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Predicting the products of reaction between ammonia and hydrogen peroxide
Like we do in organic chemistry is there any way to write the mechanism of the reaction between $\ce{NH_3}$ and $\ce{H_2O_2}$?
Can the products be predicted?

Majority of answer content came from this link and this link.
Strong solutions of $\ce{H2O2}$ with a few drops of $\ce{NH4OH}$ (aq. ammonia) is left for 24 hours without any nitrite formation occurring. But upon longer standing, even with a small amount of hydroxide then nitrite forms. Nitrite also forms when a dilute solution of $\ce{H2O2}$ is mixed with $\ce{NH4OH}$  and is evaporated over pure conc. $\ce{H2SO4}$ with a bell jar. $\ce{H2O2}$ forms (even in very dilute solutions) nitrite very rapidly, if the $\ce{H2O2}$ solution is mixed with a few drops of $\ce{NH4OH}$ and a little $\ce{NaOH}$  and this then boiled in a retort to a very small volume.
$\ce{2NO2- +  2H+(aq) <=> H2O + NO + NO2}$
$\ce{NO2- + NO3- + 2H+(aq) <=> H2O + NO2}$
Nitrogen dioxide is then able to attack ammonia.
$\ce{2NO2 + 2NO + 4NH3 -> 2NH4NO2 + 2H2O + 2N2~~~~~~~~~~~~~~~~~~~~(1)}$
In any case, solutions of ammonium nitrite decompose on heating to 60 to 70C.
Perhaps the very slow spontaneous reaction of aqueous ammonia with $\ce{H2O2}$ takes place the small equilibrium with amide anions, $\ce{NH2-}$.
$\ce{NH2- + H2O2 -> NH2- + 2OH-}$
Even in $\ce{NH4OH}$, there is a very slight equilibrium with amide ions.
$\ce{2NH3 <=> NH2- + NH4+}$
Amide rapidly and vigorously reacts with water, so the equilibrium of amide existence in water must be very low.
$\ce{NH2-+ H2O -> NH3 + OH-}$
In the base catalysed decomposition of hydrogen peroxide, the mechanisms is presumably:-
$\ce{H2O2 + OH- -> HOO- + H2O}$
$\ce{HOO- + H2O2 -> HOOOH + OH-}$
$\ce{HOOOH --> HOOO- + H+(aq)}$
$\ce{HOOO- -> OH- + O2}$
Also, ammonium nitrite decomposes into nitrogen and water.
$\ce{NH4NO2 → N2 + 2 H2O~~~~~~~~~~~~~~~~~~~~~~~~~~(2)}$
A combination of ammonia and hydrogen peroxide is often used in the process of bleaching hair. Hydrogen peroxide has the necessary properties to remove the color from hair - it oxidizes one of hair’s melanin pigments to a colorless substance. This results in the hair becoming lighter. However, it is not recommended by hair professionals (those who don't understand chemistry but want your cash), and can cause extensive damage to hair (even causing it to fall out).
Therefore, the overall reaction is:-
$\ce{2(NH3•H2O) + 3H2O2 ->[\Delta] N2 + 8H2O~~~~~~~~~~~~~~~~~~~~~~~(1) + (2)}$

As far as I can tell, the prior answer while posting several well known reactions, did not actually detail how nitrite is introduced.
To be honest, this is not an easy question to answer and as far as I know, even  the pundits have not agreed on the precise pathway. I suspect the actually kinetics starts with the action of light on an alkaline solution introducing the hydroxyl radical and possible solvated electrons:
OH-(aq) + hv --> •OH + e-(aq)  (see ‘Flash photolysis in the vacuum ultraviolet region of sulfate, carbonate, and hydroxyl ions in aqueous solutions’ by Elie Hayon, and John J. McGarvey, in J. Phys. Chem., 1967, 71 (5), pp 1472–1477, DOI: 10.1021/j100864a044 , link: https://pubs.acs.org/doi/abs/10.1021/j100864a044?journalCode=jpchax )
This could be followed by some radical reactions:
NH3 + •OH --> •NH2 + H2O  (See Eq (1) at "Kinetics and Mechanism of the Reaction of •NH2 with O2 in Aqueous Solutions", by B. Laszlo , Z. B. Alfassi , P. Neta , and R. E. Huie, J. Phys. Chem. A, 1998, 102 (44), pp 8498–8504, DOI: 10.1021/jp981529+, July 15, 1998, at https://ws680.nist.gov/publication/get_pdf.cfm?pub_id=831494 )
•NH2 + O2 --> •NH2O2  (Same source, Eq(2))
O2 + e-(aq) --> •O2- (Same source, Eq(3))
•NH2O2 + •O2- --> O2 + NH2O2- (Same source, Eq(7))
And also, some standard chemical reactions:
NH2O2- --> NO- + H2O (Same source, Eq(8))
NO- + H2O2 = NO2- + H2O  (Speculation, as NO- with O2 (not H2O2) forms peroxynitrite, but NO2- + H2O2 --> peroxynitrite, see http://www.jbc.org/content/289/51/35570.full.pdf, so NO- + 2 H2O2 = (NO2- +
H2O2) + H2O = aqueous peroxynitrite is a supportive argument for the speculated reaction as it suggests a consistent reaction system!)
NH4+ + NO2- = NH4NO2 .
The source of oxygen above follows from the instability of alkaline H2O2 to self-decompose through various pathways or from the presence of air in an open vessel. If dust (which can contain transition metals like iron,..) is allowed to enter the solution, a possible Fenton (or Fenton-like with Cu, Mn,..) reaction could proceed producing a source of hydroxyl radical (in place of the action of light) although this suggested path would be hindered by the addition of OH-, so not my preferred likely explanation.


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What is the mole fraction of dihydrogen in a mixture, given the relative rates of effusion of its components?

A mixture of $\ce{H2}$ and $\ce{He}$ at $\pu{300 K}$ effuses from a small hole in a gas chamber. What is the mole fraction of $\ce{H_2}$ in the original mixture, if $3.0$ times as many $\ce{He}$ atoms as $\ce{H2}$  molecules escape from the hole in one minute?

The answer apparently relies on the fact that $\ce{H2}$ collides with the walls $\displaystyle\frac{1}{\sqrt2}$ as often as $\ce{He}$, and we therefore need $\sqrt{2}$ as many $\ce{He}$ atoms as $\ce{H2}$  molecules to yield an equal amount of substance escaping. Considering that the ratio is in fact 3, the final answer is then that the mole fraction of $\ce{H2}$ is
$$x(\ce{H2}) = \frac{1}{3\sqrt{2} + 1}.$$
Can anybody please explain how they got the $\displaystyle\frac{1}{\sqrt2}$ part?

If you are familiar with Graham's Law, you will know that the ratio of the rates of effusion of two gases is inversely proportional to the square root of the ratio of their molar masses.  That is:
$$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$$
Here, $M_1 = M(\ce{H2}) = \pu{2 g//mol}$ and $M_2 = M(\ce{He}) = \pu{4 g//mol}$, so the lighter of the two, hydrogen, effuses at $\sqrt{2}$ the rate of helium, or conversely, the heavier, helium, effuses at $\frac{1}{\sqrt{2}}$ the rate of hydrogen.  That's where the fraction came from.
Since we know that helium is escaping three times faster than hydrogen, there must be three times more helium than would product an equal rate of flow for the gases.  An equal flow would occur when there is $\sqrt{2}$ times as much helium, and triple flow would occur with $3\sqrt{2}$ times as much helium.
The total amount of gas present is then:
$$\ce{H2} + \ce{He} = 1 + 3\sqrt{2}$$
and the mole fraction of hydrogen is:
$$x(\ce{H2}) = \frac{n(\ce{H2})}{\text{total amount of substance}}
= \frac{1}{3\sqrt{2} + 1}$$

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By which mechanism would azide react with alcohols or alkyl halides?

$\ce{NaN3}$ produces $\ce{N3-}$ ions. So the reaction with an alkyl halide will either be $\mathrm{S_N1}$ or $\mathrm{S_N2}$.  After some reading, I found that $\ce{N3-}$ is a good nucleophile. Hence the $\ce{S_N2}$ mechanism is followed.
But if I use $\ce{HN3}$ on an alcohol, the $\ce{H+}$ ions produced will help in removing $\ce{OH-}$ as water forming a carbocation. So either $\mathrm{E1}$ (if the alcohol is tertiary) or $\mathrm{S_N1}$ (if the alcohol is primary or secondary) can take place.

Is this correct?

SN1 on a primary alcohol? Why can't you just have SN2 of azide ion with water as the leaving group?

This is what my Teacher told. I have seen reaction in SN2 where H2O leaves since OH- is a weak leaving group.

The first point seems correct though.

No primary carbocations!!! I really want to shout this at every first year organic chemistry student. You could get $\mathrm{S}{N}1$ or $\mathrm{E}1$ for a tertiary alcohol. The primary alcohol might undergo $\mathrm{S}{N}2$, but there's a good chance that it will be a bad reaction. Your best bet would be to convert the primary alcohol into a tosylate.

After some reading, I found that $\ce{N3-}$ is a good nucleophile. Hence the $\mathrm{S_N2}$ mechanism is followed.

Unfortunately, it is not as simple as that. Whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ is followed, depends on both partners but is more a function of the electrophile than one of the nucleophile. Typically, primary halides will undergo $\mathrm{S_N2}$ substitutions while tertiary ones will undergo $\mathrm{S_N1}$. Secondary ones can do both. However, note that secondary halides that are not 2-halopropanes will undergo Wagner-Meerwein rearrangements if a carbocation is formed.
In an $\mathrm{S_N1}$ reaction, the rate-determining step is the formation of the carbocation, hence weaker nucleophiles can react equally rapidly as stronger ones. In the $\mathrm{S_N2}$ reaction, the nucleophilicity of the nucleophile directly determines the rate of reaction, i.e. both partners are important. Hence these tend to prefer better nucleophiles.

But if I use $\ce{HN3}$ on an alcohol, the $\ce{H+}$ ions produced will help in removing $\ce{OH-}$ as water forming a carbocation. So either $\mathrm{E1}$ (if the alcohol is tertiary) or $\mathrm{S_N1}$ (if the alcohol is primary or secondary) can take place.

Again, primary alcohols will never react by an $\mathrm{S_N1}$ mechanism and secondary ones are very unlikely to do so. The $\mathrm{p}K_\mathrm{a}$ value of $\ce{HN3}$ is $4.72$ — which means that alcohols will practically not be protonated (you need strong acids for that, e.g. $\ce{HBr}$).
I would assume that $\ce{HN3}$ will not react with alcohols at all since it is too weak to protonate them. There will also be no elimination side reactions neither by $\mathrm{E1}$ nor by $\mathrm{E2}$ — a hydroxide ion will never be a leaving group in itself.

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Does resorcinol give positive iodoform test?
It's given in my book that other than the usual compounds that answer iodoform test, resorcinol (benzene-1,3-diol) also gives positive iodoform test.
Is this because of tautomerization which gives a diketone with an active methylene group? But why should a ring lose its aromaticity by doing so?

