What is the difference between "reaction in both directions" and "equilibrium"? According to the Wikipedia page on Chemical Equations:

  Symbols are used to differentiate between different types of reactions. To denote the type of reaction:
  "$=$" symbol is used to denote a stoichiometric relation.
  "$\rightarrow$" symbol is used to denote a net forward reaction.
  "$\rightleftarrows$" symbol is used to denote a reaction in both directions.
  "$\rightleftharpoons$" symbol is used to denote an equilibrium.

How is a "reaction in both directions" and an "equilibrium" any different? Aren't they supposed to be identical?

As far as I know, they are used indistinguishably and interchangeably.
One or another could be chosen to emphasize whether you are talking about chemical equilibrium or just trying to show that a reaction can take place in both directions depending on the conditions (which can be understood as completely the same...).

I saw the $⇄$ symbol used to specify some conditions for the reaction to take place in one direction and other kind of conditions for the reaction to occur in the other direction, and $⇌$ if conditions are the same for both reactions

For example:

First case is not an equilibrium, but it's true that the reaction can go both directions using the appropriated reagents in each case. But that could not be expressed using $⇌$.

The second reaction is an equilibrium, both reactions are happening continuously under the same conditions.
It also could be expressed using $⇄$.

That is the only difference I can tell. Like Martin pointed out, IUPAC does not give more hints about it, so the question cannot be answered for 100% sure.


Reimer–Tiemann-like reaction of catechol with diiodomethane I have studied the Reimer–Tiemann reaction of phenol with chloroform in presence of aqueous strong alkali. It takes place with intermediate dichlorocarbene.

I have also studied the Reimer–Tiemann "carboxylation" reaction, which requires carbon tetrachloride, and proceeds with intermediate carbocation.

I am however unfamiliar with what the following question requires from me:

$$\ce{Catechol + CH2I2 + NaOH -> ?}$$

Can anyone please state the step-wise mechanisms through which this reaction occurs? Thanks!

This is not really an answer to your question, but I do not think that it is possible to answer it. From one of the many comments we learn that the reference is actually the "preparation of piperonal". It uses 3-4-dihydroxy-benzenecarbaldehyde instead of catechol, but the reaction is same, with $\ce{CH2I2}$ and $\ce{NaOH}$.
I was able to find this reaction described in Chemistry of Natural Products
by Sujata V. Bhat, B.A. Nagasampagi, Meenakshi Sivakumar on pages 307f.

4.22.B. Synthesis of Piperine
Piperine was synthesized (Ladenburg, 1894) by the reaction of the piperic acid chloride with piperidine, which confirmed the structure of the molecule (Figure 4.57). The synthesis of piperic acid was achieved starting from piperonal 4.288, which was obtained from catechol using Reimer-Tiemann reaction followed by the condensation with diiodomethane in the presence of a base. Piperonal was condensed with acetaldehyde in the presence of sodium hydroxide and the product obtained was then heated with acetic anhydride and sodium acetate to yield piperic acid.

I find this a lot more in accord from what you would expect. I cannot imagine any way in which a Reimer-Tiemann (RT) like reaction would occur given the starting reagents. Instead the first step from catechol to 2,4-dihydroxy-benzenecarbaldehyde occurs via an RT reaction. This is very well discussed by H. Wynberg (Chem. Rev. 1960, 60 (2), pp 169–184, DOI: 10.1021/cr60204a003). The mechanism can also be found on Wikipedia.
The second step then involves diiodomethane and forms the acetal. I suppose after deprotonation of the phenolic hydrogen, it can attack at the diiodomethane in nucleophilic substitution fashion, further following another deprotonation and an intramolecular SN step to close the ring.

Unfortunately I could not find anything to further back up my suspicion, since the original synthesis only starts at 4.288. See here (in German): A. Ladenburg,  and M. Scholtz, Ber. Dtsch. Chem. Ges. 1894, 27, 2958–2960. DOI: 10.1002/cber.18940270356.
However, since there is proof that the reaction with 2,4-dihydroxybenzenecarbaldehyde forms the acetal, it would be highly suspicious if it were not doing a similar reaction with catechol.


How can placing a lone pair in axial position differs from that in equatorial in number of electron domains? I need help understanding the following statement in my book (regarding trigonal bipyramidal geometry):

  Placing a lone pair in an axial position would put it at 90 degree to three other electron domains. Placing it in an equatorial put is at 90 degree to only two other domains, thus minimizing the number of strong repulsive interactions

What does it mean by "90 degree" to two/three "domains"?

Here's a drawing of the situation.

In the example with the axial lone pair of electrons, when they say, "90 degree to three other electron domains", they just mean that the axial lone pair is situated 90° relative to 3 other substituent bonds - the 3 equatorial substituent bonds in this case.  When the lone pair is located in an equatorial position, then it has a 90° interaction with only 2 other bonds - the 2 axial substituent bonds in this case; the lone pair is 120° away from the 2 other equatorial substituents.


What type of glass are Pasteur pipettes made of? What type of glass are Pasteur pipettes made of? And at what temperature do they need to be heated to create a micro-pycnometer?

This page from the glass shop in the Department of Chemistry and Biochemistry at the University of Delaware lists the properties of several common glasses that are used in laboratory equipment.

The two most common types of glass for less expensive applications are soda lime glass and borosilicate glass. The anneal point (or glass transition temperature) can be thought of as the temperature above which the glass can start to have some molecular motion, but it is not a liquid. The soften point is probably what you are looking for.

$$
\begin{array}{lcc}
\hline
\text{Glass} & \mathrm{Anneal\ Point\ ^\circ C} & \mathrm{Soften\ Point\ ^\circ C} \\
\hline
\text{Soda Lime} & 545 & 724 \\
\text{Borosilicate} & 560 & 821 \\
\hline
\end{array}$$

Soda lime glass is cheaper, and is generally used when you want something very inexpensive or disposal. Borosilicate glass has a lower thermal expansion coefficient so it is less likely to break. If you want to know which type of glass your pipets are made from , look at the wide end from an angle. If there is a slight greenish cast to the curve of the glass, you have soda-lime glass. Or you could look at the box they came in.


Why only F, O and N form Hydrogen Bonds? Why only fluorine, oxygen and nitrogen can form hydrogen-bonds with the hydrogen of another molecule?

No, that's not necessary. Amongst the strongest hydrogen bonds are formed by $\ce{ N, O, F}$ because of their high electronegativity.

There's evidence that the ions $\ce{Cl-, I-, Br- }$ form hydrogen bonds that are much stronger than those of the covalently bonded atoms

$\ce{Cl}$ can form weak hydrogen bonds, but $\ce{Br}$ and $\ce{I}$ form very weak bonds if at all.

A sulfur atom can also function as weak acceptor., but the $\ce{SH-}$ ion forms much stronger bonds.

Hydrogen bonding has been directly observed between a negatively charged carbon and an $\ce{-OH}$ group in the same molecule

Isocyanides' ($\ce{R- N+ #C- }$) carbon atom can also act as an acceptor (forming a rather strong Hydrogen bond).

Reference: March's Advanced Organic Chemistry, Chapter 3: Bonding weaker than Covalent


Salt and sugar surfactants? How do daily compounds such as table salt ($\ce{NaCl}$) and sugar ($\ce{C6H12O6}$) affect adhesion or surface tension of water? Considering their polarity, I guess they both increase it.

The surface tension of water is increased by the dissolution of either glucose ($\ce{C6H12O6}$) or $\ce{NaCl}$.  The following are excerpts from this table:

Solution$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $  Temp ($\mathrm{^oC}$)$\ \ \ \ \ \ \ \ \ \ \ $  Surface Tension (mN/m)
Water$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $25$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 71.97
Water$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $0$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 75.46
6M aqueous $\ce{NaCl}$$\ \ \ \ \ \ \ \ $ 20$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 82.55
Sucrose (55%) + water$\ \  $20$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ 76.45
(Water is included at two temperatures in order to bracket the temperature of the mixtures.)

According to this Wikipedia article:

  Surfactants are compounds that lower the surface tension (or
  interfacial tension) between two liquids or between a liquid and a
  solid. Surfactants may [emphasis mine] act as detergents, wetting agents, emulsifiers,
  foaming agents, and dispersants.

The article further states that "surfactant" is a concatenation of the term "surface acting agent". Furthermore, according to the Encylopaedia Britannica :

  Surfactant, also called surface-active agent, substance such as a
  detergent that, when added to a liquid, reduces its surface tension,
  thereby increasing its spreading and wetting properties.

So, by these definitions and the data above, the compounds in question do not qualify as surfactants as they raise the surface tension of water.  Additionally, as they have no amphiphillic (having both hydrophilic and lipophilic properties) nature, they would not be considered surfactants in the common context of chemistry jargon.


N-fluorosuccinimide: any scope? All halogen derivatives (except fluorine) of succinimide exist and are used in various chemical processes. What makes N-fluorosuccinimde difficult to prepare? Still chemists have not been able to figure out the preparation of N-fluorosuccinimde and don't even bother to synthesize it since electrophilic/radical fluorination is accomplished by  N-fluoro-o-benzenedisulfonimide (NFOBS), N-fluorobenzenesulfonimide (NFSI), and Selectfluor. Moreover, even if NFS is successfully synthesized, can it be used as a fluorinating agent in electrophilic/radical fluorination?

According to Wikipedia, NBS is made by reacting sodium hydroxide and bromine in an ice-water solution of succinimide. The NBS product precipitates and is collected by filtration. I am assuming that NCS and NIS are made in the similar fashion (using appropriate halogens). But, why NFS has still not been made through a similar way (using fluorine)? Is the reaction condition unsuitable for fluorine or is there any other reason?

I was reading a forum discussing the scope of NFS as a fluorinating agent. It says:

    Does anyone know why is N-fluorosuccinimide (201281-58-5) such
    uncommon reagent for fluorination? I checked some major suppliers and
    none of them sells it.

  In my knowledge, N-Halogenesuccinimide are obtained by reacting of Succinimide with the halogen itself. This is valid for Chlorine, Bromine, and Iodine. Fluorine is too reactive. For the fluorine derivative,  iodinesuccinimide is treated with silver fluoride to exchange the halogen.

  The other option is to treat N-Hydroxysuccinimide with Thionyl
  fluoride or Phosphorous trifluoride.

Is this true? I couldn't find any source confirming the above synthesis and I am skeptical about the preparation process. Can NFS be made through the above process?

