Mechanisms for Reimer-Tiemann and Kolbe-Schmidt reaction on deuterium labelled phenol I was solving a bunch of reactions when I encountered the following:

So for the first question:

This is a Reimer-Tiemann reaction. I felt that the tautomerism would result in the deuterium getting attached to the phenolic $\ce{O^-}$. As it turns out, that was wrong. Oh well. Then I proceeded with the second question. Now this one is a Kolbe-Schmidt reaction, where the $\ce{CO2}$ liberated will react with the phenol. Going by the previous reaction, this time I decided not to add the $\ce{D}$ atom to the phenolic $\ce{O-}$. Consequently, this is how it panned out:

All of this has left me quite confused. So here's my question:

Why is it that the $\ce{D}$ atom during tautomerism adds on to the phenolic $\ce{O-}$ in the Kolbe-Schmidt reaction, but does not proceed to do so in the Reimer-Tiemann reaction?

Image credit: Created using Marvin JS

This is the mechanism of the Reimer-Tiemann reaction (source Wikipedia here)

The polyhaloform reacts with strong base to create a carbene, a highly reactive species which reacts with the phenoxide anion through the alpha carbon to give an anionic intermediate with the -ve charge on the halogenated carbon. It is this species that picks up the deuterium.
In Kolbe-Schmidt reaction, the process does not proceed through a carbene. It is nucleophilic attack (through carbon) of the phenoxide on the polyhaloform (or carbon dioxide) and the deuterium shifts to the phenol oxygen as the product re-establishes aromaticity. Mechanism from Wikipedia here


Understanding "Gay-Lussac's law of Gaseous volumes/combining volumes" Understanding "Gay-Lussac's law of Gaseous volumes/combining volumes". So I was reading Wikipedia and stumble across this. It defines it as follows "The ratio between the volumes of the reactant gases and the gaseous products can be expressed in simple whole numbers" and then give an example of
$$\ce{2H2 + O2 -> 2H2O}$$
The problem is that example does not fit the definition. The volume of reactants is $3x$ where $x$ is a variable. The volume of product is $2x$. The ratio is $\frac 3 2$, which is not a whole number. Clearly, something is wrong here or something is wrong with me. Either way, can anyone help?

The ratio between the volumes of the reactant gases and the gaseous products can be expressed in simple whole numbers

Wikipedia is correct.

Expressed in, not expressed as.

$3/2$ can be expressed in simple whole numbers $3$ and $2$.


What are the maximum number of electrons in each shell? In my textbook, it says that the maximum number of electrons that can fit in any given shell is given by 2n². This would mean 2 electrons could fit in the first shell, 8 could fit in the second shell, 18 in the third shell, and 32 in the fourth shell.

However, I was previously taught that the maximum number of electrons in the first orbital is 2, 8 in the second orbital, 8 in the third shell, 18 in the fourth orbital, 18 in the fifth orbital, 32 in the sixth orbital.  I am fairly sure that orbitals and shells are the same thing.

Which of these two methods is correct and should be used to find the number of electrons in an orbital?

I am in high school so please try to simplify your answer and use fairly basic terms.

Shells and orbitals are not the same. In terms of quantum numbers, electrons in different shells will have different values of principal quantum number n.

To answer your question...

In the first shell (n=1), we have:

The 1s orbital

In the second shell (n=2), we have:

The 2s orbital
The 2p orbitals

In the third shell (n=3), we have:

The 3s orbital
The 3p orbitals
The 3d orbitals

In the fourth shell (n=4), we have:

The 4s orbital
The 4p orbitals
The 4d orbitals
The 4f orbitals

So another kind of orbitals (s, p, d, f) becomes available as we go to a shell with higher n. The number in front of the letter signifies which shell the orbital(s) are in. So the 7s orbital will be in the 7th shell.

Now for the different kinds of orbitals
Each kind of orbital has a different "shape", as you can see on the picture below. You can also see that:

The s-kind has only one orbital
The p-kind has three orbitals
The d-kind has five orbitals
The f-kind has seven orbitals

Each orbital can hold two electrons. One spin-up and one spin-down. This means that the 1s, 2s, 3s, 4s, etc., can each hold two electrons because they each have only one orbital.

