the most likely reason nobody is using n-butanol is because nobody has ever tried. Few people have sufficient knowledge of chemistry to even think of your question and then when you multiply that by the percentage of those people with manufacturing skills and an entrepreneurial personality, the chances are very very small of it happening.
Manufacturing a new product requires making thousands of decisions and solving hundreds of problems. How will you deliver the n-butanol? In a beverage? Which one? How will it taste? How will you manufacture it? How pure do you need it? Will it be regulated by the government differently than ethanol? Is it a drug subject to FDA (probably not)? ....and there are many many more issues to be resolved.
If you want to invent a new alcoholic beverage, just start experimenting. Put some n-butanol in beer or wine and see what it tastes like. Or if you are concerned with safety, create your mix and then kill some mice with it to see if it has the same LD50 as straight n-butanol.

Isn't alcoholic beverage industry a huge multi-trillion conglomerate with virtually unlimited powers? Think Bacardi and Philip Morris who owns a huge share of alcohol industry. They run sophisticated labs with multi-billion dollar budgets. And then think government agencies like BATFE etc with their own multi-billion dollar budgets. There should be at least some consideration given between those of what sorts of alcohol are more or less toxic per drink.


text:
Electrolysis of water in 0.1 M hydrochloric acid
For water electrolysis in acidic media ($\ce{TiO2}$ anode and $\ce{Pt}$ cathode) the half reactions for water splitting and $\ce{O2/H2}$ evolution are as follows:

Anode: $\ce{H2O + 4 H+ -> 4 H+ + O2}~~(E = \pu{-1.23 V}\text{ vs NHE})$
Cathode: $\ce{4H+ + 4 e- -> 2 H2} \quad\quad~~(E = \pu{0.00 V}\text{ vs NHE)}$

If I use $\pu{0.1 M}$ $\ce{HCl}$ as my acidic electrolyte, what reaction will occur at the anode (the standard oxidation potential for $\ce{2Cl-}$ to $\ce{Cl2}$ is $\pu{-1.36 V}$ vs NHE). In $\pu{0.1 M}$ $\ce{HCl}$ electrolyte, there should be $\ce{H_2}$ evolution at the cathode. Will the anode maintain $\ce{O2}$ or $\ce{Cl2}$ evolution (or maybe both?)

Possible reactions :
At Cathode:
$\ce{2H2O{(l)} + 2e- -> H2{(g)} + 2OH-{(aq)}}$
$\ce{2H+{(aq)} + 2e- -> H2{(g)}}$
At anode:
$\ce{2H2O_{(l)} -> O2_{(g)} + 4H+_{(aq)} + 4e-}$
$\ce{2Cl-{(aq)} -> Cl2{(g)} + 2e-}$
General rules of electrolysis of aqueous solution using inert electrodes :-

Reduction potential of $\ce{H2O}$ > reduction potential of cations of 1, 2 and 13 groups of periodic table. So reduction of $\ce{H2O}$ takes place at cathode.
Reduction potential of $\ce{H2O}$ < reduction potential of other cations (than cations of 1, 2 and 13 group). So reduction of $\ce{H2O}$ does not occur at cathode.
Oxidation potential of $\ce{H2O}$ > oxidation potential of fluoride, sulphite, sulphate, nitrate, etc. So oxidation of $\ce{H2O}$ occurs at anode.
Oxidation potential of $\ce{H2O}$ < oxidation potential of chloride, bromide and iodide, etc. So oxidation of these ions takes place.

Keeping these rules in mind, at cathode reduction of hydrogen ions will take place and hence $\ce{H2}$ gas will be evolved. At anode, oxidation of chloride ions will take place and chlorine gas will be evolved and $\ce{O2}$ won't be evolved.
Although the oxidation potential of $\ce{H2O}$ > oxidation potential of chloride ion, chloride ion gets oxidized because of over potential of $\ce{O2}$ i.e. extra potential is required to initiate a reaction at a required rate. And formation of $\ce{O2}$ from $\ce{H2O}$ is kinetically very slow.
Therefore oxidation of chloride ion takes place.