Maybe NFS doesn't work well?  Bear in mind that bonds from a second-period element to fluorine are generally stronger and less labile than the corresponding bonds to heavier halogens. If that's true, getting fluorine off a nitrogen-fluorine bond requires more work than, say, getting bromine off a nitrogen-bromine bond.

In https://en.m.wikipedia.org/wiki/Electrophilic_fluorination#Mechanism_and_stereochemistry several common fluorinating reagents are given, these being the compounds mentioned by the OP.  The listed compounds all withdraw electrons from the nitrogen-fluorine bond more strongly than the carbonyl groups would in NFS (they use sulfonyl groups or a cationic function), so it seems this attribute is necessary to get the fluorination to go.


How does oxygen fight odors? (Deodorant ad) I saw an ad on tv last night for Mitchum deodorant that showed big text on the screen that said "OXYGEN eliminates all odors!" or something similar. Their website indeed claims "Our exclusive formula releases pure oxygen, a powerful odor fighter...".

How does oxygen fight odors, and how can this aspect of the deodorant be more effective than the abundance of pure O$_2$ in the air?

In my estimation, this is basically marketing nonsense. If you examine the list of ingredients, the primary active agent is invariably some aluminum or aluminum-zirconium salt, which act as anti-perspirants primarily by the mechanical actions of shrinking pores (due to their astringency) and clogging ducts in the apocrine glands.

Listed among the inactive ingredients is hydrogen peroxide, $\ce{H2O2}$. At sufficient concentrations, $\ce{H2O2}$ can be a potent oxidizing agent, and probably the source of the oxygen the ad refers to. It might indeed be able to neutralize some odors by oxidizing the molecules which cause them, thereby transforming those molecules into odorless ones. However, as it's listed among the inactive ingredients, it's almost certainly present in very low concentration. This is to be expected, since prolonged skin contact will invariably cause chemical burns unless the mixture is very dilute. Furthermore, much of the odor of sweat is the result of metabolic byproducts from skin-dwelling bacteria, chief among them certain short-chain carboxylic acids that are already maximally oxidized. The only malodorous substances in sweat that can be neutralized by oxidation, as far as I can tell, are certain amines and thiols. That said, I doubt the concentration of $\ce{H2O2}$ is sufficient to be of any use here. $\ce{H2O2}$ does also have anti-septic properties, so perhaps its purpose is bacteriostatic, but there I'm just speculating. Ultimately, I'm uncertain of its purpose in the formulation, but I'm extremely dubious that it helps substantially in eliminating odor.


Calculate partial pressure of gases at equilibrium I have the reaction $\ce{CO_{(g)} + Cl2_{(g)} <=> COCl2_{(g)}}$ where $K=4.5\times10^9$ at $100\ \mathrm{^\circ C}$

The first question was to find $K_p$ which I know is $K_p=K(RT)^{\Delta n}$ where $R=0.08206$, $T=273+100=373$ and $n=1-2=-1$. Put in the numbers and get $K_p\approx1.47\times10^8$.

Now comes the second question and this is where I get stuck:

  Equal moles of $\ce{CO}$ and $\ce{Cl2}$ are reacted at $100\ \mathrm{^\circ C}$. If the total pressure at equilibrium is 5.0 atm calculate the equilibrium partial pressures of the gases.

I have tried several solutions, where I either end up with a wrong answer or hit a dead end. I just don't understand how to solve this.

The equations you need are
$$K_p=\frac{p_{\ce{COCl2}}}{p_{\ce{CO}} p_{\ce{Cl2}}}=1.47\cdot10^8$$
and
$$p_\text{tot}=p_{\ce{COCl2}}+p_{\ce{CO}}+p_{\ce{Cl2}}=5.0~\text{atm}$$
Additionally, $n_{\ce{CO}}=n_{\ce{Cl2}}\implies p_{\ce{CO}}=p_{\ce{Cl2}}$, making the system solvable.


Heats of combustion and stability of rings Which isomer would have the largest heat of combustion?
A) Propylcyclopropane
B) Ethylcyclobutane
C) Methylcyclopentane
D) Cyclohexane
E) Since they are all isomers, all would have the same heat of combustion.

Can anyone explain why?
If you think about it, without having the data of the heat of combustion of each type of bond, you can't really answer this question. I know that cyclopropane has a higher heat of combustion per bond than cyclobutane. However, that, in my sense of viewing the problem, does not prevent the fact ethylcyclobutane could have a higher heat of combustion as it is a cycloalkane with 4 bonds rather than 3.
It could happen that the three bonds in cyclopropane and the three bonds in the alkyl branch are lower in energy that the four bonds in cyclobutane and the two bonds in the alkyl branch.
For example:
Let's say a cyclopropane C−C bond is 10 J.
A cyclobutane C−C bond is 9.5 J.
A regular alkane bond is 8 J.
For propylcyclopropane:
Total energy is 54 J
For ethylcyclobutane:
Total energy is 54 J
The same! I know this sounds really simplistic and dumb, but I just don't understand how you can assume the answer is A.
How do you analyze this situation? Please help.

Compounds A-D all have the same molecular formula, $\ce{C6H12}$.  We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation

$$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$

Any differences in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).

This link provides the heats of combustion for some useful model compounds.  Look at the last column ("Total Strain") in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained.  The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.

Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is "A".


What are known examples of drugs that racemize/stereoconvert in vivo, and how are they converted? It is known that although only the (S)-enantiomer of the infamous sedative thalidomide possesses teratogenic properties, it is not very useful to administer the pure (R)-enantiomer since it is racemized within the body. Are there other known examples of drugs that are stereoconverted in vivo? How does the body perform the conversion?

Although I can't think of any drug examples other than thalidomide, here's information on thalidomide's mechanism:

The chiral carbon of thalidomide can tautomerize in basic conditions into an enol, which is achiral.  A reversal back to the ketone results in a mix of (R) and (S) enantiomers.

In the body, this tautomerization is generally catalyzed by basic amino acids.  Specifically, albumin is the main catalyst in humans.

While this is beyond the scope of your question, the reason that only (S)-thalidomide causes birth defects is that it can insert itself into DNA and suppresses certain genes necessary for embryonic development.

Reddy - Chirality in Drug Design and Development
http://www.ncbi.nlm.nih.gov/pubmed/9860497
http://www.ncbi.nlm.nih.gov/pubmed/9499573


Chemical based Water Purifier I am currently researching as too which is better for purifying a water sample. I have obviously tried the standard Google search for 'Aluminium Sulphate or Ferric Chloride' and have come up with lots of results. The only problem, is that most of these results are based more around peoples opinions or psuedo-scientific observations. So, to the question:

What chemical reasons are there for choosing Aluminium Sulphate over Ferric Chloride to refine a water sample?

On addition of inorganic coagulants like aluminium sulfate(alum) or Iron salts causes several chemical and physical interaction take place among particles.

This inorganic coagulants neutralize the negative charge on particles. Also metal hydroxide precipitates of the aluminum and iron salts ions begin to form.

This precipitates combine to form larger particles under natural processes such as Brownian motion

Aluminum hydroxides are formed within a fairly narrow pH range, typically: 5.5 to about 7.7. Iron (III) hydroxides can form over a larger pH range including pH levels lower than are effective for alum, typically: 5.0 to 8.5

Source wikipedia


Alkane nomenclature - choosing the parent chain - maximize or minimize substituents? Given structure:

Here's my go at it. The longest carbon chain has 10 Carbon atoms, so the root -- decane. Next: let's take the substituent in position 3, we get 3 ethyl. And finally, we get 5-(2, 2-Dimethylbutyl). Adding them all up:

5-(2, 2-Dimethylbutyl)-3-ethyldecane.

However, the solution in my book is given as follows:

And the answer given is: 5-(2-Ethylbutyl)-3, 3-dimethyldecane

What I don't get is: Why is the main chain taken by me incorrect?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44 SENIORITY ORDER FOR PARENT STRUCTURES

  A parent structure is defined (P-15.1) as a parent hydride, for example, methane, a functionalized parent hydride, for example, cyclohexanol, or a functional parent compound, for example, acetic acid. For choice of the preferred IUPAC name based on a senior parent structure, see P-45.

  (…)

  P-44.3 SENIORITY OF ACYCLIC CHAINS (THE PRINCIPAL CHAIN)

  In an acyclic compound consisting of individual atoms, alike or different (an acyclic chain), the chain on which the nomenclature and numbering is based is called the 'principal chain'. When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached.

  The principal chain:

  (…)

  (b) has the greater number of skeletal atoms;

  (…)

  P-45 SELECTION OF PREFERRED IUPAC NAMES

  P-45.0 INTRODUCTION

  Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. (…) Application of
  the criteria in this section will generate the preferred IUPAC name.

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has
  the maximum number of substituents cited as prefixes (other than 'hydro/dehydro') to
  the parent structure.

In accordance with P-44.3 (b), the principal chain is a decane.
It could be either “5-(2-ethylbutyl)-3,3-dimethyldecane” or “5-(2,2-dimethylbutyl)-3-ethyldecane”.

Because of P-45.2.1, the preferred principal chain corresponds to “5-(2-ethylbutyl)-3,3-dimethyldecane”, which has a principal chain with three substituents (5-, 3-, and 3-), whereas the principal chain of “5-(2,2-dimethylbutyl)-3-ethyl” has only two substituents (5- and 3-).


Silicon–Boron Covalent Bond This is a semiconductor-related question, but in order to understand the acceptor level energy state of boron in bilicon, I figured I'd ask here for hints.
Boron is a p-type material when introduced in a semiconductor like silicon. Boron forms three covalent bonds with silicon, leaving one silicon atom frustrated, not forming a bond. This doping process introduces the idea of the hole, that is, absence of electron.
However, my chemistry knowledge is limited, and I don’t understand what kind of bond exists when an adjacent silicon valence electron jumps in this electron vacancy. Is this new electron now shared by both silicon and boron, or does it belong to silicon, since silicon complies to the octet rule, whereas boron does not?
Why does the acceptor level reside above the silicon valence band edge and what does this level exactly represent? Does it represent the energy needed for an adjacent silicon valence electron to jump into this vacancy?