The 2p, 3p, 4p, etc., can each hold six electrons because they each have three orbitals, that can hold two electrons each (3*2=6).

The 3d, 4d etc., can each hold ten electrons, because they each have five orbitals, and each orbital can hold two electrons (5*2=10).

Thus, to find the number of electrons possible per shell

First, we look at the n=1 shell (the first shell). It has:

The 1s orbital

An s-orbital holds 2 electrons. Thus n=1 shell can hold two electrons.

The n=2 (second) shell has:

The 2s orbital
The 2p orbitals

s-orbitals can hold 2 electrons, the p-orbitals can hold 6 electrons. Thus, the second shell can have 8 electrons.

The n=3 (third) shell has:

The 3s orbital
The 3p orbitals
The 3d orbitals

s-orbitals can hold 2 electrons, p-orbitals can hold 6, and d-orbitals can hold 10, for a total of 18 electrons.

Therefore, the formula $2n^2$ holds! What is the difference between your two methods?

There's an important distinction between "the number of electrons possible in a shell" and "the number of valence electrons possible for a period of elements".

There's space for $18 \text{e}^-$ in the 3rd shell: $3s + 3p + 3d = 2 + 6 + 10 = 18$, however, elements in the 3rd period only have up to 8 valence electrons. This is because the $3d$-orbitals aren't filled until we get to elements from the 4th period - ie. elements from the 3rd period don't fill the 3rd shell.

The orbitals are filled so that the ones of lowest energy are filled first. The energy is roughly like this:

$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s$$

An easy way to visualize this is like this:


Is there an agent that changes colour when exposed to air? And if so, does it come in a liquid form?
Thank you for your time

Many compounds change colour when exposed to 'air', however, this is usually due to the presence of either water or oxygen, rather than 'air'.

To give a few examples:

*when distilling a solvent such as THF over sodium/potassium, benzophenone is used as an indicator. In the absence of air, the indicator leaves the solution a deep purple colour, however if an excess of air is present, the indicator turns a clear/brown colour.

*drierite is an indicator often used in inert gas drying tubes. In its dry form, it is either deep blue or orange, however when wet it turns white/pale orange.

Several other indicators exist, however its worth noting that air is merely a mixture of multiple things, and as such any indicator will indicate for a specific component of air rather than air as a whole- with water and oxygen being the most common things to indicate for.


Percent composition of nitrogen in fertilizer The question was posed as follows:

  Fertilizers supply the essential elements, nitrogen, phosphorus and potassium for plant growth. A bag of fertilizer contains 500g of ammonium sulfate, $ \small \ce{(NH4)2 SO4}$, and 500g of potassium nitrate, $\small \ce{KNO3}$. Calculate the percentage by mass of nitrogen in the bag of fertilizer.

This came in a test of mine and I got the answer wrong, now I did a redo and here's my solution:

(42 * 2)/227  * 100 = 37%

Is this answer correct?

I got 17.542% by mass nitrogen.

An explanation for beginners, a refresher for the rest of us:

To make sure anyone reading knows what molar mass means: The mass of one mole of something, in our case the mass of a mole of nitrogen, or the mass of a mole of ammonium sulfate, etc. A mole is just a way of saying a really big number which is actually $6.022 * 10^{23}$. It is a word we use instead of saying that big number, like when we say a dozen of eggs, but we know it means $12$ eggs. The same concept applies to a mole of eggs (that's a lot of eggs). To calculate the molar mass you need to add the atomic masses of each atom present in the molecule. If it is water, then we must count hydrogen's atomic mass twice (two hydrogen atoms per molecule), oxygen only once, since the formula is $H_2O$. Atomic mass can be found on any respectable periodic table, be sure not to confuse the atomic mass with the atomic number! The atomic mass always represents the weight in grams of a mole of that element. This convention makes our calculations much easier.

Ultimately we need to know how many $mol$ of nitrogen exist in this bag of fertilizer, so we can find out the mass of nitrogen in said bag (atomic mass can be used both ways because it is a ratio of grams per mole). To find that we need to do this process for both of the compounds separately because they have different molar masses. If there are two $chickens$ per $hen$ and we have $4_{mols}$ of $hens$ we should be able to tell that we have $8_{mols}$ of $chickens$. The same concept is used to determine how many $mols$ of nitrogen we have, based on how many nitrogens we expect per molecule of each substance.

