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Help understanding the following conversion

I am given the substrate and reagents in steps (1,2,3). I am asked to figure out the product! I am not sure how the reagents in 2 (aq. Na2CO3) and 3 (NaH, MeI/DMF) work to give that answer. Why is the answer what it is?

You can work backwards here. Step 3 converts cyclohexanol to the product by substitution reaction of methyl iodide and alkoxide. Sodium hydride is the base that deprotonates the alcohol.
Step 2 is a bit dubious. They want to substitute a secondary chloride with hydroxide. In principle this will work, though I am suspicious about whether or not it will actually take place. For cyclohexyl chloride (product from step 1), the substitution is probably only going to happen from the axial conformer; the equatorial conformer has the flag pole hydrogens blocking the approach to the $\sigma^{*}$ orbital. But from the axial conformer, I suspect that elimination is also a competitive pathway. Hard for me to say exactly what the relative rates are for sure.
Step 1 you've likely figured out. It's radical chlorination of an alkane.


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What is the surfactant's role in the non-coalescence of water droplets?
I have recently seen many videos that talk about the noncoalescence of water droplets; however, I do not understand the purpose of adding surfactants into the water. From my knowledge (high school), objects will stay on top of the water due to water's cohesion, resulting in a high surface tension. This is why in an experiment of a paperclip floating on water, when you add soap, the paperclip sinks - due to the decrease in surface tension. What I am really confused is that the noncoalescence of water droplets works BETTER when surfactants are added. It would be great if anyone has an answer, I am really puzzled...

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Why does burnt hair smell bad?
When I use hot stuff like hair straightener on my hair, my hair begins to smell bad, which is very different from smell produced from burning other things. So what's the gas produced that is responsible for this smell?

Two words: Keratin. Sulfur.

I'd say that the smell is "bad" precisely because it is the smell of burning hair. There probably has been some really messy natural selection been going on in the past, ingraining the badness of this smell into our collective subconsciousness. ("Selection" is not to be taken too literally here... ;) )

^ I think @AlexanderKosubek is trying to say:  It smells bad so that you will notice when your hair is burning ...

The hair straighteners may also be gassing out and getting converted to smelly compounds by the heat.

Hair is largely (~90%) composed of a protein called keratin, which originates in the hair follicle.
Now, keratin is composed of a variety of amino acids, including the sulfur containing amino acid, cysteine. All these amino acids are joined to each other by chemical bonds called peptide bonds to form these long chains that we call polypeptide chains. In the case of human hair, the polypeptide that we're talking about is keratin. The polypeptide chains are intertwined around each other in a helix shape.
The average composition of normal hair is 45.2 % carbon, 27.9% oxygen, 6.6% hydrogen, 15.1% nitrogen and 5.2% sulfur.

(I got that diagram off of Google Images)
Now, there are a whole bunch of chemical interactions that maintain the secondary and tertiary structures of proteins, such as van der Waals forces, hydrophobic interactions, polypeptide linkages, ionic bonds, etc. But there is, however, one additional chemical interaction in proteins that contain the amino acids cysteine and methionine (both of which contain sulfur) called disulfide linkages. You can see that in the diagram above (it's been marked in yellow, which is fortunately, a very intuitive color when you're dealing with sulfur).

When you burn hair (or skin or nails... anything that has keratin in it for that matter) these disulfide linkages are broken. The sulfur atoms are now free to chemically combine with other elements present in the protein and air, such as oxygen and hydrogen. The volatile sulfur compounds formed as a result is what's responsible for the fetid odor of burning hair.
Quite a few of the "bad smells" we come across everyday are due to some sulfur containing compound or the other. A great example would be the smell of rotten eggs, which can be attributed to a volatile sulfur compound called hydrogen sulfide. Yet another example (as @VonBeche points out in the comments) would be that of tert-butylthiol, which is the odorant that is used to impart the characteristic smell of Liquefied Petroleum Gas (LPG).

Quite a few of the "bad smells" we come across everyday is due to some sulfur containing compound or the other. One word: Flatulence.

Another nice example would be https://en.wikipedia.org/wiki/Tert-Butylthiol, or "gas smell".

Yes, breaking of disulphide bridges is largely responsible for the bad smell. However, I think there is also some additional bad smell coming from ammonia, which is also a part of all amino acids.

I don’t think ammonia is formed when burning hair — it should oxidise on to $\ce{NO_x}$.

@Jan The oxidation to $\ce{NO}_x$ requires high temperature ($700$–$1000\ ^\circ \mathrm{C}$), higher than atmospheric pressure (up to $10\ \mathrm{atm}$), and transition metal catalysts (e.g., $\ce{Pt}$) to occur in a significant way. The more likely end product is $\ce{N2}$; burning in air instead of pure oxygen will also leave a reasonable amount of ammonia unburnt. (H. Karik, K. Truus. Elementide keemia. 2003)

It could be 2 things;
1: your hair straighteners could be coated in formaldehyde - something that wierdly is allowed although many studies have shown this to cause cancer, and it smells awful warm.
2: for me me personally, when i have accidentally caught my hair in a hairdryer - the smell most prominent is ammonia, with a hint of sulphur, so when i saw your question i looked at it from another perspective - that some industries specifically burn hair on purpose - abbotoirs and human waste disposal companies, and i looked at their hazardous waste regulations and any research that had occured in that field, rather than trying to extrapolate it from the building blocks, because I really am not that good a chemist to be sure, and found that Ammonia is indeed a key gas released when burining hair, here is an excerrpt;
"in waste managment they have protocols for dealing with the gasesses emitted from the burning of hair, that produces foul odor and toxic gases such as ammonia, carbonyl sulphides, hydrogen sulphides, sulphur dioxide, phenols, nitriles, pyrroles, and pyridines."
taken from;
M. Brebu and I. Spiridon, “Thermal degradation of keratin waste,” Journal of Analytical and Applied Pyrolysis, vol. 91, no. 2, pp. 288–295, 2011.
So yes to sulphur compounds, yes to ammonia, and also phenols which have an unpleasant smell as well.
I hope this helps.

Lots of interesting points here. I agree that ammonia would probably be produced in much more minute quantities than nitrogen oxides, but I wouldn't rule out the formation of amines, which are associated with smells like rotting fish.
Also:

Degradation started with formation of ammonia and CO2 (from 167 and 197 °C respectively and with maximum evolution at 273 and 287 °C respectively), continues with formation of sulphur-containing inorganic compounds (SCS, SCO, H2S and SO2 at 240, 248, 255 and 253–260 °C respectively) and of water (255 °C). Thiols are formed in two stages (257 and 320 °C) while the evolution of nitriles is maximum around 340 °C and continues up to about 480 °C. Phenol and 4-methylphenol are the most important degradation compounds, formed at 370 and 400 °C respectively. Nitrogen was present mainly in aliphatic/aromatic nitriles, pyrroles, pyridines and amides while sulphur was found mainly as sulphides, thiols, thiazoles and thiophenes. — http://www.sciencedirect.com/science/article/pii/S0165237011000490?via%3Dihub

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Nitroxyl anion electron structure
Is the $\ce{NO-}$ a radical species?  If so, can someone please explain why because $\ce{NO}$ has an odd number of electrons (11), so adding an additional electron would give 12 valence electrons and a formal charge of -1 on the nitrogen. Is my general chemistry failing me or?

Your title doesn't match your question. Hyponitrite is $\ce{N2O2}^{2-}$. As for the short-lived intermediate $\ce{NO-}$, I don't know whether or not it is a radical. Note, however, that it is formally isoelectronic to $\ce{O2}$, which surely is.

"nitric oxide anion in its triplet ground state" http://scitation.aip.org/content/aip/journal/jcp/114/18/10.1063/1.1364679

The nitroxyl anion is isoelectronic to oxygen. It has pretty much the same orbitals as oxygen does and it also features the same number of electrons. The principal difference is the orbital shape: since oxygen and nitrogen are two different atoms with a different number of protons, there is no plane of symmetry bisecting the $\ce{N=O}$ bond axis (in group theory terms: the point group is $C_{\infty \mathrm v}$ rather than $D_{\infty \mathrm{h}}$). This difference means that bonding orbitals will be more oxygen-centred while antibonding ones will be more nitrogen-centred.
However, if you take a look at the molecular orbital scheme, the key orbitals are identical to those in oxygen. The highest occupied molecular orbitals are two symmetry-equivalent π* orbitals and the unoccupied σ* orbital is higher in energy. Since the two π* orbitals are symmetry equivalent, there is no way the molecule can pair the spins; the highest energy electrons will be parallel and occupy one of the π* orbitals each.
Considering that we have exactly the same situation as oxygen in the key orbitals, this also means that $\ce{NO-}$ features exactly the same states. Most notably, the ground state is a triplet $\ce{^3NO-}$ state, while a higher-energy singlet $\ce{^1NO-}$ state also exists — entirely analogously to the singlet $\ce{^1O2}$ and triplet $\ce{^3O2}$ states of dioxygen.

When calling the nitroxyl anion a radical, one should remember that it is, like the oxygen molecule, more accurately described as a diradical. Its properties are distinct from those of monoradicals such as $\ce{NO}$, which feature a doublet state.

I agree with the answer above (the foundamental state of nytroxyl anion is a triplet diradical).
Chemically, though, as shown by its strong tendency to dimerize in the absence of coordination centers, it behaves more as a nitrenoid synton  (-)O-N:
As such it tends to couple with itself to form hyponitrite (-)O-N=N-O(-)
In this aspect it recalls somehow the double nature of CO : an hybrid of a oxonium carbanion and a carbene carbonyl.


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For molecules to be enantiomers, do their bonds have to point the same way?
Enantiomers are non superimposable mirror images. What is bothering me is:
If the bonds point different ways are they still considered superimposable and hence enantiomers?
To illustrate, comparing X and XI:

If I flip XI (180 degrees around the vertical axis that runs along the $\ce{C-C}$ bond), I get Y:

Is Y considered non-superimposable on X (and hence is Y e.g. XI an enantiomer of X)?
I would say yes they are non-superimposable, as in X the horizontal groups are pointing out whereas in Y they are pointing into the page (other than this, the groups are in "matching" positions). I don't think that is "superimposable", as the groups would not overlap.

You think right, they are not superimposable.

Place your two hands face down on a table. They are non-superimposable mirror images of one another (chirality does come from the Greek kheir, meaning "hand", after all). Now flip one of your hands over so that one palm is facing up. Are your hands any different now than before you flipped one over?
No. Your hands' orientations in space have no effect on their superimposability, as is the same for molecules.


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For molecules to be enantiomers, do the groups have to be non-superimposable mirror images even when rotated around C-C single bonds?
For example, if I rotate the Carbon-Carbon bond in B around 120 degree anti-clockwise, I would get the molecule shown in image C. C looks to me like a non-superimposable mirror image of A, hence A and C are enantiomers.
However, as drawn, A and B do NOT seem to be mirror images.
Are A and B enantiomers considering what happens after a bond rotation?