Image source: National Schools' Observatory

Using your diagram, the boron-silicon bond with only the silicon electron has only half of the possible electron content.  A "hole" as you said. The hole is not confined to the one $\ce{B-Si}$ bond, but can be in any of the 4 positions. It's not like the boron sits there and the silicon pops its electron into the potential bond; the bond is there and it's only half filled.
Now this isn't a molecule, but you can focus on the boron part of the crystal and say that all four adjacent silicons are bonded to the boron, but you have only 7 electrons where the "molecular" orbitals could contain 8.
Electrons don't really "belong" to one atom or another; they comprise some sort of sea of electrons, and the sea above this boron is a little lower than elsewhere. I think of the sea of electrons in a copper-boron alloy as being very fluid, and having a level surface everywhere, but the sea in a p-type semiconductor is more like mud (quicksand?) and can have some dips that could catch and puddle some water (~electrons). The sea is still controlled mostly by the silicon, just modified by boron (submerged holes).


Can carbon monoxide contaminate water? Just wondering if you fitted a cold water tank overflow pipe into a car park at the height of a car exhaust , if the carbon monoxide fumes went back up the oveflow pipe into the tank would this contaminate the water?

Contamination would occur if a sufficient quantity of carbon monoxide wouldn't just bubble through but

would react with water to form an unhealthy product or
is dissolved in water

Both isn't really true:

$\ce{CO}$ does not react under these conditions.
The solubility of $\ce{CO}$ in water at 20 °C is as low as $\mathrm{30\,mg\cdot L^{-1}}$ (GESTATIS database)


Why is the Still-Gennari reaction Z-selective? Why is the Still-Gennari reaction Z-selective?

I guess dipole interactions have the biggest influence here. They somehow destabilize the reversion of the oxaposhpetane to the betaine and prevents formation of the more stable E isomer but I am not sure.

As you hopefully realise, the stereochemical outcome of the HWE reaction (and variations) is due to the relative orientation of the R-groups in the phosphaoxetane intermediate.
If the R groups are syn as in 5A then the (Z)-olefin is formed, with an anti arrangement 5B favouring the (E)-olefin as the product.

Source: Modern Carbonyl Olefination, Wiley, 2004

The pertinent question when looking at the selectivity is why do we get dominant elimination from the syn phosphaoxetane (leading to the correct olefin), rather than dominant elimination from what is arguably the sterically most favourable anti phosphaoxetane (in which the R groups are pointing away from one another).
As it turns out, the initial 'aldol' addition into the aldehyde is rate determining, with the anti addition being favoured over the syn addition on steric grounds.

Source: Stereoselective alkene synthesis, Springer-Verlag, 2012   (a really useful book, FYI)

Once the aldol addition has taken place, the subsequent steps are rapid due to the high electrophilicity of the Still-Gennari phosphonate (with the additional electron withdrawing groups).
This means that the entire process is essentially controlled by the initial addition to give a syn or an anti adduct. Once formed, the adducts are sufficiently short lived that despite the cis phosphaoxetane being less favoured due to the clash between R-groups, the (Z) olefin is formed via quick and irreversible elimination.
The HWE therefore is (E)-selective, in part, due to the fact that elimination is less rapid and as such there is equilibration to the more favourable phosphaoxetane from which elimination may occur.


In the last step of Kjeldahl Method, is the reaction of ammonium to NaOH not taken into account? In the last step of Kjeldahl method, the unreacted $\ce{HCl}$ in the solution is back titrated with a standard $\ce{NaOH}$ and the concentration of $\ce{HCl}$ that has been used to neutralize the ammonia is then calculated. However, other than $\ce{HCl}$, there also exist an acid in the solution that can be titrated with $\ce{NaOH}$ which is the protonated form of $\ce{NH3}$. How is this not considered when calculating the nitrogen presence in a sample?

The Kjeldahl method has two versions, using $\ce{HCl}$ as trapping solution or a weak acid like boric acid. Depending which one is used, the nitrogen content is either determined by a back titration with $\ce{NaOH}$ (your case) or a direct titration with $\ce{HCl}$.

The goal of the titration is to figure out how much ammonia (gas) has been trapped in the acid. The equation for the trapping step is:

$$\ce{NH3(g) + HCl(aq) -> NH4+(aq) + Cl-(aq)}$$

So the trapping solution will get a bit less acidic because some of the $\ce{HCl}$ is neutralized by the weak base ammonia. To quantify the ammonia, you now titrate the remaining $\ce{HCl}$ with the strong base $\ce{NaOH}$:

$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + Na(aq) + Cl-(aq)}$$

The solution is titrated to about $\mathrm{pH} = 5$ as endpoint (at the equivalence point, the strong acids and bases are neutralize, and $\ce{NH4Cl}$ solution remains). At $\mathrm{pH} = 5$, very little ammonium is deprotonated to form ammonia ($\mathrm{p}K_\mathrm{a}$ is about 9.3). In other words, ammonium/ammonia does not act as a buffer, so it does not affect the titration. You might even call it a spectator ion at this $\mathrm{pH}$.

After reaching the endpoint, the amount of $\ce{HCl}$ originally present in the trapping solution is equal to the sum of the amounts of $\ce{NaOH}$ and ammonia added. The overall reaction is (non-standard notation):

$$\ce{HCl(aq) + x NH3(g) + (1-x) NaOH(aq) ->}$$
$$\ce{Cl-(aq) + x NH4+(aq) + (1-x) H2O(l) + (1-x) Na(aq)}$$

From that, you can calculate the amount of nitrogen in the sample.

  However, other than $\ce{HCl}$ there also exist an acid in the solution that can be titrated with $\ce{NaOH}$ which is the protonated form of $\ce{NH3}$.

As explained above, if the $\mathrm{pH}$ does not approach the $\mathrm{p}K_\mathrm{a}$ of ammonium, ammonium will not react with $\ce{NaOH}$ to an extent that matters. If you choose to end the titration when the solution reaches a $\mathrm{pH}$ of 11 (to give an example), however, the ammonium will be deprotonated to ammonia, invalidating the result. Worse, it might come out of solution, so it becomes a problem that can't be fixed.

Sources:

http://www.brooklyn.cuny.edu/bc/ahp/SDKC/Chem/SD_KjeldahlMethod.html
J Martín, Fernández Sarria L and Agustín G. Asuero (2017) DOI: 10.5772/intechopen.68826

Wikipedia on the other hand I think got it wrong (https://en.wikipedia.org/wiki/Kjeldahl_method, retrieved before I edited it):

  Ammonium ion concentration in the standard acid solution, and thus the amount of nitrogen in the sample, is measured via titration. If boric acid (or some other weak acid) was used, direct acid-base titration is done with a strong acid of known concentration. $\ce{HCl}$ or $\ce{H2SO4}$ can be used. Indirect back titration is used instead if strong acids were used to make the standard acid solution: strong base of known concentration (like $\ce{NaOH}$) is first added in excess, and then the excess is titrated with a strong acid of known concentration. Titration of ammonia absorbed in boric acid solution thus has an advantage that only one standard solution is needed. One of the suitable indicators for these titration reactions is Tashiro's indicator.

They say to add an excess of $\ce{NaOH}$ and then titrate with $\ce{HCl}$. This would drive ammonia out of solution. Also, this is not a back titration but a double-back titration (which is not a thing, I think).


Adiabatic isobaric irreversible process I cannot understand that

  in an irreversible adiabatic isobaric process the enthalpy change is $ \Delta H <0$.

Could anyone please help me how to prove it.
Any help is appreciated.

For an adiabatic isobaric process the work done by the system during a spontaneous expansion is $w= -p_{\text{ext}} \Delta V $.  Since the heat $q=0$ the work is also equal to the change in internal energy, ie $w= \Delta U$. The associated enthalpy change is then

$$\begin{align} \Delta H &= \Delta U + \Delta (pV) \\ &= w + p_fV_f-p_iV_i \\ &=  -p_{\text{ext}}\Delta V+ p_{\text{ext}} V_f-p_iV_i \\ &=  p_{\text{ext}}V_i-p_iV_i\\&=(p_{\text{ext}}-p_i)V_i \end{align}$$

In order for the expansion to proceed spontaneously the initial pressure of the system $p_i$ must exceed that of the surroundings $p_{\text{ext}}$, which means $\Delta H<0$.

Since you are computing $H$, a property of the system, when computing the change in $H=U+pV$ you should use the pressure of the system, not of the surroundings. Before the expansion the pressure of the system (and surroundings) is $p_i$, while at the end of the expansion it is $p_f=p_\text{ext}$, since the system is then again in mechanical equilibrium with the surroundings. Note that previous to the expansion the system and surroundings are also in mechanical equilibrium. The pressure of the surroundings is then suddenly dropped to a lower constant value. The following figure illustrates the process:


How to find the molar mass of a compound given mass, vol, temp, and pressure? A $2.1\ \mathrm g$ sample of a liquid vaporizes and exerts $120\ \mathrm{mmHg}$ pressure at $1.5\ \mathrm L$ volume and $80\ \mathrm{^\circ C}$ temperature. Find the molar mass of the gas.

I first get the equation $PV=nRT$ and replace $n$ with $m/M$ since $M=m/n$
Then the equation becomes this $$PV=\frac{mRT}{M} \Rightarrow M=\frac{mRT}{PV}$$

$$\begin{align}
M&=\frac{mRT}{PV}\\[6pt]
\Rightarrow M&=\frac{(2.1\ \mathrm g)\times(62.364\ \mathrm{L\ mmHg/(mol\ K}))\times(80+273.13)\ \mathrm K}{(120\ \mathrm{mmHg})(1.5\ \mathrm L)}\\[6pt]
\Rightarrow M&=257\ \mathrm{g/mol}
\end{align}$$

But the answer key says $150\ \mathrm{g/mol}$. What am I doing wrong?

Sometimes (too often) answer keys are wrong. In this case we've now triple checked the calculation, and you, Jan (as implied in his comment) and myself have all come up with a molar mass of $\pu{257 g/mol}$. At this point, we have to say the answer key is wrong and you are correct.

Your answer is $\pu{1.71}$ times that given by the key, and neither that nor any other comparison appears to correlate to any obvious error on their part. It simply appears to be a misplaced or incorrectly entered answer in the key, end of story and move on. And good job getting the correct answer ;)


How do you figure out the dissociation energy for ions? Specifically $\ce{KF}$ into the $\ce{K+}$ and $\ce{F-}$ ions. I can find tables on the DE into atoms, but where would I go to find the DE between ions?

I am also given that the bond length is 2.17e-10m. Can I use that at all?

If you are trying to find the heterolytic dissociation energy for a gaseous $\ce{KF}$ molecule then as a rough estimate you can model the ions as point charges and calculate the energy required to separate them from their equilibrium separation (the bond length) to infinity, using Coulomb's Law: $$E = -\int_{2.17\times10^{-10}}^\infty\frac{Qq}{4\pi\epsilon_0r^2}dr$$

Note that modelling the ions as point charges is a major simplification and so this is probably not a very accurate estimate. Unfortunately I can't find any data for the actual energy so I can't say how accurate this is.


