Why is BCl3 a monomer whereas AlCl3 exists as a dimer? What makes dimerization possible in $\ce{AlCl3}$? Are there 3c-2e bonds in $\ce{Al2Cl6}$ as there are in $\ce{B2H6}$?

Introduction

The bonding situation in $\ce{(AlCl3)2}$ and $\ce{(BCl3)2}$ is nothing trivial and the reason why aluminium chloride forms dimers, while boron trichloride does not, cannot only be attributed to size.

In order to understand this phenomenon we need to look at both, the monomers and the dimers, and compare them to each other.
Understanding the respective bonding situation of the monomers, is key to understand which deficiencies lead to dimerisations.

Computational details

Since I was unable to find any compelling literature on the subject, I ran some calculations of my own. I used the DF-M06L/def2-TZVPP for geometry optimisations. Each structure has been optimised to a local minimum in their respective symmetry restrictions, i.e. $D_\mathrm{3h}$ for the monomers and $C_\mathrm{2v}$ for the dimers.
Analyses with the Natural Bond Orbital model (NBO6 program) and the Quantum Theory of Atoms in Molecules (QTAIM, MultiWFN) have been run on single point energy calculations at the M06/def2-QZVPP//DF-M06-L/def2-TZVPP level of theory.
A rudimentary energy decomposition analysis has been done on that level, too.

Energy decomposition analysis

The dissociation energy of the dimers $\ce{(XY3)2}$ to the monomers $\ce{XY3}$ is defined as the difference of the energy of the dimer $E_\mathrm{opt}[\ce{(XY3)2}]$ and double the energy of the monomer $E_\mathrm{opt}[\ce{XY3}]$ at their optimised (relaxed) geometries $\eqref{e-diss-def}$.
The interaction energy is defined as the difference of energy of the relaxed dimer and double the energy of the monomers in the geometry of the dimer $E_\mathrm{frag}[\ce{(XY3)^{\neq}}]$ $\eqref{e-int-def}$. That basically means breaking the molecule in two parts, but keeping these fragments in the same geometry.
The deformation energy (or preparation energy) is defined as the difference of the energy of the optimised and the non-optimised monomer $\eqref{e-def-def}$. This is the energy required to distort the monomer (in its ground state) to the configuration it will have in the dimer.
$$\begin{align}
E_\mathrm{diss} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{opt}[\ce{XY3}]
  \tag1\label{e-diss-def}\\
E_\mathrm{int} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{frag}[\ce{(XY3)^{\neq}}]
  %\ddag not implemented
  \tag2\label{e-int-def}\\
E_\mathrm{def} &=
  E_\mathrm{frag}[\ce{(XY3)^{\neq}}] - E_\mathrm{opt}[\ce{XY3}]
  \tag3\label{e-def-def}\\
E_\mathrm{diss} &=
  E_\mathrm{int} + 2E_\mathrm{def}\tag{1'}
\end{align}$$

Results & Discussion

The Monomers $\ce{XCl3; X{=}\{B,Al\}}$.

Let's just get the obvious out of the way: Boron is (vdW-radius 205 pm) smaller than aluminium (vdW-radius 240 pm). For comparison chlorine has a vdW-radius of 205 pm, too. That is pretty much reflected in the bond lengths and the chlorine-chlorine distance.
\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.0 &173.6&--\\
\mathbf{d}(\ce{Cl\bond{~}Cl'})&/\pu{pm} & 356.8 & 300.6 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}

From this data we can draw certain conclusions without further looking. The boron monomer is much more compact than the aluminium monomer. When we compare the bond lengths to the covalent radii (Pyykkö and Atsumi) we find that the boron chloride bond is about the length that we would expect from a double bond ($\mathbf{r}_\mathrm{doub}(\ce{B}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 173~\pu{pm}$). While the aluminium chloride bond is still significantly shorter than a single bond ($\mathbf{r}_\mathrm{sing}(\ce{Al}) + \mathbf{r}_\mathrm{sing}(\ce{Cl}) = 225~\pu{pm}$), it is still also much longer than a double bond ($\mathbf{r}_\mathrm{doub}(\ce{Al}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 191~\pu{pm}$).

This itself offers compelling evidence, that there is more π-backbonding in $\ce{BCl3}$ than in $\ce{AlCl3}$. Molecular orbital theory offers more evidence for this. In both compounds is a doubly occupied π orbital. The following pictures are for a contour value of 0.05; aluminium (left/top) and boron (right/bottom)

In numbers, the main contributions are as follows (this is just a representation, not the actual formula):
$$\begin{align}
\pi(\ce{BCl3}) &=
   21\%~\ce{p_{$z$}-B} + \sum_{i=1}^3 26\%~\ce{p_{$z$}-Cl^{$(i)$}}\\
\pi(\ce{AlCl3}) &=
   13\%~\ce{p_{$z$}-Al} + \sum_{i=1}^3 29\%~\ce{p_{$z$}-Cl^{$(i)$}}
\end{align}$$

There is still some more evidence. The natural atomic charges (NPA of NBO6) fairly well agree with that assesment; aluminium is far more positive than boron.
$$\begin{array}{lrr}
                    & \ce{AlCl3} & \ce{BCl3}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.4       & +0.3     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
%\mathbf{q}(\ce{X})~\text{[QTAIM]}  & +2.4       & +2.0    \\
%\mathbf{q}(\ce{Cl})~\text{[QTAIM]} & -0.8       & -0.7     \\\hline
\end{array}$$

The analysis in terms of QTAIM also shows that the bonds in $\ce{AlCl3}$ they are predominantly ionic (left/top) while in $\ce{BCl3}$ are predominantly covalent (right/bottom).

One final thought on the bonding can be supplied with a natural resonance theory analysis (NBO6). I have chosen the following starting configurations and let the program calculate their contribution.

The overall structures in terms of resonance are the same for both cases, that is if you force resonance treatment of the aluminium monomer. Structure A does not contribute, while the others contribute to about 31%. However, when not forced into resonance, structure A is the best approximation of the bonding situation for $\ce{AlCl3}$. In the case of $\ce{BCl3}$ the algorithm finds a hyperbond between the chlorine atoms, a strongly delocalised bond between multiple centres. In this case these are 3-centre-4-electron bonds between the chlorine atoms, resulting from the higher lying degenerated π orbitals.

This all is quite good evidence that the monomer of boron chloride should be more stable towards dimerisation than the monomer of aluminium.

The Dimers $\ce{(XCl3)2; X{=}\{B,Al\}}$.

The obvious change is that the co-ordination of the central elements goes from trigonal planar to distorted tertrahedral. A look at the geometries will give us something to talk about.

\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.7 &175.9&--\\
\mathbf{d}(\ce{X-{\mu}Cl})&/\pu{pm} & 226.1 &198.7&--\\
\mathbf{d}(\ce{Cl\bond{~}{\mu}Cl})&/\pu{pm} & 354.1 & 308.0 & --\\
\mathbf{d}(\ce{{\mu}Cl\bond{~}{\mu}Cl'})&/\pu{pm} & 323.6 & 287.3 & --\\
\mathbf{d}(\ce{B\bond{~}B'})&/\pu{pm} & 315.7 & 274.7 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}

In principle nothing much changes other than the expected elongation of the bonds that are now bridging. In case of aluminium the stretch is just below 10% and for boron it is slightly above 14%, having a bit more impact. In the boron dimer also the terminal bonds are slightly (> +1%) affected, while for aluminium there is almost no change.

The charges are not really a reliable tool, especially when they are that close to zero as they are for boron. In both cases one can see that charge density is transferred from the bridging chlorine to the central $\ce{X}$.

$$\begin{array}{lrr}
                    & \ce{(AlCl3)2} & \ce{(BCl3)2}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.3       & +0.2     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
\mathbf{q}(\ce{{\mu}Cl})~\text{[NPA]} & -0.4       & +0.1     \\\hline
\end{array}$$

A look at the central four-membered-ring of in terms of QTAIM offers that the overall bonding does not change. In aluminium they get a little more ionic, while in boron they stay largely covalent.

The NBO analysis offers a maybe quite surprising result. There are no hyperbonds in any of the dimers. While a description in these terms is certainly possible, after all it is just an interpretation tool, it is completely unnecessary. So after all we have two kinds of bonds in the dimers four terminal $\ce{X-Cl}$ and four bridging $\ce{X-{\mu}Cl}$ bonds. Therefore the most accurate description is with formal charges (also the simplest). The notation with the arrows is not wrong, but it does not represent the fact that the bonds are equal for symmetry reasons alone.

To make this straight: There are no hyperbonds in $\ce{(XCl3)2; X{=}\{B,Al\}}$; this includes three-centre-two-electron bonds, and three-centre-four-electron bonds. And deeper insight to those will be offered on another day.

The differentiation between a dative bond and some other for of bond does not make sense, as the bonds are equal and only introduced by a deficiency of the used description model.

A natural resonance theory for $\ce{(BCl3)2}$ gives us a overall contribution of the main (all single bonds) structure of 46%; while all other structure do contribute, there are too many and their contribution is too little (> 5%). I did not run this analysis for the aluminium case as I did not expect any more insight and I did not want to waste calculation time.

Dimerisation - yes or no

The energies offer us a clear trend. Aluminium likes to dimerise, boron not. However, there are still some things to discuss. I am going to argue for the reaction
$$\begin{align}
\ce{2XCl3 &-> (XCl3)2}&
\Delta E-\mathrm{diss}/E_\mathrm{o}/H/G&.
\end{align};$$
therefore if reaction energies are negative the dimerisation is favoured.

The following table includes all calculated energies, including the energy decomposition analysis mentioned at the beginning. All energies are given in $\pu{kJ mol^-1}$.
\begin{array}{lrcrcrcrr}
 \Delta & E_\mathrm{diss} &(& E_\mathrm{int} &+2\times& E_\mathrm{def}&)& E_\mathrm{o} &H &G\\\hline
\ce{Al} & -113.5 &(& -224.2 &+2\times& 55.4&)& -114.7 & -60.4 & -230.4\\
\ce{B}  &   76.4 &(& -111.2 &+2\times& 93.8&)&   82.6 & -47.1 &  152.5\\\hline
\end{array}

The result is fairly obvious at first. The association for aluminium is strongly exergonic, while for boron it is strongly endergonic. While both reactions should be exothermic, stronger for aluminium, the trend for the observed electronic energies ($E_\mathrm{o}$ including the zero-point energy correction) and the (electronic) dissociation energies reflect the overall trend for the Gibbs enthalpies.

While it is fairly surprising how strongly entropy favours association of $\ce{AlCl3}$, it is also surprising how it strongly disfavours it for $\ce{BCl3}$.
A look at the decomposed electronic energy offers great insight into the reasons why one dimer is stable and the other not (at room temperature).
The interaction energy of the fragments is double for aluminium than it is for boron. This can be traced back to the very large difference in the atomic partial charges. One could expect that the electrostatic energy is a lot more attractive for aluminium than it is for boron.
The deformation energy on the other hand clearly reflects the changes in the geometry discussed above. For aluminium there is a smaller penalty resulting from the elongation of the $\ce{Al-Cl}$ bond and pyramidalisation. For boron on the other hand this has a 1.5 times larger effect. The distortion also weakens the π-backbonding, which the additional bonding would need to compensate.
The four-membered-ring is certainly not an ideal geometry and the bridging chlorine atoms come dangerously close.