Here is how that calculation might look in dimensional analysis in case it surprises you:

$81_{grams} * \frac{1_{mol}}{12_{grams}}$

Notice units of grams cancel each other out, and the remainder is in moles. This is how you would find the number of moles of carbon there exist in a pure sample of 81 grams.

Notice we used the ratio $12$ $\frac{grams}{mol}$ which is carbon's atomic mass.

I highly recommend writing out the entire problem using dimensional analysis so that you can be absolutely sure you have the answer correct for whatever problem you are solving. You will know this is true when all the units cancel correctly.

To solve the problem it may be good to take a step back from the calculator and consider what we know. What is the molar mass of these two substances? It should be the summation of the atomic masses of all the atoms in the molecule. Just like a water balloon's mass is not only the mass of the water, but also the rubber. Similarly the mass of water is not just the mass of the hydrogen, but the oxygen as well.

There is $500_{grams}$ of ammonium sulfate, but if we are trying to figure out the moles based on this information, we will need to do some dimensional analysis. Use the molar mass of ammonium sulfate which has units $grams/mol$ to multiply this ratio we found by the total mass of ammonium sulfate $(500_{grams})$ in such a way that the $_{grams}$ cancel each other, leaving $mol$ units. This is how many moles of that molecule exist in a 500g sample. We need to know this because the relationship of nitrogen per molecule is not a gram to gram relationship. You could of course, go another route and figure out the percent by mass of nitrogen per molecule and your final calculation should be equivalent. You should get about $21$% nitrogen by mass in a molecule of ammonium sulfate, and therefore 21% of the mass of that substance should equal the mass of nitrogen in that substance. It will be different for the other substance in the bag.

Once you know the total moles of nitrogen, convert back into grams of nitrogen, make sure your units cancel. From there you know the mass of nitrogen in the fertilizer bag, and the total mass of the fertilizer bag.

find the answer: what percent mass is the nitrogen atoms compared to the total mass of the fertilizer bag (all the atoms/molecules)?


Predicting Ka/Kb values using values for solubility in water Given that $K_\mathrm{a}$ values are determined by the position of equilibrium for a species such as $\ce{HCl}$ where $K_\mathrm{a} = \ce{\frac{{[H^{+}]}{[Cl^{-}]}}{[HCl]}}$, I was wondering whether it would be possible to work out $K_\mathrm{a}$ values using 'Solubility in Water' values.

For example: $\ce{HCl}$ has a solubility of $\pu{720g/L}$ in water at 20 degrees Celsius. Is this possible or is there no direct correlation between solubility in water and $K_\mathrm{a}$?

Your example, $\ce {HCl}$, is considered to be a strong acid. Thus, it it dissociate $100\%$ in water. Thus, the molarity of the given solution (if it exists):

$$\frac{\pu{720 g}}{\pu{1.0 L}}\times\frac{\pu{1.0 mol}}{\pu{36.5 g}} = \pu{19.7 mol L^-1}$$
That means, since $\ce {HCl (aq) -> H+(aq) + Cl-(aq)}$, $[\ce {H+}]$ of the solution is $19.7$, and $[\ce {Cl-}]=19.7$ as well so you can find $\mathrm{pH}$ of the solution. But, since no $\ce {HCl}$ remains in the solution, $K_\mathrm{a}$ has no meaning.


How do the two half cells of Daniel cell know when they are connected? I'm currently studying electrochemistry in school, and there is something I don't can't quite understand about Daniel cell.

When you put $\ce{Zn}$ solid into $\ce{ZnSO4}$ solution, does the electrode become negatively charged even before connecting it with the $\ce{Cu}$ half cell?

I searched but I wasn't able to form a clear understanding, I read that the answer to the previous question is no, the the two cells need to be connected for the redox reaction to happen.
If that was the case, how do the half cells communicate, how do they know when they are connected even when there is distance between them?

If the electrodes were charged before connecting them that would make perfect sense for me, the electrons repel each other and when connecting the two half cells with a wire, the electrons move because of the negatively charged electrode, but from what I've read that is not the case?

I'm really sorry if that is a stupid question but I really can't find an answer.

No, electrodes do not know about each other.

