Why is NaCl3 possible? There. And there. Almost a year ago, a group of scientists claimed to have reached compounds of $\ce{Na}$ and $\ce{Cl}$ with weird stochiometries ($\ce{NaCl3, Na3Cl, NaCl7, Na3Cl2}$ and $\ce{Na2Cl}$).
What one of the articles says:

These compounds are thermodynamically stable and, once made, remain indefinitely; nothing will make them fall apart.
For example, $\ce{NaCl7, NaCl3, Na3Cl2}$, and $\ce{Na2Cl}$ are metals (that explains the apparent violation of electroneutrality since charge balance rules are inapplicable to metals), and only one semiconducting phase of $\ce{NaCl3}$ is stable in the pressure range between 250 and 480 thousand atmospheres.

At first, I thought this whole thing was bogus. But then, I faced another, apparently convincing, article and even one from RSC. Then the questions popped up:

Why does this happen? (Why is such a compound more thermodynamically stable at high pressure?)
How can the bonding in these be explained?

(Source)

Yes NaCl3 is possible.

Think of the more common iodine analog.

$\ce{KI + I2 -> KI3}$

For a comprehensive analysis see:

Unexpected stable stoichiometries of sodium chlorides

  At 25-48 GPa, NaCl3 is stable in the Pnma structure, which has 4 formula units in the
  unit cell. Unlike all the other new phases predicted here (which are metallic), this phase is
  a semiconductor. Its structure (Fig. 2(B)) contains almost linear asymmetric Cl3 groups.
  Bader analysis (19) shows that the middle atom in the Cl3-group is nearly neutral, with
  most negative charge on the side atoms (Table 1) and the total charge of this anion group
  being ~-0.8. This result is reminiscent of well-known triiodides and azides with I3
  - and
  N3
  - ions, and of recent calculations on hypothetical H3
  - ions (20), which were found to
  have charge configuration [H-0.81H+0.72H-0.81]-0.9. Pnma-NaCl3 still can be viewed as an
  octet compound, with covalent bonding within [Cl3]--groups and ionic bonding between
  Na+ and [Cl3]-. At 48 GPa, with the formation of a new phase of NaCl3, the octet rule
  breaks down.

This research group compressed chlorine gas between NaCl plates, while heating with lasers and observed formation of $\ce{NaCl3}$.


Reactions of diborane, silane, methane, aluminium, boron, sodium, phosphorus trichloride, white phosphorus and iron with water at RT How many of the following will evolve $\ce{H2}$ gas when dissolved in water under room conditions?
$\ce{B2H6},$ $\ce{SiH4},$ $\ce{CH4},$ $\ce{Al},$ $\ce{B},$ $\ce{Na},$ $\ce{PCl3},$ $\ce{P(white)},$ $\ce{Fe}.$

Initially, it seemed easy to me and I got 7 (all except $\ce{PCl3},$ $\ce{CH4}).$ Correct me if I am wrong here.
However, the answer given was 3, which given were most likely $\ce{B2H6},$ $\ce{SiH4}$ and $\ce{Na}.$
What did I do wrong here? Does the temperature condition hold that much difference here? How do you predict products like these?

$$\ce{B2H6 + 6H2O -> 2H3BO3 + 6H2 \\
SiH4 + 4H2O -> Si(OH)4 + 4H2 \\
2Na + 2H2O -> 2NaOH + H2 }$$

The rest do not react with water under ordinary conditions. $\ce{Al, Fe, B}$ react only with steam or when they are red hot. White $\ce{P}$ undergoes easy hydrolysis when coordinated with ruthenium, as illustrated by Vaira et al. [1]. $\ce{CH4}$ does not undergo hydrolysis with water and $\ce{PCl3}$ does not give hydrogen on hydrolysis.

Reference

Vaira, M. D.; Frediani, P.; Seniori Costantini, S.; Peruzzini, M.; Stoppioni, P. Easy Hydrolysis of White Phosphorus Coordinated to Ruthenium. Dalton Transactions 2005, 0 (13), 2234–2236. DOI: 10.1039/B504795A.


Why do we add copper when testing a salt for nitrate ion? In our chemistry curriculum we are doing these lab experiments where we are supposed to test salts for certain anions,cations and other stuff.

I was told that if I dissolved the salt in concentrated $\ce{H2SO4}$ and then heated it for a while over a flame, the evolution of light brown vapours would indicate the possibility of $\ce{NO3-}$. To confirm its presence, however, I would have to repeat the test but this time also add some copper bits to the solution and check for dark brown vapours.

I figured the reason was that the copper reduced the $\ce{HNO3}$ gas to $\ce{NO2}$ which increased the amount of $\ce{NO2}$ gas produced, resulting in a darker shade of brown.

My question is that why copper, which is a very weak reducing agent, used here? Why not a real reducing agent like zinc which will give more $\ce{NO2}$ even with lesser salt?

This is a preliminary test, which is neither particularly selective nor sensitive.
With nitrate, brown vapours of $\ce{NO2}$ are released.
However, bromine or iodine could also release brownish or violet vapours.
Copper (or silver) is used to remove bromide and iodide by precipitation.


Temperature dependence in absorption spectroscopy What are some qualitative guidelines one should know regarding the temperature dependencies (of the gas and maybe the radiation, if that makes sense) for an absorption spectroscopy measurement?

How does, for example, a rise in temperature influence the absorption lines?

I am aware of one effect: doppler broadening. When the temperature increases, the speed distribution spreads out. That spreads out the distribution of the velocity component in the direction of your light source. The light the particles see from the light source then get red- or blue-shifted relative to what actually comes out of the source. So the particle population ends up absorbing a broader range of wavelengths than it did at a lower temperature. The effect is  $\Delta\lambda/\lambda_0 \propto \sqrt{T/m}$


Why elemental carbon is solid? Carbon with 6 protons, the first of its group, is a solid, while the very next elements, nitrogen and oxygen, are gaseous in their elemental form.

Why carbon is a solid? If it's not the molar mass that defines carbon phase (carbon's molar mass is lesser)

In pnictogen and chalcogen group only the second element, phosphorus and sulfur, is solid.

Carbon forms large molecules such as the ring structure of graphite, or tetrahedrons of diamond. Boron, like carbon, forms relatively large covalently-bonded molecules.

Nitrogen, oxygen and fluorine are diatomic, so molecules are less massive. The inert gases He and Ne are monatomic and therefore have lower b.p. yet.

In a similar fashion, metallic bonds hold together Li, Be and other metals.


Wave function for particle in a infinite well located at -L and +L The wave function for a particle in a infinite potential well located at $0$ and $L$ is $$\psi_n=\sqrt{\frac2L}\sin\frac{n\pi x}L$$

I want to find our wave function for the same situation where the walls are located at $-L$ and $+L$

I can write the Schrödinger equation, but I am unable to apply the boundary conditions, so I am unable to find the wave function.

How can I find the wave function in case of walls at $-L$ and $+L$?

It is very simple, there is no need to solve it again since you already know the solution of infinite square well. The idea is very simple, the trap looks the same, but wider now $2L$. So first you should extend it by a factor of two and then move it back to the center, it is given by the rescaling and translation:
$$L \mapsto 2L \tag{1}$$
$$x \mapsto x-L \tag{2}$$

So the solution is:
$$\psi_{n}=\sqrt{\frac{1}{L}}\sin\left(\frac{n\pi x}{2L}-\frac{n\pi}{2}\right) \tag{3}$$
$$E_{n}=\frac{n^2\pi^2 \hbar^2}{8mL^2}$$
As expected. Note that $\sin(\pi/2-x)=\cos(x)$ so that you can split it into odd and even series, but the Eq (3) given above is more succinct.


Why do metals form ions in solution? I'm looking at redox half-reactions, such as the one below:
$$\ce{Zn(s) -> Zn^2+(aq) + 2 e-}$$
Does the zinc react with the water and dissociate into ions? Why doesn't solid zinc have a charge?

Solids can't have a charge (neither can liquids or gases). The electrostatic forces are immensely great, way greater than you seem to think right now. Just imagine the forces that would be tearing apart a droplet of water (say, 36 microliters) if 0.001 of its molecules (not all of them, God forbid) would suddenly become positive ions. If you are aware of the Coulomb's law, you can help your imagination by actually calculating these forces.

(By now you may wonder how electrical phenomena are possible at all, but that's another story.)

Just in case if you don't like that answer, here is another:

Solid zinc, like any other metal, does have a charge. It consists of $\mathrm{Zn}^{2+}$ ions. These ions are immersed in a sea of electrons, which balances their charge, holds them together, and also is responsible for such properties typical of all metals as electrical conductivity, metallic lustre, etc.


Steroid nomenclature: β-sitosterol Can someone tell me why β-sitosterol has the beta prefix? What is it denoting? When comparing its structure to other similar phytosterols I see no need for it.

While one might think that β-sitosterol is an epimer of α-sitosterol, differing by configuration at C-3 (with $\ce{-OH}$ group), it does not seem to be the case.

There are various products and their isomers of phytosterol sitosterol biosynthesis, and their trivial naming seems to be somewhat rudimentary.

All of the sitosterols are 3β alcohols. The Greek letter (beta) designation in ‘β-sitosterol’ seem to be for more general, non-systematic isomers designation. In the field of natural compounds, it has historical reasons – a compound was isolated, described and named, and later was found that it in fact is a mixture of different isomers. (E.g. “α-caryophyllene” and “β-caryophyllene” do even have different number of cycles).

Sitosterols mentioned the Merck Index (14th ed.) are as follows ("natural product" IUPAC name according to last steroid nomenclature shown as well, might slightly differ from those from the M.I.).

(I) α1-sitosterol, (24Z)-4α-methyl-5α-stigmasta-7,24(241)-dien-3β-ol
      (or (24Z)-4α-methyl-5α-poriferasta-7,24(241)-dien-3β-ol)
(II) β-sitosterol, stigmast-5-en-3β-ol
(III) γ-sitosterol, poriferast-5-en-3β-ol

So, β- and γ- sitosterols (II, III) are diastereomers, epimers at C-24 (those differently named parent structure names stigmastane and poriferastane are in fact just (24R)-24-ethylcholestane and (24S)-24-ethylcholestane, respectively; these ethyl carbons are nowadays numbered 241 and 242 instead of older 28, 29). α1-Sitosterol (I) has the double bond not at C-5 but at C-7, additional methyl at C-4, and additional double bond at C-24 (ethylidene instead of ethyl). (And I guess that that ‘1’ subscript might be because of E/Z isomerism if this double bond.)


The product of a reaction series involving 1,3-dithiane, 1-chloro-3-iodopropane and butyllithium I encountered this question while solving an objective paper.