Thanks for your comment. The oxidation potential of H2O is -1.23 V. The oxidation potential of Cl- is -1.36 V. The oxidation potential of SO42- is -2.01 V. Surely then H2O oxidation potential of H2O is less than Cl- and SO42- and so H2O oxidation should be favoured?

Modified my answer.


text:
Why is cupric oxide more basic than cuprous oxide?
I have learnt earlier that metal oxides with higher metal oxidation states are more acidic than those with less oxidation states, but copper disobeys this rule. Why?

There are many factors that affect basicity of a substance-electronegativity, size, resonance, etc. Its the cumulative effect of these factors that decide the relative order. What you learnt about the higher metal oxidation states being more acidic is more of a "shortcut" than a fact. Don't rely on them too much.

Just stumbled upon this while searching...It'd be great if anyone wrote an answer to this question. Perhaps, an explanation of why the "higher oxidation state is more acidic" rule even worked in the first place would also help.

How do you find Cu(II) hydroxide more basic to start with?  Reference?

I'm sorry. This is a question I asked a very long time back that I'm not able to recall the source


text:
What is spin spin coupling?
I am studying nmr and the concept of multiplicity is very hard to undersand. Specifically I wanted to know where this arises form. Can anyone explain the concept of spin spin coupling in detailed and in simple terms?

This question would benefit from an answer you understand IF, and only IF, you elaborately explain why you didn't understand the tutorials you have available or ones on the Net. What if an answerer repeats the same words that made you uneasy and baffled?

The reason why I am asking this seemingly difficult question is because I am not sure I understand the concept of spin properly (And i think the answer to this question is related to spin of adjacent hydrogens). Most important no one seems to answer the question as to why there are two spin value when a external magnetic field is applied. What I mean by this is why do some proton align with and some against the magnetic field applied? Being opposed to external field takes energy, so why are some protons have spin against the field?

Understanding this first then would make it easier for me to understand how does coupling occur and all.

Not all protons align with the field, because of thermal energy and the Boltzmann distribution. Simply speaking, the gap between the two energies is very small, and therefore as long as there is some heat in the system, that heat is enough to excite some protons from the lower energy state to the upper energy state. (Therefore, at absolute zero where there is no heat energy, every single proton will be aligned with the field.) However, even at non-zero temperatures, there will be an excess of aligned spins and a deficit of opposed spins.

MAR is right; these things are covered in many existing textbooks and websites, perhaps better than anybody here can explain. I would really recommend consulting one of them, or finding somebody in real life to go through this, instead of asking a large number of questions here. The issue is that we don't understand the context or the background behind these questions. You have to explain what exactly the problem is - and that's much easier in real life than over the Internet. Apart from that, Stack Exchange is not well-suited to lots of back-and-forth discussion and clarification.

As it stands, your question is very general. It is unlikely that a single answer here will describe spin coupling and multiplicity in a simple and detailed manner any better than any publicly available reference source. There are hundreds available. I suggest you read one or more of the following extremely useful introductory NMR references, and ask more specific questions that you do not understand from them.

Reich (http://www.chem.wisc.edu/areas/reich/chem605/)
Hornak (http://www.cis.rit.edu/htbooks/nmr/)
Libretexts (https://chem.libretexts.org/Core/Organic_Chemistry/Spectroscopy/Nuclear_Magnetic_Resonance_Spectroscopy
and
https://chem.libretexts.org/LibreTexts/Purdue/Purdue%3A_Chem_26200%3A_Organic_Chemistry_II_(Wenthold)/Chapter_12._Nuclear_Magnetic_Resonance)


text:
Low-spin / high-spin arguments: why overlooking pairing energy?
There are many arguments in inorganic chemistry based on a transition from a high-spin to a low-spin configuration, for example:

$K_{f,3} > K_{f,2}$ for $\ce{Fe(bipy)3}$ complex, because of "a considerable increase in LFSE" as the complex becomes low-spin (from Shriver and Atkins)