There is an equilibrium between the keto enol forms, however small the keto form may be as a contributor. The rest is as you say. The precipitation of iodoform drives the equilibrium further to the products, as per Le Chatelier's principle.Of course, it is not in all cases where the equilibrium is controlled by such factors. The net gibbs energy change for the reaction must be negative.

Speculation aside, neither resorcinol nor phloroglucinol give positive iodoform tests. Fuson, R.C. and  Tullock, C.W., J. Am. Chem. Soc., 1934, 56, 1638.

@user55119 Note that the conditions used by the experimenters in the study are in fact quite mild, i.e. 2 min at 60 deg C. The following study which was also not done under vigorous conditons, at temperature of 20 deg C. However, the experimenters waited long enough for iodoform reaction to occur for resorcinol to give a positive test. Hence, they concluded that resorcinol does give a "slow test". Seelye, R. N., & Turney, T. A. (1959). The iodoform reaction. Journal of Chemical Education, 36(11), 572. doi:10.1021/ed036p572

Tan Yong Boon: Yes, if you pound on the substrate, I'm not surprised. Marrian (1932) treated estrone with KOH/I2 to obtain the dicarboxylic acid from the D-ring. Heer and Miescher (Helv. Chim. Acta, 1945, 28, 156) treated estrone benzyl ether with methanolic KOH/I2 at 80 oC to obtain O-benzyl marrianolic acid. Although the reaction presumably goes through the diiodoketone, I would be reluctant to call this reaction or that of resorcinol "haloform tests".

Yes, resorcinol gives iodoform test.
This is due to the fact that a keto- form of resorcinol exists (via tautomerism as suggested by the original poster), which contains an active methylene group in a cyclic compound, this being a positive condition for the iodoform test.
However, the reaction will be slower than expected and is only driven forward by Le Chatelier's principle which states that precipitation drives a reaction in the forward direction.
Any question in theoretical organic chemistry asking whether Resorcinol gives the Iodoform test will always specify the rate (eg. A question with the statement: Which of the following gives iodoform test instantly? has Resorcinol being a wrong answer).

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Elimination reaction on 2-fluorobutane using alc.KOH
Question: Find the major product formed when 2-fluorobutane reacts with alcoholic $\ce{KOH}$
My answer: Elimination of $\ce{HF}$ takes place and a more substituted alkene should have formed. (Zaitsev product). I think the mechanism is E2 since strong base is used.
but answer given is a less substituted 1-butene. Is the answer wrong or am I making any mistake?

E1CB elimination takes place because of a poor leaving group. This is because of the high $\ce{C - F}$ bond strength. Stronger bonds are more difficult to break, making them less reactive.
If the reaction was E2, then zaitsev product would have been formed. But since the mechanism is E1CB, the reaction proceeds via carbanion.

From wikipedia: The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favor an E1cB mechanism.

I hope you know that for carbanions, the stability order is primary > secondary > Tertiary.
Now that you know the stability order of carbanions, the preferred position for carbanion is a primary carbon. So base abstracts a proton from a primary carbon.

This answer is, unfortunately, not correct. There is no reliable experimental evidence to demonstrate that a primary carbanion is more stable than a secondary carbanion. In fact, the opposite is true (JACS 1989, 1968). And neither carbanion would be stable enough to be formed.
So, what's the answer? I would honestly say that no-one knows for sure. High level gas phase calculations reproduce the experimental outcome, so it's unlikely to be due to solvation.
It's a little unfortunate that this has become a textbook example, since the textbook explanations do not withstand scrutiny. Please understand, I am not criticising the person who responded. It just seems that this reaction is a bit more complicated than it seems at first, and (in my opinion) it needs a bit more work to understand it.
Sorry if this response is not very helpful, but it's better to tell the unpalatable truth.


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overall energy and the boiling point
i need help with the fractional distillation method of a couple of substances at least more than two substances , just like air actually what am i looking for is that why do substances in air with a low boiling point like nitrogen reach higher the fractionating column even if it looks like that substances with higher boiling point will absorb more energy and thus eventually result in greater kinetic energy (at least to me). please help . heres how i came to know that nitrogen rises higher the fractionating column

Distillation relies on the fact that, at the same stage (tray), all the substances are approximately at the same temperature. Then, at the same temperature, the vapour will be composed preferably by substances with a lower boiling point and the liquid composition will be dominated by higher boiling point compounds.
Very quickly, distillation works as follows. The vapour flow from one stage (i) is then richer in the light compound (with lower boiling point). That vapour is enters the immediately higher stage (i-1) making the composition richer in the light compound. The vapour from stage i-1 is even richer in the light compound. This process is repeated until you reach the top stage where the light compound can be almost pure. Likewise, the liquid of stage i is rich in the heavy compound and goes down to stage i+1. The liquid leaving stage i+1 is even richer in the heavy compound. At the bottom, you can reach and almost pure heavy compound.
Formally, the fractionation of a mixture is dictated by the relative volatility of the compounds (which for ideal solutions does not vary along the column). At low/moderate pressures and for an ideal solution, the relative volatility is simply
$$
\alpha =\dfrac{\dfrac{y_1}{x_1}}{\dfrac{y_2}{x_2}} = \dfrac{P^s_1}{P^s_2}
$$
We usually take the heaviest compound as the bottom line for comparison. In your example, at the pressure and temperature conditions of the column, the relative volatility of nitrogen/oxygen is higher than argon/oxygen. Therefore, nitrogen leaves the column at the top, argon at an intermediate stage and oxygen at the bottom.

that hopefully explains why at the same stage(tray) the substance with lesser boiling point dominates . but why do the higher melting points condense at a particular tray and the lower melting points rise higher ?
thanks in advance

I have edited my answer to respond your query

thanks , it explains everything now
is this your own opinion or have you taken it from a source
please mention if there is any
i am asking you this because i had a similar post and someone posted some diffrent explanation. i would had provided th link but the post is gone now

No, this is not my own opinion. You can take any textbook on distillation and it will say the same. I would recomend "Separation Process Principles" by Seader and Henley; "Separation Process Engineering" by Wankat

one second
i pondered over your explanation and it suggests that if we achieve the boiling points of any two of three liquids in diffrent containers , then in a specific interval of time all should have achieved the same height ? but we know that this is not the case ! please explain

I don't understand your remark. Achieve boiling points? Achieved the same height?

actually what i am saying is that liquids with diffrent boiling points if kept in diffrent containers and then  beginned to heat till they convert into gases will then the gas of the liquid which had higher boiling point travel farther or the gas with the liquid having lower boiling point ?

The gases in the distillation column boil off according to their boiling point.
   Gas          Boiling point
   Nitrogen     77.355 K (−195.795 °C)
   Argon        87.302 K (−185.848 °C)
   Oxygen       90.188 K (−182.962 °C)

The distillation column shown is setup for continuous distillation. But imagine that a pot distillation was being done. Nitrogen would boil off first, then argon, then oxygen. So the various trays in the distillation column represent different temperatures. Nitrogen boils off at the top since it has the lowest boiling point, and oxygen collects in the "pot" since it has the highest boiling point.

that answered my whole question , thanks !
actually i did not knew that the trays in distillation flask are not of the same trempreature throughout .
But what about a fractionatig column with marbles in it ? it has no trempreature diffrences

The marbles would act as the trays. There has to be a temperature difference from the bottom to the top of the column or it isn't a "fractioning column."

but the laboratory setup for the marble distillation column does not show any trempreature difference !

In a typical pot chemistry experiment you heat the liquid in the pot and the vapor carries the heat up the column.

thanks !
for a second i thought that in the marble fractionating column
you need to collect the pure gas from below the container!
hopefully i was wrong , right ?


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If glucose can react with phenyl hydrazine to form osazone why can't it react with 2,4-DNP to give positive test?

If glucose can react with form phenyl hydrazine to form osazone why
  can't it react with 2,4-DNP?

Yes I know about cyclic form of glucose and its equilibrium with straight chain and all …
I read this, this and this Stack Exchange one too Does glucose react with Brady's reagent?
Still, didn't get an answer to my question. So requesting for clarification …

Glucose itself keeps interconverting between alpha beta glucopyranose and the chain form but it is more stable in this ring forms. A reagent should be strong enough to convert cyclic structure to the chain form so as to react with aldehyde group. I think due to electron withdrawing action of the Nitro groups in DNP they are less active/weaker than phenyl hydrazine and so DNP shouldnt convert glucose to chain form and react with the aldehyde. That's my explanation though.Someone might post a better answer.

Glucose converts itself to the chain form in appropriate solvents, doesn’t it?

@Jan I guess. When the reagent /solvent is highly reactive towards aldehyde group, it becomes chain.

Although glucose 1a exists principally in the hemiacetal pyranose forms (α and β), nonetheless it reacts with one equivalent of phenylhydrazine to form the hydrazone 1b via the aldehydic open form. Two additional equivalents serve to form the osazone and the products of reduction of phenylhydrazine, ammonia and aniline.The glucose 2,4-dinitrophenylhydrazone 1c has been reported1 but not the corresponding osazone. In their studies on the mechanism of osazone formation, Shemyakin and coworkers2 utilized p-nitrophenylhydrazine labeled with nitrogen-15 at the terminal nitrogen (the red N in the structures). Although glucose was not examined, the p-nitrophenylhydrazones  of fructose (forms the same osazone as glucose), benzoin and 2-hydroxycylohexanone were studied. The labelled nitrogen was located in the ammonia that was liberated during osazone formation.
Structures 2-7 represent the essentials of the mechanism.  Enolization of the hydrazone 2 produces the enol 3, which, via the zwitterion 4, undergoes oxidation of the carbon framework and reduction of the phenylhydrazine moiety. The transformation 3 --> 4 is essential. The basicity of the electron pair on the nitrogen in the 2,4-dinitrophenyl case is sufficiently reduced by delocalization into the aromatic ring, thereby repressing the proton transfer. This step must be completed before osazone formation can be accomplished.

1: Lloyd, E. A. and Doherty, D. G., J. Am. Chem. Soc., 1952, 74, 4214.
2: Shemyakin, M. M., Maimind, V. I., Ermolaev, K. M. and Bamdas, E. M., Tetrahedron, 1965, 21, 2771.


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How can the size of atom be the cause of octet expansion?
Lately, I was reading The Lewis Theory of Covalent Bonding by Peter Atkins in Appendix 4 of 'Elements of Physical Chemistry'. There he was talking about expansion of octet . As he wrote:

Many molecules cannot be written in a way that conforms to the octet rule. Those classified as hypervalent molecules requires an expansion of octet. Although it is often stated that octet expansion requires the involvement of d-orbitals, and is therefore confined to Period 3 and subsequent elements, there is good evidence to suggest that octet expansion is a consequence of an atom's size, not its intrinsic orbital structure. [...]

I had always bore in mind that octet expansion was always due to d orbitals.
But size of atoms? How can it be related to octet expansion?

Since the central atom in such compounds is coordinated to more groups than expected, but is not expanding its valence electron shell, the term hypercoordinated is generally preferred over hypervalent.