$\ce{BuLi}$ is a strong base and it can be used to abstract a proton from the substrate. Low temperature favours substitution more than elimination. So my idea was that the ring will break up $\ce{Li}$ will be substituted in place of two $\ce{S}$ atoms. The next product should involve formation of a six membered saturated ring but I am unable to proceed further. What is going to happen?

Some right ideas, but unfortunately, not correct. The correct answer is (C).
The protons in between the two sulfurs are more acidic because of hyperconjugative stabilization into the sulfur. Addition of 1-iodo-3-chloropropane results in the substitution product as you guessed. Another equivalent of butyllithium does the same thing again, except, this time, we do an intramolecular substitution for ring closure. Mercury has high affinity for sulfur, so that deprotects what is analogous to a ketal to give the ketone.

The key for these kinds of problems is to approach it from both ends. For example, the products are all carbonyl groups. Where would those groups come from? They can only be from the last step because otherwise butyllithium would add into a carbonyl. This gives you some idea for what 'C' would look, and so on.
For the forward direction, it's really important that you not kill all the reactivity by substituting off the heteroatoms. If you do that, you'll end up with an alkane, and there's no good chemistry unless you want $\ce{C-H}$ activation magic. Since butyllithium is not adding into anything, consider acid-base chemistry because it's also a very powerful base.


Is the titration of sodium carbonate with hydrochloric acid temperature dependent? I am a chemistry teacher in a Cambridge school.
In one of the titrations of sodium carbonate with HCl, there was one question asked:

  On heating the sodium carbonate solution before doing the titration, is there any change in the observation/reading?

My opinion is that the rate of reaction will increase due to more collisions among particles due to the temperature change. So the time to completion of the titration should be shortened.

However, the CIE mark scheme suggests that there is no effect, as the concentration remains same.

I think you might have a different understanding of what constitutes a 'change in the observation/reading' from the textbook, resulting in your confusion.

I am assuming that the textbook is only considering the titration graph per se, pH vs volume of acid added.

You are referring to the time to pH equilibrium during each drop of acid added. For the purposes of all titrations I have encountered, this was never a factor that was discussed. As @Ivan Neretin commented, the reaction is assumed to occur instantly.

As the concentration of the species remains the same, the titration graph will look the same. In other words: The temperature has no influence on the equivalence point.


How do you determine the pH of a solution at different volumes? For a tube with 1 mL of 0.75 % (w/v) casein in 0.10 M sodium acetate, with 0.05 mL acetic acid, and 6.95 mL H2O, how do you determine the pH?

My original thought was to use the Henderson-Hasselbalch equation with the concentration of base being 1 and the concentration of acid being 0.05, but this does not feel right.

I think my fundamental understanding of the direct correlation between concentration and volume is not correct.

With regard to your understanding of the correlation between concentration and volume, you should know that concentration $c$ is defined as
$$c=\frac nV$$
where $n$ is amount of substance and $V$ is the volume of the solution.

When you dilute a sample, you change the volume $V$, but you do not change the amount of substance $n$ in your sample. Therefore, if you know the initial concentration $c_1$ in the initial volume $V_1$
$$c_1=\frac n{V_1}$$
you can calculate the new concentration $c_2$ after dilution to the new volume $V_2$:
$$c_2=\frac n{V_2}$$
You can rearrange both equations to calculate the amount of substance $n$:
$$\begin{align}
c_1\cdot V_1&=n\\[6pt]
c_2\cdot V_2&=n
\end{align}$$
Since the amount of substance $n$ is constant, you get another useful equation:
$$c_1\cdot V_1=c_2\cdot V_2$$

In your exercise, after addition of all solutions, the new volume may be estimated as
$$V_2\approx1\ \mathrm{ml}+0.05\ \mathrm{ml}+6.95\ \mathrm{ml}=8\ \mathrm{ml}$$

The initial concentration of the base (sodium acetate) is $c_{\text{base},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{base},1}=1\ \mathrm{ml}$.

Therefore, the new concentration of the base is

$$\begin{align}
c_{\text{base},1}\cdot V_{\text{base},1}&=c_{\text{base},2}\cdot V_2\\[6pt]
c_{\text{base},2}&=\frac{c_{\text{base},1}\cdot V_{\text{base},1}}{V_2}\\[6pt]
&=\frac{0.10\ \mathrm{mol\ l^{-1}}\times1\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt]
&=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$

The initial concentration of the acid (acetic acid) is $c_{\text{acid},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{acid},1}=0.05\ \mathrm{ml}$.

Therefore, the new concentration of the acid is

$$\begin{align}
c_{\text{acid},1}\cdot V_{\text{acid},1}&=c_{\text{acid},2}\cdot V_2\\[6pt]
c_{\text{acid},2}&=\frac{c_{\text{acid},1}\cdot V_{\text{acid},1}}{V_2}\\[6pt]
&=\frac{0.10\ \mathrm{mol\ l^{-1}}\times0.05\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt]
&=0.000625\ \mathrm{mol\ l^{-1}}
\end{align}$$

Finally, you can use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of the acetate buffer from the new concentrations $c_{\text{acid},2}$ and $c_{\text{base},2}$.


Is the Beckmann Rearrangement technically an "SN2" reaction at an "sp2" center? Below is the mechanism from this link of the Beckmann Rearrangement.

In the alkyl migration step, the alkyl group migrates simultaneously as the $\ce{H2O}$ leaving group is expelled. This seems as though it is an $\mathrm{S_N2}$ reaction at the $\mathrm{sp^2}$-hybridized nitrogen atom, but I thought that it was geometrically impossible to have an $\mathrm{S_N2}$ reaction at an $\mathrm{sp^2}$ center, per this link.

What am I missing in my reasoning?

When you were taught that we don't see $S_{\mathrm{N}}2$ reactions at $sp^{2}$ centers, what evidence or rationale was given? In organic chemistry, very few things are definite. When they are, it's usually because of being forbidden by quantum mechanics (e.g., symmetry in pericyclic reactions). Otherwise, they are just very unlikely, unless you somehow magically make the conditions right.

Greg Fu used to give this example: J. Am. Chem. Soc. 1998, 120, 2275-2282.

In it, a trans-substituted alkene undergoes nucleophilic substitution to become a cis-substituted alkene. Using that fact and other data, the authors conclude that the mechanism most consistent with the observed reaction data is a vinylic $S_{\mathrm{N}}2$. If you have a better mechanism, consider writing a paper and submitting it to JACS.


Why is tetramethylsilane (TMS) used as an internal standard in NMR spectroscopy? I understand that TMS has a large proton number, giving a strong peak in the NMR spectrum. But apart from that is there any other reason why just TMS is used and not any other molecule?

TMS was first proposed as a reliable internal chemical shift reference in 1958 by Tiers. Back in them good ol' days, 1H NMR was called proton nuclear spin resonance, or nsr, and the tau scale was  used for reporting chemical shifts (10ppm in the delta scale was set to 0 and positive values were read to the right. So TMS came at t10.0), and CCl4 was the one of the most common solvents used for running dilute solution samples.

TMS was proposed as a reference at the time becasue it offered the following advantages:

the chemical shifts are largely temperature and concentration independent
it offered a single phase method for referencing (That is, it was an internal standard. Most common methods at the time used an external reference, such as a coaxial insert; a method which is still acceptably used today)
The TMS peak lies outside the 'usual spectral region and is readily identifiable'
It is largely inert, and is not likely to react with most samples
It can be used for almost amy solvent (except H2O, D2O)

Today, TMS is still accepted as the single universal reference standard for NMR. For over a decade, the IUPAC recommendations have stated that, other than for 13C, all nuclei are referenced to the frequency of the 1H signal of 1% TMS solution in CDCl3. Yes, your 31P and 77Se peaks should actually be referenced to 1H of TMS. This unified reference scale is very easy to implement on modern instruments that incorporate a digital 2H lock (ask your friendly NMR custodian if they use it!). It is in fact arguably more accurate that referencing to internal solvent residues which are much more likely susceptible to concentration and temperature factors. Internal solvent residues are called secondary references, and are less accurate than the primary reference. However, they are accurate enough for the very vast majority of work done in the laboratory.

It has long been shown that the chemical shift of TMS is, in fact, both temperature and concentration dependent. But not much. There are actual corrections that should be applied to measured chemical shifts at extreme temperatures, but these are very small and rarely seen in the literature.

TMS is not water soluble, and in aqueous samples DSS is the recommended reference standard. It contains signals in the 1H spectrum at 3.1, 1.9 and 0.8ppm, all at a fraction of the intensity of the main reference signal, and these can interfere with analysis of other spectral regions. A deuterated version is widely used in the literature. An alternative, TSP, is sometimes used, although is more pH dependent. Other reference standards can also be used, however, and as long as the reference method and sample conditions are reported and verified, then all is good. For 1H NMR, compounds such as acetonitrile, methylsulfone, benzene, 1,4-dioxane and dichloromethane are all examples of internal standards that are sometimes used, both for chemical shift referencing and concentration measurements, as they all largely satisfy that important list of reference characteristics above. They all have very simple 1H spectra, soluble in common deuterated solvents and fairly unreactive. Plenty of other compounds with multiple peaks can be used as well, such as toluene and dmf.

So, to specifically answer your question, TMS is the adopted primary standard for 1H chemical shift referencing for the reasons stated above. Any other material can be used as a secondary reference, either internal or external, provided its chemical shift can be accurately determined against the chemical shift for TMS. Results reported should include a clear description of how chemical shifts were determined. For example, 1H chemical shifts of all reported compounds were measured relative to the methyl peak of n-dodecane, determined as 0.88ppm. The disadvantages of using other materials is that it needs to be shown that the chemical shift of these signals is independent of sample conditions (concentration, temperature etc), and that they will often have peaks overlapping regions of interest.


How do you name this compound using IUPAC system (including steps)? And including stereochemistry (cis trans or R S )

I know you start numbering from the double bond because there are no functional groups like -OH, you take the longest carbon chain which is six carbons in this question, so now it should be 3-bromo-3-methyl-1-hexene. My quarrel is with the stereochemistry and how do you name the compound regarding the widge and the dash (cis and trans or R and S)

I also know CIP priorities, it's just the widge and the dash being at the middle of the compound that confuses me.

trans-2-bromo-2-vinyl-pentane ?
R-2-bromo-2-vinyl-pentane ?

Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.
Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl
Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3R)
String it all together.

(3R)-3-bromo-3-methylhex-1-ene.


London/dispersion forces "London (dispersion) forces are responsible for the fact that non-polar substances can be condensed to form liquids and sometimes solids at low temperatures"._

I learned that what results in London fores between molecules is the "cloud of electrons" that  results in an instantaneous dipole and induced dipole, which eventually results into forming these weak London forces._

The aforementioned statement suggests that as we lower the temperature we form a solid or a liquid meaning that the inter-molecular forces (London forces in this case)became stronger._

My question is: if we lower the temperature we subsequently decrease the kinetic energy of the electrons,  i.e. we decrease the frequency/probability of these "electron clouds" moving to one side of the molecule and forming instantaneous and induced dipole, which means that London forces should be weaker and not stronger to form a liquid or a solid.