However, as the complex becomes low-spin there is also an increase in the pairing energy (or a decrease in the number of Fermi holes), so I don't see why the assumption above that low-spin complexes are always more stable than high-spin is necessarily true (it can just reflect a subtle balance between $\Delta$ and P with not indication of the overall change in energy).
Based on my knowledge, successive bipy ligands would only further enhance the mixing of $\mathrm{t_{2g}}$ SALCs and that is where increasing stabilization is coming from. However, I would predict that this factor would get weaker with each ligand substitution, as the energy match of the occupied $\mathrm{t_{2g}}$ and the π* of an incoming bipy becomes worse.
Perhaps, there is some non-linear dependence of the way the pairing energy is reduced as the orbitals get "smeared" over larger space (i.e. over both the metal and π-acceptors) on the number of $\ce{bipy}$ ligands?

Is $\ce{bipy}$ $\ce{bpy}$, i.e. 2,2'-bispyridyl?

@Jan yes, sorry for the confusion!

Regarding your question about the non-linear change with ligands, I suspect this isn't that easy of a question to answer, but the data is probably out there somewhere.
Regarding the quote and question in the title, when these kinds of arguments are made, they aren't ignoring the pairing energy, but they also aren't really making any predictions.
Essentially, Ligand Field Theory (LFT) lays out a simple way that one can rationalize the geometry of a particular transition metal complex based on the energy of the d orbitals. I can see that you know this. What this means though is that when they observe whatever change happens for the iron complex, LFT says that this must correspond to some change in the energy of the electrons associated with shifting changes in the energy of occupying orbitals in a specific way.
That is, in LFT the only two terms that describe the energy of a particular electronic arrangement are the ones describing the potential energy of the electrons while ignoring their interactions, and the increase in energy from pairing electrons in orbitals.
So, when any geometrical or similar observation is made, there's really no other explanation than to say the a high-spin complex became low-spin because the orbitals split more ("considerbale change in LFSE") which guarantees this decrease in potential is larger than the cost of pairing electrons.
I agree that it's quite naive to always explain things in this way, but that's why crystal field theory and ligand field theory are not meant as real computational or predictive tools these days. They are useful for connecting geometry, spin, and energy as well as giving a first-order explanation of observations.
It's like when you'll sometimes hear inorganic chemistry students describe the behavior of a catalyst as being due to a "combination of electronic and steric effects." That's true... but those are also the only possibilities, so it doesn't tell you much.

Thank you for your answer! But in this case, I'm surprised that such arguments work at all, and do consistently!

Well they aren't really arguments so much as posterior explanations. That's what I'm trying to say in the answer. If you really wanted to be predictive the model you use to has to be much more detailed than CFT or LFT because you don't know the splitting or pairing energies without experimental measurement. And by the time you've made an experimental measurement, you already know the answer. That's why there are other methods for explaining these things from first principles.


text:
Hydrogen covalent bond
It's stated: when two hydrogen atoms form a covalent bond, the distance between them corresponds to the lowest energy. The letters A, B and C show the correspondence between shifts in electron density and the distance apart of the atoms
I am having a difficulty understanding why the figure looks the way it does and what is the energy plotted on the y-axis, can anyone explain?