Note: ChemDoodle shows those red triangles saying "questionable stereochemistry". I'm not sure what is wrong but the molecules I drew may have some issues, but I think the question still holds.

related http://chemistry.stackexchange.com/questions/42852/is-rotation-of-a-mirror-image-allowed-in-order-to-determine-chirality-doesnt-r

Real molecules surely do rotate around their single C-C bonds all the time(*). So yes, if two models can be superimposed after an internal rotation of such sort, this means they represent the same molecule, as is the case with your B and C.
This, BTW, is the reason why you should orient all substituents in a standard manner when drawing your Fischer projections. Otherwise any molecule would have just too many variants, and you'll get bogged down in them. Think of rotating the upper group the third way (the first two being B and C). Think of rotating the lower group too...
Then again, I've rarely seen Fischer projections used for anything other than sugars. Typically people just determine the absolute configuration (R or S) and go with that. It gives you a definitive answer, after all.

(*) That is, most of them do. When you have bulky substituents, rotation may pretty well get blocked, and then you may be able to separate the isomers that are (nominally) produced by this rotation - they are no longer equivalent. Look at BINOL or BINAP, for example.

Hi Ivan.  Thank you for clearing that up. Could you please explain what you mean by drawing Fischer projections "standard manner"? Do you mean e.g. Hydrogens should always be drawn as horizontal groups?

Yes, exactly. Carbon chain goes vertically and all those H or OH side groups are outstretched horizontally towards us.

Great, thanks for the tip.  I wasn't aware about that.

Ivan, in cases where the barrier to rotation is sufficiently high, is there some sort of "thermal enantiomerism" that could develop?

Thought I'd rather edit this into the answer.

Even molecules such CBrClIH can be converted from one enantiomer to the other by inversion (like what happens to an umbrella in a storm). This would  require tunneling through a energy barrier, but the barrier is so high and the atoms are so heavy that the tunneling time is typically more than the age of the universe. The simplest "chiral" molecule that is not chiral in practice due to internal rotation is H$_2$O$_2$. The barrier to internal rotation here is so low that the molecule remains only chiral for a few picoseconds.


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Reasons for existence of eutectic point for alloy systems
Why does a eutectic point exist for a system of alloy ? Why is it that there is such a definite ratio of composition at which this occurs ?

You might look at my answer over on Physics, http://physics.stackexchange.com/questions/196686/what-causes-a-eutectic-point and see if that helps.

First, there are several other answers that you might be interested in looking at first (full disclosure - they include some by me).
A question on Physics SE on eutectics
Ideal vs regular solutions and eutectics on Chemistry SE
These together go in to some detail on why the eutectic forms (when the liquid free energy drops down and touches the tie-line between the two solid phases - the Physics one), and the interactions one needs to get the liquid phase to drop below the tie line between the solids (the Chemistry one). To get a eutectic you need an interaction term in the liquid Gibbs free energy.
As for why it happens at such a definite composition, the glib answer is the Gibbs phase rule. In a system at fixed pressure, three phases can coexist at only one point.
Given that this question is, in general, a duplicate of the ones listed above, particularly of the Chemistry SE one, I have no particular need of the bounty. Go read those and see if they help.


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Numbering cholesterol structure
What could be the reason for clockwise numbering in c ring of cholesterol?

Related answer: What is the origin of the naming of deoxycorticosterone acetate?

(note that this structure is incomplete cholesterol structure, some stereochemical information is missing)

When two German chemists H.O. Wieland and A.O.R. Windaus were awarded Nobel prize in two consecutive years 1927 and 1928 for their research of bile acids and sterols, despite the enormous effort of structure determination, their alleged cholesterol structure was not yet correct at a time, but already had similar numbering:[1,2]

It's difficult to answer why this one exactly was chosen. But see, here you have counter-clockwise numbering in first three rings, however clockwise in the last one, and with methyl and ethyl intermission. I don't think that it's an important factor anyway. In fact, fused ring systems nomenclature usually numbers the ring system peripheral atoms continuously (through the peripheral carbons, if including the fusion carbons), and in the clockwise sense, when the skeleton is in preferred orientation. However, for natural compounds, especially terpenes, there is large variety of different parent structures with various numberings.
So, when the correct structure was determined in 30’s (I think I was R. Robinson; and later confirmed by X-ray analysis in mid 40’s), it adopted similar numbering scheme (changes were made in methyls / side chain numbering), and the first systematic nomenclature of steroids[3] (pre-IUPAC though) from early 50’s also uses it, without explanation, rather referring to tradition:

STEROID NOMENCLATURE.
The following proposed rules for steroid nomenclature are based on agreements reached at a Conference held at the CIBA Foundation, London, on May 30th to June 1st, 1950. It is hoped to submit them for approval to the International Union of Pure and Applied Chemistry.
NUMBERING.
Rule 1.—Steroids shall be numbered as in formula (I). If one or more of the carbon atoms shown in (I) is not present, the numbering of the remainder shall remain undisturbed.

Remarks. The numbering in (I) is that generally accepted in steroid chemistry, although it does not conform to that given in the “Ring Index” for cyclopenta[a]phenanthrene (No. 2561). It has been the general, but not invariable, practice to assign the number 18 to the carbon atom of the methyl group attached to C(13), as in (I), because this methyl group is present in all natural steroids whereas the methyl group (involving C(19)) attached to C(10) is absent in members of the œstrogen series.
(…)

…and it's used still (that's for the ring numbering. There are some changes – C-28 and C-29, not present in cholesterol, now have numbers 241 and 242).

If chemists were to decide for steroids numbering revision, an interesting option would be to choose numbering based on the biosynthesis facts. The numbering would be changed only in ring A (starting with today’s C-4, clockwise), and would be entirely continuous:

(or opposite direction)
… Based on squalene (farnesyl, or farnesene units joined tail to tail depicted in different colors, their isoprene units bold) cyclization (with a couple of methyl shifts, and also hydride shifts (not depicted)) after squalene epoxidation (bottom leftmost double bond) to lanosterol (30 carbons), which then losts three methyls via oxidation and decarboxylation, and double bond is shifted. (The correct topology of of squalene cyclization was first correctly proposed by R.B. Woodward, if I'm not mistaken, as an alternative to R. Robinson proposal[4]).

References:

Wieland Nobel lecture, PDF
Windaus Nobel lecture, PDF
Steroid nomenclature. Chem. Ind. June 23, 1951, SN 1-11.
Woodward, R. B.; Bloch, K. THE CYCLIZATION OF SQUALENE IN CHOLESTEROL SYNTHESIS. J. Am. Chem. Soc. 1953, 75 (8), 2023–2024.


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How do lone electrons impact the charge of an ion?
This is a quite elementary question, but I still can't wrap my head around it.
I know how to calculate the formal charges of an ion or a molecule, but I don't understand what happens when oxygen (when sharing a single covalent bond with sulfur in a sulfate ion) has a lone electron that doesn't bond nor pair. I don't understand why this puts a -1 charge on the atom. What happens?
I've attached an image which explains what I'm trying to figure out:

See those two red dots? What do they mean? I understand that they symbolize the lack of a single electron needed to complete an octet. But why does it give the atom a negative charge?

The two red dots indicate the presence of two electrons, not the absence, if there were no electrons there it would be balanced. The colours of the electrons in the image indicate their origin: blue electrons originate with oxygen, black from sulfur and red from an outside source (eg. Potassium or H2).
Ions have a charge because they have more or fewer electrons than the number of available protons. Sulfate has gained two extra electrons.
The arrows in the second image indicate a 'dative covalent bond' where the electrons in the covalent bond all come from one atom rather than being shared, while the lines indicate a normal covalent bond (ie. an electron from each atom. The extra electrons added to the oxygen atoms to form a full orbital level in the normal covalent bond are responsible for the extra charge.
There's a nice visual depiction here: http://chem-net.blogspot.co.uk/2013/02/electron-dot-structure-of-sulfate-ion.html

The way this is normally taught (introduced) is as the concept of formal charge.
Think about the neutral oxygen atom, how many valence electrons does it have? Now look at the oxygen bonded to the Sulfur, how many electrons does it have if you consider the bonding ones to be shared?
-lone oxygen = 6
-bonded oxygen = 7
As it has one more electron than the neutral atom this gives it a negative charge. Each oxygen is -1 and if you do the same process on the sulfur you get +2 and the total (+2 + -4) is -2 which is the charge on the ion.


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Is there a sixth type of d-orbitals with index 2z²-x²-y²?
There are 5 types of d-orbitals: $\mathrm{d}_{xy}$, $\mathrm{d}_{zx}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{x^2-y^2}$, and $\mathrm{d}_{z^2}$.
I encountered a question which asked about the various d-orbitals, but I came across one that I'd never heard of before: $\mathrm{d}_{2z^2-x^2-y^2}$.
the answer key says that it is a d-orbital.
How is that possible? Is it sixth d-orbital?

There are 5 linearly independent solutions to spherical harmonics for $\ell = 2$.
We generally pick $\mathrm{d}_{z^{2}}$, $\mathrm{d}_{x^{2}-y^{2}}$, $\mathrm{d}_{xy}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{xz}$.
Ignoring, $\mathrm{d}_{z^{2}}$, notice that it's really easy to see what the $\mathrm{d}$-orbital looks like figuring out the nodal planes. The lobes of the $\mathrm{d}$-orbitals go in between, and the sign of the function tells you how to switch the sign of the wavefunction. So, in your diagram for (a), (b), and (c), I would take off points for the reason that there is no indication that the diametrically opposite lobes have the same shading.
Here's the question that you want to wrap your head around. Why is there no $\mathrm{d}_{x^{2}-z^{2}}$ and $\mathrm{d}_{y^{2}-z^{2}}$ in analogy to $\mathrm{d}_{x^{2}-y^{2}}$?
The answer is by choice because you can add orbitals together to get new orbitals. The orbitals we write are actually linear combinations of the complex solutions we got from solving the spherical harmonics for the Schrödinger equation. Now, the full name of $\mathrm{d}_{z^{2}}$ is $\mathrm{d}_{2z^{2}-x^{2}-y^{2}}$. This is not hard if you think about the where the node is (a nodal bi-cone) and the signs, the torus is different sign than the lobes. With this and $\mathrm{d}_{x^{2}-y^{2}}$, you can construct the two missing ones (being extremely loose with notation):
$$
-\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y{2}} + \mathrm{d}_{x^{2} - y^{2}})
= -\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2} + x^{2} - y^{2}})
= -\frac{1}{2}\mathrm{d}_{2z^{2} - 2x^{2}}
= \mathrm{d}_{x^{2} - z^{2}}\\
-\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2}} - \mathrm{d}_{x^{2} - y^{2}})
= -\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2} - x^{2} + y^{2}})
= -\frac{1}{2}\mathrm{d}_{2z^{2} - 2y^{2}}
= \mathrm{d}_{y^{2} - z^{2}}
$$
So, all you need is 5.