This is how I thought of it, can anyone suggest what's wrong with my way of thinking of it? Is it that what we relate to the strength of the London forces is how long the electrons cloud stay on one side rather than how often that happens?

When we lower the temperature of a substance, we are decreasing the average molecular kinetic energy (they're slowing down). As mentioned in the comments on the question, this doesn't really affect the electron cloud.

So, the substance has been cooled down and molecules are moving more slowly. At this point, it's not that the dispersion forces because stronger, but rather that they become comparatively stronger. While in the gaseous phase, the molecules have velocities of several hundred meters per second. At those speeds, London forces don't really have much of an effect. But, as the molecules are cooled and slowed down (this seems to suggest this is around 10m/s), London forces can cause more interactions between molecules. If the substance is cooled far enough, those forces are able to overcome the kinetic energy of the molecules and they condense into a liquid. As the cooling continues, the molecules move around less and less until they take on a definite shape, at which point it is considered a solid.

The primary issue with your intuition is merely understanding what cooling the substance down really does.


Range of distance for Van der Waals force Is there any range of distance between the nuclei of the atoms (in Angstrom, say between 4 to 12 Angstrom) within which there will be an occurrence of Van der Waals force (attraction) between them?
For example, say within a distance of 4 A between two atoms of same type, there will van der Waals attraction present. Can such a certainty be obtained? If two atoms are very far apart, then there will be not Van der Waals attraction between them, this is known. So my question is how close should the nuclei of two atoms be so that one can say with certainty that there will be Van der Waals attraction?

Van der Waals force is due to interaction of two dipoles. It is never strictly zero, but it diminish very fast with distance.

It is common to look at VdW interaction at Van der Waals radius. The largest radius should be around 2.60 for Fr making the total distance between nuclei more than 4 A. Heavier elements will have larger radius, but heavier elements are not stable. Are we really interested in VdW force between two atoms that will decay in 10^-15 sec?

Also, by "4-12A" do you actually mean distance between nuclei or the surfaces of atoms (VdW radius A + VdW radius B + "4-12 Angstrom")? In the first case iodine has VdW = 2.15A, so nuclei interact on distances more than 4A. This is the distance between atoms of neighboring I2 molecules in the crystal. In the second case you will see some interaction, but it will be quite small.


Organizing my "chemical" cabinet I'm reorganizing my garage (I do metalworking and some machining), and now I have an empty white cabinet that I'd like to dedicate to chemical stuff. I'm not a chemist, and I don't have the typical assortment of chemicals that a chem lab would have; the stuff I'm referring to is typical stuff you'd find in a garage: adhesives, lubricants, solvents, paint, etc. (household cleaners are stored elsewhere).

I'd like to organize things by functional groups (like those listed above). Is there a convention/standard for organizing shop chemicals?

As mentioned in the comments, safety is the main priority.  Take particular heed to the warning labels and information on the chemical label - if unsure, look up the chemical details MSDS (or SDS) = (Materials) Safety Data Sheet.

One of the most important thing to do (and is often overlooked) is to keep an inventory of what chemicals you have.  The safety data sheets can be obtained from the manufacturers or by looking up the chemical SDS itself, or from general websites such as MSDS.com - an example is this one for mineral turpentine - these contain considerable information about storage, disposal and first aid.  (It is always a good idea to have this readily available and updated).

Personal protective gear is a must - gloves, safety glasses etc - usually the label or the SDS will have this information.

According to the document NERC Guidance on Safe Storage of Laboratory Chemicals (2010), has information that is pertinent to a workshop environment.  There are 3 main principles in regard to storage of chemicals explained in the document:

Segregation, from the article:

  The key incompatibles to segregate from each other are strong acids from
  strong bases and strong oxidisers from organic or flammable materials.

The NERC document has a list in its appendix (too long for this reply)

Separation, related to the segregation,

  In a laboratory situation adequate separation can be achieved by means of
  storage cupboards which physically divide incompatible classes of hazardous
  chemicals. The cupboards may need specific properties or provide
  separation by means of distance. They will also need to provide secondary
  containment (eg spill trays or bunded shelves) and security (eg locks / bolted
  to wall).

Ventilation (from personal experience, this one is often overlooked).

  Ventilation is often an essential requirement for safe storage of hazardous
  chemicals. Its main function is to allow dilution and extraction of vapours or
  gases that may escape / seep out from containers during storage so they no
  longer present problems from the viewpoint of noxious smell, hazardous
  personal exposure or creation of an explosive atmosphere.

Also consider an appropriate fire extinguisher, note that water is probably the worst one.  Here is an example of a fire extinguisher type guide - this guide suggests that powder based extinguishers may be best (but, make sure you study it and seek advice if unsure).

Some of this may seem 'overkill', but it is better to have some preparedness for if anything goes wrong, than be unpleasantly surprised by it (from personal experience), also, depending on where you are, there could be a legal requirement for a degree of proper storage, handling and protection.


What is chemically different about tea for hard water? At the supermarket they sold tea for hard water. (Hard water is water with limestone/calcium carbonate in it).

There appears to be a chemical modification for foods using this type of water.

My question is: What is chemically different about tea for hard water?

Edit: here is the list of ingredients: tea (only)

Edit 2:
- it appears hard water is more basic so a tea mix that is more acidic is required (perhaps)

Hard water tea

By definition hard water is defined as water that has more than 0,6mg/l of dissolved CaCO3. Therefore soft water that doesn't have that much minerals has a greater dissolubility and can dissolve more tea. It can also dissolve the tea quicker. So the tea companies need to find what tea blend dissolves quickly and efficiently in hard water.

  A hard water blend needs lots of quality Assams and East African teas, with plenty of “guts” and strength. It needs flavour too, of course, but this flavour will be the rich, malty type that punches through the hard water. -https://allabouttea.co.uk/tea-news/hard-water-tea/ (as Bdrs commented)


Why is picric acid more acidic than carbonic acid? I just had a question in a test asking to compare the relative acidity of picric acid (TNP) and carbonic acid, $\ce{H2CO3}$ (not phenol). It turns out that TNP is more acidic. I have no idea why? Is there some rule I'm missing or is this one of those exceptions that I just have to remember?

I have been told that alcohols are generally less acidic than carboxyl groups. But in this case, doesn't the hydroxyl group form a chelate ring already? If the stability of the reactant is much greater than the the stability of the product, why would the reaction even take place? Sure, there is cross conjugation in carbonic acid, but shouldn't it be extremely hard to break the chelation in picric acid?

So you want to compare the acidity of different Brønsted–Lowry acids. This means ultimately you want to compare the equilibrium constants for reactions of the type $\ce{HX <=> H+ + X-}$. To do so, you'll have to analyse what stabilisation/destabilisation effects are present in each acid $\ce{HX}$ and each conjugate base $\ce{X-}$. This is easier said than done, especially quantitatively. There are many sorts of possible effects, often acting in a compounded fashion. Fortunately, there are some basic principles to help guide us.

Your question specifically asks to contrast carbonic acid, $\ce{H2CO3}$ and picric acid, more descriptively called 2,4,6-trinitrophenol. This already starts off as being a bit tricky, because they are not very similar substances; of course, the more similar the two substances are when being compared, the easier it will be to pick out the differences in stabilisation/destabilisation effects and the safer it will be to extrapolate from qualitative reasoning. With this in mind, it is best if we build up to the answer by making comparisons between several molecules, making small changes at each step.

First off, let us start with alcohols. These are substances containing hydroxyl groups ($-\ce{OH}$) linked to carbon atoms which are saturated (carbon atoms with four single bonds, i.e. $\rm{sp^3}$-hybridised carbon) and whose three other bonds are only to carbon or hydrogen atoms (in other words, linked to no heteroatoms other than the hydroxyl oxygen). This class represents neither of your substances, but is simple and is related to phenols, so it will be a useful benchmark. A simple alcohol is methanol, $\ce{CH3OH}$, which can act as an acid by the equilibrium:

$$\ce{H3C-OH_{(aq)} <=> H+_{(aq)} + H3C-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=3\times10^{-16}$$

The small equilibrium constant establishes that methanol is a very weak acid in water, and exists almost completely as the unionised reactant. Values of $\mathrm pK_\mathrm a$ around 15 are typical for simple alcohols.

Now let us take a step towards the structures we want, by examining at a hydroxyl linked to an unsaturated carbon atom, specifically a $\rm{sp^2}$-hybridised carbon atom linked to no heteroatoms other than the hydroxyl oxygen. What we have here is an enol. Simple enols are not very stable molecules, but for simplicity we shall disregard that. For the simple molecule propen-2-ol, the equilibrium is:

$$\ce{H2C=C(OH)-CH3_{(aq)} <=> H+_{(aq)} + H2C=C(O^{-})-CH3_{(aq)}}\ \ \ \ \ K_\mathrm a=1.3\times 10^{-11}$$

Evidently, enols are quite a bit more acidic than alcohols. There are two new effects at play here. First, $\rm{sp^2}$-hybridised carbon atoms show an electron-withdrawing effect compared to $\rm{sp^3}$-hybridised carbon. This ability to "suck away" electron density helps stabilise molecules when close to electron-rich regions such as the oxygen atom. This effect stabilises the reactant, but it stabilises the enolate product more (as it has extra electron density to be pulled), which drives the equilibrium to the right. A second effect is resonance. Some of the electrons from the oxygen atom are said to conjugate with the carbon-carbon double bond, creating resonance or canonical structures, which contribute to the true description of bonding in the molecule. Once again, spreading the electrons out contributes to decrease the energy of a molecule. The resonance structures for the reactant and the product are similar, but one of the resonance structures in the reactant contains a formal positive charge on oxygen (a very electronegative atom) and a formal negative charge on carbon (a not very electronegative atom), so this form of conjugation is not quite as favourable. In the product, though there is still a formal negative charge on carbon in one of the canonical structures, there is no formal positive charge on oxygen, making conjugation more favourable. Thus the resonance argument is also a part of the explanation for the acidity.

Next up is phenol, the parent compound of 2,4,6-trinitrophenol. The effects present here are much the same as in the enol above, except that conjugation happens in a larger extent in both the reactant and the product. Experimentally:

$$\ce{C6H5-OH_{(aq)} <=> H+_{(aq)} + C6H5-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=1.1\times10^{-10}$$

This represents only a small increase in acidity compared to propen-2-ol, which suggests conjugation is not the main factor in phenol acidity, but rather the inductive effect of the $\rm{sp^2}$ carbon.