The energy on the y-axis is the total energy of the system. It uses a reference of E = 0 for when 2 H atoms are infinitely far apart.
As to why the plot looks the way it does...
When the H atoms are infinitely far apart, they are not interacting at all. When brought together, you get a covalent bond which forms. This can be treated at various levels - In the simple classical picture, the 2 electrons are "shared" between the two atoms, leading to lots of electron density in between the atoms, which the 2 protons are then attracted to, and this is what holds the 2 H atoms together. Now we have a slight problem - why do they stop? You could say due to the nuclear repulsions, but if you have a simple situation, using  $\ce {H2+}$:
$\ce{H+}$---e---$\ce{H+}$
With the electron just sat halfway between 2 nuclei. Now let's analyse this situation with just classical mechanics, all units set = 1 for convenience. q = Charge on electron and nucleus. d = Distance from nuclei to nuclei:
Force on each $\ce{H+}: $
$\ F = q^2/d^2 - q^2/(d/2)^2 =  -3q^2/d^2  $
We have $\ce{q^2/d^2}$ term for the repulsion between the 2 nuclei and the negative $\ce{-q^2/(d/2)^2}$ term is due to attraction between nucleus and electron. In this situation the covalent bond results in an attraction of both nuclei together for all values of d - IE eventually we get a singularity. This is why the covalent bond can only really be understood in a quantum mechanical framework.
Taking again the situation of $\ce{H2}$ but this time in a quantum mechanical way.
We start off with 2 hydrogen atoms infinitely far apart. They each have an electron in H 1s, with OPPOSITE spins. When they are brought closer together the wavefunctions change. Now what happens is this. The electron in H 1s of one atom is able to tunnel into the potential well of the other hydrogen atom. By doing this the total volume in which the electron resides is larger - Obviously the electron can go anywhere in the universe, as per usual QM, but the H atom is localised less. This leads to a decrease in the electron's kinetic energy - It is spread out more, and so uncertainty in position is larger, so uncertainty in momentum can be less, and so leads to a lower KE. The potential energy which the electron experiences won't change a huge amount - Yes there are 2 potential wells but it doesn't get closer to them. This is the story at first glance.
However there is the virial theorem which states:
$2KE+PE+R(dE/dR)=0$
So the decrease in KE by bringing 2 H atoms closer together must be balanced out somehow - Either an increase in KE by another means, or an increase in potential energy (PE). If we shrink the 1s orbitals contributing to the bond, this in turns increases KE and decreases PE - PE is negative, and so becomes more so. Thus we find the orbitals shrink in bonding. This all leads to a picture like this:
As 2H atoms draw closer together, electron on one can tunnel onto the other. This leads to a decrease in KE, and as at the equilibrium bond distance $2KE+PE = 0$ this decrease in KE must be balanced out by a contraction in the H 1s orbitals contributing to the bond, leading to both a larger KE term and a more negative PE term. In other words a covalent bond forms because it reduces the kinetic energy of the contributing electrons and increases the potential they feel, due to the contraction of orbitals via the Virial Theorem pathway.
Now back to your plot. At far distances, E=0, because no interaction, and this is taken as 0 point - IE referenced relative to infinite separation. As 2 H atoms brought together, get interaction described as above. At around 2 bond lengths, the KE term reaches a minimum. It then rises - the 2H atoms are so close now that more overlap - IE more tunneling of electron from well to the other well - has no more effect. Instead the wavefunction of the electron is found to contract. This raises the KE term - electron now confined to a smaller space = Larger KE term - but the PE term becomes more negative by a larger amount. At the equilibrium bond length, the virial equation applies as written above:
$2KE+PE=0$
now applies. Thus the KE term is positive with value x, and the PE term is negative, with value 2x. Thus if
$E=PE+KE$
we have a negative binding energy, as expected.
Interestingly the involvement of KE in bonding leads to a new class of bonds, called charge shift bonds - the prime example of which is difluorine, $\ce{F2}$. This molecule has an overall bond order of 1 according to MO theory, but has a large antibonding effect due to filling of antibonding pi orbitals. Calculations using VB theory indicate that the covalent contribution to the structure never actually has a minimum - IE antibonding at all distances. In this case adding in ionic structures is entirely responsible for the bond observed.
This stackexchange answer explains it well, if you're interested, and there's a number of interesting papers listed there as well.
What is charge shift bonding?

It is the bonding energy of the two atoms. Remember that in the quantum mechanic model the electrons can be found infinitely far away from the nucleus.


text:
Can iodine hydrocarbons be used as fire extinguishing agent? What are their properties?
In the German Wikipedia the following claim about fire extinguishing agents is made:

Iodkohlenwasserstoffe haben zwar eine hervorragende Löschwirkung,
  erlangten wegen ihres exorbitant hohen Preises (in den 1950ern etwa
  Faktor 10 gegenüber Bromkohlenwasserstoffen) jedoch niemals Bedeutung.

Translated:

Though iodine hydrocarbons have excellent fire extinguishing properties
  they never became important because of their exorbitant prices (during
  the 1950s they were approximately 10 times more expensive than bromide
  hydrocarbons).