When an electrode is inserted to an electrolyte, the electrochemical reaction is ongoing in both directions. If reduction direction overruns oxidation, the potential of the electrode is increasing ( or vice versa ) until the rate of both reaction gets equal, the net reaction rate is zero and the electrode reaches the equilibrium potential.

That may be quick or slow, depending on the electrode and respective reaction kinetics.

When both electrodes are galvanically connected, the electron exchange due the potential difference disbalances the equilibrium potentials of both electrodes and the net electrochemical reactions get the steady non zero rate.

Note that the electrode potential is conventionally referred to the Standard hydrogen electrode (SHE), which itself has potential $\pu{+4.44 \pm 0.02 V}$ ( Wikipedia ) wrt the potential of a free electron in vacuum. So saying $\ce{Zn}$ electrode potential is negative rather means it is less positive.

You may want to try searching various terms in wikipedia and following relevant internal/external links. The great source of compactly explaned topics is hyperphysics, but with limited scope. It is a kind of hyperlinked quick cards/cheat sheets, very handy, if you want to understand the basics.  This link is particularly for Electrochemistry


How to extract pure potassium carbonate from ash? In our village, lots of ash is produced which is thrown off. I want to extract pure potassium carbonate from that ash. I am able to extract the potassium compound and a soluble compound by recrystallizing and evaporating but I found impurities. The first test I performed is electrolysis. This test confirmed the presence of chlorine gas which is able to change the litmus paper to white. The second test I performed is flame test. During the flame test, I got a yellowish orange color while my expectation was bluish violet. So, I assume there is sodium. How can I extract the pure compound?
To conclude, there is a small amount of potassium chloride, sodium chloride, sodium carbonate which are the impurities.

The composition of the wood ashes varies with the nature and the origin of the wood. In the average, it is a mixture of $40$ % - $70$ % $\ce{CaCO3}$, $5$% - $10$% $\ce{MgCO3}$, $5$% - $10$% $\ce{K2CO3}$, $5$% - $10$% $\ce{Na2CO3}$, $2$% - $5$% $\ce{SiO2}$, $2$% - $5$% $\ce{Ca3(PO4)2}$, $0.5$% -$2$% $\ce{Al2O3}$, <$1$% $\ce{NaCl}$, <$1$% $\ce{KCl}$, plus traces of manganese and iron. So it is difficult to purify, and/or to extract any of these compounds, as a lot of these compounds are insoluble in water: $\ce{CaCO3, MgCO3}$, $\ce{SiO2, Ca3(PO4)2, Al2O3}$. The only soluble compounds are $\ce{K2CO3}$ and $\ce{Na2CO3}$, plus traces of $\ce{NaCl}$ and $\ce{KCl}$, and they form less than $20$% of the total mass of the ashes. The solution obtained from washing the wood ashes is a mixture of sodium and potassium carbonate. The problem is how to separate those two carbonates. Fractional crystallization does not seem easy. I would rather transform them into nitrates by adding nitric acid : at $0°$C, $\ce{KNO3}$ is  not very soluble ($5$ g/$100$ mL), which is much less than $\ce{NaNO3}$ ($73$ g/$100$ mL). So obtaining $\ce{KNO3}$ from the mixture by cooling a hot solution should be relatively easy.
Ref.: J. Hartwig, Ann. der Chem. u. Pharm. XLVI, p. 97. A. Wurtz, Dictionnaire de chimie pure et appliquée, Vol. 1, p. 641 (1891).


Can reactions produce microwave or radio wave radiation? I know that many reactions produce visible light,  heat and sound,  but is there a reaction (preferably one example for an inorganic and organic one)  that could produce electromagnetic waves in the microwave and radio band?

An ammonia maser qualifies as an inorganic RF-emitting reaction. As the nitrogen changes position in the $\ce{NH3}$ molecule, ~24 GHz radiation is emitted.

Similarly, organic methanol, $\ce{CH3OH}$ makes a maser found in nature, emitting at ~36.2 GHz and also at ~6.7 GHz.

BTW, if the reactants are  moving away rapidly (e.g. interstellar gases), visible radiation is red-shifted to microwaves... or lower, so wavelength depends on your frame of reference. So, as Brumley and Stevens state, listen in to radio astronomy.