Finally, we've built up to our first structure, 2,4,6-trinitrophenol. The new bit here is the addition of nitro groups ($\ce{-NO2}$), which are strongly electron-withdrawing both by induction and mesomerism (conjugation). Again, these effects decrease the energy of both the reactant and the product, but stabilize the product more because there is extra density to be spread around. As a quick interesting aside, the acidities of 2-nitrophenol and 4-nitrophenol are similar, and both are somewhat larger than 3-nitrophenol, which suggests most of the increase in acidity comes from electron-withdrawing by mesomerism rather than induction, represented by canonical structures similar to these. You also make a good point in mentioning the intramolecular hydrogen bond present only in 2,4,6-trinitrophenol and not its ionised form. Indeed, this helps stabilize the reactant a slight amount, and decreases its acidity. Let's see how all these effects add up:

$$\ce{(O2N)3C6H2-OH_{(aq)} <=> H+_{(aq)} + (O2N)3C6H2-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=0.4$$

The acidity increase from the electron-withdrawing effects of the three nitro groups far overwhelms the acidity decrease due to the formation of the intramolecular hydrogen bond. How could you have known this beforehand? Strictly speaking, you couldn't have deduced this from thin air (not without a computer, at any rate!), but generally you are expected to know the situation in similar cases and work out the answer based on previous knowledge. Intramolecular hydrogen bonding usually doesn't play a huge role in increasing or decreasing acidity, at least in simple molecules, as they are relatively weak interactions and so don't contribute massively to increasing/decreasing the energy of a structure. Furthermore, nitro groups are among the most electron-withdrawing there are, so if pulling electrons is going to have any role, it'll show some of its strongest effects with nitro groups.

To prove this answer comes to an end, we reach our final structure, carbonic acid. First, we notice it is related to carboxylic acids, and its acidic group is the $\ce{-COOH}$ in the structure. If you think about it, the carboxyl group looks like an enol where a $\ce{=CH2}$ group is replaced by an oxygen atom. Thus the analysis is similar to the enol. However, there is an increase electron-withdrawing effect due to induction (oxygen being more electronegative than carbon) and mesomerism (in the canonical structures, the negative formal charges are now on an oxygen atom, rather than a carbon atom as was the case for the enol). The instability of carbonic acid interferes, but leaving that aside, its dissociation would be something like:

$$\ce{H2CO3_{(aq)} <=> H+_{(aq)} + HCO3^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=2.5\times 10^{-4}$$

Comparing everything, the acidity increases in the order:

Simple alcohols < Simple enols and phenols < Carbonic acid and simple carboxylic acids < Picric acid

How would you know carbonic acid is a weaker acid than picric acid? Once again, it's not something you can deduce by pure logic; it requires some benchmarking and comparison with previous data and knowledge. The analysis above would have been helpful, but has its limits. In Chemistry, it is often far easier to explain than predict. Whether you should have been expected to know the answer to this question, I'm not sure. It would depend on how your chemistry course is conducted.


Relative acidity of p-chlorobenzoic acid and p-flurobenzoic acid I want to compare the relative acidity of p-chlorobenzoic acid and p-flurobenzoic acid. Both of the facts below point to p-chlorobenzoic acid being more acidic:

The carboxylic acid group is too far away from the halogen for inductive effect to have any significance. The +M effect of fluorine is greater than chlorine (due to more effective orbital overlap). +M effect is not distance dependant and hence p-fluoroobenzoic acid  should be less acidic since the negative charge on its conjugate base is destabilized.
There is resonance of the negative charge of the carboxylate ion with the empty d-orbitals of chlorine which would stabilize it.

However, in reality, p-chlorobenzoic acid is less acidic than p-fluorobenzoic acid . Why?

$\mathrm{p}K_\mathrm{a}$ of para chlorobenzoic acid $\ce{-> 4.03}$.Source
$\mathrm{p}K_\mathrm{a}$ of para fluorobenzoic acid $\ce{-> 4.14}$.Source

For acidity of para substituted benzoic acids, -I effect is helpful and +R effect decreases the acidity.

In 4-fluorobenzoic acid, the +R effect is prominent in $\ce{F}$ due to stronger $\ce{2p}\pi - \ce{2p}\pi$ overlap with nearest carbon. Thus, due to stronger +R effect of $\ce{F}$ than -I effect, it creates a negative charge on your starred carbon. This charge delocalises in the carbonyl carbon  and the electrophilicity is thus decreased of that carbon. Releasing hydrogen also doesn't help much as the negative charge created over $\ce{O}$ of $\ce{-OH}$ group can only delocalise a little.

Whereas in the $\ce{Cl}$, there is much weaker $\ce{3p}\pi-\ce{2p}\pi$ overlap. So, +R effect of $\ce{Cl}$ is much weaker and due to a relatively higher -I than its +R, there is more positive character on the starred carbon due to $\ce{Cl}$.

And, thus, in reality, acidity of 4-chlorobenzoic acid is a little higher than that of 4-fluorobenzoic acid.


Ambimodal transition states in some organic reaction pathways In many computational studies, mechanisms appear to have so-called ambimodal transition states, i.e. a transition state which can lead to multiple products, in conflict to the common undergraduate idea of a reaction pathway.
To give an example, Houk, in Mechanisms and Origins of Periselectivity of the Ambimodal [6 + 4] Cycloadditions of Tropone to Dimethylfulvene,[1] presents the following scheme:

Reaction pathway for [6 + 4] cycloadditions of Tropone to Dimethylfulvene. Taken from ref [1]
In the scheme, TS-1 may either give 7 (the 'expected' product) or go directly to 6, completely skipping TS-Cope-1 (which would allow 7 and 6 to interconvert) — Houk describes this as an ambimodal transition state, that is one transition state that can lead to two possible products.

Although what he says makes sense, it seems to me that, given 6 is the lower energy product out of the two, it would be favoured thermodynamically anyway.
Upon further reading, I discovered the concept of valley ridge inflection, which appears to describe the phenomenon observed above, in which we completely 'miss' the  intermediary transition state:

Why do these ambimodal transition states occur, and how does the situation differ from normal transition state theory?

Notes & References
[1]:Houk, K.N., J. Am. Chem. Soc. 2017, 139, 8251. DOI: 10.1021/jacs.7b02966

Why do these ambimodal transition states occur?

I don't know that there is a general reason for that. Potential energy surfaces are complicated. Very often, there is an abundance of local minima, separated by potential energy barriers. It is similar to what happens with mountain chains and valleys, but with many dimensions.
One could argue that the opposite could be asked: why would there be a single minimum associated with each and every barrier? Why would there be exactly a single downhill valley behind each mountain chain, instead of, sometimes, two different valleys being similarly accessible, once you climb over the top of a hill?

How does the situation differ from normal transition state theory?

Transition state theory is a useful simplification, assuming a simple potential energy surface. It is useful because, at room temperature, the energy differences in the different possible paths are very often sufficient for a single reaction pathway to predominate, so that others effectively do not happen. When the temperature is comparable to the difference between the energies involved in different possible reaction pathways, the situation is more complicated.


Why are inert gas (especially Xenon) compounds powerful oxidizing agents? I am curious as to why compounds with inert gases, such as $\ce{XeF4}$, $\ce{XeF2}$, and $\ce{XeO3}$ are considered powerful oxidizing agents. I would attribute the phenomenon to the highly oxidized state of the noble gas which would be unstable but I am not too sure about it.

Let's be clear here. Xenon by itself is not an oxidizing agent. Instead, certain compounds of xenon tend to be strong oxidizing agents.

First, let's consider an analogous case - the nitro group. Molecules containing nitro groups ($\ce{-NO2}$) tend to be explosive - the nitro groups contain oxygen which facilitates combustion. Remember, combustion is just a form of oxidation.

Back to xenon!

Xenon, being a noble gas, shouldn't like bonding too much. It already has a complete octet of outer shell (valence) electrons.

So it is likely the case that compounds of xenon are only weakly held together. Compounds of xenon can therefore said to be relatively thermodynamically unstable. Therefore, it is also likely that compounds of xenon tend to react in the presence of other molecules - especially ones containing atoms to which fluorine can form stronger bonds.

We verify the above by examining bond enthalpy data: the Xe-F bond has a bond energy of about 130 kJ/mol. Compare this to a typical C-F bond with a bond energy of 485 kJ/mol.

It is no wonder therefore that compounds such as xenon difluoride are used in fluorinating (oxidizing) organic molecules.

Here are some examples from Wikipedia. Why the nitro group in the second photo has been reduced to an amino group - I have no clue - and I suspect it's just an error. Further updates soon ...


Visual explanation between Molecule vs Compound vs Element vs Atom vs Substance? Im a visual learner and am trying to find a visual explanation between the meaning of those words.

Is this Venn diagram accurate?

Can you draw a better Venn diagram, perhaps one that would also include other words like homogeneous, heterogeneous, and mixture?

Even if you draw it in Paint, I can re-do it and see if you like it.

Or is this one better?

Here's a cone view.

Everything from large trees to atoms are viewed by our eyes in the cone view (just my perceptions...). This is a very interesting concept. It's like shooting an electron (into or out) of an element, that changes an physical & chemical properties of the atom, compound and substance etc..

Think about a pencil, and what it's made of.

This may be quite far from a deep chemistry expertise. So take this as a simple representation to accommodate the diagrams that you began with and looking from a different perspective. We aren't trying to fit a facts/theory into our understanding and visualisation but other way around.

An important distinction is the clarity of molecule, compound and mixture.
Molecules are constructed of atoms of same species. A compound is formed by atoms of different species. Compounds are of 2 different types mainly,

Covalent
Ionic
Intermetallic
Coordinate

Each of this bonds, origin goes back to a single atom and involvement of its orbitals energy and electrons.

For an example,

The water that we drink is a molecule and also a compound, made of two H atoms and oxygen atom covalently bonding. Isn't it odd to read in your paper or an article, "the water compound..." or wouldn't it feel fit to read as "a water molecule...".

The salt we add to balance flavour is a compound, made of one sodium atom (+charged ion) and chloride atom (-charged ion) ionically bonding.

The air we breath is a mixture of various molecules : $H_2O, H_2, O_2, N_2, CO_2$ etc (not atoms).

  Put it this way, all compounds could be molecules, but (not) vice
  versa.

Here's the top view (transverse plane from the very top) of the cone.

In physical sciences the matter is classified in pyramids like the following. Reference is in the image itself.

There are more advanced and slightly different version of discussions on the same topic here on pubchemblog : "difference between xyz..."..

The starting cone isn't pretty as your images. But I will make them more user-friendly and eye-candy if you can comment. This almost feels like a drawing session now, apart from the conceptual outlook. I am happy to read more about intermetallic and coordinate bonding as well to address that in the answer. It is not something I come across on daily basis as I am not in electronics or soldering/alloy analytical chemistry. Apart from the cone, other picture references are taken from here.