Now I am curious because I never heard of iodine hydrocarbons. What are their properties and what advantages would they have over the traditional halogen (fluoride, chloride, bromide) components if price is irrelevant?

Funny, Professor Knochel stated in a lecture that if $\ce{Ar-Cl}$ costs $1 per gram, $\ce{Ar-Br}$ will cost $10 per gram and $\ce{Ar-I}$ $1000 per gram resulting in a factor of $100$.


text:
What determines whether a step in a reaction is reversible?
What makes a step in a reaction not reversible? I'm learning all these mechanisms and having to memorize which steps are reversible. Is there a way to figure out whether it's reversible or not?

This strikes me as a good question. It is a simple concept, but pragmatically I don't think there is, in general, any really good way to just look at the reaction equation and tell.

Agreed with @MaxW. There is no one-size-fits-all qualitative approach as far as I'm aware. Theoretically, every reaction will reach equilibrium, and is therefore reversible at some point. Whether this equilibrium has to be taken into account is what we mean by adding two-way arrows $$\ce{<=>}.$$The first estimates are whether there is a gas among products (and system is thermodynamically open), then it's "not" reversible. Same when a non-soluble compound or stable complex forms. More generally, think in terms of stability and rel. likelihoods (hence speed). Hunches gets us only so far though

see also here

I think the gist is that every reaction will have a forward rate and a backwards rate. If $\ce{K_f  >> K_b}$ then we say the reaction is irreversible.

There's no way of figuring that out "from scratch", it is a lot of experience and knowledge. There is no general answer to this but there are some steps which are basically always seen as reversible or irreversible*.
protonation/deprotonation:
unless an extremly strong acid/base (like BuLi) is involved those are always in some kind of equilibrium and are reversible.
exchanging a good leaving group with a bad one:
That can be seen as irreversible.
gas evolution:
as @LinearChristmas pointed out: that can be seen as irreversible (if you are working in solution)
in general: producing a stable structure:
for example aromatization, that's irreversible

*theoretically every reaction is reversible, but if the equilibrium is far on one side we call it irreversible.
For an explanation what makes a reaction reversible see: Is every chemical reaction in equilibrium?


text:
What is the exact reason that compounds such as carbon dioxide cannot interact with other compounds through dipole-dipole interactions?
Whilst I understand that $\ce{CO2}$ has a linear structure, resulting in a vector sum of the dipoles of 0, I do not understand why the dipoles formed at the two oxygen atoms as a result of their increased electronegativity cannot interact with other molecules via dipole-dipole forces. Do the vectors equaling 0 somehow negate the present dipoles?

Do realize that there are other attractive forces besides dipole-dipole interactions. So two carbon dioxide molecules could be attracted to each other by a London dispersion force.

The attractive induced-dipole, induced-dipole interaction exists even when the molecule has no overall dipole. It is proportional to the product of average polarisabilities and mean of  ionisation potentials and inversely as the sixth power of separation of centres of mass. ($\ce{CO2}$ has an asymmetric vibration and two bending vibrations that produce transient dipoles.)

Of course it depends whether you consider the chemical reactivity of carbon dioxide as something that can be explained by dipole dipole interactions when other molecules are very close. Most people think of the chemical reactivity as something different even though the reactions are explained by the movement of electrons.

As the other answers have indicated, CO2 has no NET dipole moment. However it does have two dipoles pointed in opposite directions (as OP keeps mentioning). This means that CO2 can possibly interact through higher moments, such as the quadrupole moment. The mathematical procedure behind this is known as the multipole expansion. It is important to note however, the higher order multipoles get weaker and weaker by a factor of 1/R approximately, so quadrupole interactions are usually ignored for example.

Carbon dioxide is more than linear. It's symmetric, and the axis of symmetry perpendicular to the bonds also applies to whatever dipole moment it has. The only vector that looks the same after being rotated 180 degrees is the null vector, so the molecule has zero overall dipole moment.
Therefore, dipole-dipole interactions are not possible because carbon dioxide does not have a dipole.