Why can't sulphate ion be discharged in the anode of electrolysis? Given that there is an aqueous solution of sulfuric acid ($\ce{H2SO4}$), $\ce{2H+}$ is reduced in the cathode and $\ce{OH-}$ is reduced in the anode. Why is $\ce{OH-}$ preferentially discharged over $\ce{SO4^2-}$ in the anode?

One explanation I've found is that because $\ce S$ in $\ce{SO4^2-}$ has an oxidation number of $+6$ which is a maximum and thus cannot be oxidized any further. However, can't the $\ce O$ in $\ce{SO4^2-}$ be oxidised instead as it has an oxidation number of $-2$ (just like the case in $\ce{OH-}$ where $\ce{O^2-}$ is oxidised and $\ce{H+}$ is not oxidised according to the half equation $\ce{2OH- <=> 1/2 O2 (g) + H2O (l) + 2e-}$)? Therefore, the argument that $\ce S$ cannot be oxidised because it has an oxidation number of $+6$ doesn't seem to be valid.

In aqueous solution the anode will oxidize the water instead, as this reaction can be done at much lower potential than oxidizing the sulfate ion.

Sulfates can be oxidized if you take the water away.  Potassium peroxydisulfate can be prepared by electrolyzing a solution of potassium bisulfate ($\ce{KHSO4}$) in sulfuric acid solvent.


Grasping the concept of Electronic Spin, Effective Spin and Fictitious Spin Trying to learn alone some aspects of quantum mechanics is, sometimes, a struggle. Reading the excellent paper by Piwowarska [1] I was hoping to, finally, understand what is the origin of the so-called “fictitious spin $1/2$” in a $\mathrm d^7$ cobalt ion.

Fig. 2 Visualization of distinction between the electronic, efective, and fctitious spins and respective spin Hamiltonians in terms of the spin levels of $\ce{Co^2+}(\mathrm{3d^7})$ ions

If my cobalt is high spin $(3/2$ is the electronic spin $S),$ then the effective spin $(\widetilde{S})$ is also $3/2,$ but where is the fictitious spin $({S}')$ $1/2$ arising from if the molecule is, in fact, high spin?
Reference

Piwowarska, D.; Gnutek, P.; Rudowicz, C. Origin of the Ground Kramers Doublets for $\ce{Co^2+}(\mathrm{3d^7})$ Ions with the Effective Spin 3/2 Versus the Fictitious ‘Spin’ ½. Appl Magn Reson 2019, 50 (6), 797–808. DOI: 10.1007/s00723-018-1080-4. (Open Access)

Here's what I believe they're trying to say.
First, note that the discussion is mostly limited to high spin complexes (3 unpaired electrons out of the 7 d electrons), so the figure you reproduced is only representative of high spin states.
Second, recall that when D > 0 (ie left side of figure), the five d orbital energy levels split so that we have three t2g orbitals that are lower in energy than two e2 orbitals because of crystal field splitting. When D < 0 (right side of figure), the reverse is true, two e2 orbitals are lower in energy than three t2g orbitals.
In the absence of a crystal field, there is no difference in energy across the five orbitals, so for the case of 7 d electrons, we have a 4A state, which is the pre-split state in both cases in the figure.
In typical simple crystal field models, D>0 is associated with octahedral (i.e. OH) coordination and D<0 with tetrahedral (TH) or cubic coordination, but the authors give examples of TH with both D>0 and D<0, so the sign of D should not be assumed based on geometry here (nor should geometry be assumed based on sign).
Looking at the figure, then, we can start with the right side (because it's easier to understand) and see the transition from 4A to a state of D<0. That is the application of the crystal field. Since D<0, we know that the two e2 orbitals are lower in energy than the three t2g orbitals. Since we have high-spin d7, our electronic state is e24t2g3, with all three t2g electrons unpaired.
What we see in the figure is that there are two pairs of energy levels, with the states with spin $\pm \frac32$ at lower energy than those with spin $\pm \frac12$. The spin $\pm \frac32$ and $\pm \frac12$ values refer to the sum of the spins of the three unpaired electrons. The $\pm \frac32$ states are those in which all three t2g electrons have the same spin (either all $+\frac12$ or all $-\frac12$), while spin $\pm \frac12$ means that one of the three has a different spin than the other two. We know from Hund's rule that unpaired electrons in degenerate orbitals have the lowest energy when they all have the same spin direction, due to the exchange energy term. That is the source of the splitting in energy between the two possible total spin states.
We now get to the "fictitious spin" part. This is actually relatively simple. It has nothing to do with actual spin and simply indicates the number of experimentally relevant spin states, but is expressed in $\frac 12$ increments to have the familiar form of a spin value. Thus, fictitious spin of $\frac12$ just means two possible states, as one would observe for an electron with possible spins $\pm \frac12$. A fictitious spin of 1 would mean three possible states, as we would have for the case of two unpaired electrons.
In the figure, the fictitious spin is $\frac12$ because under the experimental conditions, only the two lower energy levels (ie actual spin of $\pm \frac32$) are observed. The two states with spin $\pm \frac12$ are higher enough in energy that they do not occur.
The left side can be interpreted similarly, except that here, when D>0, the three t2g orbitals are lower in energy than the two e2 orbitals, and the electron configuration is t2g5e22, with one of the t2g and both of the e2 electrons unpaired. Now, since the unpaired electrons are not in three degenerate orbitals, the lower energy states have the unpaired t2g opposite spin to the unpaired e2 electrons (which by Hund's rule again are the same spin).
The fictitious spin is interpreted in the same way on this side of the figure, simply meaning that only two states are accessible under the conditions of interest.


In Finkelstein reaction, how does precipitate formation favor forward reaction? In my textbook it's stated that:

  $\ce{NaCl}$ and $\ce{NaBr}$ thus formed is precipitated in dry acetone. It facilitates the forward reaction according to Le Chatelier's principle.

Is this related to common ion effect in any way? I'm not familiar with Le Chatlier's principle. Please explain this in simple words as I'm in high school.

How does formation of $\ce{NaCl}$ and $\ce{NaBr}$ (both insoluble) prevent backward reaction from taking place? Does it mean $\ce{NaI}$ will be produced again?

$\ce{NaCl} $ and $\ce{NaBr }$formed as products are polar compounds which are vey less soluble in non- polar solvents like dry acetone, but $\ce{NaI } $due to high polarizability of $\ce {I^-} $ is soluble in acetone. So the products precipitate.  If a reaction is in equilibrium in constant temperature  and it is distorted somehow, it goes in the direction to nullify the effect of distortion.  This is the Le-Chattelier's principle.  If some of the products are removed to keep the $\ce{K_c}$ constant , some more reactants will react and convert into products. Thus constant removal of products by precipitation helps to dissociate reactants into products and thus favour the forward reaction.


The color of the most sulfides of p- and d-block elements My teacher said that many of the sulfides of d-block and p-block are black. Why?

The band gap between conduction band and the valence band plays a role in color of these compounds. The good example is $\ce{SnS2}$, which is bronze color (Image $\bf{A}$; $\ce{SnS2}$ is used in decorative coating where it is known as mosaic gold) and have a band gap of $\pu{2.18-2.44 eV}$ (Image $\bf{B}$; Ref.1). When an electron in the conduction band falls down to the valence band, it loses energy and the lost energy is emitted as a photon ($h\nu$). This photon has an energy equal to the band gap (Image $\bf{C}$). In mathematics, it is: $E_\mathrm{gap} = h\nu$. The crystals of $\ce{SnS2}$ and its calculated band gap is shown below:

The compounds is colored when the band gap is between $1.8$ and $\pu{3.0 eV}$ (visible range). The compounds with larger band gaps $(\gt \pu{3.0 eV})$ are either white or transparent crystals.  Those with smaller band gaps $(\lt \pu{1.8 eV})$ give  black color crystals. For example, the band gap of $\ce{ZnS}$ is $\pu{3.54 eV}$, the color of which is white. Meantime, the color of $\ce{CdS}$ (Pigment name: Cadmium yellow) is yellow because its band gap is $\pu{2.45 eV}$. On the other hand, the colors of $\ce{CdSe}$ and $\ce{CdTe}$, which have the same structure of $\ce{CdS}$, are black because their band gaps are $\pu{1.74 eV}$ and $\pu{1.49 eV}$, respectively. Also, the color of $\ce{PbS}$ is black because its band gap is $\pu{0.37 eV}$.
For some crystals, the band gap change with temperature (Ref.1). Hence, for those, the color depend on temperature.
References:

Lee A. Burton, Thomas J. Whittles, David Hesp, Wojciech M. Linhart, Jonathan M. Skelton, Bo Hou, Richard F. Webster, Graeme O'Dowd, Christian Reece, David Cherns, David J. Fermin, Tim D. Veal, Vin R. Dhanak, Aron Walsh, “Electronic and optical properties of single crystal SnS2: an earth-abundant disulfide photocatalyst,” J. Mater. Chem. A 2016, 4, 1312-1318 (https://doi.org/10.1039/C5TA08214E).


Recoil velocity of atom when electron is excited If an electron is excited from the $n = 1$ level to the $n = 2$ level in a hydrogen atom, how can we find the recoil velocity of the atom?

There are three things to conserve during an electronic transition: energy, momentum, and angular momentum. As always, these things are related to some extent. Let's forget about the angular momentum because it isn't relevant to this question.

The energy of a photon is given by,$$E_\gamma=h\nu.$$This energy is related to the momentum of a photon through the mass-energy equivalence,$$E^2=m^2c^4+p^2c^2.$$ Remembering that the mass of a photon is zero, we have,$$P_\gamma=\frac{E_\gamma}{c}.$$

The energy of an electronic transition between two states in a hydrogen atom must equal the energy of the photon being absorbed, which conserves the energy upon either absorption or emission. Thus, we have,$$E_\gamma=\frac{m_ee^4}{8h^2\epsilon_0^2}\left (\frac{1}{n_1^2}-\frac{1}{n_2^2}\right ),$$where we have used the solution to the energy states of a hydrogen atom coming from using a Coulomb potential to describe the potential energy of the system.

Then, we impose the conservation of momentum condition that $P_\gamma=P_{atom,f}-P_{atom,i}$. When I write $P_{atom}$ we have to note that this is given by the sum of momenta of the nucleus and electron.