Your revised version of Atkins’ statement is correct, no hypervalent main group compounds require the involvement of d-orbitals. Rather, all of those molecules formerly explained by d-orbital participation can just as well be explained with 4-electron-3-centre bonds (including resonance structures) and charge separation (e.g. in sulphate; no need for resonance), giving rise to all-octet Lewis structures. In fact, calculations have shown the d-orbital participation to be minimal and neglegible at best.
A much more relevant question is whether more than $x$ substituents actually fit around the central atom. This is especially notable for the difference between the second and third periods: The reason why nitric acid is $\ce{HNO3}$ while phosphoric acid is $\ce{H3PO4}$ is likely due to the much smaller size of nitrogen, not allowing for tetracoordination by oxygens in the $\mathrm{+V}$ oxidation state while the larger phosphorus atom can support four oxygen atoms. If I remember correctly, that was also discussed as a reason why $\ce{NF5}$ hadn’t been found at the time of Klapötke’s article while $\ce{PF5}$ has been known for quite a long time.
It gets more pronounced when you go even lower in the periodic table, i.e. to higher periods. Antimony is known to be coordinated by six oxygens in the $\mathrm{+V}$ oxidation state ($\ce{H7SbO6}$, if you wish, although $\ce{H[Sb(OH)6]}$ is a more accurate description) — something phosphorus could technically do ($\ce{PF6-}$ with small fluorine atoms is known) but usually doesn’t. It is mainly the greater atom size of antimony that allows the coordination of two additional oxygen atoms.


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'In MO theory, electrons are treated as spreading throughout the entire molecule'- isn't it exhibited in VB theory also?

First statement:

In the molecular orbital theory, the valency electrons are considered to be associated with all the nuclei in the molecule. - Concise Inorganic Chemistry by J.D.Lee.

Second statement:

In molecular orbital theory, electrons are treated as spreading throughout the entire molecule: every electron contributes to the strength of every bond. - Elements of Physical Chemistry by Peter Atkins.

Two different sources are actually saying the same thing in regard to MO theory: electron density is spread over the entire molecule, okay that is quite right & agreeable.
But the way the fact is presented as if it were the sole property of MO theory. Doesn't the electron density span over the entire molecule in VB theory also?
Check this pic taken from Peter Atkin's above mentioned book:

This is the electron density in $\ce H_2$ as predicted by VB theory; notice how the resultant wavefunction after bonding is spread over the entire molecule & this implies the electron density is spread over the entire molecule. So, even in VB theory, the electron density is spread over the nuclei of the molecule. So, why those special writings above?
Why is it that the spreading of electron density over the nuclei of the molecules specially in MO theory? Is not this exhibited even in VB theory also? Or am I missing something?
Can anyone please help me whether the spreading of the electron density is the sole property of Mo theory as is put up by the two authors? Is not it exhibited in VB theory also?

Aren't your other questions asking the same thing? http://chemistry.stackexchange.com/q/33879/4945 http://chemistry.stackexchange.com/q/33951/4945

@Martin-マーチン: Same thing? How can they be?? Of course, they are related to each other, but how can it be the same thing?

@Martin-マーチン: At one, the difference between the two theories is highlighted while the later deals why resonance isn't required for MO theory. Here, I'm asking whether the property of spreading the electron density over the entire molecule is solely exhibited in MO theory or as I am saying, also exhibited in VB theory also. Of course, I would appreciate if you elaborate how the previous queries have been reflected again here.

To be clear: the electron density does spread over the whole molecule in both models.  As ifilot's answer describes, the difference is in whether all AO's potentially contribute to a given MO, or whether just AO's on neighboring atoms are used to construct that MO.

In MO theory, you construct a molecular orbital by a linear combination of atomic orbitals. Because those atomic orbitals are part of different atoms, in this way, the valence electrons are associated with all the atoms in the molecule. So even if you consider a molecular orbital that is bonding between, let's say, atom 1 and atom 2 in the molecule, that molecular orbital has also contributions from all the other atomic orbitals (besides those that arise from atom 1 or atom 2).
In contrast, in VB theory, a bond between atom 1 and atom 2 can only have atomic orbitals of atom 1 and atom 2. The bond between two atoms is then considered to arise from two weakly coupled atomic orbitals. I have to point out though that modern valence bond theory now complements molecular orbital theory and no longer adheres to the valence bond idea that electron pairs are localized between two specific atoms in a molecule but that they are distributed in sets of molecular orbitals which can extend over the entire molecule.

Given the discussion below, I would like to add to my answer a simple example. Let us consider the $\ce{H2O}$ molecule. Within the framework of Hartree-Fock and using a STO-3G basis set, the $2\sigma$ orbital of water can be expanded in terms of the atomic orbitals as follows:
$$
\psi_{2 \sigma} = (-0.4330)\; \phi_{\textrm{O},1s} + (-0.4686)\; \phi_{\textrm{O},2s} + (0.2369)\; \phi_{\textrm{O},2p_y} + (0.2685)\; \phi_{\textrm{H1},1s} + (0.2685)\; \phi_{\textrm{H2},1s}.
$$
From this expansion you can see that multiple atomic orbitals contribute to the same molecular orbital. This is a molecular orbital that forms, in a way, a multi-atom bond between O and the two H atoms. You could also say that this molecular orbital contributes to both of the O-H bonds (in fact, more molecular orbitals than just the $2\sigma$ do).

which do not arise from atom 1 or atom 2- what? Could you please clarify that?

I was not very clear here. I have edited my answer. I meant to say: "besides those that arise from atom 1 and atom 2"

You meant to say there are contributions of atomic orbitals which are not of 1 & 2?

Exactly! :-) A good book that treats electronic structure calculations within the framework of MO is Modern Quantum Chemistry from Szabo and Ostlund (http://store.doverpublications.com/0486691861.html). If you read through chapter 2, I think my point becomes clear.

I've not enough money to buy this book now. Could you please elaborate it a bit here? I never thought molecular orbital is formed by atomic orbitals not from the participating atoms!

Any molecular orbital within MO theory is a linear combination of atomic orbitals. In mathematical terms: $\psi_{i}=\sum_{j} c_{j} \phi_{j}$ where $\phi_{j}$ are the atomic orbitals, $c_{j}$ the corresponding coefficients and $\psi_{i}$ the molecular orbital. $j$ runs over all the atomic orbitals and not only the orbitals of the atoms to which the bond adheres. So for the description of a molecular orbital, you use the complete basis set that consists of all atomic orbitals in that molecule. Hence, all these atomic orbitals contribute.

Okay. But so far, I have read the examples presented by the above books I mentioned, only the atomic orbitals that are present in the atoms participate only in the formation of  molecular orbitals. Why is it so? Examples include: $\ce H_2, C_2, O_2, O_2^{-1},F_2$ etc.

@user36790 Because these examples are diatomic molecules--the only atoms the AO's can come from are necessarily participating in a bond. For something larger,  say, ethane, a given $\sigma$ bond MO may have contributions from AOs centered on atoms all over the molecule.

@ifilot Probably worth up-editing this discussion into your answer.

Indeed, I have expanded my answer. I  hope it becomes clearer this way.

Isn’t LCAO theory a subtheory of MO theory?

I am not quite sure. I think it is a matter of semantics. LCAO is an approximation used in MO theory, whereas MO theory by itself is a method to calculate electronic structure.

authors: Макс Харитонов


text:
Is rotation of a mirror image allowed in order to determine chirality? Doesn't rotation alter the mirror image of the molecule?
So, chiral molecules are those which have non-superposable mirror images.
trans-1,2-Dimethylcyclopentane is chiral; 2-butanol is chiral; both don't superpose on their mirror image.
Then my book cited 2-propanol as achiral. How?
The mirror image doesn't seem to superpose with the molecule. How can it be achiral then?
Then the book provided a pic where the author showed how 2-propanol is achiral; he rotated the mirror image & then superposed! Done! 2-propanol is achiral!
But really how can rotation be allowed? After all the mirror image is a stereo-isomer; if it is rotated, wouldn't it be changed?
I'm not getting, why rotation is allowed before superposition. Not every-time, the mirror image is rotated before superposition.
Can anyone please explain how rotation is allowed before superposition?

superposition is pertinent only to the chiral centers.

Molecules in the nature do rotate all the time, come to think of it.

As a rule of thumb, in order for a molecule to be chiral, it must have a chiral center. There are some exceptions. A chiral center is a generalized extension of an asymmetric carbon atom, which is a carbon atom bonded to four different entities.
Carbon-2 in 2-propanol is not a chiral center because it is only bonded to 3 unique substituent groups: ($\ce{-H}$), ($\ce{-OH}$), and 2 ($\ce{-CH3}$).

he rotated the mirror image & then superposed! Done!! 2-propanol is achiral!!

You seem to misunderstand. Mirror images of chiral molecules are non-superposable, and mirror images of achiral molecules are superposable (because they are the same molecule).
Rotating the molecule only means orienting a molecule differently in 3-D space, not changing anything about the molecule itself. I'd recommend playing around with a molecule building set to solidify this in your mind.

so we can rotate the mirror image at any angle and it won't make a difference,right?

https://chemistry.stackexchange.com/a/64170/15410

When checking out chirality or achirality you are always allowed and supposed to rotate the molecule freely in all dimensions. Essentially, what you are trying to prove, is that the mirror image of a molecule can be achieved by rotation alone.
The thought behind this is simple. Chirality is macroscopically proven by optical activity. If you reduce the experiment to its bare basics, you have a polarised light beam that passes through a solution and hits a single molecule to be deflected either to the right (dextrorotatory) or to the left (levorotatory). A second molecule it hits which has the same chirality will move it in the same direction by the same amount. If, however, it hits a molecule which is exactly the mirror image of the first, the effect will be cancelled and the plane of polarisation is moved back into the vertical. In a solution containing only one enantiomer of a chiral substance, the beam will never hit a mirror image and thus always be deflected in a certain direction.
If we continue this thought experiment, we realise that molecular motion in solution is random. Thus, a molecule can — if we give it conciousness — move around freely and let it try to assume exactly the mirror image of the previous one. Therefore, we are allowed and encouraged to rotate (but not to use another mirror image or inversion!) our molecule until it ‘fits’.
By the way, most if not all achiral compounds can also be rotated before you apply the plane of symmetry. Oftentimes if not always, you can rotate them in a way that you see the plane of symmetry present in the molecule itself. If you managed to do that, you won, because just use a plane parallel to that one for mirror imaging and the image will automatically be identical to its template without further rotation.
The plane of symmetry in isopropanol cuts through the $\ce{CHOH}$ bit of isopropanol, dissecting each of those four atoms into two half-spheres. One methyl group will be transformed exactly onto the other by this plane.

I know that this post is like 2 years old now, but can we rotate about a sigma bond before superimposition? I don't want to ask a separate question as it might be marked a duplicate...

@Abhigyan Yes you can. Both before and after. If rotation around sigma bonds will determine that your molecule is achiral, it does not matter whether you rotate until you find a plane of symmetry or if you create the mirror image first and check by rotation if it fits later.