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Why is the boiling point of stibane higher than that of ammonia?
I recently came across the fact that the boiling point of $\ce{SbH3}$ (stibane) is greater than that of $\ce{NH3}$ (ammonia). I was expecting $\ce{NH3}$ to have a greater boiling point as a consequence of intermolecular hydrogen bonding. I understand that as molecular mass increases, van der Waals interactions between molecules also increase, but believing that hydrogen bonding will be a more dominant factor than van der Waals interactions, I expected $\ce{NH3}$ to have a greater boiling point.
Now, I already know that:

$\ce{NH3}$ and $\ce{SbH3}$ are gases at room temperature.
Gases including $\ce{NH3}$ do not showing hydrogen bonding in gaseous phase, which I also don't understand why.

But I want to know:

Why does $\ce{NH3}$ have a lower boiling point than $\ce{SbH3}$ even though it shows hydrogen bonding?
Why does $\ce{NH3}$ not show hydrogen bonding in gaseous phase?

Though it does go against your intuition, you've actually mentioned the answer in your question. Stibane has a higher boiling point than ammonia/azane on account of van der Waals interactions (owing to the larger size of the antimony atom).
Our teacher had actually posed this question to us during my first year of high-school. All of us were incredulous when we were told that stibane had a higher boiling point than ammonia, and like you, we were fixated on H-bonding and chose to dismiss van der Waals' interactions as insignificant in this regard.
First, a look at the H-bonding argument:
At school we're taught that H-bonding only occurs in molecules where you find hydrogen atom(s) bound to either nitrogen, oxygen or fluorine ($\ce{N,O,F}$). This is because $\ce{N, O}$ and $\ce{F}$ are sufficiently electronegative (about 3.0, 3.5 and 4.0 on the Pauling scale respectively) to polarize the $\ce{N/O/F-H}$ bond. Do note, that nitrogen is only barely able to establish H-bonding, while it's much easier for oxygen and fluorine to polarize the $\ce{O/F-H}$ bond on account of 1) their higher electronegativites and 2) their smaller sizes.
As antimony is much larger than nitrogen, and since the hydrogen bonding due to the nitrogen in ammonia is pretty weak, the van der Waals interactions in stibane 'outdoes' the H-bonding in ammonia.
However the same argument does not follow for water and hydrogen fluoride, which have higher boiling points than tellane and hydrogen iodide respectively.
$$\textbf{Melting } (\vartheta_\mathrm{m})\textbf{ and boiling } (\vartheta_\mathrm{b}) \textbf{ points of}\\\textbf{Group 15, 16, and 17 hydrides  in K}$$
\begin{array}{lrrlrrlrr}\hline
&\hspace{-2em}\text{Group 15}\hspace{-1em} &&& \hspace{-2em}\text{Group 16}\hspace{-1em} &&& \hspace{-2em}\text{Group 17}\hspace{-1em}&\\ \hline
\text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b} & \text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b} & \text{Hydride} & \vartheta_\mathrm{m} & \vartheta_\mathrm{b}\\ \hline
\ce{NH3} & 195.5 & 239.6 & \ce{H2O} & 273.0 & 373.0 & \ce{HF} & 180.7 & 292.4\\
\ce{PH3} & 138.0 & 185.0 & \ce{H2S} & 190.3 & 211.0 & \ce{HCl} & 161.0 & 189.4\\
\ce{AsH3} & 159.0 & 218.0 & \ce{H2Se} & 209.0 & 231.0 & \ce{HBr} & 184.5 & 206.0\\
\ce{SbH3} & 184.0 & 256.0 & \ce{H2Te} & 222.0 & 271.0 & \ce{HI} & 222.2 & 237.0 \\ \hline\end{array}
Now here's where your logic comes into play; the H-bonding's pretty strong in case of water and hydrogen fluoride (after all, oxygen and fluorine are the two most electronegative elements we know) and this time 'round the H-bonding outdoes the van der Waals interactions.
So you see, you should've made a comparision between the higher and lower hydrides (for groups 15, 16 and 17, to really be able to spot the exception).
As for your spin-off question:

Why does $\ce{NH3}$ not show hydrogen bonding in gaseous phase?

Well, the H-bonding here is strong enough to keep the ammonia molecules together at normal temperatures. But you must never forget that H-bonding is not a true chemical bond, but an intermolecular interaction. If you provide sufficient energy to a sample of ammonia (i.e. by heating it) the ammonia molecules wiggle and wiggle and wiggle, until at some point they wiggle so rapidly, that they overcome the attractive forces brought about by hydrogen bonding and become a gas. This is why you don't observe H-bonding in gaseous ammonia: because the particles are too energetic to be held down.

True, ammonia shows no hydrogen bonding in the has phase.  But some other compounds ... see https://www.jstor.org/stable/77864?seq=1#page_scan_tab_contents.

A molecule's boiling point is directly proportional to molecular mass of the compound!
In general, melting and boiling points of compounds increase down the group, but because of hydrogen bonding, $\ce{NH3}$ has a higher boiling point than $\ce{PH3}$ and $\ce{AsH3}$, but the atomic mass of $\ce{Sb} \gg \ce{N}$. Hence, the boiling point of of $\ce{SbH3}$ is greater than that of $\ce{NH3}$.


text:
Lewis structure of nitric oxide
These are the two resonating structures of $\ce{NO}$ given in my textbook:

$$\ce{:\!N=\overset{\Large .\!\!.}{\underset{\Large .}{O}}\!: <-> :\!\overset{\Large .}{N}=\overset{\Large .\!\!.}{O}\!:}$$

In the first resonating structure, oxygen is shown to have 9 electrons. How can that be possible since oxygen does not have the third shell so it has no orbital available to accommodate any more electron?
Also, why don't elements donate only one electron in such cases? I know unpaired electrons cause repulsions but in cases like $\ce{NO}$ where it's not possible to avoid an unpaired electron. Cannot oxygen donate only one electron to nitrogen? If it does that, we could have another structure. Is this structure possible?
$$\ce{:\!\overset{\Large .\!\!.}{N}=\overset{\Large .}{O}\!:}$$
Also, just out of curiosity, between oxygen and nitrogen, on whom will be one electron pair more stable? I am not able to decide this on the basis of electronegativity.

There is actually an erroneous explanation in your textbook. There is how the nitrogen monoxide molecular structure is explained in school: as there are no extended methods known by pupils or moreover the teacher, it's ok children to think that third unpaired electron on nitrogen atom is resonating between two atomic kernels as it's presented on picture upstaris. But it's wrong. Of course, that third electron cannot entirely belong to an oxygen atom (look to the right structure of first picure), because it (ox. atom) simply doesn't have enough free "quantum cells" to take it up to.
Things became much easier ever since people could describe the situation using molecular orbitals. Consider the molecular orbital diagram below:

It's to say, that

As the modern theory of molecular orbitals considers, there is some inter-atomic space between two atoms, where is the bonding electron allocation is the most likely situated in by the moment. There is exacly where our "third" elecron situated towards to that molecule. It doesn't ever belong to oxygen atom, but, technically, we can consider that structer as a resonant. It would me more correct, if to represent it's structure like that:

As that molecule has one antibonding electron π*2px it is a molecular diradical and it has an unstable structure. That also follows because of it's real physical-chemical properties (semi-stable only under low temperatures, and is very chemically active molecule).

On the assumption of last conclusion (nitrogen dioxide as a molecular diradical), we can redraw our structure like that:

link

So will this unpaired electron will be drawn more toward oxygen atom because of the higher electronegativity of oxygen!!??

As I already said, no because it (oxygen atom) simply doesn't have enough free "quantum cells" to take it up to.

Since we have not been taught molecular orbital theory in detail, it might take some time for me to fully understand your answer. There is a question in my textbook : On which atom is the unpaired electron in more delocalized? The ànswer is oxygen, so how else will we justify that?

@Blue, yes, we can actually say, that this electron is more delocalized under oxygen atom. Talking simplier, we can admit, that there is the only oxygen's higher electronegativity factor (incorrect in real!). Unfortunately, I can not explain things simplier than it's described above. But, the thing I can tell you definitely is that the formula in the top of that page (resonant structures transformation) is absolutely incorrect. It should be drawn in other site.

Are you sure the order of the orbitals in your diagram is right? As far as I know, the pi orbital should be below the sigma orbital in terms of energy.

@ShoubhikRajMaiti, not in this case. Orbitals' order depends on concrete molecule.

Look at page 290, Principles of Inorganic Chemistry, Brian W. Pfennig, Wiley, 2015. The $\ce{\pi_{2p}}$ orbital is drawn below the $\ce{\sigma_{2p}}$ orbital in the energy level diagram.


text:
Why bath sonicators don't come with adjustable power setting?
All the bath sonicators I saw online don't have a power level setting.
(Probe sonicators, on the other hand, do have a power setting.)
Why is that?

All the bath sonicators I saw online don't have a power level setting.

This depends on the product lines.
I've mostly used BANDELIN SONOREX SUPER ultrasonic baths, and indeed, the most simple models of this line just have a (mechanical) timer.
However, models of the SONOREX DIGIPLUS line do come with power control.
In the past, the ultrasonic baths were mostly used for cleaning purposes, while the probes, which typically had a much higher power, were used for sample homogenisation and similar purposes. Here, power regulation was apparently more important to avoid degradation of the same due to local overheating.
As usual, exceptions prove the rule:

I've used ultrasonic baths to support oxidations with $\ce{MnO2}$ or PCC on silica
Cleaning of modern (composite) materials seemingly requires power control, e.g. to avoid ablation

Why is that? Is there some sort of a standard/uniform power setting across vendors of such low end models?

@Sparkler I don't know. I only remember using BANDELIN baths. I might have used others, but I don't recall the manufacturer.


text:
Is it correct to say that SOME iodine undergoes sublimation at 1 ATM
The triple point of Iodine occurs at 0.12 atm and at standard temperature it is a solid on the phase diagram.

I understand that therefore, when heating iodine at 1 atm, it (at least some of it) will pass through the liquid phase, and it is also obvious from observing heated Iodine that it yields liquid when heated gradually. Also, just as liquids can evaporate below the boiling point, I understand that some Iodine can escape from a sample in the vapor form when the vapor pressure is lower than the atmospheric pressure.
My question is 2 part:
1) Is it correct to call the change form solid to gas below equilibrium vapor pressure "sublimation".  In other words, is sublimation equivalent to the term "boiling" or is it equivalent to the term evaporation in the change of phase from liquid to vapor, or both?  (Is there a separate term, even if out of use,  for the solid to vapor transition below equilibrium pressure?)
2) At 1 atm, when Iodine molecules in a solid sample escape into the vapor form, either at standard temperature, or as the sample is heated, do they all briefly pass through the liquid phase first, either at a vapor pressure below 1 atm or as vapor pressure is raised above 1 atm.  .
Iodine's crystal vapor pressure will of course be <0.12 atm below the triple point pressure as can be seen on both graphs.  A converse but equivalent question would be whether vaporous Iodine at 1 atm can crystalize and deposit without liquifying.