Though these might be due to physical excitation, and not an exact answer for your question, an interesting experiment would be to form $\ce{NH3}$, e.g. with the Haber process, and try to detect the 24 GHz radiation, perhaps due to initial formation of the molecule in a metastable state, or due to rearrangement. A similar experiment could be done with methanol, forming it from $\ce{CO}$ and $\ce{H2}$, looking for 36.2 GHz or 6.7GHz signals. I'd be very interested to know if you or anyone else tries that!


Strange Sticky Substance on Digital Camera I purchased a used Canon Powershot A1100 IS camera online. When I received it in the mail it had a very strange sticky substance that looked like a thin film.

It resisted attempts at cleaning. Water, water + soap, baking soda + vinegar, baking soda only, vinegar only and rubbing alcohol all did not resolve the issue. Only vigorous scrubbing kind of helped.

In addition I could not get it off my hands at all even after about 10 hand washing attempts and soaking my hands in soap and water. After my hands dried the areas where my hands touched the camera were still sticky. The only thing that helps was after it dried rubbing my hands with a paper towel. When wet it was not sticky then when it dried it came back.

The battery terminal was completely corroded with a blue color and it seems like there was some but possibly less of the substance internally.

This is part of the battery door. It is used to complete the circuit between the two AA batteries that the camera takes.

Scraping away the blue crystal only results in more of what you can already see in the picture. Metal that looks grey in some places and brown/red in others.

The closest thing I can think of is petroleum jelly but I can't be sure that, that is it. I don't have any and don't work with it much so I can't compare what it looks like.

Based on my description do you have any possible ideas on what it is?

If you do have something that matches these symptoms would you happen to know what its toxicity is and proper cleanup procedure?

Also in general how should one handle touching something they don't have knowledge of but that could be dangerous and when hand-washing is not working/removing it.

Thank you in advance for your questions and answers.

After the lengthy back and forth perhaps an official answer is due: the blue salt deposit on the camera battery holder door is most likely copper(II) hydroxide by reaction of leaked alkaline battery electrolyte (likely $\ce{KOH}$) with copper metal in the battery door contact. It could possibly also consist of copper(II) carbonate hydroxide formed by reaction of the copper(II) hydroxide with carbon dioxide, but the deep blue rather than green color in the picture suggests it is the previous. Copper hydroxide is regarded as moderately hazardous, but you are unlikely to be at risk unless you ingest it. Feel free to consult a MSDS for this compound (such as here) for more information.

The sticky substance coating the camera is possibly a degradation product of plastic components in the camera housing. It is not likely to be toxic unless ingested or if you are in contact with it persistently.

Without more information about how the camera was maintained it is difficult to say much more.


Why is up spin in the px orbital and down spin in the py orbital considered an excited state? Why is this form of p orbital occupation an excited state? I understand that there should be one up spin and one down spin in one orbital box by Pauli's exclusion rule. But in the p orbitals, it doesn't matter, whether we put up spins first or down spins first according to Hund's rule. I thought unless up and down spin are in same box, they don't affect each other. But as what I thought is wrong, I would like to know the exact reason for this. Why is it considered to be an excited state?

Hund's rule of maximum multiplicity states that for a given electron configuration, the term with the lowest energy has a maximum multiplicity $2S+1$. $S$ is the total spin angular momentum of all electrons in the partially-filled subshell.

For the application of the rule, we consider the 2 electrons in the $p$ subshell and add up their spin quantum numbers to obtain $S$ and the multiplicity. For the ground state configuration, with both electrons "spin-up", we get

$$S=+\frac{1}{2}+(+\frac{1}{2})=1$$
$$2S+1=3$$

For your proposed configuration with one $p$ electron "spin-up" and the other one "spin down":

$$S=+\frac{1}{2}+(-\frac{1}{2})=0$$
$$2S+1=1$$

Since the multiplicity value is smaller, this configuration is not the one which is lowest in energy, and therefore cannot be regarded as the ground state.


What would be the correct mechanism for hydration of ethene? Given above is the mechanism that I drew but my textbook puts it a bit differently. It says that the ethene is dissovled in the concentrated $\ce{H2SO4}$ to form ethyl hydrogen sulphate $\ce{CH3-CH2OSO3H}$ which when diluted with water reacts to form ethanol and the acid
$$\ce{CH3CH2OSO3H + H-OH -> CH3CH2OH + H2SO4}$$
So which mechanism is true?