Conclusion, Summary and TL;DR:

The distortion of the geometry of the monomer $\ce{BCl3}$ cannot be compensated by the additional bonding between the two fragments. Therefore the monomers are more stable than the dimer. Additionally entropy considerations at room temperature favour the monomer, too.

On the other hand, the distortion of the molecular geometry in $\ce{AlCl3}$ is less severe. The gain in interaction energy of the two fragments well overcompensates for the change. Entropy also favours the dimerisation.

While size of the central atom is certainly the distinguishing factor, its impact is only severe on  the electronic structure. Steric crowding would not be a problem when the interaction energy would compensate for that. This is quite evident because $\ce{BCl3}$ is still a very good Lewis acid and forms stable compounds with much larger moieties than itself.

References

The used van der Waals radii were taken from S. S. Batsanov Inorg. Mat. 2001, 37 (9), 871-885. And the covalent radii have been taken from P. Pyykkö and M. Atsumi Chem. Eur. J. 2009, 15, 12770-12779.

Computations have been carried out using Gaussian 09 rev D.01 with NBO 6.0. Additional analyses have been performed with MultiWFN 3.3.8. Orbital pictures were generated with the incredible ChemCraft.

BH3 and AlCl3 each form "dimers" but by different processes.

AlCl3 is trigonal planar but in vapor phase at certain temperatures, it can dimerize to Al2Cl6 with a dH of ~ 46-63 kJ/mol. A few degrees above the boiling point, and the monomer is predominant. This happens by change of geometry from trigonal planar to tetrahedral and backbonding of Cl pi electrons to fill the octet of Al giving a dimer. This is possible due to the low relative energy difference between Cl and Al valence orbitals in the MO diagram. See http://pages.swcp.com/~jmw-mcw/Parsing%20the%20Bonding%20in%20the%20Aluminum%20Trichloride%20Dimer.htm for a good overview of bond angles and lengths.
BH3 is tetrahedral and forms a dimer  B2H6 but through a different process. The bonding is described by Cotton&Wilkerson as: "each boron atom uses two electrons and two roughly sp3 orbitals to form 2 center 2 electron (covalent) bonds to two hydrogen atoms. The boron atom in each BH2 group still has one electron and two hybrid orbitals for use in further bonding. The plane of the two remaining orbitals is perpendicular to the BH2 plane. Thus when two BH2 radicals approach eachother with the with the hydrogen atoms also in the plane, a B-H-B 3 center 2 electron bond is formed." The MO orbital diagram for the 3C2E bond requires close energy of the three atomic orbitals which is met between boron and hydrogen.
BCl3 does not form dimers as others have mentioned above, the difference in atomic radii between B and Cl explain the instability of the potential dimer. Additionally, larger difference in orbital energy would make such bonding unfavorable.


Compound having the highest enol content will be? The answer written in book is 3 but no solution is given.
After referring to some other question I think that hydrogen bonding can take place in compounds like 3 but am not sure about it. Am I right that it is the only reason why it has the highest enol content?

I would like to ask one more thing that, while comparing the keto and enol content, if in a compound more than one ketone is present, than do we enolize any one of them only to check?

If you can form an enol where the carbon-carbon double bond is conjugated to one or more other double bonds, the enol is more stable and the equilubrium mixture will therefore have more.  With that in mind, try seeing what happens to (3) if you tautomerize with the proton on the bridging carbon between the two carbonyl groups.

For a further exercise: render the structure for cyclohexa-2,4-dien-1-one where the carbonyl group is conjugated to two carbon-carbon double bonds in the ring.  Now tautomerize to the enol and see what "enol" you get.  When the conjugation of the enolic double bond forms an aromatic ring, the stabilization effect is so strong that the enol takes over completely.

(This is too elaborate to fit into a comment.)

Although one would, at first blush, immediately think of the β-diketone (3) as quite acidic, a moment's pause of what the putative enol should look like will confront you with a violation of Bredt's rule. Thus, it would prefer to enolize at the non-bridgehead hydrogens, and the acidity would likely not be quite different from the other three choices.


When to use Da or u for mass units Following up on the question Units of mass on the atomic scale,
Daltons (Da) and unified mass units (u) are equivalent, but which is used seems to depend on context.  For example, when discussing single atoms one typically sees the mass expressed as u, but when discussing large biomolecules one typically sees Da or kDa.

Is there a rule about whether to use Da or u? Or, is this simply a matter of convention or left to the writer's choice?

It doesn’t matter. Both units are basically identical, much like there is no everyday difference between the units mbar and hPa. There may be situations where one is preferred — e.g. hPa are, in my opinion, rarely used outside weather reports — but they should be understood nonetheless.

If there is anything to be said about the two, it seems using dalton is more popular in biologic contexts. But that wouldn’t be a tendency to base anything on.

Being consistent within one’s lab, within the journal one is publishing in or throughout one’s thesis is far more important.

Dalton (Da) is used to characterize polymers. For example, biochemists and molecular biologists specifically to discuss biopolymers such as nucleic acids or proteins.

Usually Da appears in the context "a 60 kDa protein band was observed on SDS-page" or "10 kDa cut off membrane was used for dialysis".

same applies to synthetic polymers.


Computational Chemistry Modelling Redox Potentials How could I accurately determine the Fe3+/Fe2+ redox potential in an aqueous solution? What would be a good computational model(s) for this?

Here is a non-electrolysis scenario that could be of interest. Add deoxygenated water to FeCl2 in a closed vessel filled with oxygen or air. Record the oxygen consumption.

Chemistry:

$\ce{4 Fe(2+)  + 2H2O + O2 (g) -> 4 Fe(3+) + 4 OH- }$

The wet ferrous chloride should be presented in a fashion to maximize oxygen exposure from air.

[EDIT] I would also place the system (till the reaction is completed) under pressure to limit adjustments owing to the speed of oxygen assimilation.

To be clear, as apparently required by a comment, the purpose of this experiment, per my understanding of the Question, is to serve as a frame for real world validation of a sophisticated theoretical/computational based model.


Geometries of SCl2(OCH3)2 and SF2(OCH3)2 I am not able to understand the difference between structure of $\ce{SCl_{2}(OCH_{3})_{2}}$ and $\ce{SF_{2}(OCH_{3})_{2}}$.

I am already well versed with Molecular Orbital and VSEPR theories. I found the geometries,
but don't understand why they are:

Why is the $\ce{F}$ at axial positions and $\ce{Cl}$ on the other hand in equatorial positions? (I am not able to understand Bent's Rule to be straight-forward, I have read Wikipedia already.)

Check the bond angles. The larger the angles around a bond, the more S character.
Electronegative elements prefer higher P character(Therefore less the bond angles).
Axial position has lesser bond angles.Hence higher electronegative element takes axial position.( F>O>Cl)

This concept however may not be applicable to every compound.

According to the Bent's rule the axial orbitals have pd hybridisation and the equatorial orbitals are sp2 hybridized. In the first example oxygen has more electronegativity than chlorine  hence it occupies the orbitals with lesser electronegativity. Similarly second example can be explained


Difference between an absorbance, emission, and excitation spectrum for uv-visible spectroscopy From the list below, circle the wavelength that would be the best setting for the UV-visible spectrophotometer used to determine the FITC concentration in the sample.
\begin{array} {}
\mathrm{wavelengths:} & \pu{400 nm}& \pu{450 nm}& \pu{500 nm}&\pu{550 nm} &\pu{600 nm}
\end{array}

The exam paper answer is $\pu{500 nm}$, because we need the wavelength of "maximum absorbance." But this graph shows two curves, and the absorbance curve is omitted.
But what is the difference between an emission, absorbance, and an excitation spectrum? What are they measuring? How do you interpret/read these graphs - those with multiple curves.

When you have an excitation spectrum and an emission spectrum, it is for fluorescence spectroscopy. To obtain these spectra, you hold either the emission wavelength or the excitation wavelength and scan the other to gather the information. The excitation spectrum shows at what wavelengths the solution uses to produce its fluorescence. The emission spectrum shows what wavelengths are given off from the solution.

For absorbency spectrum, the solution has all the wavelengths shining on the solution, and you monitor the wavelengths of interest. The absorbency shows which wavelengths are absorbed by the solution.

The excitation spectrum has the same peaks as the absorbency spectrum; although the absorbency spectrum may have more peaks.

The energy gap of absorption is greater than that of emission. The gradient in between is termed stoke's shift. This because when the excited electron looses energy, it first gets to zero point level of excited state, emits and hits the potential well of the ground state after which it looses energy again to retire to the zero point level of the ground state. Emission takes longer time than absorption


Hydrochloric acid vs hydrogen chloride? If one of them is an aqueous solution, in this case hydrochloric acid, why doesn't it have a different chemical formula ? How would I explain the reaction between hydrogen and chlorine without the use of water to make it acidic ?

Hydrochloric acid is an aqueous solution of hydrogen chloride.

You better know that the main difference is the state they are in. Hcl(g) is obtained without water, and it is highly corrosive-it even reacts with water vapor in the air. I don't know why they do have the same name.:)


Predict the products of the controlled oxidation of 2-methylpentan-2-ol Predict the products of the controlled oxidation of 2-methylpentan-2-ol.

From what I know, since 2-methylpentan-2-ol is a tertiary alcohol, there should be no reaction. However, my textbook states that the product includes 2-methylpentanone.

Edit: In response to the comments, my textbook neither specifies the isomer nor mentions any given conditions. I hate Nelson

Tertiary alcohols are resistant to oxidation in neutral or alkaline solution, but are readily oxidized by acidified oxidizing agents.

In your case two products will be acetone and propionic acid.

http://chemistry.tutorcircle.com/organic-chemistry/tertiary-alcohol.html


Organic Chemistry - Reaction scheme and structures of molecules Compound P and compound Q are straight-chain constitutional (structural) isomers with the
molecular formula C5H12O. Compound P can form optical isomers, whereas compound Q
cannot.
When reacted with concentrated sulfuric acid, compound P forms two products, compounds
R and S; compound Q forms only one product, compound S.

How come compound Q is pentan-1-ol and not pentan-3-ol?

Q is pentan-1-ol and not pentan-3-ol as on dehydration, of pental-1-ol, only one product (alkene) is formed. Whereas on dehydration of pentan-3-ol, we get two alkenes, as the alkene can show geometrical isomerism.


Carbocation stability order I was wondering whether the following order (from here) is correct:

In the first inequality, why did 9 hyperconjugations dominated resonance effect of benzene ring?
In the last inequality doesn't the 2 hyperconjugations of vinylic carbocation overpower the methyl carbocation?

In middle there is:
allyl = secondary

Why is there an equality here when resonance dominates inductive effect?

About the first two, it's an exception to the rule that resonance dominates hyperconjugation.

About the last inequality, the vinyl cation is sp hybridized which means it is highly electronegative, and a positive charge on it will be very unstable.

About the middle inequality, the benzyl carbocation (with five resonance structures) is more stable than allyl carbocation (two resonance structures). You may verify this here.

As for the 1st two:

Consider the 3 R's as independent(I am just explaining) ,each pushing a part of the electron cloud in its bond towards the central carbon through inductive effect as well as Hyper-conjugation. So in a way each R is reinforcing each other and an overall combined stability is achieved.