Related: http://chemistry.stackexchange.com/questions/37413/sublimation-of-iodine/37416#37416

So trying to distill out the essence of my question, is the heat of sublimation going to be equal to the heat of fusion+ heat of vaporization at 1 atm?
Anyway, when heating Iodine in a flask, it is very easy to see liquid Iodine in the bottom of the flask.  It can be swirled and will "freeze" to fill the shape of the container.  When the vapor deposits on the sides of the flask, it looks to be atomized, not crystals formed from condensed droplets of liquid.

But it looks like what may be the case is that melting does not happen spontaneously below the melting point because the liquid is not free to escape contact with the solid, but vaporization can happen because the energy can be carried away.

To answer part 1 of your question:
It is correct to call the change form solid to gas below the equilibrium vapor pressure sublimation. And there is no separate term for vaporization of a solid above the equilibrium vapor pressure; it is still sublimation.  Similarly, a boiling liquid is still perfectly correctly said to be evaporating.
To answer part 2 of your question:
At temperatures below the melting point of iodine, it will sublimate until the equilibrium vapor pressure is reached without going through a liquid phase in that process.
Once the iodine is heated to it's melting point, it will continue the same process of vaporizing to it's equilibrium vapor pressure. Because it is now a liquid, we just call that evaporation rather than sublimation.
If the iodine is further heated to it's boiling point at 1 atmosphere, then it will continue to rapidly evaporate as it boils away.
Regarding the thermodynamic relationships between these processes as mentioned in one of your comments (which I think you should just edit into the question), what you have said is correct. This is a path-independent thermodynamic process, so with respect to the ΔH of the phase-change processes, it doesn't matter how you get from the solid phase to the vapor phase. ΔH for the direct solid-vapor transition (sublimation) is equal to the ΔH of melting plus the ΔH of evaporation.
I hope I have addressed all your points, please ask for clarification in the comments if I've left anything unclear.

I'd quibble a tiny bit. At the melting point you could have both solid and liquid iodine. It all has to go to liquid before you could get above the melting point of course.

The "Once the iodine is heated to it's melting point..." statement could be interpreted  as saying that the vapor pressure would continually increase with increased heat input, which as you say is not the case. Once the solid starts to melt, the additional heat will go into melting the solid and neither temperature nor vapor pressure will increase again until all of the solid is melted. Of course at the melting point both the liquid and solid have the same vapor pressure, and the rate of sublimation of the solid is equal to the rate of evaporation of the liquid. Hope I got the right quibble ;)

Paragraph starting "Once the iodine is heated to it's melting point, ..." then the second sentence starts off "Since it now all a liquid ..." // My quibble is that it doesn't all have to be liquid at the melting point.


text:
What is the "smell" of solid sulfur
A solid piece of sulfur has a characteristic smell.
Is the smell of solid sulfur the result of the release of compounds like sulfur dioxide or hydrogen sulfide from a reaction with air or moisture, or is it from elemental sulfur particles that are being released by sublimation?
If it is from pure Sulfur particles (at least in part), what form is it released into the air, ie $\ce{S8}$ or $\ce{S2}$ or something else?

Sublimation would be a long shot, I doubt it. It can't be S2, since I hear it's pretty unstable and exists only at high temperature ._. The smell is most likely due to the formation of sulfur oxides.

The primary candidates for volatilization of odorous sulfur gases are $\ce{SO2}$ and $\ce{H2S}$.  Unless the elemental sulfur has been physically ground and dispersed into the air it is not likely the direct source of the smell.
According to the International Volcanic Health Hazard Network, who of course study gaseous volcanic emissions with respect to safety issues:

"Sulfur dioxide ($\ce{SO2}$) is a colourless gas with a
  characteristic and irritating smell. This odour is perceptible at
  different levels depending on the individual's sensitivity, but is
  generally perceived between 0.3-1.4 ppm and is easily noticeable at 3
  ppm."

Regarding $\ce{H2S}$, according to the US Occupational Safety and Health Administration (OSHA), $\ce{H2S}$ has a perceptible smell at between 0.01 and 1.5 ppm.
The two gases have different smells, $\ce{H2S}$ being that of rotten eggs and that of $\ce{SO2}$ being described as "pungent and irritating".  So that would be one way to differentiate which gas you are smelling.
The most likely candidate is $\ce{SO2}$ as our air is generally oxidizing, and to me elemental sulfur smells more like $\ce{SO2}$ than rotten eggs.

Pure elemental sulfur is completely odorless and tasteless.
The characteristic smell you observed is due to impurities, probably $\ce{H2S}$, sulfur oxides, and organic sulfur compounds, some of which can be perceived in very low concentrations.

Very true, though if elemental sulfur were aspirated/inhaled it would rapidly react with the moisture present to produce sulfur dioxide, so the "odorless" elemental sulfur would appear to smell like sulfur dioxide.


text:
More than one way of balancing a chemical equation
My textbook had this question on balancing chemical reactions

Ozone reacts with nitric oxide to give nitrogen dioxide and oxygen gas

Here's how I balanced it: $$\ce{O3 + NO -> NO2 + O2}$$
However I realised that the equation is also balanced if I write it as: $$\ce{2O3 + 4NO -> 4NO2 + O2}$$
and when it is written as: $$\ce{3O3 + NO -> NO2 + 4O2}$$
This seems wrong as I am sure that there cannot be more than one way of balancing a reaction but I do not have a satisfactory explanation. What is going wrong here?

This reaction is a combination of two different chemical reactions so we cannot balance it simply ....http://chemlab.truman.edu/CHEM130Labs/AlumFiles/ChemEqnNotes.asp

This happens when you have a sum of two (or more) independent reactions. You can balance each one, and then add them together in an arbitrary proportion. People usually run into this when trying to come up with an equation for combustion of gunpowder, where the oxidation of carbon and oxidation of sulfur are pretty much independent.
Now, in your case the reactions are:
$$\ce{2O3 -> 3O2}\tag A$$
and
$$\ce{2NO + O2 -> 2NO2}\tag B$$
Your first reaction is a combination of ${1\over2}A+{1\over2}B$, the second is $A+2B$, and the third is ${3\over2}A+{1\over2}B$. There are infinitely many possible combinations.
Ozone, being an extremely powerful oxidant, is known to participate in certain reactions where it spends one atom of oxygen and the rest falls off as the less reactive $\ce {O2}$, but this is not the case here. $\ce {O2}$ reacts with $\ce {NO}$ instantly, as well as $\ce {O3}$.

I am still confused. You said that I could add the equations in any 'arbitrary' proportion. That would deem the second and third questions correct, wouldn't it?

Sorry, I didn't get it. What questions? You've balanced the reaction in three different ways. Yes, each of them is correct.

Sorry. I meant the second and third eqautions

Like I said, they are as good as the first one.


text:
Help balancing reaction between Iron (II) chloride and potassium phosphate
Very rusty with chemistry and looking to review some basics. Came across the following:

Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.

What I have so far (guessing, basically):
$$\ce{FeCl2 + K(H2PO4) -> KCl + Fe(H2PO4)2}$$   [still need to balance but correct right hand side?]
I can't remember the exact rules for forming the right hand side. I just put this in place to match ion charges. What prevents me from writing the right hand side as something else entirely (not sure what but I'm asking, "well, couldn't I figure something out if I break everything apart and re-arrange into another option?").
EDIT: Additional Question Below
Ok, so assume the above reaction was balanced (it's not) and valid. Then, my question is does the reaction go both ways ($\ce{<-->}$)? If so, when/why/how? I'm trying to think back many years ago and it has something to do with "reactivity" and the row of the elements on the periodic table, correct? I'm going to look into it more myself but if someone has a quick "30 second" answer I'm sure it'd be of help to me!
[Assuming this is OK and valid in the forward direction. Again, needs to be balanced.]
$$\ce{FeCl2 + K(H2PO4) -> KCl + Fe(H2PO4)2}$$
[Is this ever ok/possible?]
$$\ce{KCl + Fe(H2PO4)2 -> FeCl2 + K(H2PO4)}$$

See, what charge does Fe have?

@IvanNeretin Since it's iron II, it's charge is 2+, correct?

@jon $\ce{FeCl}$ if it exists has iron (I) because chloride cannot be (-II). So if you are speaking about iron (II) chloride then it must be $\ce{FeCl_2}$ ;)

@jon Right! Now what is the charge of Cl?

@IvanNeretin Ah, yes 1-. So it'd have to be FeCl2. Ok, I'll make the edit and update my question accordingly. Thanks!

This question was the continuation of your question and since both question are related,  I am answering the question by emphasizing the mistake you are doing and adding a relevant portion to explain it.

Mistake:
Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.
You need to consider potassium phosphate to be $\ce{K3PO4}$ not $\ce{KH2PO4}$ according to IUPAC's Red book (see sections IR-8.2 and IR-8.4 and the anions in question are explicitly named in table IR-8.1 therein.)
If $\ce{KH2PO4}$ is meant to be added, the correct term should be potassium dihydrogenphosphate.

Extra info.:
Iron hydrogen phosphate are new in this world and has been recently synthesised and as not as stable as iron phosphates. So, only iron (II and  III) phosphate and potassium chloride forms from the reaction.
$$\ce{3FeCl2 + 2K3PO4 ⟶ Fe3(PO4)2 + 6KCl}$$
$$\ce{Fe3(PO4)2 ->[air oxidn⋅,400C] FePO4}$$
or actually
$$\ce{4Fe3(PO4)2 + 3O2 ->[400 C] 8FePO4 + 2Fe2O3}$$

Iron hydrogen phosphate can be made by reacting iron(II) phosphate with dilute hydrochloric acid.
$$\ce{Fe3(PO4)2 + 2HCl → 2FeHPO4 + FeCl2}$$

The molecular formula: $$\ce{ 2 K3PO4 (aq) + 3Ni(NO3)2 (aq) -> Ni3(PO4)2 (s) + 6KNO3 (aq) }$$
The complete ionic formula is: $$ \ce{6K+ (aq) + 2PO4^3- (aq) + 3Ni^2+ (aq) +2NO3-  -> Ni3(PO4)2 (s) + 6K^+(aq) + 2NO3^-(aq) }$$
The net ionic formula is: $$ \ce{2PO4^3- (aq) + 3Ni^2+ (aq) -> Ni3(PO4)2 (s) }$$

Be welcome to ChemSE. Please familiarize yourself mathjax (for mathematical equations) and mhchem (about chemistry like reaction equations). Especially the later will both facilitate (and thus, speed up) editing reaction equations in questions / answers / comments, and offer a nicer representation, too. This may be  a starter on both, yet mhchem equally may be used with the LaTeX markup, too (https://mhchem.github.io/MathJax-mhchem/).  But don't use it in question titles.