A chiral centre has 4 different atoms or groups attached to it. 2-propanol has two $\ce{CH3}$ groups attached to the second carbon, therefore it does not exhibit stereoisomerism.
Rotating a molecule about an axis so that it can be superimposed on its mirror image does not change its chemistry. If a molecule can be superimposed on its mirror image without having to make/break bonds to reform it, then it does not exhibit stereoisomerism.


text:
How to determine gerade & ungerade symmetry of a MO orbital?
J.D.Lee writes in his book Concise Inorganic Chemistry:

[...] An alternative method for determining the symmetry of the molecular orbital is to rotate the orbital about the line joining the two nuclei and then rotate the orbital about the line perpendicular to this. If the sign of the lobes remains the same,  the orbital is gerade, and if the sign changes, the orbital is ungerade.

Now, it can be detected easily which MO, formed due to the overlapping of s-s or p-p orbitals, is gerade or ungerade as shown in these pics:

But what about the molecular orbitals formed due to the overlapping of s-p orbitals or p-d orbitals?
mo due to to overlapping of s-p orbitals:

mo due to overlapping of p-d orbitals:

If I rotate about the first orbital about an axis perpendicular to the inter-nuclear axis, the smaller lobe goes to the right only; the sign always remains same; just the position of the lobe has been changed. But in the first picture, only the bonding mo is gerade while the anti-bonding mo is ungerade; I'm not getting how by rotating along the axis perpendicular to the inter-nuclear axis, the sign of the antibonding mo changes making it ungerade.
Can anyone please help me how to apply the procedure for the molecular orbitals formed due to the overlapping of s-p & p-d orbitals?

Those MOs simply cannot be labelled as g or u. MOs of non-centrosynmetric molecules (i.e. molecules that do not possess a centre of inversion) cannot be labelled g or u. Look at the MO diagram of water (C2v) - none of the orbitals have g or u labels.

@orthocresol: So, do you mean to say those in the pic cannot be labelled g or u?

That's right, they cannot be labelled as such.

@orthocresol: But the author did label those orbitals as g & u :/

I think the point of all this is about MO orbital symmetry, nodes, bonding and antibonding. For these asymmetric examples, consider the orbital lobes between the nuclei if they are sigma- or pi-overlap, and if the orbital formed between them has genade or ungerade symmetry.

Related : How derive g and u symmetry labels for orbitals?

The better way to do it is to check what happens under inversion ($i$ or $\bar 1$). If the orbital stays the same, it is g, otherwise u.
However, as Orthocresol mentioned in the comments, checking that is only possible if the entire molecule contains inversion symmetry. Not all point groups do, and the vast majority of molecules does not (partly because $C_1$ is likely the most prevalent point group out there).
For example, consider transition metals’ d orbitals. In octahedral complexes, they are labelled $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In tetrahedral complexes, which do not have inversion symmetry, they are $\mathrm{t_2}$ and $\mathrm{e}$.

Hmmm... could you tell what point group is? Also, if there is no inversion symmetry in those orbitals, why did Mr. Lee labelled those orbitals as $g$ and $u;?$

@user36790 If you check out which symmetry features a molecule has, you realise that these happen to form a mathematical group, i.e. there’s identity, there’s an inverse element to each element and when adding two operations together, they exhibit commutativity. Only a handful of different groups actually exist, and since they help so much with symmetry considerations, they are taught in second or so year of chemistry studies. I’ll check out if I find a question here on chem.SE about them ^^

So, should I say the author is not right in marking those orbitals as $g$ and $u$?

Unfortunately, there is no ‘basic’ question about group theory here as far I can tell. Feel free to ask one ^^'

@user36790 Yes, it does not make sense labelling something as g or u if there is no inversion present in the molecule.

Thanks for the confirmation. He confused me for 2 days! I'm going to read about point group in order to understand it better. Thanks, again.

If opposite lobes have inversion centre i.e. the centre of symmetry with respect to  phase of wave functions then the orbitals are gerade otherwise ungerade.


text:
Reaction of Hydrogen and Dinitrogen Tetroxide?
What might be the possible products of a reaction between hydrogen and dinitrogen tetroxide ($\ce{N2O4}$)? They are fuel and oxidizer respectively in rockets.

And what do you think they are?

I have no Idea.

In future please try to have any idea, otherwise question can be downvoted and/or closed.

A likely reaction between dinitrogen tetroxide and hydrogen would be as follows with nitrogen and water being the products
$\ce{N2O4 + 4H2 → N2 + 4H2O~(I)}$
$\ce{\Delta H_{rxn}=[0+(4 x -241$.$ 8)-9$.$2 - (4 x 0)] ~=~ -976$.$4~ kJ}$
For the record, dinitrogen tetroxide and hydrazine (usually in combination with methylated hydrazines) is currently used as a rocket fuel.  The equation describing this reaction is
$\ce{N2O4 + 2N2H4 → 3N2 + 4H2O~(II)}$
$\ce{\Delta H_{rxn}=[(3 x 0) +(4~ x ~-241$.$ 8) - 9$.$2 - (2 x 50$.$6)] ~=~ -1077$.$6~ kJ}$

are traces of NOx possible in either case?

Sure, $\ce{N2O4}$ can produce $\ce{NO}$ directly.

@F.Sherrif or redissociates into $\ce{NO2}$.


text:
What is the reason behind neglecting & putting the overlap integrals $0$ in Simple Hückel Theory?
This is an excerpt about Simple Hückel Theory from Elements of Physical Chemistry by Peter Atkins:

The first step that Hückel took was to ignore the $\sigma$-bonding framework and focus solely on the $\pi$ electrons. That is, he assumed that the atoms had taken up the positions they have in the actual molecule, then calculated the properties of the $\pi$ orbitals that matched that framework.
[...] we show in Derivation 14.3 that the Schrödinger equation for the orbitals $\hat H \psi = E\psi\;,$ then becomes the following pair of simultaneous equations for the coefficients: $$(H_{AA}- E)c_A- (H_{AB}- ES)c_B= 0\\ (H_{BA}- ES)c_A+ (H_{BB}- E)c_B= 0$$ ... . These equations are called the secular equations.
Hückel then made further approximations. First he neglected all the overlap integrals and set $S= 0$ wherever it appear. .....

I didn't quite get why he made this approximation. There must be some reason for neglecting the overlap integral, I suppose. Also, Atkins didn't mention why Hückel only made calculations for $\pi$ orbitals only; does his equations not work for $\sigma$-bonding?
So my questions are:
$\bullet$ What is(are) the reason(s) behind making the approximation of putting all the overlap integrals $0\;?$ After all, making $S= 0$ would mean there is no overlap.
$\bullet$ Why did Hückel work only on $\pi$ orbital? Doesn't his calculation work for $\sigma$-bonding?

What is(are) the reason(s) behind making the approximation of putting
  all the overlap integrals 0?

I am not 100% sure here, but using such a simplification, the mathematics becomes significantly easier. In principle, Hückel theory would still work if there would be overlap. In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix equation can be directly interpreted in terms of the $\pi$ orbitals in the system. You do not have to back-transform your solution as is the case in the Hatree-Fock method.

Why did Hückel work only on π orbital? Doesn't his calculation work
  for σ-bonding?

Only $\pi$ electron molecular orbitals were included because these determine the properties of the molecules that Hückel was studying (conjugated molecules). The $\sigma$ electrons were ignored. This is known as $\sigma$-$\pi$ separability and is justified by the orthogonality of $\sigma$ and $\pi$ orbitals in planar molecules. As a consequence though, the Hückel method is in principle limited to planar systems.

Taking $S=0$ implies an orthogonal basis set of AOs & this indeed makes the maths easy. But if there is no overlap between any of the constituting AOs, then how can MO form?

Even without overlap, by the diagonalisation of the Hamiltonian matrix you obtain eigenvectors and -values. Those vectors represent the molecular orbitals which are linear combinations of the atomic orbitals. For example, in the ethylene system you obtain $\psi_{0} = \frac{\phi_{0} + \phi_{1}}{\sqrt{2}}$ and $\psi_{1} = \frac{\phi_{0} - \phi_{1}}{\sqrt{2}}$ for the HOMO and LUMO, respectively.
This is for instance demonstrated on the Wikipedia page concerning the Hückel  method: https://en.wikipedia.org/wiki/H%C3%BCckel_method#H.C3.BCckel_solution_for_ethylene.

so, what does resonance integral being zero imply?

The resonance integrals are the (off-diagonal) integrals of neighbouring atoms. If those are zero as well, then you were correct that you cannot find any MO because every MO will then simply be an AO and all these MO/AO will have the same energy. (and your system is no longer conjugated)

In those examples, $S_{12}= S_{21}= 0$ meaning there is no overlap between atom 1 and 2; still $H_{12}= H_{21}\ne 0= \beta$; I'm not getting that- if there is no overlap which is evident from $s_{12}$ being zero, how can there be non-zero $H_{12}$ as it would mean there would be overlapping? I'm really not understanding how $S_{12}= 0$ and $H_{12}\ne 0$ go with each other.

Indeed, this seems rather counterintuitive, but it is not. Another way of looking at $S_{ij} = \delta_{ij}$ is saying that all atomic orbitals are orthonormal to each other. So if you would evaluate the overlap integral of two different orbitals, it would result in zero. This does not necessarily mean that evaluating the Hamiltonian integral $\langle \phi_{i}|\hat{H}|\phi_{j} \rangle$ results in zero, because first applying the Hamiltonian operator on the wave function and then integrating might result in a nonzero outcome.

Use \langle \rangle to get $\langle ;\rangle$ rather than <> to get $<;>;.$

(thanks for mentioning about the mathjax commands) To give an example of my previous comments; have a look at this website: http://www.ivofilot.nl/posts/view/29/Introduction+to+Electronic+Structure+Calculations%3A+The+variational+principle. If you look at example 2, you see that $\psi_{1}$ and $\psi_{2}$ are orthogonal to each other so they have zero overlap. Yet the off-diagonal elements of the Hamiltonian matrix are nonzero.

You are right at all your points but still I'm not getting the physical insight; I'll post it as a different question where I can reflect my problem more clearly. Thanks, nevertheless for the answer; +1.

It does not really have a physical interpretation. You should see it as something transcendental. Like a decoherence of the wave function if you will. That is the reason why diagonalization of the Hamiltonian matrix results in the correct energies of the electronic states: because you get rid of the off-diagonal elements. It is actually a very good question and deserves a more elaborate answer than what I can give in this comment.

It's probably worth mentioning that there is an extended Hückel method that includes both $\sigma$ and $\pi$ electrons.


text:
Is Hydrolysis same as dissolution?
I have book that contains the following paragraph:

Due to high dielectric constant, it (water) has a very strong hydrating tendency. It dissolves many ionic compounds. However, certain covalent and some ionic compounds are hydrolysed in water.

Is this statement self-contradictory? Doesn't 'hydolysed in water' mean dissolved in water?

Absolutely not. $\ce{NaCl}$ dissolves, but never hydrolyzes. $\ce{Al2S3}$ hydrolyzes, but never dissolves.

What's the difference ?