However consider  the resonating structures of Benzyl carbocation. If you research a bit deeper , you will find that of all the 4 resonating structures possible, the form that is given ( + on the external carbon) contributes to almost 90% of the original stable structure. This is because, if you take any other resonating structure( say + charge on a carbon inside the ring ), you will find that the structure is quite unstable and very highly tends to resort back to the + charge outside the ring ( after all, breaking a Benzene ring isn't very favorable) . So in a way , the resonating structures do not 'reinforce' each other so "Resonance" does not stabilize the way it is expected to do (as in case of similar resonating structures)

Talking of the last inequality:

The "Hyper-conjugation" that you mentioned wouldn't play the role here as you imagine. You might- have overlooked the fact that vinyl cation is least stable MOSTLY because it has a very high tendency to get converted to another structure, namely the left carbon(in your diagram) with a + charge and the right one with a lone pair. This is because ,then it will have many more hyper-conjugations to get stable. This is what makes vinyl carbocation least stable, it very very rarely ever exists or tries to exist in this particular form.

As for the middle equality,

Yes, in a way your doubt is right. (Since there is an equivalent resonance structure in the left one (as given) and that is a powerful one!!).However the thing you have missed while comparing the 2nd one is the fact that "Inductive" effect plays a 'minor' role in contributing to the stability of the secondary carbocation.The 'Major' role is played by the 7 "Hyper-conjugations" which is no less than a single resonance. Of course 1 hyperconjugation is less than 1 resonance but basic intuition will tell you that 7 hyperconjugations will have a fair chance to cope with 1 resonance.


If i keep adding sulfuric acid into water, will it keep dissolving forever? If i keep adding sulfuric acid into water, will it keep dissolving  forever?

If you remove the last two or so percent of water from concentrated $\ce{H_2SO_4}$, it starts fuming, giving off SO3 until it again contains enough water to be stable.

$\ce{H_2SO_4 <-> SO_3 + H_2O}$

Pure sulfuric acid does not exist, so there is no clear yes/no answer to your question. Also $\ce{H_2SO_4}$ (like all other mineral acids) does not just mix with water, but strongly reacts. So it's clear there can be no miscibility gap like with water/ether or similar pairings.

Yes, there is no chemical reaction between H2S04 and water. it is just a dilution of H2S04. If u want to add 1000 liter water in 1ml H2SO4 , U can add and there will be clear solution without an solid or layer


Is it valid to calculate concentrations given equilibrium constant and initial concentration? Suppose we have the following reaction:
$$\ce{2A <=> B + C}\tag{1}$$
which can be thought the sum of the following reactions:
$$\ce{A <=> B}\tag{2}$$
$$\ce{A <=> C}\tag{3}$$
with equilibrium constants $K_1$ and $K_2$ respectively. The $K$ for the first reaction will be the product of $K_1$ and $K_2$ i.e. $K=K_1K_2 $. Now if we are given that initial we have $x$ mol of $\ce{A}$ and no amount of $\ce{B}$ and $\ce{C}$, one can calculate the amount of B and C present at equilibrium using the equilibrium constant $K$:
$$K=\frac{y\cdot y}{(x-2y)^2}\tag{4}$$
If we solve (4) for a given $x$ (initial amount) then concentrations of $\ce{B}$ and $\ce{C}$ must be equal. But how can this be possible if $K_1$ and $K_2$ have different values? I mean the following equality must holds:
$$[\ce{C}]= [\ce{B}] \Longleftrightarrow K_1{[\ce{A}]} = K_2\ce[{A}] \tag{5}$$
Edit I am asking if it is valid to use the equilibrium constant $K$ to find the concentrations of $\ce{B}$ and $\ce{C}$ because every reaction can be thought of as sum of other reactions. It is a common exercise in many general Chemistry textbooks where you are given a reaction with an equilibrium constant and you must find the concentrations at equilibrium. Well the substitutions into equilibrium constant that they make is the one that I have also did (assuming 0 initial concentration for both $\ce{B}$ and $\ce{C}$. So what is going wrong?

The equation
$$\ce{2A <=> B + C}$$
implies a fixed stoichiometry between B and C. If the two separate reactions (2) and (3) were happening, the correct net equation would be
$$\ce{(x + y) A <=> x B + y C}$$
Here is an example that would be well-described by reaction (1):
$$\ce{2 H2O <=> H3O+ + OH-}$$
One water molecule cannot make a hydronium ion (goes against the conservation of mass and charge). The same goes for a water reacting to form a hydroxide. Instead, for every hydronium that forms, exactly one hydroxide is formed. So the separate reactions do not exist, and the separate equilibrium constants do not either.
Of course, there are other reactions where two independent products exist, such as linear glucose forming the alpha or beta anomer of the pyranose ring. In this case, there are two equilibrium constants, but there is no combined equilibrium constant because the two reactions are not the steps of an overall reaction, but rather two distinct reaction paths.

You have stated that the reaction:

$$\ce{2A <=> B + C}\tag{1}$$
[...] can be thought the sum of the following reactions:
$$\ce{A <=> B}\tag{2}$$
$$\ce{A <=> C}\tag{3}$$

However, this is not always the case. In particular, your assertion that the overall reaction (1) is a linear combination of reactions (2) and (3), is based upon the assumption that the rate of forward reaction ($r_f$) and rate of reverse reaction ($r_r$) are equivalent.
I have re-written your reactions (2) and (3) in more detail below (including the forward and reverse reaction rates):
$$\ce{A <=>[\ce{r_1}][\ce{r_2}]}B\tag{2*}$$
$$\ce{A <=>[\ce{r_3}][\ce{r_4}]}C\tag{3*}$$
Chemical systems are dynamic. At equilibrium, even though the concentrations specified by the reaction equation may not be changing, it does not imply that reaction is NOT occurring. In actuality, at equilibrium, all reactions are occurring, just at equal and opposite rates. If one looks at the equation for $K$, the equilibrium constant, one would see that this can also be expressed as the ratio of the forward and reverse reactions for the reaction system, such that:
$$K = \frac{r_f}{r_r} $$
In your case, you are assuming that in equations (2*) and (3*), that $r_1=r_3$ and $r_2=r_4$. Assuming this allows equations (2*) and (3*) to be "added" together. As such, these equations revert to the forms you've outlined in (2) and (3) and when combined they constitute equation (1).

Now, regarding the portion of your question:

The $K$ for the first reaction will be the product of $K_1$ and $K_2$ i.e. $K=K_1K_2 $. Now if we are given that initial we have $x$ mol of $\ce{A}$ and no amount of $\ce{B}$ and $\ce{C}$, one can calculate the amount of B and C present at equilibrium using the equilibrium constant $K$:
$$K=\frac{y\cdot y}{(x-2y)^2}\tag{4}$$

The equilibrium constant, $K$, is specified in terms of the equilibrium concentrations of each reaction constituent. For the reaction system below, we wouls specify $K$ as being,
$$\ce{2A <=> B + C}\tag{4*}$$
$$K = \frac{[B][C]}{[A]^2}$$
In particular, $[A]$, is the equilibrium concentration of A. Modifying $[A]$ to be dependent on the initial amount of A is contradictory, as $[A]$ should not be dependent on the initial amount of itself.


Preparation of picric acid My textbook says "Nowadays picric acid is prepared by treating phenol first with concentrated sulfuric acid which converts it to phenol-2,4-disulfuric acid and then with concentrated nitric acid to get 2,4,6-trinitrophenol"

My question: Why is this a better method than simply treating phenol with concentrated nitric acid to obtain picric acid?

What I think is that maybe because $\ce{NO2}$ is a stronger deactivating group than $\ce{SO3H}$, it is more difficult to add 3 $\ce{NO2}$ to a benzene ring directly.

But we would also have to substitute $\ce{SO3H}$ for $\ce{NO2}$. Wouldn't that be hard?

The benzene ring in phenol is highly activated toward electrophilic substitution and hence attempts to directly nitrate it result in charring and copious evolution of oxides of nitrogen. The reaction is highly exothermic and difficult to control.
To reduce the reactivity, the phenol is first mono-sulfonated ( some of the product which is substituted may also be used). The products are ortho- and para-isomers. The para-isomer is separated and then nitrated. The nitration is comparatively far smoother (easier to handle).
Ipso substitution of $\ce{SO3}$ groups occur.

Links: Wikipedia, Research paper
Link: (For ipso substitution): PDF, Wikipedia (see subsection: Ipso substitution)

Here is the small scale manufacturing process we are giving to our students in their first organic chemistry laboratory, for their synthetizing and destroying picric acid.
Take $1$ mL concentrated nitric acid $\ce{HNO3}$ ($65$%) in a $\pu{20 mL}$ test tube. Add with care, and in several portions, a tiny amount of phenol (maximum $0.02$ g). The reaction is highly exothermic ! Wait two minutes, then cool down the mixture. Add enough water to nearly fill the test tube (about $20$ mL). The picric acid is an oil, that soon gets solid after one minute or two. Throw the liquid away. Half fill the tube with $\pu{10 mL}$ water. Heat to $\pu{100°C}$ : the picric acid gets dissolved. Wait : nice yellow needles are produced in the tube by slow cooling. It is picric acid, which can be filtered in case of necessity. But don't do it presently. Decant the aqueous solution, and keep the solid in the tube
As picric acid is an explosive stuff when dry, don't dry it and don't save it. Instead, add $\pu{1 mL}$ water to the whole sample of picric acid from the preceding experiment. Add $\ce{1 mL NaOH 1 M}$. This produces a yellow solution of sodium picrate. Add one grain $\ce{Na2S}$. Heat for $2$ minutes. The solution darkens and becomes red-orange. The picric acid has been reduced into picramic acid ($\ce{2-amino-4.6-dinitrophenol}$), which is no more an explosive substance. $\ce{Na2S}$ has been oxidized into $\ce{Na2S2O3}$.


Is there any formula that can be used to find loss of mass due to binding forces in atomic and sub atomic particles? Atomic weight of Br-79 is 79.641 if you add the masses of protons and neutrons. However, in periodic table, it is less than the value given here. How is the difference arrived at for all the elements listed in the periodic table?

Yes, there is the Semi-empirical mass formula which will give you the binding energy/mass of the nucleus.

Once you subtract the binding energy/mass from that of the nucleons, you have the mass of the nucleus.  Then you should add the mass of the electrons.  Finally, in principle at least, you would subtract the binding energy/mass of the electrons (in other words the ionization energies of the electrons), but this is a very minor correction.

Yes - they call it the Theory of Relativity. E=mc^2

Take all the mass of the protons/neutrons lost in the nuclear reaction (you need to know your isotopes here). That mass is a tiny one, but then multiple that by the speed of light squared, and you have the energy created from the nuclear reaction.

Bang! Thats it! [A very big bang in fact...]


Nernst equation for a fuel cell when gases are stored separately I have run some tests with a PEM fuel cell and the voltage we have found was $1.44\ \mathrm{V}$. Given that the standard potential for a fuel cell is $1.229\ \mathrm{V}$ what is causing the difference in our measurement?

Initially we provide power to the fuel cell so it can generate oxygen and hydrogen into separate containers. These containers are submerged in water so there is water pressure keeping them contained.

All measurements were done in standard conditions, except that the pressure of each container is slightly different (hydrogen displaced twice the amount of water since there are twice the moles of atoms present).

I am not sure how to calculate this using the Nernst equation since the conditions of both the oxygen and hydrogen are slightly different (each gas is of a different concentration in its respective container since there is different water pressure and water vapour pressure present).

My equation is:

$$E_\text{cell} = 1.229\ \mathrm{V} - \frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}]}{[\ce{H2}][\ce{O2}]^{0.5}}\right)$$

For $[\ce{H2}]$ I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).

I also do the same thing for $[\ce{O2}]$

I'm not sure what to use for $[\ce{H2O}]$

The answer I am getting is $1.24\ \mathrm{V}$. What am I missing?