At first I thought your mechanism was very comprehensive but upon closer inspection I think it has some flaws because in this reaction conc. sulphuric acid is used as a catalyst and it must be regenerated at the end and in the above mechanism only ethanol is generated.

This must be the mechanism shown in your book.


To extend the life of a cartridge razor blade, should the blade be stored in oil or alcohol? I've read varied recommendations about extending the life of a razor blade. Some say should stick the blade in oil when not in use, others say in alcohol, others say in a bowl of rice or silica gel.

I thought this community would be able to settle this:

What is the best (scientifically proven) manner to prevent corrosion of a cartridge razor blade?

I believe my Gillette Mach3 razor blades are stainless steel.

[The reason for this question is that cartridge blades are not cheap. Some may suggest going to a DE safety razor. I have tried it. It took me 2x as long to shave, so I returned to the cartridges.]

Most blades are dulled by either the edge folding over itself, or small amounts of rust forming. To stop the first problem, you can strop the blade, but that's not really chemistry. For the second one, anything that stops oxygen and water from reaching the blade will stop it from rusting.
Silica gel does absorb moisture,so that would prevent rust from forming, but this would be temporary. You would have to replace it, and even then it might not stop 100% of moisture from reaching the blade.
Oil might prevent water and oxygen from reaching your blade, but it would make it... oily. If you are okay with that then it would probably be a fine solution.
Alcohol would prevent water from reaching the blade pretty well, as well as disinfect it. This might be the best solution to extend your blade's life. Just make sure you use medical grade alcohol.
Finally, blow drying or towel drying your blade after every use will help a lot as well.


Oxidation to allylic alcohols without stoichiometric metal reagents Common methods of oxidation of alkenes to allylic alcohols in one step usually use stoichiometric metal oxidants, e.g., sodium dichromate.
Without using this type of reagent, are the other methods typically peroxide forming,  requiring further chemistry to adjust to the allylic alcohol, e.g., allylic hydroperoxides or t-butyl peroxides, or are there more "direct" methods?

Almost all allylic oxidation reactions have been using metal catalysts such as chromium, selenium, copper, etc. because it is relatively easy to abstract hydrogen radical from allylic position and metals are well known for doing so with or without single electron transfer (SET). This fact is well documented in the review published in 2013 (Ref.1), the abstract of which simply states that:

  Although $\ce{C–H}$ oxidation of hydrocarbons is generally difficult, allylic $\ce{C–H}$ oxidation is relatively simple and predictable, even on a preparative scale, because active species generated at the allylic position are stabilized by the double bond. Therefore, allylic oxidation has been employed in natural product synthesis, and a variety of reagents and conditions for allylic oxidation have been reported...

Some of the specific catalysts the review referred are  selenium Dioxide, diphenyldiselenide–Iodoxybenzene, chromium Trioxide–3,5-Dimethylpyrazole, pyridinium chlorochromate (PCC), chromic acid, etc., all of which are toxic to humans (e.g., selenium and chromium compounds) and fairly expensive (e.g., palladium and rhodium compounds). Recently, a paper published in Nature journal describes an electrochemical allylic $\ce{C–H}$ oxidation strategy that does not need toxic and expensive metal catalysts, but uses inexpensive and readily available materials. The method exhibits broad substrate scope, operational simplicity, and high chemoselectivity. The notable advantage of this electrochemical allylic $\ce{C–H}$ oxidation strategy is It represents a scalable allylic $\ce{C–H}$ oxidation (results have demonstrated on $\pu{100 g}$ scale reactions).