I think that under the IUPAC definition of dissolution, hydrolysis is (or at least can be) a type of dissolution.

Hydrolysis is the chemical breakdown of a compound due to reaction with water.
Dissolution is a process by which the original states of matter turn into a liquid or other solvent and become solutes - dissolved components in solution.

Thanks for the response , but I don't think it answered my question effectively. It will be a great help for me if you could explain the statement given in the book(written in the blockquotes).

OK wait for a while..

Water has high dielectric constant because due its polar nature, it has more tendency to decrease the attraction between the ions in an ionic compound. It means that water is dissociating or ionising the ionic compound into its constituent particles i.e. it is known as hydrolysis which is done on some ionic and covalent compounds (not for all ionic compounds).

Dissolution simply means mixing. As all ionic compounds can mixed in water to from a solution which my be homogenous or heterogenous.  IF you like my answer, please up vote me.

On hydrolysis , the ionic compound is split into its constituent ions but on dissolution , the ions are mixed with water so what forms is not the original ions...Am I right ?

I unable to understand your question

I have edited the question...

Yeah.. Correct. You got it..

Hydrolysis is a broad chemical term that means reacting with the solvent water in some way. So for examples esters can be hydrolysed according to the following reaction:
$$\ce{H3CCOO-C2H5 + H2O -> H3CCOOH + HO-C2H5}$$
(Ethyl acetate and water react to form acetic acid and ethanol.)
Dissolution is the process of something being dissolved in water, intermixing with the water molecules so that each atom/ion/molecule of the substance is more or less completely surrounded by water molecules. This process is usually understood to happen without any reaction, thus the term hydrolysis is not used.
However, the above is often a simplification. For example, some water-free salts such as $\ce{AlCl3}$ or $\ce{CuSO4}$ will react with the water to form the corresponding hexaaquacomplexes $\ce{[Al(H2O)6]^3+ + 3 Cl-}$ or $\ce{[Cu(H2O)6]^2+ + SO4^2-}$. These are technically reactions, best exemplified in the aluminium example where the metal was originally coordinated by chlorido ligands and after dissolution is coordinated by aqua ligands.
The only thing that changes is the immediate vicinity of the metal ion in these cases, so speaking of dissolution rather than hydrolysis (essentially the metal cation remains the same as do the corresponding anions) is often tolerated.

So, putting it in a simple language, dissolution involves no reaction, just the constituent atom/molecule/ion of the substance is surrounded bu water molecules, whereas, hydrolysis involves the reaction. This is what I got from your answer. Am I correct ?

@user3503781 yeah.


text:
How can off-diagonal Hamiltonian matrix elements be nonzero when the overlap matrix is diagonal?
While studying about Hückel theory, I got accustomed to the approximation of making the overlap matrix an identity matrix; that is making the off-diagonal elements zero as $S_{AB}= S_{BA}= 0\;;$ this implies the use of orthogonal base states of AOs. The off-diagonal elements of the Hamiltonian matrix are still taken as constants that may be non-zero: $H_{AB}= H_{BA}= \beta_{AB}$, where $\beta_{AB}$ is a negative quantity.
Then I wondered why it isn't the case that $H_{AB}= H_{BA}= 0$ strictly, as it represents the expectation value - the average energy contribution of the overlapping region of AOs $A$ and $B$.  But, it would seem that overlapping is not actually possible, as is evident from the fact that $S_{AB}$ is zero.
In this question, when I asked about this, ifilot replied:

Indeed, this seems rather counterintuitive, but it is not. Another way of looking at $S_{ij}=δ_{ij}$ is saying that all atomic orbitals are orthonormal to each other. So if you would evaluate the overlap integral of two different orbitals, it would result in zero. This does not necessarily mean that evaluating the Hamiltonian integral $\langle ϕ_i|\hat H|ϕ_j\rangle$ results in zero, because first applying the Hamiltonian operator on the wave function and then integrating might result in a non-zero outcome.

I know he is right in this point; but I'm still having trouble seeing how this is possible.
As Peter Atkins in his book Elements of Physical Chemistry wrote:

[...] The integral $H_{AB}$ depends on both $\psi_A$ and $\psi_B$, and we can interpret it as the contribution to the energy due to the accumulation of electron density where the two atomic orbitals overlap, including, for instance, the Coulombic attraction between the extra accumulation of electron density and both nuclei.

Evidently, this phrase makes clear that $H_{AB}$ is non-zero iff there is overlap between the AOs.
So, how can $H_{AB}\ne 0$ and $S_{AB}= 0$ both be true at the same time? And, what are the physical implications? While the former means there is overlap, the latter means the opposite; it seems really contradictory.

Please don't use Math Jax in the title. It does not render in the mobile app and it comes with multiple more problems. I can't provide the link at the moment, but there are recent discussions on Meta. (Cc @brian)

@Martin-マーチン Meant to change the title when I edited; forgot.  Will tweak here shortly.

I believe the confusion lies in some ambiguity in the use of the term overlap.
For the $S_{AB}=0$ approximation of Hückel theory, overlap is used in the specific mathematical sense of the "overlap integral":
$$
S_{AB} = \left<\varphi_A|\varphi_B\right> =
\int_{\vec r}{\varphi_A^*\!\left(\vec r\right)\varphi_B\!\left(\vec r\right)\mathrm{d}\vec r}
$$
However, this does not mean that at each point in $\mathbb{R}^3$ only one of $\varphi_A$ or $\varphi_B$ is allowed to bear a non-zero value.  In fact, at most points in $\mathbb R^3$, $\varphi_A$ and $\varphi_B$ will both bear non-zero values. Since $\varphi_A$ and $\varphi_B$ will in general have "co-located regions of non-zero values," the two functions can be described qualitatively as overlapping, in the same sense as in your second quote. It is the overlap of the functions in this more general, qualitative sense, that allows $H_{AB}=\left<\varphi_A^*\middle|\hat H\middle|\varphi_B\right>\neq0$.
(To note,  in order for Hückel's approximation to be exact, the orbitals would have to be selected such that the specific integral $S_{AB}$ over all $\mathbb R^3$ equals zero$^\ddagger$. I believe Hückel did not actually do this, but instead just coerced to zero any nonzero $S_{AB}$.)
To help explain more explicitly why this is, I'll first write out the integral implied by the Dirac notation for $H_{AB}$:
$$
H_{AB}=\left<\varphi_A^*\middle|\hat H\middle|\varphi_B\right>=
\int_{\vec r}{\varphi_A^*\!\left(\vec r\right)\hat H\!\left[\varphi_B\!\left(\vec r\right)\right]\mathrm{d}\vec r}
$$
Then, I'll introduce two (real-valued) one-dimensional functions defined on the closed domain $x=[0,1]$ to use for demonstration purposes:
$$
\varphi_A\!\left(x\right) = \sin\!\left(2\pi x\right) \\
\varphi_B\!\left(x\right) = \sin\!\left(3\pi x\right)
$$
$\hspace{10mm}$
It is straightforward to show that $S_{AB}=0$ for these two functions.  One way is to plug the overlap integral into Wolfram Alpha. Another option is to do it by hand.  In the below figure, the green curve is a plot of the product $\varphi_A\!\left(x\right)\cdot\varphi_B\!\left(x\right)$, and the orange curve is a plot of the cumulative integral of this product:
$$
\int_0^x{\varphi_A\!\left(x'\right)\cdot\varphi_B\!\left(x'\right)\mathrm{d}x'}
$$
$\hspace{10mm}$
As can be seen, the integral over the entire domain is zero, satisfying $S_{AB}=0$.
Now, let's introduce a simple "Hamiltonian"$^\dagger$:
$$
\hat H = x
$$
The next figure is a replicate of the first one above, with $\hat H\!\left[\varphi_B\right]$ also plotted as the dashed red curve:
$\hspace{10mm}$
As in the second figure above, one last plot shows the local value of $\varphi_A\!\left(x\right)\hat H\!\left[\varphi_B\!\left(x\right)\right]$ along with the cumulative integral:
$$
\int_0^x{\varphi_A\!\left(x'\right)\cdot\left[x'\cdot\varphi_B\!\left(x'\right)\right]\mathrm{d}x'}
$$
$\hspace{10mm}$
The value of $H_{AB}$ here is clearly not equal to zero.  (Wolfram Alpha concurs.) Thus, it is absolutely possible to have $H_{AB} \neq S_{AB} = 0$.
To note, the principle exhibited here on a closed domain in $\mathbb R^1$ applies equally well to orbitals on an infinite domain in $\mathbb R^3$. Note also, if I were actually performing a quantum mechanical calculation here, I would have been a lot more careful about normalizing my wavefunctions (viz., ensuring $S_{ii}=1$).
$^\ddagger$Numerous methods are known for generating orthogonalized orbitals with $S_{AB}=0$ strictly, from a general set of orbitals where $S_{AB}\neq0$ occurs for some non-null subset of the $\left(\!A,\!B\right)$. To my knowledge, the most commonly used orthogonalization for atomic orbitals is Löwdin's method (see this PDF). The Gram-Schmidt orthogonalization (usually in its "modified" form) is used in a variety of other areas of quantum chemistry. Wikipedia also has a pretty solid article on orthogonalization methods in general.
$^\dagger$You'd never use this as a Hamiltonian operator. Further, I'm not sure if it technically even is a Hamiltonian operator. (I think it'd work as a dipole moment operator?) But, whatever, leave me alone, it's for purposes of illustration.

Okay, this means at most points $\varphi_A$ and $\varphi_B$ are non-zero; but still the sum of them over all the volume is $\approx 0,$ isn't it? Then how it makes the difference; overlap region is still on the verge of null; how could there be $H_{AB}$ as it is also the integral over all the volume? If the overlap is zero, then why should $H_{AB}\ne 0$,irrespective of how the overlap integral became zero-either by one of $\varphi$s zero at a point or the net sum being zero?

Cool! I am looking forward to seeing your example. In my previous answer, I already referred to my website (http://www.ivofilot.nl/posts/view/29/Introduction+to+Electronic+Structure+Calculations%3A+The+variational+principle) where you can see in Example 2 a scenario where an orthonormal basis gives off-diagonal Hamiltonian integrals. But perhaps this example is too complicated and you can come up with something simple and really to the point. :-)

I agree with the general principle illustrated by your example, i.e., that for two arbitrary orbitals $A$ and $B$, $S_{AB}=0$ doesn't necessarily imply $H_{AB}=0$. However, it should be emphasized that for two atomic orbitals $A$ and $B$ belonging to different atoms, it is almost invariably (if not always) true that $S_{AB}=0$ iff $H_{AB}=0$.

We still ignore $S_{AB}$ because most qualitative features of the solution don't change by this approximation (a counterexample: http://chemistry.stackexchange.com/a/42636/7841). But it would be a poor approximation if we really wanted to get accurate numbers (e.g., bonding energy) from Hückel theory.

@higgsss Mmm, so actually $S{AB} \neq 0$ holds generally, it's just that the Hückel approximation treats them as equal to zero?  Hadn't appreciated that.  Good points.