Why does monoammonium phosphate (MAP) have 52% of phosphorus (P₂O₅)? The formula of monoammonium phosphate (MAP) is $\ce{NH4H2PO4}$. Manufacturers say that MAP has 10% of nitrogen (N) and 52% of phosphorus (P). However, its molecular mass is 115 u and the atomic mass of P is 31 u $(31/115 × 100\% = 27\%)$. How is there 52% of P?

When it comes to fertilizers, usually what is advertised by general name monoammonium phosphate 10-52-0, in reality is ammophos, which is a mix of monoammonium and diammonium phosphates, $\ce{NH4H2PO4}$ and $\ce{(NH4)HPO4}$ with some water, so the brutto formula is

$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}$$

where $0 ≤ x ≤ 1$.
To find out the range for $y$, let's find the average molecular mass from the given nitrogen ($\ce{N}$) and phosphorous ($\ce{P2O5}$) content $α_i$:

$$\bar{M} = \frac{1}{2}\cdot\left(\frac{M(\ce{N})}{α(\ce{N})} + \frac{M(\ce{P2O5})}{2α(\ce{P2O5})}\right) = \frac{1}{2}\cdot\left(\frac{\pu{14 g mol-1}}{0.10} + \frac{\pu{142 g mol-1}}{2\cdot 0.52}\right) \approx \pu{138 g mol-1}$$

Using the expression for molecular mass of brutto formula and its average value:

$$x = 1.35 - 1.06y$$

Since $0 ≤ x ≤ 1$, the following holds true for $y$:

$$0.33 ≤ y ≤ 1.27$$

resulting in formula with the following ranges for the variables:

$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}, \quad x \in [0; 1], y\in [0.33; 1.27]$$

Since monoammonium phosphate usually prevails in this mix, there is a substantial amount of water; e.g. one of the possible brutto-formulas would be the following monohydrate:

$$\ce{(NH4)_{1.3}H_{1.7}PO4 * H2O}$$

Obviously, all the above is valid in the absence of third-party contaminations, e.g. if pure phosphoric acid and ammonia were used for the synthesis.


Which material to use for athletic accessories? I'm looking for some guidance in selecting the right material for dance shoes heels and women's self-defense accessories.

I'm looking for a material that 1) has a superb strength-to-weight ratio, can withstand punishing and frequent impact, and very light (like plastic).

The part for the shoe will be very, very slender (stiletto heel), but will have to be able to endure hours of jumping, sliding, stomping, spinning, and skidding dancers… is compression strength the right term?

2) The material must also be safe to use as jewelry, so nothing toxic. Also, I am looking for something that tends to snap, rather than shatter, when it fails under pressure.

3) Lastly, it would be great if this material can be 3D printed, although that is not a must.

An inexpensive material would be ideal, of course, but I am open to learning about a fuller range of materials, including the more expensive ones. I am willing to consider a higher priced material if it will offer substantial value to the accessories in terms of strength, quality, and safety.

For plastics, my first instinct for the first two properties would be a polycarbonate plastic. Polycarbonates are known to be very strong plastics (tensile strength up to ~70 MPa) and deform without breaking under many conditions, which is why they're commonly used in things like safety glasses and bullet-resistant glass. It also looks very interesting as it's quite optically clear. PC is 3D printable, I've seen a few people using it in hobby printers (http://www.protoparadigm.com/blog/2011/12/printing-polycarbonate/) and in very high end StrataSys printers, but it's not a super common material so it might take some experimenting.

Another option is a nylon. There are a number of different types available and many have similar properties to polycarbonates (minus the optical clarity). Nylons seem to be a bit more popular for 3D printing and I've heard good things about the Taulman filaments.

Of course, you're not limited to a single material and there are many things that could potentially be used in this kind of application. Often, the combination of multiple materials makes sense (e.g. a metal insert in a plastic part).

Something like carbon fibre-reinforced epoxy construction might be appropriate. It's not 3D printable, but can be constructed in layers around a lightweight (expanded polystyrene or the like) form. This approach is often used for parts that need to be very light while having high tensile strength (think bicycle frames, ski poles, etc.) I could see a carbon fibre-wrapped plastic insert making a very strong and light shoe heel.

Ultimately, these things are balancing games between strength, mass, fabrication cost, aesthetics, etc. but these are a few potential materials that come to mind.


How to make γ-MnO2 in a lab? I am working on Li-ion batteries, and I need to use γ-$\ce{MnO2}$ (electrolytic manganese dioxide). I have googled and read quite a few papers, but I am not very sure of those methods. I'd love to know if you are aware of some! The ones which I encountered, again and again, was the one involving precipitation of $\ce{MnO2}$. But is it possible to precipitate the γ form? Also, I think the commercially available $\ce{MnO2}$ is the α form. Is there any way to convert it to γ form? I am particularly looking for a way to synthesize  $\ce{MnO2}$ using electrolysis.

I know that I can also break a few batteries to get my $\ce{MnO2}$, but it would be heavily contaminated by carbon. Commercially available  $\ce{MnO2}$ is very expensive.

A quick search yields this.

  [Electrolytic manganese dioxide] is produced in a similar fashion as
  electrolytic tough pitch (ETP) copper: The manganese dioxide is
  dissolved in sulfuric acid (sometimes mixed with manganese sulfate)
  and subjected to a current between two electrodes. The $\ce{MnO2}$ dissolves,
  enters solution as the sulfate, and is deposited on the anode.

Unfortunately, there is no source for the information.

Some further searching yields this article, with the following information:

  EMD ($\gamma$-$\ce{MnO2}$) can be prepared from the direct electrolysis
  of an aqueous bath of manganese sulfate and sulphuric acid. The
  electrochemical deposition of $\ce{MnO2}$ on an inert electrode from
  an electrolyte containing $\ce{Mn^2+}$ occurs as per the reaction
  noted below:

  $\begin{alignat*}{3}&\textrm{anode: }\quad&&\ce{Mn^2+ + 2H2O} &&\ce{->
MnO2 + 4H+ + 2e-} \\ &\textrm{cathode: } \quad&&\ce{2H+ + 2e-}
&&\ce{->H2} \end{alignat*}$

Note that the linked article is a review, with a lot of information, both about the actual structure of $\gamma$-$\ce{MnO2}$ as well as the reactions that lead to the production of the correct crystalline structure in the presence of acid. There are a number of references that you may find useful as well.


Deflection of Alpha Particles in Rutherford's model of atom Rutherford stated that deflection of alpha particles is due to repulsive positive charge of nucleus. Why can't alpha particles be deflected by attraction of so many electrons revolving around nucleus?

See, what the Geiger-Marsden-Rutherford experiment achieved was the following: by bombarding (with alpha particles) a one-atom thick gold sheet and counting how many alpha particles passed through, they were able to relate the already known atomic radius with the actual area that could get collided by alpha particles.
Figure 1 shows a sketch of the apparatus.

Figure 1: depiction of the apparatus (courtesy of Wikipedia).
Now follows a simplified rationalisation.
Assume you have an atom-thick sheet of some material.
By knowing the element's molar mass and the mass of the sheet you can calculate how many atoms $n$ are there, and knowing the sheet's surface area $A$ allows you to calculate the atomic surface density $\sigma = n/A$.
Now if the atoms have radius $\rho$, an area $A_\rho = n \pi \rho^2$ of the sheet should be covered by atoms.
This means a ratio
$$r = \frac{A_\rho}{A} = \pi \rho^2 \sigma$$
is actually covered by atoms and $f = 1 - r$ is the ratio of "free space".
Rutherford actually tested if this made sense by throwing particles at the sheet: those scattered simply hit something, the rest passed through.
Imagine $M$ particles were thrown and $m$ of those passed through.
If $M$ is big enough we can set $f_R = m/M$ as Rutherford's "free space" estimate.
Going backwards we may set the coverage ratio
$$r_R = 1 - f_R = 1 - \frac{m}{M} = \frac{A_R}{A} = \pi \rho_R^2 \sigma$$
were $\rho_R$ above is the measured radius of the collision cross section of an atom.
The experiment has shown that $r_R \ll r$ and thus $\rho_R \ll \rho$, which implies that most of the atomic mass is buried inside the atom.
In fact (Wikipedia),

The diameter of the nucleus is in the range of $1.75$ fm ($1.75 \times 10^{-15}$ m) for hydrogen (the diameter of a single proton) to about $15$ fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about $23,000$ (uranium) to about $145,000$ (hydrogen).

In Rutherford's own words (p. 68),

It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive centre, carrying a charge.
--- Ernest Rutherford

Thus, nuclei are simply very small, but very dense, as stated by Enigma and Ilmari Karonen.


What is the hybridisation of carbon in transition state in SN2 mechanism? Normally, a carbon atom is $\ce{sp^3}$ hybridized, whereas in a transition state five bonds are present, then what's the hybridisation of carbon?

In transition state of SN2 mechanism, the carbon atom is $\ce{sp^2}$ hybridised due to its planar structure. At this point, carbon almost acquires 'pentavalency' with three full bonds and two 'partial' bonds, and a planar complex is formed, which is $\ce{sp^2}$ hybridised.


Connection between Gibbs Energy of Formation and Gibbs Energy of Mixing Recently I found the following formula, which I cannot verify to be correct. Since I found this formula in several publications I assume that it is correct but I don't know why.

First there is the Gibbs Energy of Formation $$\Delta_fG$$ and second the Gibbs Energy of Mixing $$\Delta_mG$$
The formula I found says, that $$\Delta_mG = \frac{\Delta_fG^{\circ}}{N}$$ where N is the number of particles. I am not sure if this is really N or n (amount of substance). Have anyone seen this formula and/or can explain why it is valid/how it can be derived?

Thanks!

Edit:

Indeed, my problem is also in the frame of binary alloys. Consider the reaction $$2A + B = A_2B$$ which is performed in an electrochemical cell (emf measurement). Then $$\Delta_fG^\circ = -zFE$$ because the reactants are elements and A is the active species. So far this should be correct. In this context the formula in question reads as follows $$\Delta_mG = \frac{\Delta_fG^\circ}{3}$$

The full definition of the Gibbs Energy of Mixing is $$\Delta_mG = x_A RT\ln{a_A}+x_B RT\ln{a_B}$$

Unfortunately, I cannot give more detail, because I don't understand the formula. :)

Thanks for the links in the comments above. The second one J Chem Thermo, 37, 97 (2005) seemed to be clearer. Eqn (12) reads
$$\Delta_M G=\Delta_f G^\circ /4
$$
but is preceded by the sentence

  According to the definition of the molar free energy of mixing,
  $\Delta_M G$ for Ni$_3$Al would refer to the formula
  Ni$_{3/4}$Al$_{1/4}$, and thus the equation ... was obtained.