The General Procedure for electrochemical allylic oxidation is described as follows in the paper:

  With no precautions to exclude air or moisture, a test tube is charged with a stir bar, $\ce{Cl4NHPI}$ $(\pu{0.2 equiv})$ and $\ce{LiClO4}$ ($\pu{0.6 equiv}$, $\pu{0.1 M}$). Acetone ($\pu{6 mL/mmol}$ substrate) is added, and the resulting suspension is stirred until most of the solids dissolve. Substrate ($\pu{1 equiv}$) is added, followed by pyridine ($\pu{2 equiv}$) and $70\%$ aqueous tert-butyl hydroperoxide ($\pu{1.5 equiv}$). The anode and cathode (RVC, separated by a glass slide) are placed in the solution, and the reaction mixture is electrolyzed at a current of $\pu{10 mA/mmol}$ substrate until the reaction is complete as judged by TLC or GCMS analysis. The reaction mixture is transferred to a separatory funnel, and the electrodes are washed thoroughly with hexanes, diethyl ether, or ethyl acetate. The organic layer is washed twice with $\ce{H2O}$, once with brine, dried, and concentrated. The resulting crude product was then purified on $\ce{SiO2}$ to furnish the desired enone product. Note: Pyridinium perchlorate can be added to the reaction ($\pu{0.1–0.2 M}$) to improve solubility of the reaction components. This co-electrolyte has negligible effect on product yields and may improve the reaction if the substrate contains any easily reduced functional groups since this increases the concentration of protons available to be reduced at the cathode.

Note that the mediator, $\ce{Cl4NHPI}$ is N-Hydroxytetrachlorophthalimide, which was synthesized from tetrachlorophthalic anhydride and $\ce{H4N-OH/HCl}$ in pyridine using thermal or microwave technique for this research. However, it was available in Sigma-Aldrich for $\pu{62 USD}/\pu{10 g}$.

The schematic diagram of the general procedure is depicted in above diagram (Figure A).  The variety of mediators used during the progress of the process are listed in Figure B. Among them, N-Hydroxyphthalimide ($\ce{NHPI}$) and $\ce{Cl4NHPI}$ have given the best results (have given total yield of $56\%$ and $77\%$, respectively in otherwise optimized conditions). They are drastically efficient than others, e.g., next best has been TEMPO, which has given only $26\%$ total yield under identical conditions. Among the co-oxidants tried, $\ce{t-Bu-OOH}$ has given the optimal results ($51\%$). The next best is $\ce{PhC(CH3)2-OOH}$ with $43\%$ yield. Acetone ($56\%$) and $\ce{CH3CN}$ ($51\%$) have given equally impressive results as choice of solvent, and acetone was chosen for optimal conditions as the solvent for this research. Hence optimized electrochemical parameters were: $(\pu{0.16 M})$ of substrate with $\ce{Cl4NHPI}$ $(\pu{0.2 equiv})$, pyridine $(\pu{2.2 equiv})$, $\ce{t-Bu-OOH}$ $(\pu{1.5 equiv})$, $\ce{LiClO4}$ $(\pu{0.6 equiv})$ in acetone.

The tentative mechanism is illustrated in Figure C. The formation of tert-butyl-peroxy-intermediate (III) was confirmed. In a separated experiment, one of corresponding tert-butyl-peroxy-intermediate was synthesized in a different route and reacted in identical condition used in electrolytic procedure ($\ce{Cl4NHPI/t-BuOOH/LiClO4/pyridine/acetone}/RT$). The reaction has completed both with and without electrolysis within $\pu{6 h}$, confirming requirement of electrolysis during initial tert-butyl-peroxy-intermediate formation steps (Ref.2).

References:

Akihiko Nakamura, Masahisa Nakada, “Allylic Oxidations in Natural Product Synthesis,” Synthesis 2013, 45(11), 1421-1451 (DOI: 10.1055/s-0033-1338426).
Evan J. Horn, Brandon R. Rosen, Yong Chen, Jiaze Tang, Ke Chen, Martin D. Eastgate, Phil S. Baran, “Scalable and sustainable electrochemical allylic C–H oxidation,” Nature 2016, 533, 77–81 (https://doi.org/10.1038/nature17431).


Which conformations of 2,3-dimethylbutane are more stable? I am not sure what the proper arrangement of these conformations are from least to most stable. I got the wrong answer, but my thinking was this:
 First, I thought that because 2 and 4 are in eclipsed conformations they are least stable.  From there I thought that 4 was less stable than 2 because of the steric interactions of the methyl groups.  Then of the two staggered conformations 1 and 3, it appears to me that 1 has more steric interaction of the methyl groups than 3.  So my ordering from least to most stable was 4,2,1,3.  Is there an error in my line of reasoning?

The correct arrangement (from least stable to most stable) is probably: $4 > 2 > 3 \gtrapprox 1$ based on my calculations for the four gas phase rotamers you can see in the figure below.