Right. It is hard to make $H_{AB}\neq 0$ with $S_{AB}=0$ with two atomic orbitals localized about different atoms. On the other hand, Wannier functions (https://en.wikipedia.org/wiki/Wannier_function) have the nice property that they are localized about different atoms as well as being orthogonal to one another. But finding such functions is another complicated problem.

@higgsss . I was picturing Loewdin-orthogonalized orbitals as a good example of an $H_{AB} \neq S_{AB}  = 0$ function space, but you're right, they're not single-center atomic orbitals.

@Brian:+1;  Really thanks for those plottings; I got the point that though $S_{AB}= 0,;;H_{AB}\ne 0$; I could extrapolate these graphs to 3D domain at ease. So, does that mean even if there is no overlapping, there can be accumulation of electron density in the internuclear-region?

@user36790 I think you should ask that as a new question, linked to this one, with a more full explanation of what you're asking. It'll be too much to try to answer in this comment thread. Also, make sure you read higgsss's comments above, in particular the one emphasizing that in actuality $S_{AB} \neq 0$ in Hückel theory,  but the values are taken as equal to zero as a key part of Hückel's approximation.

Maybe a little example from linear algebra (just imagine it to be spin space and we're back in the quantum world) can also add something to the discussion.
Suppose you have two wavefunctions
\begin{equation}
\psi_A = \frac{1}{\sqrt{2}} \begin{pmatrix}
1 \\
1 \end{pmatrix} \qquad \psi_B = \frac{1}{\sqrt{2}} \begin{pmatrix}
1 \\
-1 \end{pmatrix}
\end{equation}
Then obviously the overlap of the two is
\begin{equation}
S_{AB} = \langle \psi_A | \psi_B \rangle = \frac{1}{2}(1 \cdot 1 + 1 \cdot (-1)) = 0
\end{equation}
Nevertheless, if you consider the Hamiltonian
\begin{equation}
H = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
\end{equation}
then you get $H_{AB} = \langle \psi_A | H | \psi_B \rangle = 1 \neq 0$.
My suggestion is to forget about the interpretation of these matrix elements given in Atkins, since it is very difficult to imagine how quantum-mechanical operators act (he also seems to forget that there is not only Coulomb interaction but also kinetic energy present in the Hamiltonian, the latter acting as a derivative operator on orbitals).


text:
If a gas absorbs heat, then how can its temperature remain constant?
My book has this line:

In an isothermal expansion, the gas absorbs heat and does work

But, if a gas absorbs heat, then how can its temperature remain constant? (as the process is isothermal)

From the first law, $\Delta U=Q-W$.  If the process is isothermal, then $\Delta U=0$ (assuming an ideal gas), so Q = W.  The heat energy is consumed by your gas doing work, so there is no change in internal energy (temperature).

Hopefully I can provide you with some physical intuition on this phenomenon through a simple example.
Imagine a gas heated by a flame in a closed cylinder with a piston on the top balanced by the same pressure as the gas inside. As you heat the gas, the particles move faster (gain kinetic energy) and collide with the walls of the container harder, increasing the pressure of your system. This pressure increase is then used to drive the piston out. As the particles of your hot gas collide with the piston wall to move it, they necessarily dump some of their kinetic energy into it, slowing down (i.e. cooling) in the process. On the other hand, kinetic energy is not lost when the gas particles bump into the other rigid walls of the container, since those collisions are basically elastic.
So overall you have a heat source dumping kinetic energy into the gas particles from one side of the cylinder, and the gas particles dumping their kinetic energy into the surroundings while pushing a piston on the other side. Now run this process very carefully and slowly, and the temperature of the gas can be held more-or-less constant!

This answer is basically the same as the shorter answer by Chet Miller on Dec. 30 '15.

I was attempting to provide some physical insight into why the other answer makes sense, but I guess that isn't appreciated here.


text:
Is it possible to have accumulation of electron density in the inter-nuclear region even if the overlap integral is zero?
Recap:
Lately, I've been studying Hückel theory where I learned about the approximation of neglecting all overlaps and making overlap matrix an identity matrix that is $S_{AB}= 0\;_; \;S_{AA}= 1$ or $S_{ij}= \delta _{ij}\;.$ He then replaced the off-diagonal Hamiltonian elements with parameter viz. $H_{AB}= \beta_{AB}\;_,$ where $\beta_{AB}$ is a negative quantity.
Question then arose if the overlap is zero as is evident from $S_{AB}$ being zero, how can there be a non-zero value of $H_{AB}$. It was known to me that $H_{AB}$ is the

contribution to the energy due to the accumulation of electron density where the two atomic orbitals overlap, including, for instance, the Coulombic attraction between the extra accumulation of electron density and both nuclei.

So, it is contradictory to simultaneously have the condition $H_{AB}\ne 0$ with $S_{AB}= 0.$
I asked this querying for the same; there I got a nice answer from Brian where he showed that even if $S_{AB}=0$; it need not be necessary to have $H_{AB}= 0$; the graphs he provided in his answer prove this otherwise a counterintuitive fact.

Present thinking & associated query:
Brian's answer confirms the fact that $S_{AB}= 0$ and $H_{AB}\ne 0$ can go with each other. But doesn't that mean there is no need to have a net overlapping of AOs between the internuclear region in order to have accumulation of electron density in the internuclear region?
For example, while using Simple Hückel Theory for ethene, Atkins put all $S= 0$ as usual per the approximation and then wrote:

[...] Finally, as explained in the derivation, he also supposed that the terms $H_{AB}$ and $H_{BA} ,$ which represent the energy of the interaction of the two nuclei with the accumulation of electron density in the internuclear region .......

Really, if he writes this after considering $S_{AB}= 0\;_,$ doesn't that mean there can be accumulation of electron in the internuclear region even if the the participating AOs are orthogonal and there is no net overlapping? He is considering the case of ethene $\pi$-network which means there is electron density that leads to $\pi$-bonding even if $S=0$.
So, is it right to conclude that there can be accumulation of electron density in the internuclear region even if there is no net overlap in the atomic orbitals?

Yep, definitely better as a new question. :-)

@Brian: Also, I did ponder a while & wondered how Hückel actually made this approximation of $S= 0$; making this would make the basis sets orthogonal. But orthogonality of AOs depends on the geometry of the molecule, isn't it? I googled a bit & didn't found anything regarding this. Nevertheless I do have a book Computational Chemistry by Errol Lewars which briefly mentioned it citing  'a twisted allyl species'. So, definitely, Hückel Theory is really eerie: he didn't consider those overlaps & didn't pay heed to the geometry of the molecule:/

Oh, certainly Hückel's bases had $S_{AB}\neq 0$ in actuality -- this was higgsss's point in the comments on my answer at the other question, I believe.  In order to make for a tractable calculation, though, I think Hückel just made his approximation by editing the matrices to coerce $S_{AB}=0$. Absolutely, this makes the mathematical problem diverge from the physical system, but in many cases it still gives qualitatively reasonable results.

Using approximations to simplify calculations and having the given parameters physically zero are two completely independent problems. If neglecting S makes no big difference in the results, or simple parametrisation can mostly compensate this simplification, it is justified irrespective if the actual S is zero or not.

If you define overlap as given by the overlap integral,
$
S_{ij} = \langle \psi_{i} | \psi_{j} \rangle
$
than you can have an accumulation of electron density in a specific part in space even though $S_{ij} = 0$.
Of course, if the complete wave functions of the two AOs are completely non-overlapping in all space, for instance because the centers of the two AOs are seperated by a long distance (say more than 5 A), then there can be no accumulation of electron density anywhere. This is of course logical. There is hardly any accumulation of electron density between two atoms that are quite far apart.
An interesting scenario exists where two wave functions have opposite sign and an overlap of these wave functions actually gives a decrease of electron density. This is termed deconstructive interference opposite to constructive interference when there is an accumulation of electron density.

Thanks for answering the question; so that means even if the net overlap is zero, if there is, at some place, non-zero $\psi_i\psi_j$, it can have a net  Hamiltonian $H_{ij},$ is it so?  Also, in ethene, the $p$ orbitals do overlap to give $\pi$ bonding; although Hückel put $S=0$.

Another thing I want to ask you is: if the the sum(integral) of $\psi_i\psi_j$ is zero, even though there may be places where $\psi_i\psi_j$ is non-zero; does that mean even $\psi_{i} \hat H\psi_j$ is non-zero at some places, but the net sum is zero since the net overlap is zero? Brian in his answer has shown that this doesn't necessarily happen. So, can I conclude that $H_{ij}$ gives the average energy of those $\psi_{i}\psi_{j}$ which are non-zero?

There are two things to keep in mind here; you have the product of two wave functions $\psi_{i} \psi_{j} (\vec{r})$ and you have $S_{ij} = \langle \psi_{i} | \psi_{j} \rangle$ which is the integral over all space of this product. Even if this $S_{ij} = 0$, the former can have a non-zero value at a specific $\vec{r}$. Obviously, if the value of $\psi_{i} \psi_{j} (\vec{r}) = 0$ at all $\vec{r}$, then the integral will be zero as a consequence of that.

Furthermore, if $\psi_{i} \psi_{j} (\vec{r}) = 0$ at some $\vec{r}$, then  $\psi_{i} \hat{H} \psi_{j} (\vec{r}) = 0$ as well. Because if there is no overlap density at some point in space, then there can be no energy density at that point. This is a consequence of the following: If $a \cdot b = 0$, then either $a=0$, $b=0$, or $a = 0$ and $b=0$.


text:
Breakdown of butyl rubber
I'm in my high school's theatre crew and of course we build a lot of props and such. However I just opened (yesterday) an old drawer of screwdrivers and they smelled like puke! I know that the chemical responsible for that smell is butyric acid, so the rubber must have been a butyl rubber. Could someone point out the chemical equation that shows the reaction for this?

If ever, it would be rather isobutyric or isovaleric acid, as butyl rubber is copolymer of isobutene with few percents of isoprene.

One really does not know much until a chemical analysis is done.


text:
Why is conjugation not possible if there is not present alternate $\pi$ molecules according to MO theory?
I was reading this page at Chemwiki; there they were discussing conjugated-$\pi$ bond.
In order to show that conjugation requires  alternate double bonding, they cited the example of 1,3,6-heptatriene:

To be considered conjugated, two or more π bonds must be separated by only one single bond – in other words, there cannot be an intervening sp$^3$-hybridized carbon, because this would break up the overlapping system of parallel 2p$_z$ orbitals. In the compound below, for example,  the C1-C2 and C3-C4 double bonds are conjugated, while the C6-C7 double bond is considered to be isolated, due to the effect of the sp$^3$-hybridized C5.

I've a problem with their explanation; firstly are they following MO theory or VB theory? While describing molecules with MO, they, suddenly from nowhere, used the concept of hybridisation. Why?
Can anyone explain only with he help of MO theory why, there needs to be alternate $\pi$ bonding in the molecule for conjugation to happen without use of hybeidisation?

I have a feeling that you don't exactly understand what MO theory is so you may don't know what your asking for - there was no MO stuff in what you mention.