Let's define the recoil velocity as, $v_{recoil}=v_f-v_i$

I don't wanna deal with the special relativistic correction to the momentum, so I'll leave that to you if you wanna figure it out.

Now, we can write,$$P_{recoil}=\frac{E_\gamma}{c},$$which can be written in terms of the recoil velocity according to,$$v_{recoil}=\frac{E_\gamma}{(m_e+m_p)c}.$$In the last expression, $m_p$ is the mass of a proton (i.e. the mass of the nucleus).

The expression would get uglier if you used $P=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$ instead of $P=mv$.

Using the above expression, the recoil velocity ends up being,$$v_{recoil}=\frac{10.2\cdot10^{-5}MeV}{938.272\frac{MeV}{c}}=3.26\ \frac{m}{s}.$$

So that's a reasonable velocity I think.

It's totally possible I made a mistake somewhere because I've never thought about this question before, but I think that is all correct. It's a little bit weird to think about whether the momentum is being transferred to the electron or the nucleus, but it seems like it must be transferred to both somehow?

One way to check the work (which I don't wanna do) is to try to get the same expression by only balancing energies and using $\frac{1}{2}mv^2$.


Does exchange energy affect completely filled orbitals? In my textbook, it's written that exchange energy stabilises half filled and completely filled orbitals. But my teacher said electrons can't be exchanged in full filled orbitals because of pairing up of electrons. So which explanation is correct? My textbook has an explanation for half filled electron orbitals, by it had no explanation given for completely filled orbitals.

It seems like you are just being introduced to quantum mechanics and is confused with the concept of exchange energy. The concept can be confusing at first, since it neither involves exchanging electrons as classical particles nor has a classical analogue. I will try explaining the problem with a simple two electron system.

Before we discuss, here is a short introduction. In quantum mechanics, one solves the Schrodinger's equation and tries to estimate the wave function of the system. We are talking about time-independent case here, so we solve time independent Schrodinger's equation to obtain the wavefunction.
$$\hat{H}\psi=E\psi$$
Here $\hat{H}$ is the Hamiltonian (energy operator), $\psi$ is the wavefunction and $E$ is the energy. This equation is an eigenvalue equation, which essentially means any valid wavefunction should have a fixed energy irrespective of coordinates. In the equation, the $\hat{H}$ is the only known quantity (operator), $\psi$ and $E$ has to be estimated based on boundary conditions.

Let us consider a system with two electrons, with wavefunction $\psi_1(r)$ and $\psi_2 (r)$, where $r$ denotes the four dimensional space spin coordinates. Now, we know that electrons are indistinguishable and two electrons (fermions) cannot have same space-spin coordinates (Pauli's exclusion principle). Two electrons not having same space-spin coordinates is same as saying that the electrons should not have all same quantum numbers.
So, in order to consider the indistinguishability  of the electrons, we can have two cases of the total wave function: (1) $\psi_1(r_1)\psi_2(r_2)$ or (2) $\psi_1(r_2)\psi_2(r_1)$. On a side note, these are known as Hartree products and are valid for bosons. However, in case we wish to include the Pauli's exclusion principle, we write the system as a Slater determinant, i.e.
$$\psi(r_1,r_2)=\frac{1}{\sqrt{2}}\begin{vmatrix} \psi_1(r_1) & \psi_2(r_1)\\\psi_1(r_2) & \psi_2(r_2)\end{vmatrix}=\frac{1}{\sqrt{2}}\left[\psi_1(r_1)\psi_2(r_2)-\psi_1(r_2)\psi_2(r_1)\right]$$

Now let us talk about the Hamiltonian (energy operator). The Hamiltonian for an atom/molecule can be divided into three parts:

Constants: This include any constant terms, for example, nuclear-nuclear repulsion for poly-atomic molecules.
One electron operator: This part contains terms dependent on single electrons, for example, kinetic energy and nuclear-electron attraction.
Two electron operator: This part contains terms dependent on two electrons, and usually contains only electron-electron repulsion terms. This part gives rise to the exchange energy. We shall expand this part.

Before we go into details, if one is provided with a operator $\hat{A}$ and a wavefunction $\psi$, the expectation (average) value of the observable corresponding to the operator is given as $\langle\psi|\hat{A}|\psi\rangle$, where $\langle\cdot\rangle$ implies integral over all coordinates.

In atomic coordinates, the electron-electron repulsion is given as $1/r_{12}$, where $r_{12}$ is the distance between the electronic coordinates. Now, if we use the wavefunction in the Slater determinant, we get the total two electron interactions as:
$$\left\langle\psi(r_1,r_2)\left|\frac{1}{r_{12}}\right|\psi(r_1,r_2)\right\rangle \\= \frac{1}{2} \left[ \left\langle\psi_1(r_1)\psi_2(r_2)\left|\frac{1}{r_{12}}\right|\psi_1(r_1)\psi_2(r_2)\right\rangle + \left\langle\psi_2(r_1)\psi_1(r_2)\left|\frac{1}{r_{12}}\right|\psi_2(r_1)\psi_1(r_2)\right\rangle - \left\langle\psi_1(r_1)\psi_2(r_2)\left|\frac{1}{r_{12}}\right|\psi_2(r_1)\psi_1(r_2)\right\rangle - \left\langle\psi_2(r_1)\psi_1(r_2)\left|\frac{1}{r_{12}}\right|\psi_1(r_1)\psi_2(r_2)\right\rangle \right]$$
Now, in this cases the first two terms are called Coulomb terms and are easy to imagine classically, i.e. what is the repulsion of electron 1 if present in coordinate $r_1$ with electron 2 if present in coordinate $r_2$ for the first term and vice-versa for the second term. These are repulsive in nature.
The third and fourth terms are attractive terms, and are known as exchange terms. There is no classical analogue of these, and appears purely from the indistinguishability of electrons and Pauli's exclusion principle.

Now remember we discussed that $r_1$ and $r_2$ are space-spin operators, and $r_{12}$ depends only on space? This essentially means that we can separate the space and spin parts while calculating the terms. So the exchange terms vanish while the two electrons have different spins, and remains only when electrons have same spins. This is the origin of exchange energy and IT DOES NOT INVOLVE PHYSICALLY EXCHANGING THE ELECTRONS.

I hope I could explain you in brief the origin of exchange energy. In reality, it is very difficult to calculate the exact energy of a system with N-electrons and say half filled/full filled systems are more stable due to "exchange interactions" simply because:

We do not exactly know how the wavefunctions look. We assuming hydrogenic wave functions but that may not work out always.
Optimising the wavefunctions based on Hamiltonian is carried out self consistently and is costly.
Let us say one resolves the first two issues somehow, still total energy has other terms like kinetic energy, electron-nuclear attraction, Coulomb terms, and even nuclear-nuclear repulsion for polyatomic molecules. So it is very difficult to assert only "exchange energy" is responsible for the extra stability.

Whatever I have tried to explain in short takes up at least two quantum chemistry courses in Bachelor's degree and Master's degree. In case you have some doubts, please feel free to ask in comments. In case you wish to study more, I will suggest the following books (in the order given):

D. J. Griffiths, Introduction to Quantum Mechanics
A. Szabo and N. Ostlund, Modern Quantum Mechanics
I. Levine, Quantum Chemistry


Continuous flow mass spectrometer for isotope analysis If i took thousands of discrete samples of mineral fragments in a field study using a specialized drill, what mass-spectrometry techniques/equipment would allow me to analyse the samples as fast as possible?

Some machines can take 5000 readings per second, how can that be applied to a continuous flow of mineral fragments?

The premise is a feasibility-study of a robot archaeology tool:
Cartesian Axis Robots have an accuracy of about 0.01mm, so it would be possible to excavate sensitive sites to sample submillimeter 3d slices of the soil.
The technical challenge of the machine would be how to prepare and record tens of thousands of chemistry results every hour.

To find out about current spectrometer sampling speeds, research the term: "spectra per second"

There is alot of precis informations i.e.:

"The standard interface supplied with all nTM detectors is
12 bit USB2.0. With this interface and our SpectraArray SL software,
up to 30 spectra per second can be captured.

For even faster acquisition, a PCI bus interface is available as an
option. This  allows up to 160 spectra/second to be acquired. "
50 to 80 micrograms of mineral is sufficient for a typical analysis, although small samples on the order of 12-20 micrograms (single foraminifers) can also be analyzed

So it seems that a major challenge is the data bandwidth between the spectrometer and the SSD/HDD.

Spectra per second returned by the machine are not going to be the limiting factor, because they can go up to 10 000 spectra per second. A graph with 250 floating point humbers is 4bytes times 250, so about 1kb per graph, research is necessary to understand why USB can only handle 12 samples per second, it should be in the 1000ds.

In spectroscopy, the key is to constantly calibrate the sample against an empty control sample. If the sample is drilled, it can be ground into a find dust and then it should be possible to process it in a homogenous way similar to liquid or gas spectroscopy.

All the different spectroscopy methods can be considered, they are all very fast, and have different detection strengths. IR and Xray have different advantages, IR is easier to deal with. The selection of appropriate IR or other wavelengths to use for a specific type of study is an advanced topic.


Why do copper (II) complexes contain so many valence electrons? Consider tetraaminecopper (II) ion. $\ce{[Cu(NH3)4]^2+}$ or $\ce{[Cu(NH3)4(H2O)2]^2+}$.
The copper(II) ion has the electron configuration $\mathrm{[Ar]\ 3d^9}$. How do the 4 electron pairs from the ammonia ligands form the coordination bond with the copper ion, if the 3d subshell is almost entirely occupied?

I'll address $\ce{[Cu(NH3)4(H2O)2]++}$, but the reason is similar for $\ce{[Cu(NH3)4]++}$

The 9 d-electrons in the complex are in non-bonding and antibonding orbitals.

See pages 6-9 of Ligand Field and Molecular Orbital Theory.

  the ligand σ orbitals are derived from AOs with energies much lower than those of the metal d-orbitals. Thus, the six bonding MOs of the complex are mostly ligand-orbital in character. These six bonding orbitals can accommodate the 12 electrons provided by the six ligand lone pairs. Therefore, the electrons provided by the ligands are largely confined to the ligands in the complex.

So, to a large degree, the 12 electrons from the ligands are in bonding orbitals, but the 9 d-electrons are in nonbonding or antibonding orbitals.

  The d electrons of the metal centre...occupy the lower energy, non-bonding metal $t_{2g}$ set and the higher energy, antibonding $e_g$ set. Therefore, the metal based d electrons tend to reside largely on the metal atom. In summary, the frontier orbitals of the complex are the non-bonding, entirely metal based $t_{2g}$ orbitals and the antibonding $e_g$ orbitals, which are mainly metal in character anyway.