But assymetric stretch of $\ce{CO2}$ creates a non-zero dipole moment. So why we say that it has zero dipole moment?

Because in the ground state, the time average is zero.

You mean the time average of the excited state right?

No, I mean the ground state...

This would be the same also on excited states right? I mean for every $υ$ the time average must be zero.

If you reasonably assume symmetry, then yes.

Exactly that.
Dipole-dipole interactions are caused because the positive end of one dipole is attracted to the negative end of another dipole. This macroscopic attraction can only occur if the centre of positive charge and the centre of negative charge do not coincide.
In molecules like carbon dioxide that contain a centre of symmetry, these charge centres do coincide. They fall on top of each other. So macroscopically, a neighbouring molecule cannot see any positively or negatively charged centre and therefore sees no possibility to attach via dipole interactions.

so basically the charges are distributed equally within the molecule? But how can that be ? Aren't the 2 o atoms suppose to deprive the c atom os electrons and make it positively charged?

Well yes, each of the oxygens is negatively charged and carbon is positively charged. But these microscopic charge centres both fall together in the molecule’s centre of gravity when viewed from a greater distance. That’s the power of symmetry.

so even though the oxygen is negatively charged, it cannot interact with the positively charged C atom of another CO2 molecule

Thank you Jan. So due to the symmetry the overall dipole is 0? Wouldn't the electrons still be more attracted to the oxygen, resulting in dipoles? And wouldn't these dipoles be able to interact with other dipoles? I.e a positive carbon atom

Consider the CO2 molecule as a superposition of two dipoles, one for each C=O bond. A nearby polar molecule will feel a dipole-dipole interaction with each C=O bond, but since those two dipoles are equal and opposite, those two interactions will almost cancel out. They may not completely cancel since the interaction with the closer of the two bonds will be stronger.
The resulting net interaction will therefore be much weaker than the interaction with a polar molecule such as water.
Note that a superposition of two equal and opposite dipoles in different locations forms a quadrupole. So, instead of electing to treat the interaction of a polar molecule with CO2 as the interaction with two equal and opposite dipoles, we can treat it as the interaction with a zero dipole but nonzero quadrupole. As user157879 points out, quadrupole interactions are much weaker than dipole interactions.


text:
Cell walls in lath martensite - Meaning
What does cell walls mean in the context of lath martensite? Are they grain boundaries?
Quote (page 66 Structure and properties of engineering alloys - Smith): In the lath martensite of low-carbon steels, there is a high density of individual dislocations and many cell walls.
Sorry for the stupid question, I don't find any image to understand it.
Thank you for your time.

The general idea is the same. Grains in a metal typically refer to irregular blobs revealed by etching or some other method. I'd think of a cell wall as being more like the boundary of a crystal structure.  Lath martensite would be criss-crossed needle like crystals of  martensite. https://www.researchgate.net/profile/Emre_GULER/publication/42389007/figure/fig2/AS:276959977328642@1443043812275/Fig-2-SEM-micrograph-of-specimen-B-Lath-martensite-crystals-in-the-prior-austenite.png

According to this Wikipedia page:

Martensite, named after the German metallurgist Adolf Martens
  (1850–1914), most commonly refers to a very hard form of steel
  crystalline structure, but it can also refer to any crystal structure
  that is formed by diffusionless transformation.1 It includes a class
  of hard minerals occurring as lath- [emphasis mine] or
  plate-shaped crystal grains.

Lath basically means the shape of a woven screen when viewed close up, or like a grid of alternating appearance.

Hi @airhuff, I'd like to know the meaning of cell walls.

Though you've done a pretty good job of explaining, I wish I could see more context from your book.  I'm pretty sure that you are correct in describing it as grain boundaries in this context.  Rather than being flat and smooth, there is more of this lath-like structure due to the individual dislocations which, again, I think are essentially grain boundaries.  I apologize that I missed the crux of your question; I'm not sure if this comment answers exactly what you are looking for either.

Take a look down in my Wikipedia link for the photo of martensite steel.