Given that the standard potential for a fuel cell is 1.229 V what is causing the difference in our measurement?

Multiple sources (e.g. here, here and there) estimate the voltage to be much less than the standard potential, especially when there is a load.

Individual fuel cells produce relatively small electrical potentials, about 0.7 volts, so cells are "stacked", or placed in series, to create sufficient voltage to meet an application's requirements.

So it is surprising that the voltage is higher than the standard potential. Perhaps two cells were stacked to provide about 1.4 V, similar to the voltage of a standard alkaline battery, and about twice what you would expect from a single cell under load.

For [H2] I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).

For the standard hydrogen electrode, the standard state is defined as a partial pressure of 1 atm. This is much lower than a concentration of 1 M. In any case, the Nernst equation does not predict a voltage that is significantly higher than the standard potential.

Seems your cell is operating as,
Anode: $\ce{H2 = 2H + 2e'}$
Cathode: $\ce{2H(+) + 2e' = 2H}$, and then $\ce{2H + 1/2O2(g) = H2O(l)}$ -Heat.

E = 1.4 V gives, 1.4 = [2.303RT/F] log R, where R = ratio H activity between cathode and anode R = 10^-23.65    if your anode assumes as activity H = 1.0, at the cathode H activity is 10^-23.65 the removal H by the oxygen as, $\ce{2H + 1/2 O2(g) = H2O(l)}$.  Atomic H is highly reactive as $\ce{H + H = H2}$ or $\ce{H + O = OH}$, or $\ce{2H + 1/2O2(g) = H2O(i,g)}$, etc.

$\ce{H2O(l)}$ blocks Oxygen contacting cathode (electrode reaction), i.e., $\ce{2H+ + 2e' + 1/2O2(g) = H2O(l)}$. Actually if current is OK, it is better do Fuel Cell operation in concentration cell that the oxygen is removal hydrogen by the H2O(g) vapor which is high T fuel Cell. Therefore we need high T Fuel Cell that is more appropriate in science and technology with Fuel Efficiency, CHP module.

Currently, I am developing such as:
HECE = HFC + HTEC
[H-eChemEng = H-fuel Cell + H-Thermal to Electric converter]

you have good measurement data!


Difference in acid strength of oxalic acid and malonic acid Despite very similar skeletal structures, the difference in acid strength between oxalic acid (ethanedioic acid) and malonic acid (propanedioic acid) is quite significant. What is the reason for that?

It seems that the inductive effect plays a role, but what is the explanation for the inductive effect in oxalic acid as there is same group on opposite sides?

How does the inductive effect plays different roles in both acids?

Linear saturated dicarboxylic acids
$n$ = number of $\ce{-CH_2}-$ groups in chain
 n                        pKa1     pKa2   Ka1/Ka2
 0  Oxalic acid           1.25     4.14   776
 1  Malonic acid          2.83     5.69   724
 2  Succinic acid         4.21     5.41    15.8
 3  Glutaric acid         4.34     5.41    11.7
 4  Adipic acid           4.41     5.41    10.0
 5  Pimelic acid          4.50     5.43     8.51
 6  Suberic acid          4.526    5.498    9.4
12  Dodecanedioic acid    4.45     5.05     4.0

Dodecanedioic acid
Aromatic carboxylic acids
                                   pka1     pka2    pka3
 benzoic acid                      4.202    ----    ----
 ortho-phthalic acid               2.89     5.51    ----
 meta-phthalic acid                3.46     4.46    ----
 para-phthalic acid                3.51     4.82    ----
 o,o'-bibenzoic acid
 p,p'-bibenzoic acid
 benzene-1,3,5-tricarboxylic acid  3.12     3.89    4.70

Linear unsaturated dicarboxylic acids
$n$ = number of $\ce{-CH=CH}-$ trans groups in chain
n                            pka1      pka2
0  Oxalic acid               1.25      4.14
1  Fumaric acid              3.02      4.38
2  trans,trans-Muconic acid  3.87
3

Straight-chain, saturated carboxylic acids
$n =$ total number of carbon atoms in molecule ($n \geqslant 2$ has terminal methyl group).
n                     pKa
1  Formic acid         3.77
2  Acetic acid         4.76
3  Propionic acid      4.88
4  Butyric acid        4.82
5  Valeric acid        4.82
6  Caproic acid        4.88
15 Pentadecanoic acid  4.8

most $\mathrm{p}K_\mathrm{a}$ data from Wikipedia.

How does the inductive effect plays different roles in both acids?

First think of what happens when the carboxylic acid group of a straight-chain, saturated carboxylic acids ionizes to form the carboxylate ion. The $\ce{C=O}$ and $\ce{-C-O^{-}}$ bonds of course hybridize. This hybridization occurs faster than hydrogen bonds to the solvent form. The net result is effectively two oxygen atoms with 1/2 a charge each. This increases the ability of the  carboxylate ion to solvate, which increases its acidity.
In malonic acid the $\ce{C-C}$ bond "shares" some of that hybridization. This is the "inductive effect." The additional delocalization of the electrons from the carbon atoms makes the carboxylate ions more electronegative than they would be in a long linear saturated dicarboxylic acid. So the first ionization ,  the second ionization

Acidic nature is directly proportional to strength of conjugate base.

In oxalic acid, the hydrogen oxalate ion and the oxalate ion (Which are conjugate bases in first and second dissociations) are resonance stabilized. The negative charge on oxygen atom is in conjugation throughout the molecule (with two pi bonds). Hence it is more stable.

In malonic acid, the conjugate base is relatively less stable, as the conjugation is restricted to only one pi bond.

Resonance effect is more stronger than inductive effect and moreover, resonance effect in the conjugate base of oxalic acid takes place throughout the molecule.(there is a negative charge, pi bond, pi bond, lone pair conjugation ).

Hence oxalic acid is relatively more acidic as compared to malonic acid.


Why oxides with the element in higher oxidation state are more acidic We see this trend of acidic-basic property of oxides in most cases:
$\ce{Mn2O7}$ is acidic, $\ce{MnO}$ is basic, the intermediate oxides are progressively less acidic and more basic, as the oxidation state of $\ce{Mn}$ decreases.

What is the reason of this general trend in almost all cases?

I have seen many explanations of this, most of them involving the proton releasing capacity of $\ce{E-O-H}$ bond, in the hydrolysed oxide. But most oxides of transition metals do not hydrolyze in water. So, I am seeking an alternative explanation without having to involve the hydrolysis of oxide. Also, most explanations on the internet lack any reference, so I prefer a reference.

It does not appear to be always true.  According to Wikipedia, copper(I) becomes acidic, forming an anionic hydroxide complex, at about 1.5 pH unit lower than copper (II).  The higher oxidation state of copper qualifies as less acidic.

In the bonds $M^{z+}-O-H$, where $M$ can be $Mn$ or $Cr$, the atom $M$ is positively charged. When this charge $z$ is large, $M$ repells the $H$ atom stronger than if z is small. So the molecule containing these bonds is a stronger acid. For example, $Z$ = $6$ in $H_2CrO_4$ and $Z = 7$ in $HMnO_4$. They both contain at least one $M-O-H$ bond. As $Z$ is high in these molecules, they are strong acids. Molecules containing $H, O$ and the same atoms $Mn$ and $Cr$ at a lower oxidation state exist. But they are not acidic in water. Examples : $Mn^{2+}$ makes $Mn(OH)_2$ which is insoluble in water and not acidic. $Cr^{3+}$ and $Cr^{2+}$ both produces $Cr(OH)_3$ and $Cr(OH)_2$ which are also insoluble and not acidic.


Le Chatelier's Principle and Rate constants Consider the reaction: $$\ce{H2O2(g) <=> 2OH^.(g)}\label{a}\tag{1}$$ to be at at equilibrium ($\ce{OH^.}$ is a radical). As the pressure is increased, the reaction moves towards forming $\ce{H2O2}$ as this direction reduces the amount of substance in the system.

However, it is well-established that the forward rate constant of $\ref{a}$ increases with increase in pressure, meaning forming more $\ce{OH}$. This seems to be in conflict to what the Le Chatelier's principle suggests. Could someone help me get my head around this?

It is true that the forward reaction rate increases as you increase pressure. However, the reverse reaction rate will increase as well, and it will increase more than the forward reaction rate.

The key here is that the system is in equilibrium; the forward and backward reaction rates are equal. Altering the pressure affects both reactions.

With increasing pressure volume of system increases and thus active mass increaeas  then to keep the constant value of Keq whole setup is disturbed in direction where no of moles of reacting substance decreases. Thus increment in rate of one reaction is preferred over other.


Removal of oxygen from pyridine N-oxide Under what condition do the reaction proceed with sole removal of oxygen in contrast with the product also ending up with chlorine substituent?

I'm familiar with both reaction mechanism, but can't seem to figure out when will a certain pyridine end up with or without chlorine. I guess one factor would be electron richness of the ring system, chlorine being deactivating subsituent ends up only on the electron rich pyridines.

I'm not sure whether $\ce{PCl3}$ can lead to the installation of a chlorine atom at C-2. I performed a cursory Reaxys search and in the first page saw several examples of deoxygenation without chlorination, but no examples of concomitant chlorination.

As far as I know, treatment of pyridine N-oxides with $\ce{PR3}$ simply leads to deoxygenation, with $\ce{P(III)}$ being oxidised to $\ce{P(V)}$.

However, if you were to treat pyridine N-oxide with phosphorus oxychloride $\ce{POCl3}$, then you would see chlorination at C-2.

If you have a source that says that $\ce{PCl3}$ can deoxygenate and chlorinate, please do inform us. Otherwise as far as I can tell, you need $\ce{POCl3}$ to effect a chlorination.

References

Joule, J. A.; Mills, K. Heterocyclic Chemistry, 5th ed.; Wiley: Chichester, U.K., 2010; pp 121, 155.
Weickgenannt, A. Pyridine N-Oxides, Baran Group Meeting, June 9, 2012; p 3.

I would argue that electron withdrawing groups will favour the oxidation to POCl3 while electron donating groups will favour the POCl path.

$$\ce{HPyN+-O- + PCl3 <=> HPyN+-O-PCl2 + Cl- <=> Cl-HPyN-O-PCl2}$$

With electron withdrawing groups the oxygen atom is more positively polarized than with electron donating groups, so let the bonds flip to the left:

$$\ce{Cl-HPyN-O-PCl2 + Cl- -> Cl- + HPyN + POCl3}$$

In the opposite case let the bonds flip to the right:

$$\ce{H-ClPyN-O-PCl2 -> H+ + ClPyN + POCl + Cl-}$$


Reverse of E1cB reaction As we know that the first step of E1cB step is second of E1 and vice versa. So can we say that by reversing E1cB reaction we will get E1 reaction.

In this the main problem which I am getting is in E1cB reaction carbanion is formed while in E1 carbocation. By reversing we will convert carbanion into neutral not into carbocation. So is this possible?

I asked my teacher about it and he told me that no it is not possible as these two are totally different mechanism.

Sorry for english or improper answer( if it is so) as my teacher told me this much only so I can't say much clearly about it.


What makes diazo compounds so unstable and explosive? I once had an Orgo TA refer to a diazo compound as "diazo-boom-boom" (the technical term).  I have always been curious as to the reason behind the instability and reactivity.