Homoassociation and Acid Strength I understand that the strength of HF differs significantly depending on the concentration. I understand the basic theory behind this - homoassociation - and stabilization of the highly negative charge dense fluoride anion through hydrogen "bonds"/intermolecular attractions with undissociated solute $\ce{HF}$ molecules.

However, I am unclear on the Wikipedia explanation of homoassociation:
http://en.wikipedia.org/wiki/Homoassociation

Regarding the equations - I cannot tell if the authors meant to refer to pure, liquid $\ce{HF}$ or a solution of HF. I'm thinking that the authors are referring to liquid $\ce{HF}$ because auto-ionization of HF is mentioned. I'm thinking that auto-ionization would only be significant in liquid $\ce{HF}$. Or in a concentrated solution.

Also, regarding the second equation - the authors make it look like the fluoride anion is forming an actual, intramolecular, bond with the $\ce{HF}$ molecule. I think the authors are implying the formation of an intermolecular hydrogen bond, correct?

On the other hand the bifluoride anion seems to be a real molecular entity (http://en.wikipedia.org/wiki/Bifluoride).

This only muddies the water further. I.e. which of the following best represents how the fluoride anion is stabilized, causing concentrated solutions of $\ce{HF}$ to ionize further than one would expect from $\ce{K_a}$ values?

$\ce{HF}$ ... $\ce{F^-}$ (hydrogen bonding)

or

$\ce{HF_2^-}$ (formation of bifluoride anion)

Also, any further links to resources about homoassociation would be appreciated.

ETA: Also, can't the fluoride anion form hydrogen bonds with water? The fact that concentrated HF is stronger than dilute HF than what mere $\ce{Ka}$ values suggest implies that H-bonding with water isn't as effective as H-bonding with HF molecules. Why? Is water less polar? On the other hand water can form a maximum of two hydrogen bonds with the fluoride anion, while a HF molecule can only form a maximum of one hydrogen bond with a fluoride anion.

I think this Wikipedia article on hydrogen fluoride will answer your questions.  It discusses the acidity of $\ce{HF}$ from dilute to concentrated aqueous solution, and then pure $\ce{HF}$.  They present four equilibria to describe these situations

\begin{aligned}
\ce{H2O + HF &~<=> [H3O^+ {.} F^{-}]&&(I)}\\
\ce{[H3O^+ {.} F^{-}] &~<=> H3O^+ + F^{-}&&(II)}\\
\ce{[H3O^+ {.} F^{-}] + HF &~<=> H3O^+ + HF2^{-}&&(III)}\\
\ce{3HF &~<=> H2F^+ + HF2^{-}&&(IV)}\\
\end{aligned}

In dilute solution equilibrium (I) lies far to the right and the second equibrium (II) far to the left.  So even though the $\ce{HF}$ is extensively dissociated, the resultant hydrogen bonded ion pair is very stable and reluctant to undergo significant dissociation.  Hence the solution is weakly acidic.  As the concentration of $\ce{HF}$ is increased homodissociation comes into play [equilibrium (III)] producing the hydrogen difluoride ion as a discrete entity which is stabilized through intermolecular hydrogen bonding.  The increase in $\ce{H3O+}$ makes the resultant solution very acidic.  In pure $\ce{HF}$, autoionization occurs producing an exteremly acidic solution.  Page 4 of this link shows a graph of HF acidity from near zero to 50% concentration.


What does ozone smell when treating with household water? We live in a rural area and our well water comes from storage tanks that are continously being ozonated before passing through an infra red lamp.

The water in the house smells like it contains chlorine and so does the water in the storage tank upon climbing to the top cover and smelling.

There is no source of chlorine and want to know if the ozone can be the source of the bleach odor.

Although the primary products of chloride oxidation resulting from the ozonation of saline waters are chlorate salts ("Photochemical oxidation of chloride ion by ozone in acid aqueous solution", Environ Sci Pollut Res (2015) 22:16554 – 16569), a portion of the chloride is also known to oxidized to molecular chlorine, $\ce{Cl2}$.

Human olfactory senses can detect $\ce{Cl2}$ at levels as low as 0.2 ppm in air. This World Health Organization publication reports that some non-seawater contaminated surface waters in the UK contain greater than 40 mg/L chloride, and some seawater-contaminated aquifers used for drinking water in both the US and Philippines contain chloride concentrations on the order of a few hundred mg/L.

The degree of oxidation of $\ce{Cl-}$ to $\ce{Cl2}$ by ozone depends on several factors including temperature, lighting conditions, pH of the well water in question, and of course chloride concentration. However, based on the information given above, low-percent level conversion rates of $\ce{Cl-}$ to $\ce{Cl2}$ by ozone oxidation should be feasible, which under the right conditions, would result in the production of a noticeable chlorine odor. It indeed seems that $\ce{Cl-}$ to $\ce{Cl2}$ conversion represent the most likely scenario for the chlorine odor detected from the treated well water.

NOTE: OP mentioned an "infra red lamp" as part of the disinfectant process for the drinking water.  I suspect this really meant, or was in addition to, a UV lamp, which would further promote the oxidation of chloride per my first reference.


Computing the concentration of a reactant given formation of products At high temperatures nitrogen dioxide decomposes into nitrogen oxide and oxygen gas according to $$\ce{2 NO_2 -> 2 NO + O_2}$$

I have a question now. Let's say at a certain temperature I'm given data for the formation of $\ce{O_2}$. For example:
At $t = 0$ we have $[\ce{O_2}] = 0$ mol/L, at $t = 50$, $[\ce{O_2}] = 7.10 \cdot 10^{-4}$ mol/L, at $t = 100, [\ce{O_2}] = 1.205 \cdot 10^{-3}$ and so on until a certain time.

If I'm given that the initial concentration of $\ce{NO_2}$ is $0.006M$, how can I know what the concentration of $\ce{NO_2}$ is at certain times? I'm asked to determine the order of the reaction on the basis of the given data, and also to present graphically the rate constant.

To do that I need to know if $\ln([\ce{NO_2}])$ versus time $t$ is a straight line. But how to find $[\ce{NO_2}]$ in the first place?

In this reaction, the production of $\ce{O2}$ is tied to the decomposition of $\ce{NO2}$; for each molecule of $\ce{O2}$ that is produced, there have to be 2 molecules of $\ce{NO2}$ that have decomposed. So:

$$2 \Delta n_{\ce{O2}} = -\Delta n_{\ce{NO2}}$$

where $\Delta n$ represents a change in the number of moles. The question does not specify, but I will assume that the volume of the reaction is kept constant; since concentration is amount divided by volume, this means that the same relation holds for concentrations:

$$2\Delta [\ce{O2}] = -\Delta [\ce{NO2}]$$

Since the initial concentration of $\ce{O2}$ is zero, $\Delta [\ce{O2}] = [\ce{O2}]$, and therefore

$$2[\ce{O2}]_t = [\ce{NO2}]_i - [\ce{NO2}]_t $$

$$[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$$

To use the numbers in the question, at $t = 50$ s we have $[\ce{NO_2}] = (0.006 - 2(7.10\cdot 10^{-4}))$ M = $4.58\cdot 10^{-3}$ M.


How does a lower pH change the alcohol concentration of wine? I have been told it that when the pH of wine is lowered the alcohol concentration will decrease because of Le Chatelier's Principle. Which states; if a  equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. How can I best expand on this?

This is a slightly broad question, but ill try to assist some. Google has a large assortment of articles describing pH, Le Chatelier's Principle, and EtOH.

Im not sure if you mean altering the pH during the fermentation process or altering the pH of the final product? It also makes a difference on the types of yeast you use. Some yeasts can handle different conditions and produce a different amount of alcohol.

Here is a great response from last year on the Stack Exchange, regarding fermentation without the addition of commercial yeast, using the change in pH as a catalyst for fermentation.

With the production of wine that uses commercial yeast for fermentation (which is most commercial wines), altering the pH too far outside of it's "designated" range will cause it to produce less alcohol. Some connoisseur winemakers will alter the pH slightly sometimes, varying on the types of grapes, to alter the terpene profile present. Different variables, equals different systems of equilibrium.


Why Isodecane (B.P.: 327 *C) has a higher boiling point than decane (B.P.: 174.1 *C)? Branched chain molecules have lower boiling point than straight chain molecules because branched chain molecules have a smaller surface area where they can be attracted by other molecules. Then, why Isodecane (a branched chain molecule) has a higher boiling point than Decane (a straight chain molecule)?

The only problem here is the units - the boiling point of isodecane is given in Fahrenheit and the boiling point of decane in Celsius.


Iron Filing Catalyzed Toluene Reaction with Bromine Solution When toluene and a bromine solution react, there is no reaction. That is, the solution turns a yellow/brown colour, the same as the bromine solution.

I conducted an experiment where toluene and a bromine solution were meant to be catalysed by iron filings. However, instead of the solution losing its colour (like you'd expect in an alkene/alkyne), it became rustic in colour. If the iron filings were intended to catalyse the reaction, why did the solution become darker? The reaction that took place seems entirely counter-intuitive.

The iron and bromine must first react to form the catalyst (FeBr3) as the iron does not actually catalyze the reaction (this is known as in situ synthesis of the catalyst).  The FeBr3 catalyzes the reaction between toluene and Br2 to form ortho and para bromotoluene.  Catalysts are NOT consumed in the reaction though, which means the culprit of the rusty color is the catalyst (google iron(iii) bromide to view the color).

Also, ensure that you use a larger excess of the bromine.  The FeBr3 interacts with bromine to make it more electrophillic and able to react with the benzene ring of toluene.  If there is no bromine left after the catalyst has been formed, no reaction will occur.


Benzylic and allylic carbocation stability order Most books I have read stop at this order :
Benzylic > Allylic > Tertiary > secondary > primary
However If I want to compare tertiary, secondary and primary benzylic and Allylic carbocations what will be their stability order?

Yes tertiary allyllic carbocation is more stable than secondary because no.  of electrone donating methyl groups are greater than secondary (in tertiary 3 methyl groups  while in secondary  are 2) so they decrease  electrophilicity carbocation but it still it remain electrophillic.


Work done in irreversible isothermal expansion against vacuum Why is the work done in an irreversible isothermal expansion against vacuum equal to zero?

A system does its maximum amount of expansion work when the external pressure is equal to the pressure of the system all the way through the expansion process.