I'll try and answer when I have access to my Mac.  But in the meantime, your exaxt question is answered in organic chemistry by Warren, Clayden and Greeves...

I think some help will be to read "Orbitals are not real". Orbitals are mathematical constructs. What's more important is that MO and VB theory are not competing, but complimentary, approaches to modeling multi-electron wavefuntions. They emphasize different aspects of electron behavior and depending on what you want to know, the maths may be simpler for one over the other (or whatever approximations were needed).

And to second NotNicolaou, just about any organic chemistry textbook will answer this question using MO theory only. It has to do with accessible orbitals having appropriate symmetry.

@Mithoron: Are you saying MO is not required for conjugation?


text:
High Viscosity of Noble Gases
Why are the viscosities of noble gases higher than almost all other gases? This seems counter-intuitive since they have much weaker inter-molecular interactions and lower boiling points.
Gas Viscosity Table:
http://www.engineeringtoolbox.com/gases-absolute-dynamic-viscosity-d_1888.html

I wonder if this is at all related to the unintuitive fact that gasses increase in viscosity as temperature is increased, unlike the majority of liquids. Noble gasses and higher temperatures both provide situations which behave closer to an ideal gas.

It is very related.  Higher gas temperature means higher particle velocity.  Higher velocity means more collisions.  More collisions means higher viscosity.

In the case of temperature particle velocity is increased by higher temperature as opposed to lighter particles (He, Ne, Ar.)

Viscosity in gasses is generally different than liquids.  Like liquids exhibiting fluidic properties, the viscosity is related to internal friction; however, the friction mechanism is different than inter-molecular interactions experienced by closely packed molecules. The molecules in a gas are far apart resulting in the influence of an entirely different friction mechanism.
The friction mechanism dominating gas viscosity is related to the number of collisions between gas particles. More collisions means more internal friction. In gasses, two dominant factors affect the number of collisions: (1) the mean distance a gas particle has to travel before it collides with another particle and (2) the average velocity of the particle within the gas.
Some noble gasses (He, Ne, Ar) are monatomic particles with a lighter mass than the diatomic elemental gasses. The light mass results in a much higher velocity particles at a given temperature (consider  $E=\frac{3}{2}kT=\frac{1}{2}m \bar{v^2}$) resulting in more collisions. Xenon has a high molecular diameter, so the shear increase in size of the molecule increases the probability of a collision, thus reducing the mean distance it travels before a collision.
More information is available from the following link.


text:
Why does ozone, being heavier than air, not settle down?
Is this because of some equilibrium between diffusion and gravity ? What exactly is responsible?

Not sure why this is considered a dupe. It asks about ozone where some of the issues are specific to the gas and not adequately answered in the other questions.

The absolute concentration of ozone in the ozone layer is very, very small.

While gravity does cause some separation of atmospheric constituents at high altitudes, the reason why ozone doesn't migrate to the ground has more to do with its chemistry
Ozone is pretty unstable and is formed from some chemical reactions and from the interaction of oxygen with UV light. But it is unstable and decays fairly quickly. This means that is only appears anywhere when the reaction forming it keeps happening.  This is true in the upper atmosphere which is why we have an ozone layer. Some hydrocarbons (not just from pollution from cars but also from volatile stuff released by trees) can react to generate ozone at ground-level. But we only see a significant concentration when the reaction forming it is happening.
The effect of gravity on the separation of the components of the atmosphere is small, especially in the lower atmosphere where turbulent circulation mix things up thoroughly. In the upper atmosphere the dominant process is diffusion and gravity can have more of an effect. This paper describes the process and has some details of the calculations about the size of the effect. It summarises the situation like this:

Turbulent mixing keeps the relative concentrations of gases nearly constant in the lowest 100 km. At higher altitudes, molecular diffusion controls the concentrations, with the lighter gases becoming relatively more abundant with increasing altitude.

It is a worthwhile read if you want the detail.
But the core issue is that, in the lower atmosphere, all gasses are well mixed so there is not separation by density. Ozone only appears there if there are some very specific reaction conditions that create it. The bulk of ozone in the ozone layer occurs because there are constant reactions there driven by UV light from the sun. Separation by density isn't relevant there either.

Hi Matt: I reckon only the second part of your post is the correct answer. Up to the turbopause/homopause turbulence causes a well mixed atmosphere. Gravity can't compete with eddy diffusion. How heavy a gas is (see question), does hence not play a role. However, while ozone does have a complicated photo chemistry, the answer does not have to do with that. The atmosphere is also well mixed for O2, N2, H2O, He, Ar, Ne, ... due to turbulence, and chemistry plays no role.

@tipavi The reason why ozone is found where it is is entirely to do with photochemistry and nothing to do with its relative mass. So this is relevant to the question. And I don't disagree with you on atmospheric mixing: it is certainly irrelevant here, though it does happen for some parts of the upper atmosphere.

I agree. But while its correct, that's not what he asked for. Many other correct things could be said, but the more concise and precise an answer is, the more helpful it will be. :)

@tipavi "... that's not what he asked for" That the OP accepted this is proof enough (in addition to the upvotes, one of which I cast myself) that the answer is better off without the omission you suggested. Thank you for sharing your opinion though :-)

@paracetamol I respectfully disagree. The question whether or not an answer is correct does not depend on a single opinion (e.g., the OP's) or a popular vote (e.g., 5 upvotes). Neither is "proof enough", see discussions about "alternative facts". While the photochemistry of ozone is interesting, it's not relevant in a question about mass ("being heavy than air"), gravity, and diffusion. If you turned off all the (photo)chemistry, ozone's altitude distribution would look different, but ozone would still not settle. Molecular diffusion and turbulence would keep it well mixed up to the turbopause

"If you turned off all the photochemistry, ozone's altitude distribution would look different, but ozone would still not settle" True, but in the real world there is photolytic decomposition going on, and to me it seems prudent to mention it. As I understand it, Matt didn't form his answer solely upon the photochemistry bit, rather, the photochemistry part was an insightful addendum, because it aims to justify why ozone is localized to the upper atmosphere. Matt then builds on this  "In the upper atmosphere the dominant process is diffusion and gravity can have more of an effect." O:)

In short: This answer isn't fixated over the photochemistry of ozone. ;-)

"But while its correct, that's not what he asked for" @tipavi though Matt's already posted an answer, that mustn't stop you from posting yours. If you're willing to write up a sufficiently well-tailored answer, by all means go ahead and post it :-) If the OP finds it helpful enough, he might accept it. Since you also understand this topic well-enough (although you might want to frame the answer different), in all likelihood, your answer will get my upvote as well O:)

@tipavi If you are going to complain that ozone photochemistry isn't what the OP asked for, you might want to also read the second sentence which is "What exactly is responsible?"

If ozone wasn't so reactive that it only exists in regions where it is produced, it would still never settle down, because no gases can unmix due to gravitational fields. Not even in the complete absence of winds and turbulence.
That is exept if the flight path of molecules is significantly influenced by an accelerating field on the lenght scale of the mean free path lenght. That's why ultracentrifuges can do it, and helium and hydrogen slowy escape out of our atmosphere.
Bulk gases of course have buoyancy in the surrounding atmosphere. That's a different issue. The concentration of ozone in the stratosphere is however below 10 ppm.

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


text:
Why perborate is greener bleaching agent?
All I know is, perborate degrades into supposedly less harmful compounds.  But what exactly happens that it is considered greener? Consider say perborate vs. hypochlorite.  Why perborate is "greener"?

According to Wikipedia:

Chlorination... can oxidize organic contaminants, producing chloroform and other trihalomethanes, which are carcinogenic, and many hundreds of [other]... by-products.
[W]hen household bleach and wastewater were mixed, 1–2% of the available chlorine was observed to form [halogenated] organic compounds."

Perborates, percarbonates and similar bleaching agents, while strong oxidizers, will not create these chlorocarbon chemicals (unless the chlorine is also supplied).
However, boron and perborates are not entirely "green". Boron compounds are somewhat toxic. If the prices were the same, I'd prefer percarbonate bleach to perborate.


text:
Woodward Fieser rules for dienes - non conjugated double bond as exocyclic?
When looking for exocyclic double bonds in a diene, should a double bond not connected to the conjugated system be considered?
For example, in this compound, should the marked double bond considered as an exocyclic double bond?

Woodward's rules are an empirical method for predicting the maximum absorption of a molecule based on the functional groups present (that is, conjugated functional groups which possess a distinctive maximum absorbance such as a diene or an unsaturated carbonyl).

Should the marked double bond considered as an exocyclic double bond?

Exocyclic is defined as being outside of the ring system (or in more complex systems, outside of the ring system being considered), the bond highlighted is indeed inside the ring system and is as such endocyclic.
If we start with the alkene at the bottom left, then it is in conjugation with an endocyclic alkene (to the immediate right) and an exocyclic alkene (in the ring to the right hand side). These double bonds are all in conjugation with one another.

The bond you've pointed at isn't in conjugation with the 'triene'
The Woodward rules are specifically related to the effect of groups in conjugation, a simple test of which is to see if you can push curly arrows between the pi bonds. If you try this for your molecule you'll quickly see that the bond you've highlighted is too distant from the conjugated system to have any involvement.

Thanks for the answer. But according to the paper by Fieser the pointed double bond is exocyclic to the 5 membered ring or right. What I want to know is whether a double bond must be a part of the conjugated system for it to be considered as an exocyclic double bond in the calculation?

If you find a double bond outside a ring (as in the 4 ring system here), it will be considered as exocyclic. So, the pointed double bond is exocyclic to the adjoining cyclopentane ring.

Considering the double bond is in a six-membered ring, I have difficulties seeing it as exocyclic …


text:
Why can anthracene, but not phenanthrene, take part in Diels–Alder reactions?
Both anthracene and phenanthrene seem to be structurally similar, with three aromatic rings. The middle rings in particular seem to be similarly activated towards cycloadditions.

I wouldn't be entirely surprised if phenanthrene could undergo D-A reaction, but as the dienophile instead of the diene.

Have a look at the reaction mechanism of the Diels-Alder Reaction, e.g. at Wikipedia.
We have talked about this reaction before on this site, and concluded that a sufficient explanation is only possible with the help of molecular orbitals: How accurate is this polar mechanism for the Diels-Alder reaction?
Borrowing from Wikipedia,[1] here is the most important part of the mechanism, the HOMO of the diene and the dienophile:

Now the most important part of this picture is the symmetry of the HOMO/LUMO of the diene, since this is the subject of your question. We need either a HOMO or a LUMO of these symmetries and then could match a suitable dienophile.
In anthracene we find the appropriate symmetry of the HOMO:

The reason for the regioselectivity has also been discussed before on this site: Diels Alder with Anthracene and Maleic anhydride.
We could also tune our reaction to be of inverse demand, as the LUMO has also appropriate symmetry:

In phenanthracene we do not have the appropriate orbital symmetry in the HOMO:

The LUMO is also not suitable for an inverse Diels-Alder reaction.