In the first link J Electrochem Soc, 128, 1181 (1981) it is harder to spot your equation. However, I think I see it, in comparing eqns [7] and [10] which relate $\Delta G_m$ and $\Delta G_f^\circ$ to a Coulombic titration expression involving an integral over the compositional variable $y$. This is for the alloy Li$_y$Sn, and the expressions look identical except for a factor $1/(1+y)$. So I guess that's it. However there's no helpful explanation.

Bearing in mind the sentence I quoted from the (2005) paper, I suspect that this factor is simply a question of how these quantities are defined, in relation to the formula of the alloy under consideration (i.e. per atom or per formula unit), and not intimately connected with the expression for the free energy of mixing in terms of mole fractions and activities (except in that it is written per atom, or more accurately, per mole of atoms).

[Edit following OP comment]

I think that this is less complicated than it may have looked. $\Delta G$ is just the difference in Gibbs energy between two well defined states. I see, for instance, on p43 of Statistical Thermodynamics of Alloys by NA Gokcen (1986)
the definition

  The molar Gibbs energy of mixing refers to the process prescribed by

  $x_1$ (pure component 1) $+$ $x_2$ (pure component 2) $\rightarrow$ $1$
  gram atom of alloy

where, of course, "gram atom" is old terminology; but the meaning is clear. If the original metals are in their standard states, this is essentially the same as the definition of the standard Gibbs free energy of formation of the alloy, except that the latter is usually expressed in terms of one mole of the compound. I think this difference is what is intended by the numerical factor, irrespective of any particular ways of writing the Gibbs energy of mixing in terms of other variables (for example, one can sometimes use regular solution theory to estimate $\Delta S$ and $\Delta H$ of mixing, but I don't think that's relevant here).


Peroxide effect in reagents apart from HBr and CHCl₃ Is the peroxide effect observed only with addition of hydrogen bromide and chloroform? If so, why not with other reagents in electrophilic addition on alkenes?

Is the peroxide effect observed only with addition of hydrogen bromide
  and chloroform?

No, the "peroxide effect" or "anti-Markovnikov" addition of a free radical to an unsaturated bond is observed with other compounds too, $\ce{CCl4}$ is another  example.

Also, just to be clear, these reactions involve the addition of radicals ($\ce{X^{.}}$), not electrophiles, to alkenes (or alkynes).

Let's examine the radical addition of $\ce{HX}$ to an olefin.  I'll use a general method that you can later use to examine any radical addition to an unsaturated bond.

There are two propagation steps that are key to our analysis:
$$\ce{X* + H2C=CH2 -> X-H2C-CH2. ~~(1)}$$
$$\ce{HX + X-H2C-CH2. -> X-H2C-CH2-H ~~(2)}$$

If two halogen radicals combine, or if a halogen radical and a halo-alkyl radical (formed in the propagation steps) combine, the reaction is terminated. The propagation steps are in competition with the termination steps. If the propagation steps are endothermic, then they compete less effectively with the termination steps and the radical addition will not proceed in a useful manner.

In propagation step (1) we are breaking a pi bond and making a $\ce{C-X}$ bond, let's look at the thermodynamics of this  step.

\begin{array}{|c|c|c|c|} \hline
\ce{C-X} & \text{Bond Strength} & \text{Pi Bond Strength} & \text{Overall}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{C-F} & 116 & 66 & -50\\ \hline
\ce{C-Cl} & 81 & 66 & -15\\ \hline
\ce{C-Br} & 68 & 66 & -2\\ \hline
\ce{C-I} & 51 & 66 & 15\\ \hline
\end{array}

We see that this step is exothermic for the first 3 hydrogen halides, but is endothermic for the $\ce{HI}$ case.  Already we see that hydrogen iodide will not add to olefins via a radical pathway at room temperature.

In propagation step (2) we are breaking an $\ce{HX}$ bond and making a $\ce{C-H}$ bond.  Let's perform a similar thermodynamic analysis of this step.

\begin{array}{|c|c|c|c|} \hline
\ce{H-X} & \text{Bond Strength} & \ce{C-H}~ \text{Bond Strength} & \text{Overall}  \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline
\ce{H-F} & 135 & 99 & 36\\ \hline
\ce{H-Cl} & 103 & 99 & 4\\ \hline
\ce{H-Br} & 88 & 99 & -11\\ \hline
\ce{H-I} & 71 & 99 & -28\\ \hline
\end{array}

(source for bond strengths)

Both propagation steps (1) and (2) must occur for the reaction to proceed.  Only in the case of the addition of $\ce{HBr}$ is the reaction exothermic for both steps.  For the hydrogen halide series, only $\ce{HBr}$ will add to an olefin by a free radical mechanism.


How does one compute the number of unique 2-electron integrals for a given basis set? I have run into trouble with an exercise in Szabo and Ostlund's Modern Quantum Chemistry.

  Exercise 3.14: Assume the basis functions are real and use the symmetry of the two-electron integrals [$(\mu\nu|\lambda\sigma)=(\nu\mu|\lambda\sigma)=(\nu\mu|\sigma\lambda)$, etc.] to show that for a basis set of size $K=100$ there are $12,753,775=O(K^4/8)$ unique two-electron integrals.

I have seen answers to this on the Physics SE here and here, but the case-by-case analysis used in the answers seems confusing to me and not easily generalized to other values of $K$. The symmetry of the integrals when all the basis functions are real suggests that there should be a simpler way of determining this. How can one find the number of unique two-electron integrals in this case and, ideally, in general?

To find the number of unique $2e^-$ integrals $\left<AB|CD\right>$, it is useful to first find the number of unique $1e^-$ integrals $\left<A|B\right>$.

If $n$ is the number of real basis functions, we can find the number of $1e^-$ integrals as ${n+1}\choose{2}$. This stems from the general formula for combinations with replacement: $${n+k-1}\choose{k}$$ where $n$ is the selection pool and $k$ is how many are being chosen. Combinations with replacement are needed to ensure we count the cases where the two Gaussians are the same.

As an aside, this is only true for real basis functions, since we can only say the order doesn't matter (what we do when we compute the number of combinations) because all the basis functions are explicitly real and thus the complex conjugate equals the original function. This explains one of my own questions on the site regarding why we don't commonly use complex basis functions.

For this particular case ($n=100$), we find that the number of unique $1e^-$ integrals is $${{101}\choose{2}}=\frac{(101)*(100)}{2}=5050$$

So how does this allow us to find the number of unique $2e^-$ integrals? Well, for the $1e^-$ integrals, all we did was find the number of unique pairs of real basis functions. So, for a $2e^-$ integral, we might expect that all we need to do is find the unique pairs of basis function pairs. This winds up being exactly the case (again, only for real basis functions).

To find the pairs of pairs, we need to again find the number of combinations with replacement, but instead of sampling from the $100$ basis functions, we are now sampling from the $5050$ basis function pairs. We again need replacement to account for the fact that we can pick the same pair twice. So, we obtain: $${{5051}\choose {2}} =\frac{(5051)*(5050)}{2}=12,753,775$$ as the number of unique $2e^-$ integrals. This method avoids the checking of specific cases (i.e none of the basis functions are the same, two are the same, etc.) and gives a general way of finding the number unique integrals for any amount of real basis functions.

To show explicitly why complex basis functions aren't used, we can look at the same problem, but assume all the functions are complex instead. In bra-ket notation, the bra is the complex conjugate of the function inside, so for $\left<A|B\right>$, the order would matter if $A$ and $B$ are complex. So finding the number of pairs requires finding the number of permutations rather than combinations. This only requires a subtle change of the original calculation $$2!{101\choose2}= (101)*(100)=10100$$ So, we now have twice as many unique pairs, so there are twice as many $1e^-$ integrals.
The equation for the $2e^-$ integrals remains the same, but with the new amount of unique pairs filled in. $${{10101}\choose {2}} =\frac{(10101)*(10100)}{2}=51,010,050$$ We still need to do a combination for the pairs because regardless of whether the functions are complex or real $\left<AB|CD\right>=\left<CD|AB\right>$ via exchange of electron labels. So using complex basis functions would double the number of $1e^-$ integrals and nearly quadruple the number of $2e^-$ integrals. So even if some feature of complex basis functions made them easier to integrate, it would have to make them easier enough to warrant doing and storing 2-4 times as many integrals.


Decarboxylation / Elimination type reaction I'm quite familiar with E1/E2 reactions and usually use those ideas to explain elimination reactions, however I came across a reaction which was a bit different (in the sense it isn't E1/E2) but still an elimination.

Looking at this reaction I ruled out an E2 type elimination because there's no axial hydrogen which is trans to the bromine group, so something else must have happened here. I drew out the lowest energy chair conformation ('locked' in that conformer because of the t-butyl group) that would react to help me understand what happened and this is what I came up with (excuse the poor chair drawings):

I know hydride is a very strong base and the first step would be deprotonation of the carboxylic acid group (Due to it being the most acidic proton). However after this, mechanistically I wasn't so sure what would happen and how it would happen. My thinking was the oxygen which is now quite charged would push the electron density in towards the carbonyl carbon, and hence the bond would break (between the carbonyl carbon and the cyclohexane carbon). That would form a double bond, and $\ce{CO2}$ and bromide would leave. Is this realistically a correct explanation for what is observed? Also what kind of reaction is this? I haven't come across it before during the usual E1/E2 reactions. Are there any other examples of similar elimination reactions?

This type of reaction falls into the category of Grob fragmentation  [Wikipedia]. Conceptually, it is very similar to an E2 mechanism, where an anti-periplanar pair of electrons flows into the sigma anti-bonding orbital of a carbon-leaving group bond resulting in a new pi bond and cleavage of the carbon-leaving group bond. The primary difference between these two is that in an E2, that anti-periplanar pair of electrons comes from deprotonation, and in a Grob fragmentation, the electrons are from breaking some other sigma bond other than a C-H bond.

A syn-oriented version of this reaction can be seen in the generation of benzyne.

Also, decarboxyation is very common with a beta-carbonyl group. This doesn't exactly follow the Grob pattern, because there is no leaving group. The carbonyl plays the role of electron sink.


Is pK_{In} = pH at equivalence point still true for weak basic organic indicators? I am reading Ostwald Theory of Titration which says that indicators are organic weak acids or bases.
To prove the relation that $\mathrm{p}K_\mathrm{In} = \mathrm{pH}$, the textbook uses the following equilibrium as an example:

$\ce{HIn(aq) <=> H^{+}(aq) + In^{-}(aq)}$

which of course produces the the required result, but a thing I noted that it was for acid type indicator, so I decided to fill in the gap and do it for base type indicator myself, but it did not proceed as imagined:

Consider a weak organic base indicator $\ce{InOH}$, for it, the equilibrium will be:

$\ce{InOH (aq) <=> In^{+} (aq) + OH^{-} (aq)}$

and it's equilibrium constant then would be:

$\displaystyle{K_\mathrm{In} = \frac{[\ce{In^+}][\ce{OH^-}]}{[\ce{InOH}]}}$

then following the reasoning as done in previous case, for a sudden color change, $[\ce{In^+}] = [\ce{InOH}]$, so that,

$\displaystyle{K_\mathrm{In} = [\ce{OH^-}]}$

so that,

$\displaystyle{\mathrm{p}K_\mathrm{In} = \mathrm{pOH}}$

which inturn implies that,

$\displaystyle{\mathrm{pH} = 14 - \mathrm{p}K_\mathrm{In}}$

which doesn't seem to match with the previous result! Am I missing something here?