According to Wikipedia

  Some of the most stable diazo compounds are α-diazoketones and α-diazoesters since the negative charge is delocalized into the carbonyl.  In contrast, most alkyldiazo compounds are explosive

What is it about the alkyldiazos that makes them so much more unstable?  There doesn't appear to be any bond strain or other factors.  Naively, I can say that having a resonance structure should make it marginally more stable.

Well, your question is equivalent to “what is it about α-diazoketones that makes them so much more stable?”, which is easier to see. Compared to an alkyldiazo, the α-diazoketone has a resonance structure in which the negative charge goes to the ketone’s oxygen (and far away from the positively-charged nitrogen atom):

Because the oxygen is a quite electronegative element, the resonance form is quite stable and explains the extra stability of α-diazoketones. It is for the same reason that the protons in position α to the ketones are always more acidic than alky chain protons.

Coming back to alkyldiazo compounds, you have to realize that merely being able to write a resonance structure does not intrinsically imply stabilization: the resonance structure has to have some intrinsic stability factor. In the alkyldiazo, the resonance form you wrote is a carbanion, which is considered quite unfavourable unless it has a further stabilizing factor. Moreover, the most common reactions gives N2, which is a very stable compound… the reaction is thermodynamically very favourable.

Very short answer:

$\Delta G = \Delta H - T\Delta S$

As this is an org-chem question, no need to solve the equation numerically. As you might know, $\ce{N2_{(g)}}$'s standard formation enthalpy ($\Delta H_\text{f}^\circ$) is 0 – at standard $T$ and $p$, $\ce{N2_{(g)}}$ is gaseous. Therefore, its enthalpic contribution is 0. Its $S^\circ$ is $191.32\ \mathrm{\frac{J}{mol\ K}}$, which indicates that, at 273.15 K, $\Delta G^\circ_\text{f}$ is not only negative, but also large.

Thermodynamically, $\ce{N2_{(g)}}$ is favored. Take a look at nitrogen's phase diagram, generated by Wolfram Alpha.

EDIT

I've just noticed you're considering diazonium ions, not azo compounds. For the reasons F'x explained, the ions are more prompt to explosion than are the azo compounds. I think azo compounds are used in the fabrication of CDs, or so my teacher told us... They're supposedly light sensitive.


Products formed on cleavage of epoxides using HI There seems to be multiple forms of $\ce{HI}$ used to cleave epoxides:
conc. $\ce{HI}$, anhydrous $\ce{HI}$, cold $\ce{HI}$, hot $\ce{HI}$, cold conc. $\ce{HI}$, hot conc. $\ce{HI}$, etc.
So, does concentration of $\ce{HI}$ and temperature have any significance in determining the product?

Cleavage of epoxides can happen in 2 ways, resulting from the cleavage with the 2 carbons to which the oxygen atom is bonded.
If the medium is acidic, it will follow the SN1 pathway, proceeding via the most stable carbocation. This makes sense, since in acidic medium, the alcohol can get protonated and break away from one of the carbon atoms.
If the medium is basic, it will proceed via the SN2 pathway.
In other words, the type of cleavage doesn’t depend on the concentration of $\ce{HI}$, the only thing that matters is the acidity or basicity of the medium.

for unsymmetrical epoxides or oxiranes reaction takes place in this way:

in acidic medium- ring on reaction with HI ring opens from more hindered side i.e., the more hindered C-O bond. So the carbon in which more number of groups are attached which is bonded to the oxygen in ring gets -I.

in basic medium- the ring opens from less hindered side.

temperature is not a much important factor in this reaction.


What is the chemical structure / hybridization of sulfuric acid? I was looking at the chemical structure of $\ce{H2SO4}$.

Intuitively, I would have expected this molecule to be square planar in accordance with $p^2d^2$ or $sp^2d$ hybridization, but instead it is shown to be in a tetrahedral geometry consistent with $sp^3$ hybridization. Why is this?

I guess that an alternative way of asking this question is: What is the hybridization of the sulfur atom in $\ce{H2SO4}$?

It is much easier to explain it on simpler example – $\ce{O3}$ molecule. It has structure of resonance hybrid of $\ce{O=O+-O-}$ and its mirror. And of course, central atom has hybridization state $sp^2$. One bond here is normal covalent bond and another bond is dative: an electron pair is donated onto vacant orbital of $\ce{O}$ atom with all electrons paired. in $\ce{H2SO4}$ molecule to bonds are simple covalent ($\ce{S-OH}$ ones) and two are dative ($\ce{S-O}$ ones). A common concept of electron unpairing to my knowledge is proved brocken by quantum chemistry calculations and spectral experiments for hypervalent compounds of $\ce{P}$ and $\ce{S}$.

A little more interesting example is $\ce{XeF2}$ molecule, where three-atom four-electron bond $\ce{F-Xe-F}$ are formed, that can be think of as reasonance hybrid of structure $\ce{F-Xe+\ F-}$ and its mirror $\ce{F- \ Xe+-F}$

Of course, this scheme is still far from perfect, as reality is much more complicated, but if you do not wish to take course of quantum chemistry, it should be enough. However, I'll recommend to search for "MO LCAO model": it is quite simple and very useful. It is also often used in advanced chemistry books and articles.

Sulfur has four $sp^3$ orbitals (four $\sigma$-bonds) with oxygen and two $\pi$-bonds ($d$ orbitals).
In the case of $\ce{SO4^{2-}}$ ion, also sulfur also has four $sp^3$ orbitals and four $\pi$-bonds with $d$ orbitals. The electron from hydrogen is moves to sulfur's d orbitals


Inductive effects of halogens The Wikipedia page on inductive effects states that the order of inductive effects of halogens is as follows:

$$\ce{-F} > \ce{-Cl} > \ce{-Br} > \ce{-I}$$

From this, and from the fact that electronegativity of fluorine is higher than that of chlorine, I guessed that the acidity of p-fluorophenol is more than p-chlorophenol.

But the $\mathrm{p}K_\mathrm{a}$ values are:

p-fluorophenol: 9.8
p-chlorophenol: 9.4

Where did I go wrong? Why does this discrepancy occur? Detailed explanations are very much welcome.

This is due to the fact that we also consider +M effect . In fluorine it’s 2p orbital that participated in overlapping but in chlorine it’s 3 p orbital. 3p of chlorine does not overlap efficiently with 2p of benzene carbon so +m decreases in chlorine thus acidic nature increases

All your observations are correct.But one thing you seem to be missing is the presence of vacant d orbitals in Chlorine.I am sorry but I couldn't find pictures as such but I am sure you will understand after drawing out the resonance structures.

So when the lone pair on oxygen is delocalized and is at para position,then the d orbital of chlorine can "take up" and stabilize the compund.Flourine can't do the same.


Synthesis of ZnO by wet chemical method and reducing particle size by ball milling I synthesized bulk $\ce{ZnO}$ which is pure white in colour, but after high energy ball milling colour changes from white to yellow, why?

Figure 1: Zinc oxide (courtesy of Wikipedia).

I had a parallel experience and noted it in my thesis, 20 years ago.

Was working on an oxide ternary phase diagram (not wet chemistry, traditional shake and bake).  There were some conducting phases (dark) and some insulating ones (white).  I had a ZnO pellet (post firing) and I ground it under acetone (just mortar and pestle).  There was a definite color change to yellow.

I had a lot of other results and never ran this issue to ground.  I did note it and speculated that it had to do with the grain interiors having some conductivity (e.g. from oxygen defects or from impurities) and thus a different color, with the grain exteriors more insulating and white (not having the defects).  This would explain the color change.  There have been papers written (more than the 20 years ago) about differences in defects at grain boundaries and interiors of ZnO.

Very much a speculation at the time and never looked into it more.  But very interesting to see you have a parallel experience.  Maybe you will do more than I did and run it to ground?

P.s.  Consider the appearance of In2O3 and Ga2O3.  Indium oxide out of the jar is yellow and has semiconducting properties, from oxygen defects or the like.  Gallium oxide is much more insulating and is bright white.  Perhaps your ZnO grain interiors are a little In2O3 ish?  After all, both materials are dopable (despite the large band gap) and are TCO materials.  Ga2O3 on the other hand is extremely insulating and undopable, with a wider band gap.


Is [Co(NH₃)₄Cl₂]Cl paramagnetic or diamagnetic? $\ce{NH3}$ is known to be a strong field ligand, while $\ce{Cl}$ is known to be a weak field ligand.

Is $\ce{[Co(NH3)4Cl2]Cl}$ a high spin complex or a low spin complex?

I assumed this to be a high spin complex. Oxidation state of $\ce{Co}$ is $+3$.
And electronic configuration of $\ce{Co(III)}$ is $\mathrm{[Ar]~3d^6}$.

Since I assumed this to be a high spin complex — pairing of electrons of $\mathrm{3d}$ orbitals will not happen. The $6$ pairs of electrons from the ligands must be included in the $\mathrm{4s}$, $\mathrm{4p}$, $\mathrm{4d}$ orbitals — which leaves electrons unpaired in the $\mathrm{3d}$ orbital, making this a paramagnetic complex, with hybridisation $\mathrm{sp^3d^2}$.

Answer given to this question says otherwise. Hints are welcome!

This question is an excerpt of a question that appeared in the JEE(A) - 2016

The point is not really whether chloride or ammonia is a strong or weak field ligand, the point is $\ce{Co^3+}$ is $\mathrm{d^6}$, and virtually all "octahedral" $\mathrm{d^6}$ complexes are low spin - essentially some complexes of $\ce{Fe^2+}$  and a very small number of fluoro complexes of $\ce{Co^3+}$ are the only exceptions to the rule that all $\mathrm{d^6}$ octahedral complexes are low spin.

Why is this? Well let's use a bit of Crystal Field Theory (much of below stolen from https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy). One can consider the crystal field stabilization energy which is how stabilized an ion is compared to the free ion when in a given electronic state electronic and geometry. Here we are interested in high and low spin, and octahedral geometry. Taking a table of these values from the above link:

In the above $\Delta_\circ$ is the splitting energy of the orbitals. We can see that low spin $\mathrm{d^6}$ has the largest possible stabilization energy of any electronic configuration compared to the high spin case ($2\Delta_\circ$-P). Thus unless the splitting is very small octahedral $\mathrm{d^6}$ prefers low spin. So how do we get small splitting to see high spin $\mathrm{d^6}$? Well small splitting is favoured by

Low charge first row transition metals - for instance $\ce{Fe^{2+}}$
Very weak field ligands

And what were our exceptions above?

This is not a full answer, but you should note that halogens occupy the lower end of the spectrochemical series, and they are (almost) never able to act as strong field ligands. The magnitude of crystal field splitting energy of $Co^{+3}$ is low, such that any ligands barring the halogens do act as strong field ligands. This makes it a low spin complex.
 Therefore, in this case, $\ce{NH3}$ acts as a strong field ligand.


Why does the reaction between 2‐methylbut‐3‐en‐2‐ol and hydrogen bromide show allylic shift? In the following reaction

shouldn't the minor product be the major one as the carbocation formed before attachment of bromine is 3°? Also, can someone tell me why the product includes only nucleophilic substitution as there is no other condition mentioned to judge whether it should show elimination or substitution.

Please mention if what favours allylic shift in such type of reactions.

Even thought the immediate carbocation formed is tertiary, the stability of the double bond should also be taken into account. A more substituted double bond is also favored/stable. In this instance it seems that it's a competition between the double bond's stability and the carbocation's stability. There could be other situations where this is not the case, but you would need experimental data to back up stability (there's no universal ranking system that I know of to say when a double bond outweighs a positive charge).