At constant pressure work done is given by:
$$w=-p_\mathrm{ex}\Delta V$$
Where $p_\mathrm{ex}$ is the external pressure. Therefore the minimum work done is when the external pressure is $0$, which is the case under vacuum. This would mean that the work done under vacuum is $0$.

This is called free expansion. The greater the opposing force then the greater the amount of work done is (as work is equal to force applied over a distance). Therefore when the external pressure is $0$ there is nothing for the piston to push against, because there is no opposing force, no work is done by the system.

If you are interested to see where $w=-p_\mathrm{ex} \Delta V$ comes from then note that
$$w=Fd$$

Because pressure can be given by $P=F/A$ and volume can be given by $V=Ad$ we can see that $$w=\frac{F}{A} Ad = p\Delta V$$

This is work done on the gas. For work done by the gas, we put a negative sign in front of the pressure term.

*Note that this is at constant pressure.


Crystallization in a solution of 2 or more solutes This is a very basic question I found myself wondering. For simplicity's sake, we'll say it is a super saturated aqueous solution of an equivalent measure of sugar and salt. As the solution cools and crystals form, do you get sugar crystals, salt crystals, cocrystals or discrete crystals of both?

When the solution contains several substances, it is very often possible to achieve their separate sequential crystallization due to the different nature of the solubility of the substances as a function of temperature. In a certain temperature range, the solution will always be saturated with respect to one substance and unsaturated with respect to the other. This leads to the isolation of only the first substance from the solution, while the other remains in the mother liquor.

The difference in the compositions of the liquid (vapor) and solid phases formed during the partial crystallization of the solution, melt, and gas phase always leads to the fractions enriched with one component, and sometimes even to practically pure components. Possible degree of separation during crystallization is determined by the type of the phase diagram (composition vs temperature) in the solid-liquid(gas) system.

Obviously, salt and sugar do not form solid solutions, therefore depending on the concentration and temperature one component will crystallize first, and, when water is fully evaporated, a mechanical mixture of crystals is left.

This is a basis of various crystallization methods of mixtures (binary or multicomponent) separation, such as:

Fractional crystallization;
Crystallization in the presence of auxiliary substances;
Fractional melting and sublimation;
Multiple recrystallization.


Cis/trans Proline vs. Double bond character I am learning about proteins and we have learned that the C-N bond is rigid because of it's partial double-bond character. We have also learned that proline is commonly found in beta turns (type 1) because it can isomerize from trans to cis (which helps make the 180 degree turn).

I don't understand how these 2 things fit together--how can it change from trans to cis if it can't rotate around the C-N bond?

I found this which talks about the double bond being broken (in this case, it would be the partial double-bond, I guess) and then reforming...is that what is going on here?

The change from cis to trans (called isomerization) is not spontaneous. There are proteins called prolyl isomerases (see Wikipedia Prolyl Isomerase) that assist in the isomerization process. This happens when the protein is being folded into its final secondary and tertiary structures.


The violation of a certain rule or principle in the following electronic configuration I'm to find the violation of a certain rule or principle in the following electronic configuration:

$$\ce{1s^2 2s^2 2p^1_x 2p^0_y 2p^0_z}$$

$n+l$ rule is not violated because: for $\ce{1s^2}$, $n+l=1+0=1$ so it is filled first, for $\ce{2s^2}$, $n+l=2+0=2$ so it is filled second, and so on.
Aufbau Principle is not violated because: the orbitals with minimum energy are filled first with electrons and $\ce{1s < 2s < 2p}$ which is the sequence of increasing energy orbital judged by the $n+l$ rule.

I'm quoting here a paragraph from my chemistry textbook of the chapter Atomic Structure:

  The orbitals given by a particular value of $l$ if $n$ is same, have the same energy and such orbitals are called degenerate. Like $l=1$ gives three $p$ orbitals specified as $\ce{p_x, p_y, p_z}$ and the p orbitals are said to be three fold degenerate. These orbitals are filled according to Hund's Rule.

The numerical value of the principle quantum number $n$ for $\ce{2s}$ and $\ce{2p}$ is the same; however, they don't have a particular value of the azimuthal quantum number $l$. For $\ce{2s}$, $l=0$, and for $\ce{2p}$, $l=1$, so $\ce{2s}$ and $\ce{2p}$ are not degenerate. So none of the three principles or rules whether Aufbau's, $n+l$ or Hund's is violated in this electronic configuration.

However, my friend says that there is a violation of the Hund's Rule in this electronic configuration, and according to him the right configuration is: $\ce{1s^2 2s^1 2p^1_x 2p^1_y 2p^0_z}$, which in turn violates the Aufbau Principle, because the orbitals of lower energy level have to filled first completely and $\ce{2s < 2p}$.

  So, which one of the electronic configurations is right, $\ce{1s^2 2s^2 2p^1_x 2p^0_y 2p^0_z}$ or $\ce{1s^2 2s^1 2p^1_x 2p^1_y 2p^0_z}$?

Note: The required electronic configuration is for the electrons in their ground state.

If we talk only about electronic configurations and rules violation then you are correct. The 2nd one (excited state) is referred for hybridization that is sp2. An excited state  configuration is meant only when the particular atom gets ready or excited for forming a bond. Like boron does it with 3 fluorine atoms to form $\ce{BF3}$. So, if we talk about electronic config and rule violations we must go for ground states exceptional cases are $\ce{Cu}$ , $\ce{Cr}$ and $\ce{Mo}$.


How could I produce/rig this flame-colour lottery? Not sure if this is the wrong SE for this, happy to move it if not...

In the Netflix TV series Sabrina, there is a scene where people take part in an appropriately-gothic-and-ominous lottery, which proceeds as follows:

Fourteen participants draw, one at a time, pieces of paper from a box.  The papers are all blank and have no differences visible to the human eye in the (rather moody) lighting.
The participants burn their paper in the flame of a candle which is in front of them.  There is a separate candle for each participant, but they all appear to burn with the same yellow flame.
Twelve of the papers burn up quickly, like flash paper.  The other papers burn more slowly, and release red (for the 'runner up') or white (for the 'winner') smoke, respectively.

   The runner up waits on the winner hand and foot for three days, and then the winner is killed and eaten cannibalistically, with a strong religious 'ascension' dogma.

Depending on your point of view, this is a lottery you may or may not want to win, and also one which there would be great advantage in being able to rig to ensure a particular winner.  So my questions are:

What chemistry could you use in the papers and candles to achieve the effects shown?

Smokeless and quick burning isn't hard to achieve, just use flash paper. For white smoke, I would use flash paper again but with a white oxide powder such as titanium oxide with a flame retardant to slow the burn. For red, again flash paper but, you need a compound that releases bromine when burned. I think copper(i) bromide or silver(i) bromide would be ideal as both are easy to decompose and tolerant to moisture in air, though silver bromide is not tolerant to prolonged light exposure. These salts may discolor the paper and you may need to color the non-red smoke papers to make them all look the same.


Formulas for the number of spectral lines While reading through Narendra Avasthi's Problems in Physical Chemistry, I came across two formulas on p. 64 (Scanned page):

  When electrons de-excite from higher energy level ($n_2$) to lower energy level ($n_1$) in atomic sample, then number of spectral line observed in the spectrum is

  $$\frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$$
  When electron de-excites from higher energy level ($n_2$) to lower energy level ($n_1$) in isolated atom, then number of spectral line observed in the spectrum is

  $$n_2-n_1$$

Can anyone explain how the process of emission differs in these two cases?

To my understanding, it is simply single atom versus many number of atoms. For example, suppose one atom with an electron at energy level 7 ($n_2=7$). That electron can "de-excite" from $n_2=7$ to $n_1=6, 5, 4, 3, 2,$ or $1$. All those transitions give one spectral line for each. Thus, total of $1 \times 6 = n_1(n_2-n_1)$ (foot note 1) spectral lines would be present in the spectrum.

Similarly, when there were more than one atom in the sample, excited electrons ($n_2$) would be in different states $(n_2=2, 3, 4, 5, 6,....,\infty)$. For example, suppose we have aom population having electrons in all levels up to energy level 8 ($n_2=8, 7, 6,...$). Suppose those electrons "de-excite" to energy level 2 ($n_1=2$). Thus, electrons in $n_2=8$ can "de-excite" to energy levels $7, 6, 5, 4, 3,$ and $2$ meaning total of 6 spectral lines $(8-2=n_2-n_1)$. Some atoms with electrons in energy level $n_2-1=7$ can also "de-excite" to energy levels $6, 5, 4, 3,$ and $2$ meaning total of 5 spectral lines $(7-2=n_2-1-n_1)$, etc., etc. Thus, total numbers of spectral lines ($s$) in this case would be:
\begin{align}
s&=6+5+4+3+2+1=21=\frac{42}{2}=\frac{7\times6}{2}\\
&=\frac{(8-2+1)(8-2)}{2}\\
&=\frac{(n_2+1-n_1)(n_2-n_1)}{2}
\end{align}

Foot note 1: Total number of spectral lines for single atom where $n_2=7$ should be: $1 \times 6 = (n_2-n_1)$ in the spectrum, not $n_1(n_2-n_1)$ as I originally suggested (Thanks @porphyrin for careful reading).


What do you think is a good way to explain electronic configurations? Understanding electronic configurations is vital, but a huge number of students face difficulties in understanding the why and the how. What approach do you think I should take in explaining electronic configurations to a student who only knows the Bohr model? Thanks.

As an undergrad student that has recently learnt that the Bohr model is just an approximation at best, and mostly incorrect, I think I am well equipped to answer this.

When I first learnt about the structure of an atom, I was told that "electrons move around the nucleus in fixed paths having fixed energy, called orbits. The angular momentum of the electrons is an integer multiple of $h/2\pi$."

When I was first taught the electronic configurations and the quantum mechanical structure of an atom, my teacher explained it thus (and it was easily understood by most of the class): he made an analogy between a person's location and the location of an electronic.

Just how a person can be located if we know which country they're in, which city they're in, which locality, which room, and finally which area in the room; an electron can be approximately located if we know its principal quantum number, its azimuthal quantum number and its magnetic quantum number. Another quantum number, i.e., spin quantum number tells us which direction it is spinning in.

After this, a brief historical understanding of why the Bohr model was discarded and this model was proposed was given to us. The Heisenberg's uncertainty principle, and the Schrodinger equation were introduced, as the uncertainty principle explained why Bohr's model was incorrect, and the Schrodinger equation explained how to find the probability density of the presence of the electron.