This might also shed some light.


text:
Differentiation of H = U + pV
I was asked to use direct differentiation of "H = U + pV" to find a relationship between:
$$ \left(\frac{\partial H}{\partial U}\right)_{\!P}$$   and   $$\left(\frac{\partial U}{\partial V}\right)_{\!P} $$
So my first step was to take a differential of the equation twice, once in terms of U and once in terms of V:
$$ \left(\frac{\partial}{\partial U}H\right)_{\!P}=\left(\frac{\partial}{\partial U}(U +PV)\right)_{\!P}\rightarrow\left(\frac{\partial H}{\partial U}\right)_{\!P}= 1 + P\left(\frac{\partial V}{\partial U}\right)_{\!P}$$
and
$$ \left(\frac{\partial}{\partial V}H\right)_{\!P}=\left(\frac{\partial}{\partial V}(U +PV)\right)_{\!P}\rightarrow\left(\frac{\partial H}{\partial V}\right)_{\!P}= \left(\frac{\partial U}{\partial V}\right)_{\!P} + P$$
When I tried to substitute one equation into the other, they would simply end up all canceling out. Is there another way?

A way to do it would be:
\begin{align}
\left( \frac{\partial H}{\partial U} \right)_P \left( \frac{\partial U}{\partial V} \right)_P &= \left( \frac{\partial H}{\partial V} \right)_P = \left( \frac{\partial U}{\partial V} \right)_P + P\\
\left( \frac{\partial H}{\partial U} \right)_P &= \frac{\left( \frac{\partial U}{\partial V} \right)_P + P}{\left(\frac{\partial U}{\partial V} \right)_P}\\
\left(\frac{\partial U}{\partial V} \right)_P &= \frac{P}{\left( \frac{\partial H}{\partial U} \right)_P - 1}\\
\end{align}

Not seeing the algebra from step 2 to step 3

corrected sign error

Where does the -1 come from?

Multiply both sides by step 2 denominator. Subtract del U / del V from both sides. Factor del U / del V from left-hand side. Divide by (del H / del U - 1).

Got it. So $$ \left(\frac{\partial H}{\partial U}\right){!P} = \frac{P}{\left(\frac{\partial U}{\partial V}\right){!P}} +1$$

@Nova That is correct

Not really a full answer per se, but since you are curious about why the partial derivatives can be manipulated in the way they are:
Consider $z = z(x,y)$. We have
$$\require{begingroup} \begingroup \newcommand{\md}[0]{\mathrm{d}} \newcommand{\pdiff}[3]{\left( \frac{\partial #1}{\partial #2} \right)_{\!#3}} \md z = \pdiff{z}{x}{y}\md x + \pdiff{z}{y}{x}\md y \tag{1}$$
but also $x = x(y,z)$, so
$$\md x = \pdiff{x}{z}{y}\md z + \pdiff{x}{y}{z}\md y \tag{2}$$
Substitute $(1)$ into $(2)$:
$$\md x = \pdiff{x}{z}{y}\pdiff{z}{x}{y}\md x + \pdiff{x}{z}{y}\pdiff{z}{y}{x}\md y + \pdiff{x}{y}{z}\md y \tag{3}$$
Comparing coefficients of $\md x$ gets you to
$$1 = \pdiff{x}{z}{y}\pdiff{z}{x}{y} \tag{4}$$
This is what you need to relate the partial derivatives $(\partial V/\partial U)_P$ and $(\partial U/\partial V)_P$; as Chester Miller already noted they are reciprocals of one another.
[Incidentally: not relevant, but if you compare coefficients of $\md y$, you will get:]
$$\begin{align}
\pdiff{x}{z}{y}\pdiff{z}{y}{x} + \pdiff{x}{y}{z} &= 0 \tag{5} \\
\pdiff{x}{z}{y}\pdiff{z}{y}{x} &= -\pdiff{x}{y}{z} \tag{6} \\
&= -\pdiff{y}{x}{z}^{-1} \tag{7} \\
\pdiff{x}{z}{y}\pdiff{z}{y}{x}\pdiff{y}{x}{z} &= -1 \tag{8}
\end{align}$$
where in going from $(6)$ to $(7)$ we have used equation $(4)$. And yes, the negative sign is supposed to be there! The partial derivatives don't "cancel out" like fractions, so don't think of them as fractions. $\endgroup$

Ok, this is really cool. I guess the one thing I don't understand is steps 5/6. I took multivariable calculus years ago so I guess it must be some rule I forgot...