In short, I would say that the relative stability of a tertiary double bond (compared to a tertiary carbocation) is what would justify the major product being labeled as it is.


How are related the electrons magnetic dipole moments in atomic orbitals? I wondering, has somebody ever try to take in account the electrons magnetic dipole moment for the shapes of the probability distribution in atom? Does this magnetic dipole moment play any role in the inner bounds of molecules?

In my paper, mentioned above, I have had some doubts about the role of the electrons dipole moment distribution in chromium atom which is quite symmetrical if I use my octahedral model instead of s- and p-orbitals.

Now a have read on Wikipedia about the magnetic characteristics of pure chromium: "Chromium is remarkable for its magnetic properties: it is the only elemental solid which shows antiferromagnetic ordering at room temperature (and below). Above 38 °C, it transforms into a paramagnetic state." This matches perfect into my model abot octahedral structure of neon and argon shell electrons.


Why does the dielectric constant of a solid increase due to a Frenkel defect? In crystalline solids, an atom may be displaced from its original position in order to create a vacancy in the original location. This is referred to as a Frenkel defect. Why does this defect result in an increase of the dielectric constant of the solid?

When Frenkel defect alone is present, there is no decrease in density, however
the closeness of the charge brought about by Frenkel defect tends to increase the dielectric constant of the crystal.
Compounds having such defect conduct electricity to a small extent.


Does DIBAL-H reduce carboxylic acids? Diisobutylaluminium hydride(DIBAL-H) reduces only carboxylic acid, ester or both?

The answer given was that it only reduces ester.

Is it correct that under any condition carboxylic acids are not reduced by DIBAL-H?

Does diisobutylaluminium hydride (DIBAL-H) reduce only carboxylic acids? or  only carboxylic acid esters? or reduce both both types of compounds?

The answer is yes, it does reduce both. Not only them, it also reduces nitriles to aldehydes, and is a more selective reagent than lithium aluminum hydride (LAH) in the reduction of nitriles (Ref.1). About reduction of carbozylic acid, Miller et al. (Ref.1) found that DIBAL-H can be used to reduce benzoic acid to give 72% yields of benzyl alcohol in 1959. In same publication, they have also shown that esters can be reduced to corresponding alcohols as well.

In reduction of nitriles to aldehydes, following results were achieved and compared with the result obtained by Amundsen and Nelson (Ref.2) using $\ce{LiAlH4}$:

The summery of reduction of carboxylic acids and their esters are illustrated in following scheme, where two different groups (Ref.1 & Ref.3) achieved two different results for ester reduction under different conditions:

References:

A. E. G. Miller, J. W. Biss, and L. H. Schwartzman, “Reductions with Dialkylaluminum Hydrides,” J. Org. Chem. 1959, 24(5), 627–630 (DOI: 10.1021/jo01087a013).
L. H. Amundsen, L. S. Nelson, “Reduction of Nitriles to Primary Amines with Lithium Aluminum Hydride,” J. Am. Chem. Soc. 1951, 73(1), 242–244 (DOI: 10.1021/ja01145a082).
L. I. Zakharkin, I. M. Khorlina, “Reduction of esters of carboxylic acids directly into aldehydes with diisobutylaluminum hydride,” Tetrahedron Letters 1962, (14), 619-620 (https://doi.org/10.1016/S0040-4039(00)70918-X).

Why DIBAL is weaker than LiAlH4 . as per mechanism mentioned in the site https://www.entrancei.com/questions-lialh4-reaction-and-mechanism
LiAlH4 is H- Donor , the RDS is H- release if we take DIBAL it consist of two electron donating group makes it e rich so i think it can donate H- faster than LiAlH4 . secondly i have seen in many book it can reduce all functional group which are reduce by LAH .


Adding hydrogen ion concentration while calculating pH I have just started learning how to calculate pH and I have noticed that adding [$\ce{H+}$] makes a negligible difference in few cases whereas in other cases it makes a major change in the final value. Why is that?

pH is calculated by a logarithmic formula given by:

$$\text{pH}=-\log{\ce{[H+]}}$$

Lets say, the initial hydrogen ion concentration is $\ce{[H+]_0}$, and change in concentration is $d\ce{[H+]}$. The change in pH would be:

$$d\text{pH}=-\log{(\ce{[H+]_0}+d\ce{[H+]})}$$

This can otherwise be written as:

$$d\text{pH}=-\log{\ce{[H+]_0}}×\frac{d\ce{[H+]}}{\ce{[H+]_0}}$$

You can see that the change in pH depends on the initial hydrogen ion concentration. It's in fact inversely proportional to the initial concentration. Adding a drop of concentrated acid to a base would give a much bigger change in pH compared to adding a drop to pure water, which in turn is bigger compared to adding a drop to an acid.


How to derive composite acidity constant for H2CO3*? I see the equation (19) and (20) on Stumm and Morgan's Aquatic Chemistry book (Fig. 1 here). They said "a combination of equations 19 and 20" gives Equation 21. But I don't know how the Equation (21) is generated? Because if I make summation of (19) and (20), it will be:

$$\ce{2H2CO3 <=> H2O + CO2(aq) + H+ + HCO3-}$$

[OP] Because if I make summation of (19) and (20), it will be: [...]

They meant they combine the mathematical expressions, not the chemical equations.

To express

$$K_\ce{H2CO3^\Huge{*}} = \frac{K_\ce{H2CO3}}{1 + K}$$

in terms of concentrations, you substitute equations (1) and (2) into the expression:

$$K = \dfrac{\ce{[CO2(aq)]}}{\ce{[H2CO3]}}\tag{1}$$
$$K_\ce{H2CO3} = \dfrac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}\tag{2}$$

So

$$\frac{K_\ce{H2CO3}}{1 + K} = \frac{\dfrac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}}{1 + \dfrac{\ce{[CO2(aq)]}}{\ce{[H2CO3]}}}$$

Multiplying top and bottom of the fraction by $\ce{[H2CO3]}$ gives:

$$\frac{\ce{[H2CO3]}}{1 + K} = \frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]} + \ce{[CO2(aq)]}}$$

If you want to make the intention clearer, you define a total concentration of carbon dioxide as

$$[\ce{H2CO3^\huge{*}}] = \ce{[H2CO3]} + \ce{[CO2(aq)]}$$

and then write

$$K_\ce{H2CO3^\Huge{*}} = \frac{K_\ce{H2CO3}}{1 + K} = \frac{\ce{[H+][HCO3-]}}{[\ce{H2CO3^\huge{*}}]}$$

I take perhaps a more intuitive view of the system, and in the case of, for example, Iron bicarbonate, it intuitively explains its formation in well water under pressure and its subsequent decomposition upon exiting a faucet at ground level.

I start with a modified cited equation (19) citing gaseous and not aqueous carbon dioxide in the equilibrium reaction:

$\ce{ H2CO3 (aq) <-> CO2 (g) + H2O (l) }$

I also cite Equation (20) and introduce dissociated CO2 as represented by carbonic acid:

$\ce{ CO2 (d) = H2CO3 (aq) = H+ (aq) + HCO3- (aq)}$

Now, at high pressure, examining the modified (19) above, per Le Chatelier's Principle, the position of equilibrium will move to produce more carbonic acid, and more bicarbonate per the second equation, resulting in the creation of Iron bicarbonate.

However, with the release of pressure, the bicarbonate equilibrium concentration shifts to create more carbonate acid, which can decompose releasing CO2 gas resulting in the ultimate removal of the bicarbonate ion and the decomposition of Iron bicarbonate, as reported.

Note, to preserve the total concentration of carbon dioxide in the system:

CO2 (d) + CO2 (g) = [CO2]


Does a positive electron affinity actually correspond to the release of energy? I read about the periodic properties — ionisation energy and electron affinity — in my book. It was quite intuitive that energy is required to remove an electron from the atom but I wonder why energy is released when an electron is added to an atom? Is it actually released or is it because of our convention that energy of electron at infinite distance from nucleus is 0, that we consider it to be released?

It may help to take a look at the two processes you are comparing in terms of reactions. Ionisation follows the following reaction:

$$\ce{A ->[][$\Delta E_\mathrm{i}$] A+ + e-}\tag{1}$$

This process involves tearing charges apart, separating a positive and a negative one. Even if the original atom was somewhat unstable, it will be unfavourable to remove that electron simply due to electrostatic interaction. You write in the question that you were able to deduce this already.

The electron affinity on the other hand describes the process in this following reaction:

$$\ce{A + e- ->[][EA] A-}\tag{2}$$

Here, a charged particle is approaching a neutral particle and we have the same number of charged particles (and the same charges) on both sides of the equation. This means that we don’t have a strong, obvious physical effect that will make all electron affinities sign-equal for very simple reasons. Instead, we need to consider the details more.

The process often involves a release of energy because most atomic configurations have space for extra electrons at little or no extra cost. Except for the noble gases and the second group, atoms don’t have completely populated sub-shells so there should always be a position available in a subshell at an acceptable exothermic energy level. While adding an electron will increase electron-electron repulsion and increase all energy levels by a certain amount, this increase is typically less than the first ionisation energy which corresponds to the energy level of the highest occupied atomic orbital. Thus in electron affinity, we in principle gain the entire ionisation enthalpy minus the destabilising contribution of the new electron. (Caution! This is a very simplified picture and not suitable for generalisation or even strictly for explanation. It gets the basic point across though.)

Therefore, most electron affinities are exothermic with the notable exceptions of group 2, group 18 and nitrogen.

This is because potential energy is more in this case when electron is further from nucleus as the forces between them are attractive. As the distance is decreased between nucleus and electron, potential energy decreases. Since energy is always conserved, the net decrease in potential energy is released as any form of energy. However the energy is not released in all the cases. It depends on the net effect of the interactions of the electron with the nucleus and the electrons that are already present in the atom.


Nucleophilic nature of ions Consider the following species:

  $$
\underset{(\text{A})}{\ce{\overset{⊖}{O}H}}
\quad
\underset{(\text{B})}{\ce{CH3-\overset{⊖}{O}}}
\quad
\underset{(\text{C})}{\ce{\overset{⊖}{C}H3}}
\quad
\underset{(\text{D})}{\ce{\overset{⊖}{N}H2}}
\quad
$$

  Arrange these species in their decreasing order of nucleophilicity.

  (a) C > D > A > B
  (b) B > A > C > D
  (c) A > B > C > D
  (d) C > A > B > D

My teacher taught in polar protic solvent nucleophilicity is inversely proportional to basic character. I assumed solvent to be $\ce{H2O}$ as solvent was not mentioned in the question.
As acidity order is

$$\ce{H2O} > \ce{CH3OH} > \ce{NH3} > \ce{CH4},$$

so basicity order should be

$$\ce{CH3-} > \ce{NH2-} > \ce{CH3O-} > \ce{OH-},$$

therefore using above proportionality I arrived at the nucleophilicity order

$$\ce{OH-} > \ce{CH3O-} > \ce{NH2-} > \ce{CH3-},$$

but you can see the option is not even given. Why is it so?

What you probably missed was that the order you mention is for the SAME GROUP elements. For example, down the group basicity of halide ions is-

F->Cl->Br->I-

But the the nucleophilicity order in polar protic solvents is-

I->Br->Cl->F-

This reverse in order is due to higher charge density of smaller halide ions and thus better interaction with polar protic solvents.

But along a period the charge density is similar and thus the interaction between solvent and nucleophile is almost similar. Thus, along a period the order follows basicity order.