Reference and Notes

"FMO of Diels-Alder reaction" by Chrito23 - Own work. Licensed under CC BY-SA 3.0 via Commons.
Calculations on the DF-BP86/def2-SVP level of theory employing Gaussian09 rev. D. Contour values for the orbitals are at 0.05 e/rBohr. Occupied orbitals are depicted in blue and orange and unoccupied orbitals are depicted in yellow and red. Phases are chosen at random.

HOMOs and LUMOs don't explain everything.   Polyaromatic hydrocarbons also have their reactivities explained by Clar's aromatic sextet rule.   There is an excellent reference here from the Journal of the American Chemical Society that explains both the rule and reactivities of polyaromatics (http://pubs.acs.org/doi/abs/10.1021/ja00529a046).  In short, not all rings in an polyaromatic have equal aromatic character.   Clar's rule explains how to determine this for a given polyhex, which is a good idea since there are 22 possible structures for five ring polyaromatics and doing the HOMO/LUMO calculations can be tedious.  BTW, there are 82 possible six ring polyaromatics, and the numbers keep going up fast beyond that.


text:
How does one determine the probability of H⁺ to H₂O in pure water?
The ratio of $\ce{H+}$ to $\ce{H2O}$ is something around $10^{-8}$. How did people determine that?
You cannot use the $K_\mathrm{w}$, $\mathrm{p}\ce{H}$, $\ce{[H+]}$ ion concentration value to answer because you get all that from the probability.                                                                   Is there any independent way to find the probability. Please elaborate on the steps.
I do not want to know why the ionic product is what it is. how was the ionic product found ?

I would guess that this number comes from conductivity experiments on ultra-pure water.

Technically, the probability in pure water would be zero...

@LewisMiller:  the conductivity of a solution also depends on electrical mobility, which I'm not sure that you can measure independently the conductivity itself.

You are allowed to use the conductivity values but please elaborate on the steps

Even pure water dissociates. Pure water has conductivity of the order of 10^-9

Possible duplicate of Why is the ionic product for water $10^{-14}$?, though this appears to be more of an experimental question rather than a ~philosophical one.

I disagree, that this is a duplicate. I think this question is really about the determination of the constant rather than the impact of a proposed change.

The answer to this question was surprisingly difficult to find because all modern ways to measure pH use reference solutions of known pH and thus rely on prior knowledge of the $H^+$ activity of said reference solutions.
Finally I found this on wikipedia:

it is possible to measure the concentration of hydrogen ions directly, if the electrode is calibrated in terms of hydrogen ion concentrations. One way to do this, which has been used extensively, is to titrate a solution of known concentration of a strong acid with a solution of known concentration of strong alkaline in the presence of a relatively high concentration of background electrolyte. Since the concentrations of acid and alkaline are known, it is easy to calculate the concentration of hydrogen ions so that the measured potential can be correlated with concentrations. The calibration is usually carried out using a Gran plot.[9] The calibration yields a value for the standard electrode potential, $E^0$, and a slope factor, f, so that the Nernst equation in the form
  $$E = E^0 + f\frac{2.303RT}{F} \log[\mbox{H}^+]$$
  can be used to derive hydrogen ion concentrations from experimental measurements of E. The slope factor, f, is usually slightly less than one. A slope factor of less than 0.95 indicates that the electrode is not functioning correctly. The presence of background electrolyte ensures that the hydrogen ion activity coefficient is effectively constant during the titration. As it is constant, its value can be set to one by defining the standard state as being the solution containing the background electrolyte. Thus, the effect of using this procedure is to make activity equal to the numerical value of concentration.

This means that the precise cocentration of $H^+$ is actually unknown what is known is it's chemical activity which has been defined as equal to the concentration for a certain standard solution.

You are, I believe, referring to the auto-dissociation reaction of water:
$$2\mathrm{H_2O}\text(l) \leftrightarrow \mathrm{H_3O^+}\text(aq)+\mathrm{OH^-}\text(aq)$$
This equilibrium reaction obeys the following equation (the square bracketed quantities are concentrations):
$$K_W=[\mathrm{H_3O^+}][\mathrm{OH^-}]=10^{-14}\:\mathrm{mol^2/L^2}$$
At $pH=7.00$ and in pure water this means:
$$[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=10^{-7}\:\mathrm{mol/L}$$
$$pH = -\log[\mathrm{H_3O^+}]=7.00$$
The concentration of water in water is $55\:\mathrm{mol/L}$, so the ratio of $\mathrm{H_3O^+}$ to $\mathrm{H_2O}$ in pure and neutral water is about: $$1.8 \times 10^{-9}$$
Naked (non-solvated) protons $\mathrm{H^+}$ do basically not exist in water because $\mathrm{H_2O}$ is such a hard Lewis acid (nucleophile), which immediately bonds to any free protons.
The $pH$ of water is determined by means of a $\mathrm{H_3O^+}$-selective electrode, commonly known as a $pH$-meter.

The answer you have given is sort of circular. You cannot use the Kw, pH, [H+] ion concentration value to answer because you get all that from the probability.                                                                                Is there any independent way to find the probabitlity.


text:
Why carbon is so special?
One special branch in chemistry is allotted to compounds of only one element (carbon). Is it justified when there are more than 115 elements and their compounds but not with any special branches?

Most of the compounds that make up life on earth are based on carbon. That is why they are studied so thoroughly.

related http://chemistry.stackexchange.com/questions/43744/carbons-catenation

Carbon has a valency of four which helps in bonding with other atom or groups easier and adds to catenation. When one talks of Carbon's ability to bond, then Catenation property should come to your mind. Catenation is the linkage of atoms of the same element into longer chains. Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This is the reason for the presence of the vast number of organic compounds in nature. One such marvel is DNA! So, while studying chemistry, Carbon compounds form a class of peculiar natured compounds and the vast range of molecules and development of their own domain in their own scale might have made people feel that a separate category should be made for Carbon compounds because of their vastness and availability, also predominately manifestation of them in almost every thing, even life! In fact life became possible only because of Carbon, such is the importance. They have got their own nomenclature system in their domain.. Its really vast. The study of everything begins with Organic compounds first.. be it isomerism, medicinal reaction or anything. So, you may see, how important Carbon is. If you are a biology student, you must have learned about biomolecules.. DNA and gene pool, replication, transcription, translation, protein coding, defective proteins and some viral components and its lifecycle. Molecular basis of genetics is based on DNA and other organic compounds. That gives you thrilling experience to see how stuffs work and wonder about their importance and manifestation. Carbon is a wonder.


text:
How can borane exist?
Disclaimer: Sorry if this is a noob question, I am a high school chem student.
So I was reading my textbook in preparation for the midterm, and it talked about exceptions to the octet rule.
I understand that they exist, and what they are, but how are the chemicals structurally stable?
I also know that they are reactive, but that still doesn't help.
How do they exist?
I was thinking something like only under high pressure/heat, but I can't find anything to back that up.  Amazingly, google hasn't help.
Can somebody please explain?

Also, feel free to edit my tags.  I am new here.

Octet rule is less absolute than you seem to think. Sure, it would be nice to have a full octet, but what if the electrons are just not there? Then we'll have to do without, that's all.

Search for 'octet rule' here. There are a whole bunch of questions which will answer yours.

related http://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane

@IvanNeretin Okay.  I didn't know that, and assumed the atoms just wouldn't form a molecule in that case.


text:
Why does phosphine reduce HgCl2 to Hg but LiAlH4 / LiH reduces it to Hg2Cl2?

Why does phosphine reduce $\ce{HgCl_2}$ to $\ce{Hg}$ but $\ce{LiAlH4}$
  or $\ce{LiH}$ reduces it to $\ce{Hg2Cl2}$ ?

It seems phosphine has more reducing tendency.But why?
Moreover can it be predicted whether a reducing agent will be reduce $\ce{HgCl_2}$ to $\ce{Hg}$ or $\ce{Hg2Cl2}$ ?

Actually $\ce{PH3}$ reacts with $\ce{HgCl2}$ to form a mercury(II) salt called  chloridomercury(II) phosphide. see this:- http://chemiday.com/en/reaction/3-1-0-9509


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Nitration of Naphthalene Having Electron Withdrawing Group
In the following nitration which one out of 50 and 51 will be the major product.

In my book it is said that (51) will be major.But in this video it is said that (50) will be major.I'm confused.
Oh yes I saw this answer too Nitration in anthracene and naphthalene but did'nt help in my case.

If you look at p.450 onwards in Chemical Micro Process Engineering: Fundamentals, Modelling and Reactions by  Volker Hessel, Steffen Hardt and Holger Löwe, the major nitration product can be controlled to be mono. Conversely,  if you put this sort of system under forcing conditions,  you could favour a trinitro product. This is why these sort of set questions belong in the 19th century

@Beerhunter PLEASE read my question once again...

I think you cannot predict the outcome of 50:51 selectivity by theory and you need to look up the original paper (or try it yourself). However I wouldn't be surprised if 2 different labs report contradicting results since this kind of nitration reaction is not easy to control (violent and exothermic), and slight variations of the experimental conditions (scale, stirring, temperature control etc) can possibly lead to different results.

@K_P Ok ignoring this particular reaction.What if just mono nitro napthalene is nitrated using nitrating mixture? Even that can't be predicted?

Draw resonance structures of 48. Where are the positive charges?

@SanchayanDutta  I would expect as in your original question wrt to 50:51 selectivity that sterics would favour 50 but this would be on the assumption that there is no difference in electronics. But if your book is correct then this assumption is not right so maybe a more subtle / complex theoretic approach (computational chemistry? ) is required compared to the simpler explanation that works for 48:49 selectivity.

The reaction conditions are not plausible!  Dinitrogen trioxide $\ce{ON-NO2}$ has a "boiling" point around -3 °C, where it signficantly decomposes: $\ce{N2O3 <=> NO + NO2}$. I suspect that the reagent is in fact $\ce{N2O5}$.


text:
What is the meaning when it is said "Temperature of an Oxyacetylene Torch is 3500 degrees Celsius"?
I don't think flame is anything but the product at that temperature but I am not sure. What exactly is flame? And I think the correct statement should be that the exothermic reaction here produces heat at such level that the temperature of the product gas reaches 3500 degrees.
What will be the exact idea behind the statement?

Oxyacetylene is a mixture of acetylene (ethyne) and oxygen (not air). The mentioned temperature of about $T=3500\ \mathrm{^\circ C}$ is the adiabatic flame temperature for the combustion reaction of acetylene and oxygen.
An adiabatic process is one that occurs without transfer of heat. The adiabatic flame temperature is a theoretical temperature, which would be achieved in a combustion system in which there are no heat losses (e.g. no radiation losses from the flame). Note that this cannot be achieved in a real flame; therefore, adiabatic flame temperatures are calculated values.
The heat $Q$ released from the combustion reaction of acetylene and oxygen is the heat of combustion. If all of the energy released by this chemical reaction is used to increase the temperature $T$ of the products ($\ce{CO2}$ and $\ce{H2O}$) with no heat losses, the resulting adiabatic flame temperature can be calculated from the heat capacity $C$ of the products.
Hence, the adiabatic flame temperature represents the maximum possible theoretical temperature for a particular fuel/oxidant combustion.