Why speak of Nickel ?


text:
Does the free energy of the universe decrease for every process?
The second law of thermodynamics states that $dS_{universe}>0$ for all processes.
In addition to the second law, gibbs free energy is used to consider the feasibility of a reaction. $dG=dH_{syst}-T*dS_{syst}$. If $dG_{syst}<0$, this means that the reaction is spontaneous, but a reaction with $dG_{syst}$ can also occur if energy is input into the system (for example electrolysis).
Therefore, $dS_{univ}>0$ does not imply $dG_{syst}<0$.
However, is there such a quantity as $dG_{univ}$, and if so is the relationship $dG_{univ}<=>dS_{univ}$ true?

There are a lot of different notation you use here and also "rxn" if you can clarify this I would be glad. I'm not sure what you mean in the first sentence, what I'm sure yet is that $\mathrm{d}S>0$ always for isolated system.

You mix up things. Entropy extreme and similar principles are applied for certain systems, eg entropy max is only true for isolated systems!

@Greg Is the whole univserse not be defined as an isolated system?

Didn't you ask the last question about this, too?
The issue here is that when $\Delta G > 0$, the reaction is indeed non-spontaneous as written. If you wish to add energy to the system, for example, via electrolysis, you should include the additional changes in the computation of the free energy change. You will find that after you sum up everything $\Delta G < 0$.
Also, note that $\Delta S_{\mathrm{universe}} < 0$ does not mean that the process will not happen; it only means that it's (fairly) unlikely to happen.

It is praiseworthy that you have included the last line. Unlike thermodynamics, Statistical Mechanics has no inherent time asymmetric ingredient. So, indeed the reversal of the Second Law is , though highly unlikely, possible.

The relationship is:
$\Delta G = - T \Delta S_{univ} $
Also, if you want to consider electrolysis in this context,  your system needs to include the power input in some form.  Your entropy increase from obtaining that power is large enough to offset the entropy penalty of an unfavorable process making it spontaneous.

It seems to me now, that dG is always defined to deal with only the system. Is this the case?

That is just the equation, G is defined based on entropy of the universe. Which you use just depends on how you are thinking about the problem. You can substitute in $-T/Delta S_{univ} $ for $/Delta G$ if you prefer to think about the problem that way.

$\Delta G$ is a more convenient way for chemists to think about is since typically we only care about a system and whether the change in that system is favorable or not. In a crude practical sense who cares about the entropy of the universe, I just want to know if a process in a system is spontaneous and if not how to make it spontaneous.

I think the answer is no, there is no such thing as the Gibbs energy of the universe, and thus no way to say that it only decreases.  The reason is that thermodynamics does not take account of how things got to be the way they are.  If a raised weight happens to be conveniently waiting there attached to a pulley, which is attached to a mechanism that allows it to drive the pistons of a Van't Hoff equilibrium box, in which a chemical reaction with $\Delta G > 0$ takes place, thermodynamics does not ask how all that stuff got there.  Unless I am mistaken, the potential energy, $mgh$, of the weight does not figure into its Gibbs energy because it is not internal energy (though I would be excited to be corrected on this if I am mistaken).  Thus it is entirely possible for the weight to fall very slowly and reversibly drive the increase of free energy of the chemical system without any compensating decrease of free energy elsewhere.  In the reversible limit, the falling weight does not generate any compensating entropy, either, which is frequently claimed to be necessary to avoid violating the second law.  The position of the weight has no entropy associated with it.
So does this net $\Delta G > 0$ process mean that the total entropy decreases, violating the second law?  No, it doesn't.  The formula $\Delta G = -T\Delta S_{total}$ only applies if there is no work being done on the system other than expansion or compression work.  That formula is derived by assuming that the change in enthalpy is equal to the heat absorbed from the surroundings, or $\Delta H = Q$, which is the usual situation for a reaction in a lab.  In that case $$\Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T$$.
But if work is also done on the system, then the enthalpy change is not just the heat absorbed.  You also have to add the work done.
$$ \Delta H = \Delta(U+PV) = Q + W + P\Delta V = Q + W - W_{PV} = Q + W_{non-PV}$$.
Now when you calculate the total entropy change, you get
$$ \Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - (\Delta H-W_{non-PV})/T = (W_{non-PV}-\Delta G)/T \geq 0$$.
The last inequality comes from $W_{non-PV} \geq \Delta G$, which follows from Clausius' inequality, $\Delta S \geq Q/T$.  So the second law checks out, even though it seems like some free energy was created out of nowhere (but actually was converted from the potential energy of the raised weight).


text:
Why does elemental Iodine form a brown solution with Ethanol but not with Isopropyl alcohol?
Although it is not the standard procedure for making iodine solution, I have dissolves pure Iodine granules in Ethanol to produce a brown solution which will produce a positive test for starch by creating a purple color.
I tried to create the same solution with isopropyl alcohol but it doesn't work.  The Iodine does not dissolve noticeably, and whatever small amount may be dissolved does not complex with starch.
How does Ethanol dissolve elemental Iodine so well especially compared to Isopropyl alcohol?

I'm afraid you did something wrong or maybe used solution with high water content. Iodine dissolves well in organic compounds.

It was 70% Isoprop but absolut EtOH.  Is that enough water to do it?

While iodine is readily soluble in ethanol and isopropanol, it is only soluble to about 300 ppm in water. I think it's pretty clear that the 30% water in your isopropanol solution was plenty to thoroughly inhibit the iodine's solubility.

Diluting iodine solution with water does not reverse the dissolution of the Iodine.

I see that iodine solution is actually KI in water.


text:
How many different alkyl chlorides result from C4H10 and chlorine?
Let's say there's a reaction $\ce{C4H10 + Cl2}$ taking place under UV light. How many different alkyl chlorides do we obtain?
I'm not sure if the resulting molecule would be monochlorinated or dichlorinated? My guess was monochlorinated, as $\ce{HCl}$ is also going to be a product. And I assumed there would be 2 different alkyl chlorides as there are two different possibilities for a chlorine to be attached to a carbon of a butane.  But I guess I was wrong as the answer the book gives is 4 and not 2. How can the book’s answer be rationalised?

I guess that your "C4H10" was a mixture of n-butane and isobutane.

Well, that is probably it because I don't see how there would be 4 possibilities for an n-butane. I thought that I maybe was missing something somehow and that I was supposed to attach two chlorines instead of one, but I can't see something like that happening. Thanks for the input.

If you were to attach two Cl, that would make more than 4 isomers, even if we stick to n-butane, and even if we disregard mono-Cl stuff.

The question needs clarification. Does C4H10 mean n-butane and/or isobutane? Such questions usually state monochloro alkyl halides. What does "different" mean? Constitutional (structural) isomers or stereoisomers?


text:
Ni-Mh battery makes fizzy sounds while and after charging - why and do I need t replace it?
I have some Ni-Mh rechargeable batteries (size AA/Mignon 1.2V, 2100 mAh) and just started to charge them using an automatic charger with 450 mA output.
Now one of those batteries surprisingly started to make silent fizzy sounds, like if you're frying something in a pan. It started within seconds after putting it in the charger and keeps going for minutes after removing it from the charger again. There is nothing visible from the outside though, I don't see any bubbles or leaking fluids.

What exactly is the chemical process here that causes this sound (probably something is producing gas) and why does it happen in this one battery and not in the others I have of the same type?
Is that battery still usable (safely?) or do I need to replace it?

Update: As of now, the battery in question is replaced and returned to a recycling station. I'm still interested in the chemical process behind this noise though.

sounds unsafe. discharge and dispose of it

It's not possible to give a proper answer to (a) without examining the battery but it is likely due to a compromised casing.  The noise is likely hydrogen gas leaking out as it's produced. Nickel metal hydride (NiMH) batteries use a metal hydride (MH) for energy storage.  When this Hydride is exposed to moisture in air it reacts to form hydrogen gas: $$\ce{MH_x + xH2O \rightarrow M(OH)_x + x H2 }$$
For (b) I would say that your battery is probably not usable—safely or otherwise—anymore and may present a fire hazard. You should replace the battery as rechargeables don't cost near as much as the device they can leak inside of and destroy or the cost of fire damage to a house.

See also: http://www.ka7oei.com/nicds.html

I've heard of this before in  lead acid batteries... it's caused by hydrogen gas being produced and leaking out of the casing of the battery.This is dangerous!!! It makes the very air in the vicinity of the battery explosive. Please dispose of ASAP!!!

We're talking about a Ni-Mh cell here though, not a lead-acid battery.

That doesn't matter. Even though everything from Ni-MH to potassium hydroxide is used in batteries, the underlying principle of energy storage is the same in all commonly used batteries.


text:
When is rotation around double bonds to form cis-trans isomers allowed?
Textbook Question:

The rhodopsin molecules in the retina contain a derivative of vitamin A called cis-retinal (Figure 6). On exposure to light, cis-retinal becomes an isomer called trans-retinal.

This process helps cause an electrical impulse to be transferred to the brain. Around which numbered bonds does the isomerism rotation occur?

The answer is bond 11 (shown in red).
What I don't understand is: I thought rotation is not allowed around double bonds.
So when light hits the structure in Figure $6$, how does it cause rotation around bond $11$ to form the structure in Figure $7$?

Generally, all single bonds are said to permit free rotation while all double bonds are said to inhibit free rotation. This can actually be explained numerically as well: the rotation around the $\ce{O-O}$ bond in $\ce{H2O2}$ has an approximate rate constant of $2.0 \times 10^{8}~\mathrm{s^{-1}}$ while the rotation around the $\ce{C=C}$ bond in ethene has $1.3 \times 10^{-31}~\mathrm{s^{-1}}$. This translates to an equilibration half-life of ethene which is greater than the age of the universe.
It is, of course, helpful to understand why such a large barrier exists. This is due to the way double bonds are formed. ‘Normal’ single bonds are σ-symmetric overlaps of orbitals, i.e. they are necessarily totally symmetric with respect to rotation of the bond axis. Thus, any mechanisms that inhibit single bond rotations must derive from outside these bonds. The ‘second’ bon, however, is a π-symmetric bond, meaning that a plane of symmetry exists which is parallel to the bond axis (i.e. the bond axis is in said plane of symmetry). This automatically means that the bond is everything but symmetric with respect to rotation around the bond axis; if the p-orbitals do not align parallelly the overlap gets smaller and the bond starts to get broken. Note that triple bonds, which contain two perpendicular π bonds, are freely rotatable again since what is lost in one direction (e.g. vertical) is mathematically strictly gained in the other (e.g. horizontal). You can see the three types of bonds in the figure below.