Decomposition of alkene in acidified potassium permanganate A low boiling point liquid A, molecular formula $\ce{C5H10},$ reacts with chlorine to form B, molecular formula $\ce{C5H10Cl2},$ and with $\ce{HCl}$ (gaseous) to form substance C, molecular formula $\ce{C5H11Cl}.$ Compound A decomposes in acidifed potassium permanganate solution to form two acids D (molecular formula $\ce{C4H8O2}$) and E (molecular formula $\ce{CH2O2}$).
Find the systematic names of A, B, C, D, E.

I'd like to verify that my answers are correct and I would also like a good explanation of why the answers are what they are. My answers are:
A - pent-1-ene
B - 1,2-dichloropentane
C - 1-chloropentane (Minor), 2-chloropentane (Major)
D - butanoic acid
E - methanoic/formic acid
Currently, I doubt my answer because of C, where I got two possible compounds from applying Markovnikov's rule. Any help would be great.

gaseous hydrochloric acid or anhydrous,

we have an alkene where each side is attached to the same number of hydrogens —> both “equally substituted”.

Products in this case, 3-chloropentane and 2-chloropentane.


Is borazine aromatic? I came across this structure which has got pi bonds only in its canonical forms (which are also unstable compared to the Ist structure), is it aromatic?

Also, is it compulsory for an aromatic compound to contain a carbon ring? Can't a conjugate system without carbon be aromatic (as in this case)?

The modern definition of aromaticity from deep theoreticians is that the π-system needs to support aromatic ring currents. Borazine can support it, so it is technically aromatic.

Aromatic systems that do not contain carbon are not really all that common, but they are known. Pentazole was detected, for example. The problem is, that for aromaticity to be a thing we need a π-system, and π-systems effectively restrict us to BCNOS elements, since other elements are not eager to form π-bonds. Aromatic systems are usually 5- or 6-membered rings. Add in the fact that rings with a lot of N/O atoms often happily go kaboom after a funny look, and it's not surprising we have very few actual examples here. Still, some people insist on doing unreasonable things with results from as far back as the 50s.

Yes, Borazine is aromatic.

According to Hückel’s rules, a compound is aromatic if it satisfies three conditions:

Planarity
Complete delocalisation of π-electrons
Presence of $(4n+2)$ electrons where $n$ can be $1, 2, 3, \dots$.

Borazine is aromatic as there are 3 nitrogen atoms in the structure of borazine that donate two electrons each. Boron doesn't donate any, but it has an empty p-orbital available through which the electrons can delocalize. Hence, in total it has 6 π-electrons. Though the delocalisation is lesser than that in benzene, it still is aromatic.

FunFact: Borazine is also called inorganic benzene because of its resemblance to benzene by structure.


Can linear SOCl2 exist? Specifically, instead of its normal structure, this Lewis structure seems to be valid as a constitutional isomer:

$\ce{Cl-O-S-Cl}$

So what prevents this from existing, or if it already exists how is it made?

I highly doubt that it could exist. The co-linear arrangement, $C_{\infty\mathrm{v}}$, of $\ce{Cl-O-S-Cl}$ is a second order saddle point on the DF-BP86+D3(BJ)/def2-SVP level of theory on the $S=0$ potential energy surface. It has two imaginary bending modes, $\nu_1=371.39\mathrm{i}$, $\nu_2=108.23\mathrm{i}$. (Please note that they are actually perpendicular to each other.)

That basically means, that distortion along these modes will lead to an arrangement of the nuclei which is lower in energy. Since this is a second order saddle point, this arrangement can itself only be a transition state (first order saddle point), which also cannot be considered stable. (Compare also here.)
Therefore distortion among $\nu_1$ would probably lead to $\ce{Cl-O}$ and $\ce{S-Cl}$ fragmentation. While distortion along $\nu_2$ might preserve the $\ce{S-O}$ bond, fragmenting the chlorine atoms/ chloride ions - but that is pure speculation of a hypothetical case.

Let's try a very tempting ad-hoc explanation of why this is the case. In that case let us assume all atoms are $\mathrm{sp}$ hybridised and the bonding axis is $z$. The Lewis structure that seems to be valid has therefore three lone pairs ($\mathrm{sp}$, $\mathrm{p}_x$, $\mathrm{p}_y$) at each chlorine and two lone pairs ($\mathrm{p}_x$, $\mathrm{p}_y$) at the oxygen and sulfur. The three σ bonds are formed by the $\mathrm{sp}$ orbitals.
That would result in four lone pairs that are co-aligned in the same plane. Overlap is inevitable, and in the most simplest terms of VSEPR theory, we would call that lone pair repulsion. (Please note that for strictly VSEPR theory we would have to assume tetrahedral hybridisation.)
In other words, same spin electrons have to avoid occupying the same space and since the orbitals are so close together this will rise the energy a lot. You can probably call it Pauli repulsion.

The MO picture is too messy, so that I won't show it here. Basically most of the occupied orbitals are anti-bonding with respect to one of the bond. The HOMO is of π symmetry and anti-bonding with respect to every bond.

Not for long. Sulfur is in this case a reducing agent (love to give electrons away). And Oxygen is an odixidizing agent. It's just not stable or favorable. But sure you can make it.


What are the rate laws for each elementary step? Overall reaction

$$\ce{2NO + H2 -> N2O + H2O}$$

Mechanism

\begin{align}
\ce{NO + NO &-> NO2}\\
\ce{NO2 + H2 &-> NO + H2O}\\
\ce{N + NO &-> N2O}\\
\end{align}

I am aware that this is not a valid mechanism for the overall reaction however I am still supposed to find the rate law for each elementary step. I am having trouble determining what this rate law would be in step 3 of the mechanism.

For elementary reactions, the rate law can be directly read from the chemical equation. In your last equation this is a bimolecular reaction, i.e.
\begin{align}
\ce{N + NO &-> N2O} & r &= k\cdot c(\ce{N}) \cdot c(\ce{NO})
\end{align}

To be more precise here, one molecule of $\ce{NO}$ reacts with on atom $\ce{N}$, and in a wider sense, that is two (bi) molecules reacting. The order of the reaction is two.

That being said, there are a couple of things you need to pay close attention to. An elementary reaction is always a reaction between the number of molecules given as reactants, with exactly one transition state leading to the products. If this number or reactants is greater than three it is highly unlikely to happen this way, since a concerted transition state is statistically very unlikely. An elementary reaction therefore also has to be balanced, so your first reaction needs to be:
$$\ce{2NO -> NO2 + N}$$
In this case we could see $[\ce{ON\bond{~}O\bond{~}N}]^\ddagger$ as a transition state.

I very highly doubt that your second reaction as an elementary reaction, from my point of view, there are too many bond that would need to be broken. Instead, the following reactions might be possible and/or necessary decompositions:
\begin{align}
\ce{NO2 + H2 &-> NO + OHH}\\
\ce{OHH &-> OH + H}\\
\ce{NO2 + H &-> NO + OH}\\
\ce{2OH &-> H2O + O}\\
\ce{NO + O &-> NO2}\\\hline
\ce{NO2 + H2 &-> NO + H2O}
\end{align}
And there are more possibilities.

In all but one the elementary reactions are bimolecular - guess which one would be unimolecular.

As for the mechanism itself, I think these are reactions that occur, just not all of them, maybe not even the majority of them.

Remember that when a reactant is used in more than one step you have to use the rule of

rate of production = rate of consumption

This leads to $k1[NO]^2 = k2[NO_2][H_2]$

See where you can go from there remember in the last reaction its $k3[N][NO]$

Hope that helps.


Why does nitration of N,N-dimethylaniline occur at the meta position? The nitration of N,N-dimethylaniline with $\ce{H2SO4}$ and $\ce{HNO3}$ gives mainly the meta product, even though $\ce{-NMe2}$ is an ortho,para-directing group. Why is this so?

In the presence of these strong acids the $\ce{-NMe2}$ group is protonated, and the protonated form is electron-withdrawing via the inductive effect. This discourages attack at the electron-poor ortho position.

Under the conditions I know for that experiment, you get a mixture of para- and meta-product, but no ortho-product due to steric hindrance.

There is form of $\ce{NH3+}$ ion when sulfuric acid is in prior meta ($47\%$) but when sulphuric acid is in excess meta ($90\%$) . due to formation of $\ce{NH3+}$ ion ring become meta directing


Mechanism of hydroboration oxidation reaction I know that the hydroboration reaction is used to add $\ce{H2O}$ to the double bond in an anti-Markovnikov fashion.

I went to the Wikipedia page and found the following mechanism:

But I could not understand as to why the addition is syn.

The addition is concerted, i.e., everything happens at the same time; there aren't multiple steps to the addition. You can expect the boron to go on one face of the double bond while the hydrogen goes on the other face. There's just no way for the hydrogen to reach over to the other side for the anti-attack. It has to be syn.

It develops a cyclic transition state between C double bond:

  CH3     H
    \    /
     C--C
 H/       \H
  |         |
  |         |
  H---  ---BH2

Then replaces BH2 by OH


What does it mean that an electron pair "spends more time" nearer an atom? In a book I am reading, it is said

  In the $\ce{HCl}$ molecule, the shared pair of electrons spend more time nearer the chlorine atom. In the $\ce{HF}$ molecule, the shared electrons spend more time nearer the fluorine atom.

What does "spend more time" mean here? Has this been experimentally validated? If yes, how?

When you come to the quantum level (looking at one electron) things don't behave anymore like we normally think. To make it anyway understandable those expressions were invented such as "spending more time". Electrons are less described as a dot but as a smearing (electron density).
Chlorine and fluorine have higher electronegativities than hydrogen. The electron density therefore is not equally distributed between the two bonding partners but drawn towards fluorine/chlorine. To illustrate this, you can say "the electron spends more time at F/Cl".

This thinking was established as molecular orbital theory (s orbitals, p orbitals, ...). There have been other models before, e.g. you might know the shell model (k shell, l shell, ...). For experimental evidence you might want to take at look at this -- but most of all the citations therein: http://en.wikipedia.org/wiki/User:Chem507f10grp4/sandbox#Experimental_Evidence_Supporting_Molecular_Orbital_Theory
Basically, the shell model ran into a series of properties it did not calculate correctly or precisely. Molecular orbital theory was found to predict many properties and came to well calculated values.

On the question: 'What does "spend more time" mean here?', from a theoretical vantage, why explore approximations that are based on the time-independent Schrödinger equation?

Perhaps instead, this could be viewed as an application of the nonrelativistic version of the time-dependent Schrödinger equation for the wave function in position space of a single particle (a shared electron) subject to a potential (from an electric field from the respective cases of the chlorine and fluorine atoms).

Conceivably (for those more of an expert in this field), the product could be a theoretically based graphic depiction of the expected position over time. Visually, does the shared pair of electrons spend more time nearer the chlorine atom with HCl? In the case of the HF molecule, does the shared electrons spend more time nearer the fluorine atom?

As an example of one of the more readable discussions on the time-dependent Schrödinger equation in a position basis, complete with graphs, please see, for example,this source. Here is a good educational video describing the probabilistic nature of the quantum realm at here and also this. A related video is found here.


Chemistry underlying blacks spots in banana bread I've noticed that when I bake banana bread with certain brands of self-raising flour, the finished product looks more or less like any other cake.

However, when I use flour with sodium bicarbonate as the leavening agent, black spots appear peppered throughout the cooked cake.

What is the chemistry behind this phenomenon?

There are, at least, two reactions taking place and causing the brown colour of your bread. One is the oxidation of monophenols by an enzyme called PPO and the other is Maillard reaction, which is the same reaction that makes your steak looks deliciously brown.