This method helped me understand as it felt like I was transitioning through the same phases that all the people of the scientific community had to when the quantum mechanical model was on its way in, and the Bohr model on its way out. This really felt like a fluid transition, rather than "Bohr was wrong, this is right", which would've caused a lot of confusion (at least for me).

Finally, there isn't a one size fits all answer to this, as every student is different. I am just sharing my experience.


Does an increase in pressure always causes increase in molar concentration of each gaseous substance involved? Question

Statement :
"An increase in pressure (caused by decrease im volume) at equilibrium results in increase in molar concentration of each gaseous susbstance  involved"

This statement is given true, so i tried this reasoning:

Since decrease in volume(V) causes increase in pressure and hence increase in no. of moles(n), so n/V increase  hence molarity increases.

Is this reasoning correct? Please give the real reason for the statement.

Your reasoning is correct as long as there is a decrease in volume. Note that pressure can also increase with temperature and the volume can stay constant. In this case, the molarity will not change. Molarity is not a very useful unit for expressing gas concentrations in equilibria because you have to quote pressure, volume, temperature and of course moles to quote molarity.


What will be the major product on chlorination of m-nitrophenol? I came across a question asking for the major product on chlorination of m-nitrophenol. In my opinion the deactivating nature of nitro group should control the regioselectivity, but I found out the answer to be 4-chloro-3-nitrophenol. But why is the activating nature of the hydroxyl group dominating over the deactivating nature of the nitro group?

But why is the activating nature of hydroxyl group dominating over the deactivating nature of Nitro group?

This is the precise reason why Picric acid is formed. ( Check here )
The activating nature of the $\ce{-OH}$ group is strong enough to counter the deactivating nature of the nitro groups.
Look at Ron's comment on the question over here.
Even methyl group sustains deactivating nature of nitro group in the formation of TNT. So no wonder phenolic group, a better activator, dominates.

The phenolic group increases electron density on ortho para (making it o-p director) while nitro groups decrease e-density on the o-p positions (making it a meta director). When both these effects act simultaneously, the question is who wins and is the density on o-p carbons less than corresponding meta ones? (as that would decide where the electrophile will attack)

Resonace structures of the groups gives the solution.
In nitro group:

There are three charged atoms bonded together in the $\ce{-NO_2}$ molecule. This is less table than the positive charge bearing oxygen atom in phenolic group's resonance structure:

Hence phenolic group wins!!

Coming back to the given starting reactant:

So now we consider the dominating ortho-para effect of phenolic group over the meta-directing effect of nitro groups to get 4-chloro-3-nitrophenol:


How Is This Acid Base Map Established? In the Acid-base map -shown below- each line denoted as an isohydric line shows the pairs of $ PCO_{2}$ & $ [HCO_{3}^{-}] $ that give a certain $ pH $. So each line connects the values of $ PCO_{2}$ & $ [HCO_{3}^{-}] $ that give a value of $ pH $ that is constant across that line..
The 7.4 line shows the isohydric line of physiological pH,the central ellipse shows the normal range.
I understand the Henderson–Hasselbalch equation:
$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log _{10}\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)$
What I would like to know is how this map established either experimentally or mathematically or both preferably.

The curves are rather obtained clinically, statistically evaluating diagnose, symptoms and results of clinical lab tests.
Metabolic curve follows malfunctioning metabolism keeping too low ( m. acidosis ) or too high ( m. alkalosis ) level of bicarbonate.
Respiratury curves follows states of too intense breathing ( r. alkalosis ) or too shallow breathing ( r. acidosis ), because of too low respectively high level of carbon dioxide partial pressure, caused by abnormal breathing.
Note that the $K_{\ce{H, CO2}}$ Henry constant  or CO2 is $\pu{29.41 atm/(mol/L)}$. the value will be higher because of the body temperature, can be corrected by temperature dependent $\ce{CO2}$ solubility.
$$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)=\mathrm{p} K_{\mathrm{a}}+\log\left(\frac{[\ce{HCO3-}] \cdot K_{\ce{H, CO2}}}{p_{\ce{CO2}} }\right)$$
Where pKa of carbonic acid is 6.35, if counting the total $\ce{CO2}$ concentration.
There are 3 variables ( $\mathrm{pH}$, $p$, $\ce{[HCO3-]}$ ) and 2 degrees of freedom. $\mathrm{p}$ is controlled by breathing, $\ce{[HCO3-]}$ is controlled by metabolism. $\mathrm{pH}$ is result.


Why doesn't the viscosity of water change much with temperature like it does for other substances? I would like to know why water doesn't ever become like syrup before it freezes when many other liquids, such as hydrocarbons, seem to have a more gradual change from becoming more viscous and then turning solid.

As I understand it, the more viscous a material is, the stronger the inter-molecular forces are. When the liquid is heated up, the molecules have more kinetic energy and so they can more easily break free of these forces--the liquid will become less viscous.

Why doesn't water have the same spectrum of viscosity as other materials? It seems like it is either solid, or it has very low viscosity.

This is something I spent a lot of time thinking about during my PhD, so let me see if I can explain it.

What you are talking about is the relationship between macroscopically observable properties (like viscosity), the relaxation time spectrum (which is a way of looking at the time scales involved in molecular movements), and the structure of the molecules themselves. It's kind of complicated! But, I think I can explain it in way that will make sense.

You know that as we increase the temperature, molecules have more kinetic energy and can more easily break free of the forces holding them together. I have some other answers here and here that explain this in more detail. With something like viscosity, however, there is more to the picture.

The first thing you need to know is:

  Viscosity is a measure of momentum transfer in a fluid

In other words, it is a way of quantifying the effect that movement in one part of a fluid has on the parts immediately adjacent to it. If the movements are highly coupled (meaning that if one layer moves, the layers next to it move almost as much), the viscosity is high. If not, the visocosity is low. Wikipedia has a pretty good explanation of this on the macroscopic scale.

So what causes a fluid to have a viscosity? Since you know about intermolecular attractive forces, this is an obvious candidate. And indeed, the stronger the intermolecular forces are in a fluid, the higher the viscosity tends to be. However, there is another factor involved - intermolecular repulsive forces (sometimes called stearic effects or excluded volume interactions). The idea is that if one molecule wants to move in a dense fluid, the other molecules have to get out of the way first.

As an analogy, imagine moving through the crowd at a packed concert or club. In order for you to move, someone else needs to move, and for them to move, someone else has to move, and so on. You can be polite and wait patiently for them to move, and eventually someone will. Or you could be assertive and ask, and move faster. If you are really impatient, you could barrel through people, knocking them out of the way (and hope you don't get beat up). Your impatience in a crowd is analogous to shear stress in a fluid. At lower temperatures, the molecules are closer together (since their kinetic energy isn't high enough to move very far out of the potential energy wells of their neighbors), and they move more slowly in general. This means that the time it takes for neighbors to move goes up, and as a result, viscosity goes up too - if you force them to move, they will knock others out of the way, transferring momentum through the fluid. At higher temperatures molecules move much more frequently, and so it is relatively easy to "drag" some through the fluid - they don't have to knock other molecules out of the way, because by the time they get there, they have moved on their own. The viscosity is lower.

Now, for simple molecules, the line between "can't move" and "moves easily" is pretty distinct. Once the molecules are mobile enough that excluded volume interactions are overcome, the material melts, and further changes to viscosity are small and mostly due to intermolecular attractive forces. For example, here is a plot of the viscosity of water vs. temperature

As you can see, it decreases smoothly as temperature increases. Below the freezing point, the viscosity is harder to define (although large masses of ice do flow), but the curve is discontinuous and it "jumps up" to some higher value.

But what about complex molecules? Let's link the monomers together:

Now it becomes much harder for the molecules to move. At a given temperature, you have to wait about the same amount of time for a single monomer to move, but it takes much longer for a pair to move together. Going back to the crowded room analogy - imagine trying to move through a crowd in which everybody - including you - was handcuffed in pairs to someone else. It would be a lot more difficult! In general, it turns out that the relaxation time for linear polymers increases as (approximately) the square of the chain length (for short lengths below the entanglement length). This means that if you double the number of monomers in a polymer chain, the average time it takes for them to move out of the way goes up 4 times. This is true up to a certain length of chain called the entanglement length (which depends on the polymer), at which point the likelihood of coordinated movement of the whole chain becomes so low that each polymer is sort of "trapped" in a tube made of other polymers. The analogy often used to explain this is a nest of snakes - they have to slither around each other. As a result, this type of motion was named reptation - after reptiles.

The interesting thing is that for shorter lengths (below the entanglement length) each length scale of the chain moves essentially independently of the rest. This means that in a longer chain, on short time scales the shorter lengths act as though they were in a low-viscosity fluid - they move easily and motion is not correlated with their neighbors. For longer length scales, the chain is essentially "frozen" - it can't move until its neighbors move - and so it acts more like a solid.

The result is that short lengths of chain can "wiggle" on short time scales, but longer lengths move more slowly. This gives us a relaxation time spectrum - it means that if you apply very fast changes (high shear rates) the molecules don't have time to get out of the way, and so the viscosity is high. If you increase the temperature, things move faster, and so at the same shear rate, the viscosity will be lower (more of the chain can move out of the way). The same is true at lower shear rates for the same temperature - the shorter length scales can move out of the way, and so it becomes more liquid - the viscosity decreases. On the longest time scales (lowest shear rates, coldest temperatures) the entire length of the chain can move, and so you see the lowest viscosity.

This is called shear-thickening, and is commonly seen in linear polymer melts or concentrated solutions. It is an example of viscoelastic behavior, which means that depending on the conditions (temperature or time scale of interaction) the material will exhibit a range of liquid-like (viscous) or solid-like (elastic) behavior. An example commonly used to demonstrate this is silly putty (although silly putty is not a pure linear polymer.) On short time scales, the polymers don't have time to relax completely and so it behaves like an elastic solid. This is why you can bounce a ball of silly putty on a table. At longer time scales, the polymers can fully relax, and so the silly putty flows. This is why if you let a lump of it sit out for a while, it will form a puddle. Silly putty is unique in that the relaxation time spectrum ranges from elastic to viscous over time scales that are commonly observed at room temperature, so you can see these effects in high contrast very easily.

There are many other types of non-Newtonian behavior (which just means viscosity is not constant), but they are all fundamentally connected to the relaxation time spectrum of the fluid, which is directly related to the molecular structure.

To sum up:

Water has a narrow viscosity spectrum (as a function of temperature) because it has a narrow stress relaxation time spectrum.
It has a narrow stress relaxation time spectrum because it is a relatively simple, small molecule.