Note: I doubt that the condition $[\ce{In^+}] = [\ce{InOH}]$ might be the trouble, but I think it shouldn't be as in theory, the different color arises due to the respective indicator and their respective conjugate acid or bases.

It is true that acid-base indicators ($\mathrm{pH}$-indicators) are either weak acid or weak base. What I understood reading your question is that you have wrong impression about bases. To clear your view, not all bases contain $\ce{OH-}$ ions. Specially, most organic bases are weak and contain electronegative ion (e.g., $\ce{N}$ or $\ce{O}$) with at least one lone pair. This fact is well cleared by Poutnik's answer, thus, I'm not going to elaborate it more. Instead, I'd like to explain the action of a $\mathrm{pH}$-indicator (e.g., $\ce{HIn}$ in acidic form) to some depth.

Since Indicators ($\ce{HIn}$) are weak acids (or weak bases), in aqueous medium, it's in an equilibrium:

$$\ce{HIn + H2O <=> H3O+ + In-} \tag1$$
$$\therefore \ K_\mathrm{aIn} = \frac{[\ce{H3O+}][\ce{In-}]}{[\ce{HIn}]} \tag2$$
Taking log on both side followed by simplification gives:
$$-\log [\ce{H3O+}] = -\log K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{[\ce{HIn}]}\right) $$
$$\text{Hence}, \ \mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{[\ce{HIn}]}\right) \tag3$$

This is a resemblance of Henderson-Hasselbalch equation. The equation $(3)$ shows that when $[\ce{HIn}] = [\ce{In-}]$, $\mathrm{pH} = \mathrm{p}K_\mathrm{aIn}$.

Let's consider the indicator, Litmus, which is we commonly use in labs. Litmus is a weak organic acid (see the structure in bottom box of the diagram; $\mathrm{p}K_\mathrm{aIn} = 6.5$). The $\color{green}{\text{green circle}}$ shows the $\ce{H}$ of $\ce{HIn}$ molecule. The un-ionized litmus ($\ce{HIn}$) is $\color{red}{\text{red}}$ (below $\mathrm{pH}$ 4.5), whereas the ion part $(\ce{In-})$ is $\color{blue}{\text{blue}}$ (above $\mathrm{pH}$ 8.2):

Usually color change happens within a range of $\mathrm{pH}$. It is a rule of thumb that this range falls between the $\mathrm{p}K_\mathrm{aIn}\pm 1$. That means, acid color persists if $[\ce{HIn}] = 10 \times [\ce{In-}]$ and base color dominates when $10 \times [\ce{HIn}] = [\ce{In-}]$. For example, apply these values in the equation $(3)$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{10 \times [\ce{In-}]}\right) = \mathrm{p}K_\mathrm{aIn} + \log \frac{1}{10} = \mathrm{p}K_\mathrm{aIn} -1 $$

Similarly,

$$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{10 \times [\ce{HIn}]}{[\ce{HIn}]}\right) = \mathrm{p}K_\mathrm{aIn} + \log 10 = \mathrm{p}K_\mathrm{aIn} +1 $$

Hence, where $\mathrm{pH} = \mathrm{p}K_\mathrm{aIn}$ is the most intense point during color change. For litmus, color change happens between the range, $\mathrm{pH} = 5.5-7.5$ since $\mathrm{p}K_\mathrm{aIn} = 6.5$ for litmus.

Let's see litmus in basic solution where $\ce{In-}$ ions are predominant ($\color{blue}{\text{blue}}$ solution). Now you start adding acid to the solution and you can see what would happen to the solution (equation $(1)$) by using Le Chatelier's Principle. Since excess hydronium ions disturb the equilibrium, it'd shift to left hand side of equilibrium to reduce excess hydronium ions by reacting $\ce{In-}$ ions. Eventually, solution changes the color to $\color{red}{\text{red}}$. Opposite would happens if you added hydroxide ions to the $\color{red}{\text{red solution}}$ in equilibrium.

The scaning UV spectra in the upper box of the diagram shows how color change happens in solution with $\mathrm{pH}$. The indicator here is bromothymol blue, which is yellow in the acidic medium absorbing light at about $\pu{430 nm}$. When $\mathrm{pH}$ has changed systematically with time (by adding known aliquot of $\ce{OH-}$ solution with in fixed time), the solution changed the color accordingly, and bromothymol blue has changed to its basic color blue, absorbing light at about $\pu{620 nm}$. Note that the peak at $\pu{430 nm}$ decreases while the peak at $\pu{620 nm}$ increases while $\mathrm{pH}$ increases.


What happens to copper(II) hydroxide when it's heated? what's inside its decomposition? So just wondering, copper(II) hydroxide undergoes thermal decomposition when it's heated. But isn't it a solid by itself too? Why does it turn into another solid and water all of a sudden?

Can someone please explain to me the reason behind it, thank you!!

Well,this is a kind of dehydration reaction,and you will come across similar such reactions plenty of times as you read about the effect of heating on various compounds.

I am going to approach this in a sligthly different manner: how the copper(II) hydroxide was generated from the copper(II) oxide.
According to Wikipedia

  Copper(II) oxide belongs to the monoclinic crystal system. The copper atom is coordinated by 4 oxygen atoms in an approximately square planar configuration

So we can think of it this way,that when water was introduced in the copper(II) oxide powder,the water molecules entered into the voids(in a stoichiometric fashion)of the oxide lattice.Mind you,this is just a way of visualization,but it can help a lot. So now,when you are heating the hydroxide in a controlled manner,then we can imagine that the water molecules residing in the voids escape out,leaving the oxide in the same oxidation state. (Also,now you should keep in mind that most hydroxides lose one or more molecules of water to form their respective oxides)

Note: if heating is done excessively,then the oxide itself can decompose to a variety of products


How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution? Below was question 34 in the USNCO 2017 exam:

  If $\pu{0.10 mol}$ of solid $\ce{NaOH}$ is added to $\pu{1.00 L}$ of a saturated solution of $\ce{Ca(OH)2}$ $(K_\mathrm{sp} = \pu{8.0 \times 10^-6})$, what percentage of the calcium hydroxide will precipitate at equilibrium?

  (A) Roughly 50%
  (B) Roughly 75%
  (C) Roughly 95%
  (D) Over 99%

My solution is as follows:

Find concentration of $\ce{Ca^2+}$ $(\pu{0.02 M})$ and $\ce{OH-}$ $(\pu{0.04 M})$ ions from dissolved calcium hydroxide using $K_\mathrm{sp}$.
Add hydroxide ion concentration from sodium hydroxide (assuming full dissolution) to get total hydroxide concentration of $\pu{0.14 M}$
Find reaction quotient $Q = 0.02 \times 0.14^2 = 3.92 \times 10^{-4}$
Find amount of calcium $(x)$ and hydroxide ions $(2x)$ that will precipitate at equilibrium by using algebraic equation:
$$
(0.02 - x)(0.14 - 2x)^2 = 8.0 \times 10^{-6}, x = \pu{0.019 M}$$
Find percentage of calcium hydroxide precipitated:
$$\frac{0.019}{0.02} \times 100\% = 95\%,$$
hence (C)

I am unsure about step 4, where a cubic equation appears, and would not be able to be solved in exam conditions (use of graphing calculator is not permitted).

Is there a simpler method?

A less analytic aproach:

Initial concentrations:

$K_{\mathrm{sp}}=\ce{[Ca^{2+}][OH^-]^2}=x\cdot (2x)^2=8\cdot10^{-6}$

$\ce{[Ca^{2+}]=0.0126}$ M; $\ce{[OH^-]=0.0252}$ M.

Now 0.1 mol NaOH is added. Let $p$ be the fraction that precipitates.
The concentrations that remain in solution are:

$\ce{[Ca^{2+}}]=0.0126(1-p)$

$\ce{[OH^-]}=(0.10 + 0.0252 -2\cdot0.0126p)^2$

Since the solubility product remains the same:

$K_{\mathrm{sp}}=0.0126(1-p)(0.1252 -2\cdot0.0126p)^2$

Now, since there are only four scenarios, try them out substituting every value of $p$ in  the equation and take the closest one to $8\cdot10^{-6}$:

\begin{array}{|c|c|}\hline
    p&K_\mathrm{sp}\\\hline
    0.50&\pu{7.99E-05}\\
    0.75&\pu{3.56E-05}\\
    0.95&\pu{6.46E-06}\\
    0.99&\pu{1.27E-06}\\\hline
\end{array}

The closest one is for roughly 95% precipitation, without solving any equation.


Does an orbital have its own energy, separate from the electron? Can I picture them as a entity independent from the hosted particle? Does an orbital have its own energy, separate from the electron? If so, is it possible for an electron to have a different energy than the orbital it occupies?

An orbital of current quantum models of atoms and molecules has 3 major interpretations:

orbital(1) as  a wave function, being a solution of the quantum wave equation describing behaviour of electrons around an atomic kernels. Functions do not have energy, even if they describe it.

orbital(2) is the 3D region describing probability of occurance of an electron, related to orbital(1) . Space regions do not have energy either.
(Not exactly true, considering radiation energy and quantum vacuum energy, but nothing comparable to the electron energy at usual conditions.)

orbital(3) is a quantum state of an electron, described by orbital (1) and orbital (2), consisting of set of discrete values of physical quantities as functions of quantum numbers, including energy. In the case this quantum state is not occupied, the orbital energy means "an electron would have this energy".

orbital (4) from the obsolete Bohr model means a path of an electron, which is a crude approximation of the quantum model of an atom.

Last, but not least: Orbitals are not physical things, they are part of the quantum theoretical model of atoms and molecules. Within the given quantum model, an electron cannot have different energy than energy of it's quantum state.

As models of multielectron cases are imperfect, real electron energy usually more or less differ from its energy predicted for the given quantum state.


Why f is multiplied in solubility of salt constant I understand till we get “f” And also why f is directly proportional to PH. But I don’t understand why f is multiplied in solubility product with solubility of salt of weak acid.

(Sorry for poor english) (couldn’t find link of pdf of this book online)

Notice that $f$ is equal to the maximum concentration of $\ce{M}$, which is unchanged as part of the acid-base equilibrium. On the other hand, the pH changes the concentration of the counterion in proportion to $f$. Therefore, the second term in the solubility product is scaled accordingly.