Figure 1: Single, double and triple $\ce{C-C}$ bonds. The single bond is formed by σ overlap of $\mathrm{sp}^n$ hybrid orbitals, double and triple bonds by π-symmetric p-orbital overlap.
It may be immediately obvious that the overlap of a σ bond is generally much better than that of a π bond. Hence, the difference in orbital energies between σ and σ* is much greater than that between π and π*: π orbitals are less stabilised than σ orbitals. In turn, that means that the HOMO will generally be a π orbital (least stabilisation), while the LUMO will typically be a π* orbital (least destabilisation). Thus, the energy difference between π and π* is the smallest energy difference between an occupied and an unoccupied orbital. If a photon with the appropriate wavelength comes along, it can excite the π bonding electron into the antibonding π* orbital. (It also needs to come along from the appropriate direction — parallel to the $\ce{C=C}$ bond axis in ethene if I did my group theory correctly, but that is a minor issue.)
If we excite $1\unicode[Times]{x3c0}^2\,2\unicode[Times]{x3c0}^0 \ce{->} 1\unicode[Times]{x3c0}^1\,2\unicode[Times]{x3c0}^1$, we have effectively reduced the bond order from $1$ to $0$, creating a ‘no bond’ where there used to be a ‘second single’ bond. If there is no overall bonding interaction, the rotation restrictions no longer apply. As long as one electron is excited into π*, the molecule is freely rotatable. When the electron finally relaxes, chances are that the geometry of the double bond has inverted. You can’t observe that in (symmetric) ethene, but other systems such as stilbene or hemithioindigo have been studied extensively.

If you don’t wish to invoke molecular orbital theory, you can also explain the entire idea with Lewis structures. Under the Lewis formalism, an incoming photon ($h\cdot \nu$) will cleave the $\ce{C=C}$ double bond while leaving the underlying $\ce{C-C}$ single bond intact. It can be written as:
$$\ce{-C=C- ->[$h\cdot \nu$] -C^.-C^.-{}}$$
Here too, you are making a double bond a single bond. The former cannot rotate, the latter is freely rotatable. Upon relaxation, whichever rotamere the molecule was in will be fixed again. Since the trans rotamer is more stable then the cis rotamer in almost all linear cases, that one will be accessed by irradiation.

Tl;dr: Yes, double bonds are not freely rotatable, but by irradiation you are breaking the double bond into a formal single bond diradicalic structure.

I really appreciate the thorough explanation. It made me aware of other things I need to read up on such as triple bonds. I assumed they were just the same as double bonds and couldn't rotate.

@K-Feldspar You’re welcome. Yeah, the thing about triple bonds is interesting; you don’t really get to appreciate it with carbon because a $\ce{C#C}$ triple bond is always flanked by two single bonds. But it can be observed with metals, e.g. with $\ce{[Re2Cl8]}$ where a quadruple bond is present between the rhenium atoms. If an electron is excited from $\delta$ to $\delta^*$, the quadruple bond is broken and the resulting triple bond rotates by $45^\circ$ to prevent eclipsed chlorine atoms.

Do you have any thoughts on this animation http://demonstrations.wolfram.com/RotationAboutCarbonCarbonBonds/
You won't be able to see it as it requires the wolfram player download but here are some frames.  http://i.imgur.com/b4PAxmI.png    The double bond makes sense, as one carbon rotates the bond gets twisted and stuck.

But from the looks of things the triple bond should do the same. The issue is: it seems almost as if both carbons on either side of the triple bond are rotating (which suggests the bond itself isn't rotating but the single bond adjacent to it is?) Wouldn't the 3 lines from the triple bond get twisted and stuck the same way as in the double bond? In the animation they don't, and they freely rotate (i.e. one side of each line forming the triple bond rotates at the same rate the other side does so they never get twisted).

@K-Feldspar The issue is that you should draw it with orbitals.

In rhodopsin this normally occurs only after the 11 cis in the chromophore (a retinylidene schiff base) is placed into an electronically excited singlet state after absorbing a visible photon. This excited state has one electron promoted from the HOMO orbital into the LUMO (antibonding) orbital of the conjugated $\pi$ system. Most data has been obtained from the protein bactoeriorhodposin, which acts a  proton pump, as it is easier to use and has a similar chromophore to rhodopsin.
In the protein the rhodopsin molecule is somewhat strained by adjacent (tryptophan) and other amino acids. The isomerisation takes about 200fs so is extremely fast; among the fastest reaction ever studied. After that there is a series of at least 6 intermediates that slowly returns the trans molecule to the cis form and these thermal reactions take about a second or so.
The initial reaction is thought to proceed as follows; just after the electronically excited state is formed calculations indicate that the cis bond is rapidly stretched (~20 femtoseconds) and experiments show that a  small barrier exists between the minimum energy of the excited state and the pathway to isomerisation. At or close to this barrier rotation & oscillation of the groups either side of the 'double' bond occur during isomerisation. (Kobayashi et.al Nature, 2001, v414,p531).
Many experiments have ben performed on trans/cis stilbene isomerisation which shows similar behaviour. Trans stilbene is the most stable isomer and can be converted into the cis also by photo excitation. The cis ground state has a relatively low barrier to the trans at room temperature (the opposite is not true) and so reverts to the trans spontaneously. This does not happen in rhodopsin  so the protein has to go through a complex set of reactions.


text:
Would singlet oxygen in the engine cause a more efficient combustion?
Today someone told me about a new product — a mesh that is made from 5 different metals and when oxygen passes through it, singlet oxygen appears for a short period of time.
This mesh needs to be installed in a car air filter and the "better" oxygen creates a better combustion inside the engine. This is supposed to save fuel, give the engine more power etc.
My questions are:

Is it possible or this is just more one of the advertising lies that stores are trying to sell us?
if an interaction like this is something that can be acquired, will it really cause a better combustion inside the engine?
What are the downsides of interactions like this?
Can a mesh like that be permament without the need to replace it every couple of weeks or months?

I have this bridge in Brooklyn for sale...

Do you have a link to the claim? I'd like to see the level of technical competence in the claimant.

This hypermileage scheme is one of the more interesting ones I've come across, so I think it's worth some discussion. I don't know much at all about the engineering aspects of internal combustion engines, so I'll stick to a theoretical analysis.
Let's start by describing the difference between singlet oxygen and triplet oxygen. Both of these are the same substance, molecular (di)oxygen, or $\ce{O2}$. However, they are in different energetic states. In its lowest energy state, a molecule of $\ce{O2}$ contains two unpaired electrons (which is a very interesting and important fact in itself), and this form is called triplet oxygen. This is the regular stuff you breathe all the time.
However, if some energy is given to triplet oxygen, it can be excited into a different state, and the two formerly unpaired electrons now pair up. Since there are now no unpaired electrons, we call this singlet oxygen (technically there is more than one singlet state possible, but for our purposes that can be ignored).
So in summary, singlet oxygen and triplet oxygen are the same molecule, except that singlet oxygen has inherently more energy. So why does this matter? If singlet oxygen has more energy, then one expects it to be more reactive. It is entirely conceivable that a fire supplied with singlet oxygen will burn hotter/more completely than the same fire supplied with triplet oxygen. So that part of the scheme is plausible.
That said, you may ask yourself, if all it takes is some energy to turn triplet oxygen into singlet oxygen, then shouldn't oxygen in any fire be continuously converted from the triplet state to the singlet state, simply by the heat of the roaring flame? This would mean preparing singlet oxygen separately is entirely superfluous! As it turns out, due to the quirks of quantum mechanics it is not easy to excite triplet oxygen into singlet oxygen. Technically speaking, this is called a spin-forbidden process.
However, due to special relativity (!), spin-forbidden processes turn out to not be completely forbidden, just very unlikely. A concept called spin-orbit coupling comes into play. The more spin-orbit coupling in a chemical system, the more likely it is for spin-forbidden process to nevertheless happen. Spin-orbit coupling becomes more and more relevant as you go further down the periodic table. This means heavy atoms such as platinum can favour the transformation of triplet oxygen into singlet oxygen. So again, the necessity of a metal honeycomb as pictured in the scheme is at least conceptually plausible for this singlet oxygen strategy to work.
(Of course, it is possible to produce singlet oxygen through other means, such as the decomposition of some oxygen-containing molecules like hydrogen peroxide and ozone, but doing so would defeat the whole purpose; you'd essentially have to buy a second kind of fuel!)
At this point it seems there is actually something going for this scheme, but here is where so many others like it stumble: the laws of thermodynamics. If it takes energy to transform triplet oxygen into singlet oxygen, then that energy has to come from somewhere. If the energy comes from the metal honeycomb itself (or more precisely, the energy imbued into the metal honeycomb during its manufacturing process), then yes, it is plausible to get increased mileage, but the mesh will lose energy until it is inactive and has to be replaced or reactivated. My bet is that this would cost much more than the meagre fuel savings.
In principle, the energy to make singlet oxygen can come from the extra energy from the more efficient fuel burn, leaving the mesh intact. However, now you're dipping into the very extra energy that was supposed to make the car get more mileage. The detailed energy balance is generally very hard to calculate, and I couldn't begin to tackle all of the real-world variables in play. That said, the answer to whether a hypermileage scheme is cost-effective almost always ranges from a flat-out "no" to "yes in principle, but no in practice". Processes that are cost-effective (precise spark timings, regulation of air-fuel ratio, etc) tend to make it into standard issue vehicles in the first place. I can't help but be reminded of the joke that alternative medicine which works is just called medicine.
Lastly, regardless of energy savings, singlet oxygen may not be great inside the engine. Not only is it more chemically reactive than triplet oxygen, but it can perform reactions which are completely different from triplet oxygen. It could cause more wear of the engine components, or could generate more of some unwanted sideproducts during combustion, such as nitrogen and sulfur oxides. Any singlet oxygen that manages to escape combustion and exit the tailpipe could also pose a significant biological and chemical hazard.

I think, a catalytic transformation mechanism would make most sense, but maybe that’s my organic chemistry bias. Also, if I am informed correctly, the energy difference between singlet and triplet oxygen is not that high, so potentially warm air could supply enough — remember that the scheme says ‘partially’. And finally, singlet oxygen doesn’t pose a significant biological or chemical hazard when exiting the exhaust. It just converts to triplet oxygen given time (but nobody can see the faint red glow because it’s too faint). Other than that, you have earned +1.

@Jan Thermal energy $k_\mathrm{B}T$ at 298 K is 0.026 eV, whereas the triplet-singlet energy gap for $\ce{O2}$ is 0.98 eV. So, probably not enough. (Gosh I hate typing MathJax on a tablet...)

@orthocresol Good to know actual numbers =D Yeah, probably not enough.

I've heard of these before, and they're total rip-offs. Even if u add extra free oxygen to the intake air, there still has to be an optimum fuel air ratio, which means the oxygen sensor will sense this and direct more fuel to be added. Usually it 14:1 for gasoline.


text:
What kind of solvents favor elimination reaction?
This question is confusing and I can’t seem to find a answer to this. The question is what solvent promotes E2 and E1 reactions? My teacher says that methanol as solvent with strong base favors E2 over Sn2. But she never explained why. Also she keeps using methanol as the example which is confusing. Can anyone please elaborate on this?