Also that same rule probably explains step 4 as well. Cause earlier I was treating them as "fractions" because I remember being told it was ok to manipulate them like that even though it wasn't "technically" right.

From (3) to (5) is comparing coefficients of dy - on the LHS of eq (3), there's no dy term so the coefficient is zero. From (5) to (6) is just bringing $(\partial x/\partial y)_z$ over to the RHS. It's OK to treat them like fractions up to a certain point. Sometimes it works but sometimes it doesn't as you can see in (8).

No, I wasn't asking about the algebra, but about the calculus reasoning.

I don't know - there's no real calculus reasoning as far as I can tell, apart from the thing about comparing coefficients

Then again, I'm hardly the most knowledgeable person on this topic - I did very brief multivariable calculus a couple of years ago. I wouldn't say I have a lot of insight into the equations.

ortho, I have struggled on and off for years to understand where Eq. (8) comes from, and it makes perfect sense now. Amazing. Thank you.

The partial of V with respect to U at constant P is the reciprocal of the partial of U with respect to V at constant P.

Also, is that always true? For example, is delx/dely always equivalent to (dely/delx)^-1? Can you give an example? Cause I would think taking a derivative of an equation with respect to one variable would yield a different answer than taking the derivative with respect to another...

The thermodynamic state of a closed system is determined by specifying any two intensive variables.  So the molar U can be expressed as a function of pressure and molar V.  At constant pressure, U is a unique function of V.  This allows you to do what I've done with the reciprocal.


text:
Why are Salts Removed before Ion-Exchange Chromatography?
In my organic chemistry laboratory, we used centrifugation and salting out to begin the process of purification of Myoglobin from hamurger meat. After that process, we used to dialysis to remove the Ammonium Sulfate from the salting out process to begin ion-exchange chromatography. I was wondering why we had to use dialysis to remove the salts, and why the ion-exchange chromatography couldn't just occur in the presence of the salts?

Think about what the ion-exchange material does, and the mechanism of separation by this method. The ions of interest interact with the ion exchange sites to a different degree than do the other ions or other species in the solution, thus you get separation.  What if your solution is loaded down with ammonium sulfate?  The ion exchange resins have a limited capacity to accommodate ions and all of the sites would be swamped in ammonium ions and all of the much lower concentration ions of interest thus would not be able to be separated.


text:
Is the product of Lindlar hydrogenation of (2Z,4S)-4-methylhept-2-en-5-yne optically active?
According to me the poisoned catalyst will only reduce the triple bond so the compound produced must be optically active.. But the answer given is that the final product is optically inactive.. is it correct and if so how?

this is the final product I made:

The triple bond gets reduced to double bond as Lindlar's catalyst has been used so the reduction stops at this stage.. is this correct?

What do you think you get when you reduce the triple bond? Have you drawn the structure?

I have posted the structure I made.. Please check it @matt

And is this optically active? Does it have a carbon with 4 distinct units attached?

sorry.. I overlooked the 2 group[s are same.. Thanks.. i understood..

The two cis-configured pro-1-en-1-yl groups present on the central carbon remove the asymmetry. You can put a plane of symmetry through the $\ce{Me-C-H}$ plane of the central carbon to transform the molecule onto itself. You can confirm that by rotating the $\ce{(C=)C-C(HMeR)}$ bond.
Thus, this hydrogenation removes chirality.


text:
Reaction of benzenediazonium chloride with potassium ferricyanide

I have gone through all my notes but could not find any reaction of benzenediazonium chloride with potassium ferricyanide. What are possible products?

Google is your friend, read about the Sandmeyer reaction.

@ron .. I could not get the exact reaction on google.. but now that you mentioned Sandmeyers reaction.. Is the final product CyanoBenzene?

yes, cyanobenzene or benzonitrile