How is nomenclature decided in this case (formyl vs oxy)? I was asked the IUPAC Name of the following compound in JEE Mains (2nd September, Shift 1) 2020.

There were two names occuring to me:

5-formyl-2-methyl-hex-2-enoic acid
(2,5)-dimethyl-6-oxo-hex-2-enoic acid

I finally chose (2), but the temporary answer key says (1). Can someone please tell which is the correct one and why?
Note: As I realised after receiving a comment, I've accidentally made the double bond have (E) configuration. This is an unintended error due to the website I used to construct the compound's structure. Please note that the original compound did not have any stereochemistry indicated

According to the new recommendations of IUPAC nomenclature rules, when choosing the senior parent structure, you have to first consider the senior parent structure, which has the maximum number of substituents corresponding to the principal characteristic group or senior parent hydride in accord with the seniority of classes (functional groups). Then, if there is a choice, choose the principal chain, which has the greater number of skeletal atoms. Next in order is the chain, which has the greater number of multiple bonds, and then has the greater number of double bonds (This is according to Blue Book in 2013, which was displayed in 5 steps you have mentioned from Wikipedia).
Accordingly, your most senior function is carboxylic group, which would have the highest priority:

Then, there is a choice to choose the principal chain, which has the greater number of skeletal atoms. Both choice have six carbons, but the numbered chain indicated in the image has additional function, thus it is the principle chain. You also need to consider the stereochemistry if they are indicated. Here, there is a double bond in trans-orientation. That should be included in the name. Although, $\ce{C}$-5 is a chiral carbon, its stereochemistry is not indicated so you can leave it alone. Therefore the correct IUPAC name is: (2E)-2,5-dimethyl-6-oxohex-2-enoic acid.


If stirring negatively affects crystal purity, why is it so? My Intellectual Expectations
Intuitively, I would expect strong stirring to improve crystal purity during the process of crystalization. My rationale is that strong stirring would increase contact between each crystal and the variety of ions present in solution at each given level of solubility, and thus minimize the gap between reality and the equilibrium that would be reached if the conditions at each instant (solvent volume and temperature) would be kept equal for a very long time.
Information I have Read
I've read that as a general rule, the size of the crystals that form correlates well with the purity inside the crystals (i.e. that crystal
size in general is a good indicator of purity).
My Conflicting Observations
My observations are that under strong stirring the crystals that form are much smaller than if the crystalization is carried out in still water (at equal rate of cooling / solvent evaporation). This indicates that either crystal size is not a great indicator of purity, or that my rationale is wrong and by counterintuitive mechanisms which I ignore, crystals of greater purity would be formed without stirring.
My Question
Does anyone know 1) Wether stirring the solution during the crystalization  process positively or negatively affects crystal purity? 2) What is the physical explanation of the answer?

"Strong stirring" is a rather imprecise term.  There is such a thing as too much speed.  Rule of thumb for APIs, if you're working with a magnetic stirrer in a round bottomed flask in a lab (rather than in some industrial vessel or pilot plant setup) 350 rpm is too fast (and no, I can't quote a reference on that, it's one of those things you get told and then confirm by experiece - usually by being a wise-arse and bucking the "established wisdom" and ending up with a product that doesn't make medical grade).
Agitation for different crystallisation processes are generally kept in certain narrow bands to avoid inclusions as wel as a number of other reasons.
An example is described in Chow et.al [1] where when crystallizing paracetamol from water in a 500 mL round-bottom flask with p-acetoxyacetanilide present, increasing the stirring speed from 200 to 320 rpm increased product purity, but increasing further to 400 rpm lead to decline in product purity.
According to Meedara et.al. [2] stirrer-particle and interparticle collisions can facilitate crystal attrition  in suspensions with high energy agitation, leading to attrition-induced inclusions.
Urwin et.al. [3] did a reasonably thorough job testing various methods and reasons for presence of impurities in crystals, and is worth a read.
Another good reference is The Handbook of Continuous Crystallization, start with chapter 1 [4]
That said, in crystallisation especially of APIs, mechanisms can be compex and are still not completely understood. All kinds of workarounds have been developed, such as e.g. ultrasonics to help prevent agglomeration when crystallizing paracetamol from certain solutions.
If you have no literature on the process you are attempting, finding the "sweet spot" where you get the greatest purity of product, is only going to happen with a lot of experimentation.
[1] Chow, A. H. L.; Chow, P. K. K.; Zhongshan, W.; Grant, D. J. W. Modification of Acetaminophen Crystals: Influence of Growth in Aqueous Solutions Containing P-Acetoxyacetanilide on Crystal Properties Int. J. Pharm. 1985, 24 (2–3), 239– 258,  DOI: 10.1016/0378-5173(85)90024-9
[2] Meadhra, R.; Kramer, H. J. M.; Van Rosmalen, G. M. Model for Secondary Nucleation in a Suspension Crystallizer AIChE J. 1996, 42 (4), 973– 982,  DOI: 10.1002/aic.690420410
[3] Urwin, SJ, Levilain, G, Marziano, I, Merritt, JM, Houson, I and Ter Horst, JP
A Structured Approach To Cope with Impurities during Industrial Crystallization Development Organic Process Research & Development 2020 24 (8), 1443-1456,  DOI: 10.1021/acs.oprd.0c00166
[4] McGinty, J, Yazdanpanah, N, Price, C, ter Horst, JH and Sefcik, J  Nucleation and Crystal Growth in Continuous Crystallization , in The Handbook of Continuous Crystallization, 2020, pp. 1-50 DOI: 10.1039/9781788013581-00001


chemically inert molecule with piezoelectric properties For a research project, I'm looking into existing  chemical molecules that have the following properties:

It is mostly chemically inert, safe for animals and children
It exhibits piezoelectricity -  production of high electric oscillating potentials under pressure
paramagnetic moment a plus, but not required

Are there known candidates for this?

The ur-piezoelectric crystal is quartz. Being composed of $\ce{SiO2}$, it more or less epitomizes chemical/biological inertness. It has the additional advantage of being comparatively cheap and common. There are a number of other natural minerals which are piezoelectric (e.g., various perovskites). Most of those minerals are, as far as I know, relatively unreactive and non-toxic. Naturally, if toxicity is a concern, avoid the common lead-based compounds.

I've seen mention of materials that exhibit both ferrimagnetic and piezoelectric properties. If ferrimagnetism is acceptable to you instead of paramagentism, then you may wish to consider that as an avenue for research. If you need paramagnetic materials specifically, then it's probably not worth your time (unless your project is compatible with subjecting those materials to their respective Curie temperatures).


An alternative basis set for analytical computation of two electron integrals It is well known that the usage of the Gaussian basis set, in contrast to Slater basis set, leads relative simple semi-analytical expressions for the two electron repulsion integral $(ab|cd)$. Could you propose any other basis sets allowing simple computations of two-electron integrals? This is quite general questions meaning that these functions should not fit atomic orbitals well (the problem is formulated for wider class of potentials rather than for the Coulomb one only).

A few things come to mind: 1) my understanding is that (ab|cd) in a gaussian basis is exactly solved for arbitrary "angular momenta" using recurrence relations such as the Obara-Saika method. The O-S method also has ways of generating the requisite derivatives with respect to coefficients, without which it is difficult to minimize the energy or do anything practical.

2) On the question of alternative basis sets to the Gaussian, one simply needs a complete basis set (of which there are many), helps to be orthogonal (almost as many), has to describe the boundary conditions of vanishing at infinity or being periodic with unit cell (fewer), hopefully is cusp-like near the nuclei (very few), and integration has to be tractable without resorting to quadrature, since there are four indices (each with 3 dimensions) but also millions of these integrals to compute -- and that leaves you with the Gaussian.


How is it possible that meta xylene exists? Methyl group is ortho para directing so it should direct the
 second methyl group to ortho or para position giving rise to ortho or para xylene but how can meta xylene be formed?

Friedel Crafts alkylation of toluene is far less regioselective than you might think.
The meta-substituted products are thermodynamically favoured. At higher temperatures, the alkylation yields more 1,3-dialkyl substituted benzenes.
On a larger scale, Dimethylbenzenes are typically not synthesized by alkylation of toluene, but stem from catalytic reforming processes using alkanes as feedstock.


Why is this named 3-phenyl-2-butanol instead of 2-phenyl-3-butanol? Why is the following compound named 3-phenyl-2-butanol and not 2-phenyl-3-butanol? Does the alcohol or the phenyl group have priority?

You have already correctly identified the compound as a phenylbutanol. The only remaining problem is the numbering of locants. The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds (first to double bonds if there is a choice)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

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

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Clearly, the $\ce{-OH}$ group is the principal characteristic group in this case since you have already used it as principal characteristic group to find the correct suffix ‘ol’.

Therefore, a low locant is assigned first to the $\ce{-OH}$ group according to Rule (c), which yields phenylbutan-2-ol.

After that, a low locant is assigned to the phenyl group according to Rule (f); however, the only possible numbering left is 3-phenylbutan-2-ol.


How does absence of nd orbital in Oxygen affect its valency? The original question was 'Oxygen exhibits Oxidation state -2 to +2 but other elements of grp 16 exhibit only +2,+4,+6. Explain Why'. In the solution given i can't understand meaning of this line 'It doesn't have nd-orbitals due to which its valency doesn't increase'. Please explain how?
PS:I am a chemistry noob. So please provide as much details as possible :)

Whether or not oxygen can reach oxidation states higher than $\mathrm{+II}$ is completely independent of the type of orbitals the second shell has. If the orbitals would matter, nitrogen would not be able to reach every single oxidation state between $\mathrm{-III}$ and $\mathrm{+V}$ but would also be restricted to <insert arbitrary number here>.

The actual reason why oxygen’s accessible oxidation states are limited is its very high electronegativity, second only to fluorine among the non-noble gas elements. There is simply no partner that can draw away electrons strongly enough to access high oxidation states. Fluorine itself, while typically being able to access high coordination numbers (higher than oxygen, compare $\ce{XeF6, NF3}$ or $\ce{SF6}$ to $\ce{XeO4, NO3-}$ or $\ce{SO3}$) is not the element of choice for accessing high oxidation states — that is oxygen. But oxygen cannot oxidise itself since it has the same electronegativity. Fluorine itself, as is evident by $\ce{NF5}$ being unknown as of today, would not be able to stabilise an oxidation state of more than $\mathrm{+II}$ on oxygen given our current knowledge.

Furthermore, the statement’s claim that group 16 elements except oxygen only exhibit positive oxidation states is bogus. $\ce{Na2S}$, $\ce{Na2Se}$, $\ce{Na2Te}$ and $\ce{Na2Po}$ are all known and in all of these compounds the chalcogen has an oxidation state of $\mathrm{-II}$.