This website gives good rules of thumb: http://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/
And so does this website: http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch08substitutionvselimination.html
In addition, according to page 583 of "Physical Organic Chemistry" by Anslyn and Dougherty:

Almost all bases are also nucleophiles, and hence we expect competition between elimination and substitutions. In both $\ce{S_N2}$ and $\ce{E\text{2}}$ reactions, the nucleophile or base reacts in a single rate-determining step with the reactant. In both $\ce{S_N1}$ and $\ce{E\text{1}}$ reactions, the nucleophile or base reacts in a step after the rate-determining heterolysis...
An $\ce{E\text{2}}$ reaction requires the addition of a base, and can be performed in solvents of low or high ionizing power. Since most strong bases are also good nucleophiles, $\ce{S_N2}$ reactions are important competitors. However, elimination will dominate if the carbon with the leaving group is not susceptible to nucleophilic attack, such as a tertiary R group. Examples of this trends are given in Table 10.6.
$\ce{E\text{1}}$  reactions involve carbenium ion intermediates, and therefore are facilitated by all the factors that stabilize carbenium ions. These are the same factors that facilitate $\ce{S_N1}$ reactions. Strongly ionizing solvents and substitution of electron donating groups on the carbon undergoing heterolysis are necessary.
In highly ionizing solvents and with R groups that readily form carbenium ions, the ratio of substitution to elimination is typically independent of the leaving group. This evidence supports the notion that the substitution and elimination products are formed b branching from a common intermediate, and therefore the two reactions share a common rate-determining step. In contrast, in solvents of lower ionizing power, the ratio of substitution to elimination products does depend on leaving group, and indication that the two reactions do not share a common intermediate. Instead, contact ion pairs are formed, and either the leaving group or the solvent can remove the proton to give elimination products. The formation of a contact ion pair also has an influence on the stereochemistry of the elimination.

Table 10.6
Reactant            Solvent   Base      Elimination
Tert-Butyl Bromide  Ethanol   NaOEt     100.00%
Tert-Butyl Bromide  Acetone   NBu4Cl    96.00%
isopropyl Bromide   Ethanol   NaOEt     75.00%
isopropyl Bromide   Acetone   NBu4Cl    0.00%
Ethyl Bromide       Ethanol   NaOEt     9.00%
Ethyl Bromide       Acetone   NBu4Cl    0.00%


text:
Why is the endo product the major product in a Diels-Alder reaction?
I understand that the endo product is, at least in all cases in my textbook, the major product in a Diels-Alder reaction. However, I don't understand why this is the case.
The explanation in my textbook says that the endo product allows for more interaction between the diene and dienophile during the transition state.
I also found the following post, which suggests that secondary orbital interactions are the cause. Stereochemistry of product in Diels-Alder reaction
Then I found the following paper that indicates, at least to my understanding, that steric strain during the transition state is the main reason behind the endo rule (or at least for the example reaction in the article). https://www.ncbi.nlm.nih.gov/pubmed/24449044
So, I'm left wondering what is the reason for the endo rule?

Your textbook's description is probably a watered-down version of the secondary orbital interactions, which have traditionally been the explanation for the endo rule (see any other org chem textbook for diagrams). Good find on the paper though.

related: Endo rule and stereoselectivity in the Diels-Alder reaction

Both the explanation of secondary orbital interactions (SOI) and extra configurational strain in exo through transition states (and along reaction coordinates overall) result in endo being favored regularly. Diels-Alder's "endo rule" was born from experiment, and is also not actually a rigid "rule" (see tuning endo/exo selectivity in the two papers, and in the stack exchange related question by @ron), and is only intended as a tool in predicting expected products. Fernandez & Bickelhaupt's work is probably the most thorough description to date, supporting expt & previous theory to some degree.

What is the endo-selectivity of Diels-Alder reactions?
If the diene used in the Diels-Alder reaction has asymmetric substituents at the end carbons, and if the dienophile is unsymmetric, then two different isomers of the final adduct can form. The isomer, where the functional group(s) (usually carbonyl) on the alkene end up on the same side as the newly formed double bond, is called the endo-isomer [1]. The other one is the exo isomer. The naming of the isomers can be a little ambiguous in cases where there are multiple functional groups of different types on the starting alkene. Usually the electron-withdrawing group is considered when identifying which isomer is endo and which is exo.
 [2]
The experiments tell us that, in general, the endo-product is the one which is favoured kinetically i.e. it forms faster. This is not a rule as such, because there is no way to predict to what extent the endo product will be favoured. (In many cases, the exo product is the thermodynamic product i.e. more stable, so allowing the reaction to equilibriate will result in higher yield of exo product)
Why does this happen?
It's difficult to answer this question definitively, because there is still a lot of active research going on in the area. I will try to summarise the main effects which have been proposed as possible causes of the endo-selectivity.

Secondary orbital interaction

The interaction between the frontier molecular orbitals at the points where the new $\sigma$-bonds form are the first-order interactions. However, there are other interactions possible between the diene and dienophile, there are called second-order interactions, or secondary orbital interactions (SOI). This explanation was proposed by Woodward and Hoffmann, in the book The conservation of orbital symmetry[3].
The following diagram will be useful:
 [4]
As you can see, in the transition state leading to the endo-product, the carbonyl part of the alkene can interact with the back of the diene, and the interaction is favourable, as the symmetry matches. In the transition state leading to the exo-product, this interaction would not be available, as the carbonyl group is away from the diene. For a long time, SOI was considered as the right answer, and was extensively invoked in explaining not just Diels-Alder reaction, but other types of cycloaddition reactions as well.
However, there are a lot of evidences, both experimental and computational, which calls into question the validity of SOI. There is a good paper from Salvatella et.al. summarising them: "Do  Secondary  Orbital Interactions  Really  Exist?" Acc. Chem. Res., 2000, 33, 658.
For example, the dimerization of butadiene was shown to have slight endo preference.
 [5]
This was frequently used as an example of the SOI effect. However, computational studies at RHF and CASSCF level have found that the reaction proceeds via a stepwise mechanism. [6] If the mechanism is stepwise, not concerted, then it does not go through the constrained transition state which is required for the secondary orbital interactions to come into existence. So, we cannot invoke SOI for this endo-preference.

Steric effects

In certain Diels-Alder reactions, especially those involving the cyclopentadiene, steric effects have been proposed.
 [5]
Experimental data shows that when $\ce{R}$ = $\ce{CH3}$ and $\ce{X}$ = $\ce{CHO}$ the endo product is formed in $17.0%$ yield, i.e. the exo product now dominates. [7] This indicates that the steric repulsion of cyclopentadiene ring with $\ce{CH3}$ and the  is greater than that with $\ce{CHO}$, which is consistent with $\ce{CH3}$ being bulkier.
These observations can also be explained by SOI, but the trends in the endo-selectivity for a range of $\ce{R}$ and $\ce{X}$ groups seem to fit better to the steric argument.[5]
In a more recent study by Fernandez and Bickelhaupt, the selectivity was found to arise from activation strain, i.e. steric strain as the reactants approach the transition state.[12] They used $\mathrm{M05{\text -}2X/def2{\text -}TZVPP}$ to model the reaction of cyclopentadiene and maleic anhydride.

The above image, taken from their paper, shows the full reaction coordinate, with ZPE-corrected energies in kcal/mol and distances in Å. As you can see, the exo-reactant complex (exo-RC) is higher in energy than the endo one, suggesting that the strain comes into existence even before the TS. The extra strain in the exo-pathway is thought to arise from the steric repulsion between the methylene of cyclopentadiene and the oxygen of the anhydride.

Hydrogen bonds

Hydrogen bond-like interactions between C-H in alkene and the double bond has also been proposed as a cause of endo-selectivity in particular cases, such as the reaction of cyclopropene and butadiene.[8] Here, both steric effects and H-bond like interactions are thought to be present.
 [5]
This is somewhat different than the SOI interaction, because in SOI, only orbital overlap (delocalization) is considered. H-bond interactions are a sum of orbital overlap and electrostatic attraction.

Electrostatic forces

Electrostatic forces can be used to explain the unusual exo-selectivity of the furan+cyclopropenone addition.

In the exo-TS, there is a favourable electrostatic interaction between the electronegative O of furan and the electropositive carbonyl carbon of cyclopropenone. This makes the exo-isomer the preferred one. The exo-TS is about 1.81 kcal/mol lower in energy than the endo-TS, according to computational studies. [9]
The usual endo-selectivity of other Diels-Alder reactions can also be explained by the electrostatic repulsion between slightly electropositive terminal C-H of the diene, and the electropositive carbon of the electron-withdrawing group on dienophile:
 [5]

Solvent effects

Sustmann and others carried out computational studies on Diels-Alder reactions in both gas phase and with dielectric-continuum solvation.[10] They found that in the gas phase, the exo-TS is lower in energy i.e. kinetically favoured. Using a solvation model resulted in the endo-TS being the favoured one.
They suggested that solvent effect might be the cause of endo-selectivity. The endo-TS is generally more polar than the exo-TS, so it is stabilised to a greater extent in solvent. It is also observed that the preference for endo-TS increases the more polar the solvent is, which lends support to this argument. The gas phase exo-preference would also disprove the role of SOI effects, because they should be present regardless of solvation.
However, computational studies like these have to be considered carefully, because they were mostly carried out using HF or B3LYP methods, both of which severely underestimate dispersion forces (which are necessary to describe SOI). So, the results might be artificially skewed in the direction of exo-TS. [11]
Conclusion
All of this means that we are not really sure if there is a single cause that can explain the endo-selectivity in Diels-Alder reactions.
In an extensive experimental-computational study , Paddon-Row et. al.(2020) studied reactions of isotope labelled butadienes with mono- or di-substituted alkenes.[11] Surprisingly, it was found that mono-substituted alkenes (such as methyl acrylate) did not have a strong endo-preference when the reaction was done in benzene solution. Methyl acrylate and butadiene, for example gave 50:50 exo-endo product yield. Di-substituted alkenes showed higher endo-selectivity. In their analysis they suggested that SOI effects are important in some cases, whereas steric effects are forcing the selectivity in some other cases.
So, it might just be that different effects are working in different reactions, and somehow the end result is that endo-product is favoured in almost all cases. Then it might not be useful to look for one particular blanket cause of the selectivity, rather, each reaction has to be explained uniquely.