Reaction 1. Oxidation of monophenols.

Bananas have a lot of polyphenol oxidase (also known as phenolase or PPO), an enzyme that catalyzes the oxidation of polyphenols naturally present in the banana fruit. The product is a colored ortho-quinone, which is brown or dark reddish. When you smash bananas to make a banana bread you end up releasing a lot of PPO which will catalyze the polymerization of the phenols and, therefore, make your bred looks darker then others. Below is a representation of the oxidation of monophenols into the colored ortho-quinone.

Reaction 2. Maillard reaction

The Maillard reaction is one of the most important reactions in food chemistry. Its products can add sensory features to many preparations, from steaks to coffee. Basically, sugars from the ingredients react with an aminoacid's amine group to form a glycosylamine. This compound is then isomerized to a ketosamine, which breaks into smaller substances. These substances are the Maillard products, and one class is called Melanoidins, responsible for the dark spots in your bread.
The reason why it is formed when you add sodium bicarbonate to the mixture is because alkaline media speed up the fragmentation of the ketosamine, yielding more Maillard products. 2

I have included an infographic from CompoundChem with other information and the Maillard reaction in chemical notation.

2Owusu-Apenten R., Introduction to Food Chemistry, CRC Press, 2004

It’s because bananas are acidic and when it reacts with baking soda it’s creates the black specks in the banana bread


Need help with stoichiometry problem I'm pretty new to chemistry and I've been stuck on it for hours.

Question: $\pu{10g}$ of the hydroxide of a metal on ignition gave $\pu{8g}$ of oxide. The equivalent weight of the metal is:
a) $\pu{136g}$
b) $\pu{40g}$
c) $\pu{56g}$
d) $\pu{28g}$

I used the law of equivalence but the answer I got was $\pu{3g}$. what am I doing wrong?

How I went about it:
10/E=8/16;
E=20;
E=M(equivalent mass of metal)+16+1;
M=3

I found your numbers confusing, but I think the first equation you wrote is not correct. This is how I'd solve it:

Let's say we have $\mathrm{n}$ equivalents of metal in the sample, and let's call the equivalent weight of the metal $\mathrm{e}$. For the hydroxide sample, we have:

$n(e+17)=10$

where $\mathrm{e+17}$ is the molar mass of the hydroxide $\ce{MOH}$.

For the oxide, we have:

$n/2(2e+16)=8$

where $\mathrm{2e+16}$ is the molar mass of the oxide $\ce{M_2O}$. Note that $\mathrm{n}$ becomes $\mathrm{n/2}$ because of stoichiometry:

$\ce{2MOH->M2O + H2O}$

So you have a system of two equations with two unkowns ($\mathrm{n}$ and $\mathrm{e}$), which you can solve normally. For instance, if you divide the equations:

$\cfrac{n(e+17)}{n/2(2e+16)}=\cfrac{10}{8}$

$\cfrac{e+17}{e+8}=\cfrac{10}{8}$

$8e+136=10e+80$

$2e=56$

$e=28$

So the equivalent weight of the metal is 28g.

Makes sense?

Nothing wrong with Vic's answer. I'd solve it differently, but hopefully you are familiar enough with algebra to understand that there are various ways to correctly come up with the correct answer...
Here's how I did it: There is a loss of 2 grams of what we assume is water.
Now since we know we started from M?(OH)? we assume the reaction was 2OH → H2O + O
meaning ~34 grams of OH would produce 18 g of water and 16 g of oxide oxygen.
Since we know that ~2 grams were lost (assumed to be only water, and all of the water) then we know  2g of water must have come from (2/18)*34 grams of OH (2:18 as X:34), or 34/9 grams of OH. Since we started with 10 grams total, we started with 90/9 grams total and of that 34/9 is OH. Subtract to determine how much metal: (90-34)/9 = 56/9 grams of metal. So far, so good. But what is its oxidation state? Since the question is asking for equivalent weight, we need some way to pin down the equivalency. Well, for every 18 grams of water lost, the residue retained (we assume) 16 grams of oxide. So, 2:18 as X:16 or x= 16/9.
So now, we have the whole enchilada: 90/9ths grams = 56/9 M + 16/9 O + 18/9 OH (as H2O - meaning this treats the OH lost during ignition as H2O rather than OH ) leaving us, after ignition with 56/9 M and 16/9 O. For each equivalent of O= there should be one equivalent of M. This means that since O has an equivalent weight of of 8 that 16/9 ÷ 8 = 2/9 Eq are present (after ignition) so that there must also be 2/9 equivalents of M.  56/9 grams ÷ 2/9 equivalents = 56/2 g/eq = 28 g/eq which is the eq. wt of the Metal.  (A quick search of the periodic table indicates that Fe with an AtWt of 55.845 and assuming we're talking about Fe(OH)2 would produce 2.00 grams of water, theoretically. So, it checks out.)


Use of fractional distillation in organic chemistry Can fractional distillation be used to distinguish between:
1) d/l form of a compound
2) geometrical isomers of a compoun (Cis and Trans)

The boiling point of an (R)-isomer is the same as the one of the enantiomer, the (S)-isomer.   Since the condensor does not offer a chiral environment to the chiral molecules, there is no discern of them and consequently, neither "single" nor "fractional distillation" separates them.   (You may expose the molecules to a chiral surface to adsorb, though, and separate the enantiomers by chiral gas chromatography; a specialized approach in analytical chemistry and -- because of the scale of amount typically accessible -- much less frequently deployed on a preparative one (a review).)
A successful separation of (cis)-(trans)-isomers by distillation of course depends on how different the boiling points of the pure compounds are.  Looking up the data for (trans)-stilbene (bp 305 C reference) and (cis)-stilbene (307 C at 1 atm reference) suggests other separation techniques may be much more suitable since less demanding.  This especially becomes relevant if you need a technique easy to scale, to manage and to be run continously (e.g., distillation).

fraction distillation means seperation of compounds on the basis of specificity of boiling point.so we cannot find the geometrical and stero isomerism using fractional distillation.
But we can find the form of the substance wheather it is gas or solid or liquid using fractional distillation.


Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached? To be honest, I just don't get it.
I had this question while reading a Chemguide article, on this topic:

I think my confusion is caused by the way I look at pKa of acid-base indicator. Should you think of it more as a reaction quotient. Because technically, at equilibrium, the two concentration aren't the same so cannot be cancelled out.

Look at the following equation:

  $$\mathrm{pH} = \mathrm pK_{\mathrm a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]}$$

At the half equivalence point, say we have 10 moles of $\ce{WA},$ and so there will be 5 moles of $\ce{SB}$ as the name suggests ("half equivalence"). So we know that the 10 moles will be neutralized to 5 moles of $\ce{WA}$ by the $\ce{SB}.$ If we do this then the $\ce{CB}$ is also now equal to the $\ce{WA}$ since it will increase by 5 moles.

$$\ce{[WA]~ =~ [CB]}.$$

Thus,

\begin{align}\mathrm{pH} &= \mathrm pK_{\mathrm a} + \log 1\\  &=\mathrm pK_{\mathrm a}\;.\end{align}

So this is why the the $\mathrm{pH} = \mathrm pK_{\mathrm a}$ at the Half-Equivalence Point.

  Vocabulary:

  SB : Strong Base. (eg. $\ce{NaOH}$)
  CB : Conjugate Base. (eg. $\ce{Ac^{-}}$)
  WA : Weak Acid. (eg. $\ce{HAc}$)

NOTE: We can never be 100% accurate here, this is just used to derive that $\mathrm{pH} = \mathrm pK_{\mathrm a}$

well, weak acids ionize partially n thus we should expect a half ionization . if the acid is xmole at the start, at the end point we should expect the conjugate base to be x/2mole leaving behind x/2mole of the acid. hasselbach equation applies to weak acid n bases


Most energy-dense commercial battery available What's the most energy-dense and currently commercially available chemical battery technology? Would it be the Lithium-Ion Polymer or has it been surpassed already?

There's is a chemical non-rechargeable battery which is more energy-dense than LiPo, even if it's less power-dense?

I'm trying to have a general sense of the current state-of-the-art technology, as the Wikipedia articles and tables seems to be very dated.

Edit: I meant "dense" as in respect to mass. I believe that "specific energy" would be the right terminology.

This has a list of common battery types with energy densities.

Lithium ion primary batteries appear to have the highest specific energy at around 300 Wh/kg. Lead-acid batteries have a specific energy around 10% that of lithium-ion batteries, but are extremely reliable (depth of discharge, self-discharge, etc.), and do not have nearly the same heating/fire issues (though they can explode if you mess with them), which is one of the reasons they have endured so long in automotive applications.

More exotic chemistries have higher specific energies, but I am unsure as to whether they are available to the average consumer.

Not neccesarily truth, I don't have exhausting knowledge in the area

  What's the most energy-dense and currently commercially available chemical battery technology

Assuming we talk about energy per mass unit.

In case one doesn't care about reversibility, it is air-metal batteries (say, alumunium-air). In case one does, it is probably Lithium-sulfide batteries.

Both are inferiour to common internal combustion engine, though.

Assuming we are talking about power per mass unit

Lead batteries are close to top, AFAIK. Otherwise they would be pushed out from cars already.


How can a neutral atom attract electrons when it's supposed to have zero charge? A neutral atom is an atom with an equal number of protons and electrons

We know the force of attraction between the proton and electron is equal
and assuming that:

one proton's positive charge attracts one electron.
The "neutral" in a neutral atom means electrically neutral.

How can a neutral atom attract electrons when it's supposed to have zero charge ?

An atom is only neutral when viewed as a single object from large enough distance. But as an electron comes closer to the atom, it "notices" the electron cloud first. This cloud also "notices" the electron and deforms—the atom polarizes—so as to keep the atomic electrons farther on average from the extra electron, since like charges repel. But this deformation leads to less shielding of the nucleus than in isolated atom, so the extra electron begins to feel a non-negligible attracting electric field.

The answer lies in electronegativity. When a proton attracts an electron, the electron doesn't magically suck out the charge of the proton. The proton's charge is still distributed in all directions. The reason why 1 proton on average can attract only 1 electron is because electrons push each other out.

Now let's first take H - it has 1 proton which attracts 1 electron. If another electron jumps in, only 1 electron stays in the end. Then He - it has 2 protons, so it attracts electrons even more. So even though electrons are fighting for the place, the nucleus charge is enough to hold them.

It gets interesting with Li. It should have 3 electrons, right? But 1st shell can take only 2 electrons, so the 3d electron must go to the 2nd shell which is further away. In such case the inner shell of electrons has a much greater effect on the outer electron, this is called electron screening, not to mention that the further you are from nucleus - the weaker the attraction. So even though Li has more protons than He, it's too weak to hold electrons on the 2nd shell, so some other atom will take the electron away and Li will be ionized and become Li$^+$ with only 2 electrons.

How strongly an atom wants new electrons is called electronegativity. It increases to the right of the periodic table because nucleus gets larger and larger and can hold on more and more electrons. In the last columns atoms want electrons so much that they can mug other atoms with weaker electronegativity.

But then the row of the table finishes and new row starts. At this point previous shell is completely filled and a new shell starts, and the electron screening kicks in again.

You can see these trends here. Electronegativity is the reason why Na & Cl can't form a molecule (covalent bond) - Cl (strong electronegativity) simply takes Na's (weak electronegativity) electron and both become ions: Cl$^-$ and Na$^+$. In the end they form an ionic bond instead.