References
(1) Organic Chemistry, Clayden, Greeves, Warren, 2nd ed., Oxford University Press, p.884
(2) Image source: https://employees.csbsju.edu/cschaller/Reactivity/pericyclic/peri%20endo.htm
(3) The conservation of orbital symmetry, R B Woodward, R Hoffmann,   Weinheim : Verlag Chemie GmbH, 1971
(4) Image source: https://www.cpp.edu/~psbeauchamp/pdf/316_MO_DA_rxn.pdf
(5) Image source: Acc. Chem. Res., 2000, 33, 658
(6) Y. Li and K. N. Houk, J. Am. Chem. Soc., 1996, 118(4), 880–885
(7) Y. Kobuke, T. Fueno, J. Furukawa, J. Am. Chem. Soc., 1970, 92, 6548-6553
(8) M. Sodupe,R. Rios,V. Branchadell, J. Am. Chem. Soc., 1997, 119, 4232-4238
(9) S. M. Bachrach, J. Org. Chem., 1995, 60, 4395-4398
(10) T.  Karcher,  W.  Sicking, J. Sauer and R. Sustmann, Tetrahedron Lett., 1992, 33, 8027–8030
(11) W. J. Lording, T. Fallon, M. S. Sherburn, M. N. Paddon-Row, Chem. Sci., 2020, 11, 11915–11926
(12) I. Fernández, F. M. Bickelhaupt, J. Comput. Chem., 2014, 35, 371– 376. DOI: 10.1002/jcc.23500

This looks very promising :-) Let me nitpick, while you are still editing it, and I will read the full thing when you are done. You should use $\ce{R} = \ce{CH3}$ $\ce{R} = \ce{CH3}$ to indicate that R is equal to CH3. When you write $\ce{R=CH3}$ $\ce{R=CH3}$,  what you are saying (and what it displays as) is that R is doubly bonded to CH3.

Bickelhaupt and Fernandez certainly do excellent research, but I really cannot get behind their strain model. I never was a fan of it and none of their articles have convinced me. In any case, that is only my opinion, please add the level of theory at which they obtained their profile. I expect it to be high, but it should be included anyhow. Knowing them, they probably use the ancient kcal/mole, where 2 is somewhat significant, but it should be noted somewhere, because 2 kJ/mol is certainly within the error margins of any method.

From what I know, there reason why the endo product is preferentially formed is still not clear and subject to further enlightenment. But it seems to arise from more efficient interactions between the pi electrons of the dienophile and the pi electrons of the developing double bond in the diene during the transition state. In this website you can the both transition states in 3d
http://www.chemtube3d.com/DAendo_vs_exo,cyclopentadiene_and_maleic_anhydride.html


text:
Structure of the bifluoride anion
I came across this structure of $\ce{HF2-}$ and realized that the bond shown between H and F is a hydrogen bond.
$$\Large\ce{[F-H-F]-}$$
HF hydrogen bond are the strongest hydrogen bonds we know, but still they are not real bonds.
Writing the molecular formula as $\ce{HF2-}$ will give anyone an idea that the bonds between H and F are only covalent bonds.
Why do we write the molecular formula of $\ce{HF2-}$ as this and can you throw more light on its structure?
Have I correctly understood the structure?

How did you get the idea that those are hydrogen bonds?

Hydrogen cannot make two covalent bonds since it has only one orbital in which an electron is present. Looking at the structure, it was easy to say that the other bond must have been hydrogen bond. Hydrogen bonds don't require orbital overlap between the two concerned atoms, they are more of electrostatic interactions caused by the difference in electronegativity.

http://chemistry.stackexchange.com/questions/8549/existence-and-stability-of-bifluoride-ion

Hydrogen bonds have a significant covalent component. That's why they're so strong.

There is no such distinction as "real" vs "not-so-real" bond. Hydrogen bonds are as real as any other bonds (although weaker than many), and this particular one, as you already know, is the strongest of all hydrogen bonds.
As to why we write the formula this way, the answer is simple: because it is this way. We fool around with some fluoride compounds, and then we find out this one; we analyze it, and it seems to contain K, H, and F, and the composition turns out to be $\ce{KHF2}$. Before we know anything about hydrogen bonding, or chemical bonding whatsoever, we already have the reason to write the formula the way it is. Then X-ray crystallography comes along, so we can actually look at it, and see it is really this way; it really does contain those $\ce{HF2-}$ anions.

I thought we say hydrogen bonds are not real bonds because they are not formed by orbital overlap like other covalent bonds, they are a result of electrostatic interaction between some highly electronegative atoms and hydrogen...so i was not very convinced with the idea that these electrostatic interactions could result in real bond formation...

We also observe hydrogen bonding in HF molecule, but we dont consider that hydrogen bonding in the actual composition of HF. This does not seem intuitive to me, maybe because its the first time i am actually seeing hydrogen bond being a part of molecular formula of a compound.

Hydrogen bond, as well as covalent bond or any other kind of bond, is never a part of molecular formula. Again, we write the formula this way because the compound is composed this way, even before we know anything about bonding.

Quoting from a comment of your’s:

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

This sentence is correct for simple two-electron-two-centre bonds; i.e, if each of the two bonds going to hydrogen were to represent a 2e2c bond, hydrogen would have a double negative charge.
However, the $\ce{HF2-}$ anion does not contain two 2e2c bonds. Rather, it should be understood as a 4-electron-3-centre bond, similar to the situation in $\ce{ClF3}$. This can be understood by the following graphic:

Figure 1: Scheme of the 4-electron-3-centre bond in $\ce{HF2-}$. Image taken from Professor Klüfers’ web scriptum of his general and inorganic chemistry course.
The lowest molecular orbital, $\sigma_1$, is a fully bonding $\ce{F-H-F}$ orbital with two electrons. The second, $\sigma_2$ is nonbonding; no (accessable) hydrogen orbital exists that could complete it. The final one, $\sigma_3$, is fully antibonding. This is similar to the orbital scheme of extended linear π systems, where the number of nodal planes increases by 1 with every higher orbital. And it should be understood as such: the nodal plane of $\sigma_2$ traverses the hydrogen atom. Thus, while formally nonbonding, $\sigma_2$ is actually weakly bonding for both $\ce{H-F}$ bonds. If we calculate the overall bond order, we still count the electrons in $\sigma_2$ as nonbonding, giving a bond order of $0.5$ for each $\ce{H-F}$ bond.
The molecule can be understood in 2e2c Lewis terminology by using the following two resonance structures:
$$\ce{F-H\bond{...}F- <-> ^-F\bond{...}H-F}$$
‘Multiple bonds’ going away from a single hydrogen is not restricted to the 4-electron-3-centre bond known in $\ce{HF2-}$; they are also known and well-studied in boranes, although they are typically 2-electron-3-centre bonds in those electron-deficient compounds.

Re: "Rather, it should be understood as a 4-electron-3-centre bond, similar to the situation in $\ce{ClF3}$" But, sir, in $\ce{ClF3}$, we can "excite" the  p-electrons of chlorine atom into its vacant d-orbital. (I am using VBT) This makes three unpaired electrons in chlorine's orbitals, which can then form covalent bonds with the three fluorine atoms. I was taught that such an excitation in not possible in the fluorine atom. Then, how can you say that $\ce{HF2-}$ has a 4c-3e bond similar to $\ce{ClF3}$, when such an excitation is not possible in the former case? Thanks for your answer!

@GaurangTandon Surely you can excite an electron into chlorine’s 3d orbital—if you blast a significant amount of energy (probably visual light range) onto it. However, the d orbitals have been shown to not take part in bonding at all because of how much higher their energy is (remember that even 4s comes before them according to the aufbau principle). There are enough answers on this site that debunk this myth.

Understood. You meant the hypercoordinated valency, while I was confusing myself with the traditional VSEPR $sp^3d$ thing. Thanks!

I have drawn a (not so neat) diagram in ChemSketch - is this a valid Lewis diagram for $\ce{HF2-}$? I have based it on $\ce{ClF3}$'s structure. Three lone pairs are in the $sp^2$ hybrid orbitals of $F$ atom, and the two $H$ atoms in its pure $p$-orbital. Is it correct?

@GaurangTandon No, fluorine should be unhybridised. One s-type lone pair, two p-type lone pairs. ther than that, it’s okay.


text:
Why is the melting point of hydrogen iodide larger than that of hydrogen fluoride?

Compare the melting points of hydrogen fluoride (HF) and hydrogen iodide (HI).

I know the following things:

$\ce{HF}$ melts at $189.6~\mathrm{K}$ and $\ce{HI}$ at $222.35~\mathrm{K}$.
There are a lot of factors that have to be considered while comparing melting point, i.e., lattice enthalpy, packing fraction, intermolecular forces.
Generally, compounds that show hydrogen bonding show it in the solid and liquid phases because in the gaseous phase the molecules have enough energy to break the electrostatic interactions of hydrogen bonding.
So, HF shows hydrogen bonding in solid state. In fact, it even shows hydrogen bonding in gaseous state, though its not relevant here.
As molecular mass increases, van der Waals interactions between molecules also increase. If we only consider this factor, then melting point of HI must be definitely higher.
The strength of hydrogen bonding of HF is so much that the boiling point of $\ce{HF}>\ce{HI}$ although boiling point of $\ce{SbH3}>\ce{NH3}$, i.e. the hydrogen bonding dominates the van der Waals interactions in HI molecules because of their high molecular mass.
It's the combined effect of all the factors that determine the melting point of a substance.
Lattice energy is inversely related to the internuclear distance, it is also inversely proportional to the size of the ions. So, I expect HF to have more lattice energy.

Every factor that I can think of is in favour of HF except molecular mass. But if molecular mass of iodine is the reason that melting point of HI is higher, than boiling point of HI must have also been higher. Then why is the melting point of HI higher than that of HF?
Why is this contradiction?
Do let me know if I am wrong in the above mentioned facts or if I have any gaps in my understanding of the above points.

The increase in dispersive attraction between HI molecules is greater than the loss of dipole interactions. Only in $\ce{H2O}$ do dipole forces dominate (Intermolecular and Surface Forces, Jacob N. Israelachvili). Contribution of dispersion to intermolecular bonding for various molecules:

$\ce{CH4-CH4}$ - 100%
$\ce{HI-HI}$ - 99%
$\ce{HBr-HBr}$ - 96%
$\ce{HCl-HI}$ - 96%
$\ce{H2O-CH4}$ - 87%
$\ce{HCl-HCl}$ - 86%
$\ce{CH3Cl-CH3Cl}$ - 68%
$\ce{NH3-NH3}$ - 57%
$\ce{H2O-H2O}$ - 24%

What makes the contribution of dispersion so great is that it's non-directional and always attractive. Think of atoms as sticky spheres.

Why H20 serves as an exception?

1:1 ratio of H-bond donors and acceptors means much stronger attractive forces and with H and O being small atoms, the dispersive forces are weak.

Like you said many factors contribute to melting points. It is an interesting trend, that melting points increase moving down while boiling points increase. I'd be interested to know how the data was measured.
One thing you did not take into account is polarizability of the larger halides which will greatly influence inter and intra molecular forces. If I can think of a better answer I'll post.