Your reasoning for 4 over 2 sounds good, that would also be my point. For the two minima 1 and 3 (with Boltzmann weights of 46%:54% at the highest calc. niveau (red), 33%:66% at the lowest (blue)) you might also be right. But from the really small energy difference I don’t want to argue too much about it.

My calculation methods seem a bit random, but at first I only wanted to make that fast wB97X-D3/def2-TZVP//HF-3c scan and then decided to calculate the bigger one.


Is it possible to use diatomic carbon to utilize a new type of fuel that takes away the waste product of carbon dioxide being released into the air? These days, we as a community what to polish and make our world a greener environment. I found a way to accomplish this task, however I don't know if it is applicable. I am wondering if the use of diatonic carbon would do the trick to produce a process that could regenerate the fuel in our vehicles and eliminate the releasing of carbon dioxide into the air. The purpose of this reaction would be to remove the Carbon Dioxide waste that is created by gasoline. Gasoline waste contains one less carbon than was originally in the reaction

(UNDER CONSTRUCTION)

No doubt, diatomic carbon does exist - in blue flames and in comets.

The characteristic uv absorption pattern (Swan bands) has already been described almost 100 years ago.

At low temperature (77 K) it reacts with hydrocarbons by H-abstraction to form $\ce{HC#CH}$, and the addition to $\ce{C=C}$ double bonds has been reported too.

But I can't figure out any reasonable way for a reaction with carbon dioxide.


Pentabromophenol Mass Spectrum I am working on analyzing the mass spectrum of pentabromophenol but am struggling to determine the fragment at 60 m/z. It could possibly be 5 carbons however that is too much positive charge to exist at high abundance.

Below are the mass spec retrieved experimentally and from literature.

Literature Mass spec: NIST WebBook

Experimentally determined mass spectrum:

It seems to have the 1:1 $\ce{Br}$ ratio spaced by 1.

I think a reasonable assignment would be that it is $\ce{[BrCCOH]^2+}$

m = 120 and 122

m/z = 60 and 61


What happens to the carbon in carbon steel when it rusts? Carbon steel is iron containing some carbon in various forms of iron carbide.

When it rusts (oxidises), the iron becomes part of all kinds of crystalline and amorphous iron oxides and hydroxides, mostly as Fe3+.

Where does the carbon go in this process? Does it also oxidise to carbon dioxide or monoxide and leaves the system as a gas? Does it combine with the iron to form siderite (iron carbonate)? Does it remain as graphite or some other form of elemental carbon? Something else?

Cast iron has considerable free carbon as graphite, but carbon steel has iron carbide in cementite, $\ce{Fe3C}$ in a ferrite ($\ce{α-F}$e) matrix, along with other phases and additives. In a corrosive environment, the ferrite and cementite form a galvanic couple, causing increased corrosion of the $\ce{α-F}$e while effectively protecting the $\ce{Fe3C}$. According to Acta Metallurgica Sina,, "So, the ferrite phase diminishes gradually, and the accumulated cementite multiplies on the surface of the specimens along with the development of corrosion."

That said, what now happens to the remaining $\ce{Fe3C}$? Wikipedia states, "While cementite is thermodynamically unstable, eventually being converted to austenite [$\ce{γ-Fe}$] and graphite at higher temperatures, it does not decompose on heating at temperatures below... 723 °C..."

So, at room temperature, the remaining carbon would be specks of cementite amid the rust.


Does titration usually always involve neutralization? I've been doing titration exercises that involve figuring out the molarity of an unknown solution when you know the molarity of another (so like knowing the molarity of solution A but figuring out the molarity of solution B), and it usually always involves neutralization. Is it common that titration exercises involve neutralization?

While neutralization of an acid (or a base) is a frequent example of titration, titration is more general. IUPAC's gold book mentions  "Also contains definitions of: acidimetric titration, acid–base titration, alkalimetric titration, chelatometric titration, complexometric titration, coulometric titration, equivalence point, precipitation titration" right in the beginning.  Even the emblematic glass buret is not a must, as publications like My Dear Buret, Your Time Has Indeed Come! (Journal of Chemical Education) showcase.


Are PCC and Collin's reagent completely identical in their oxidizing action? Are PCC and Collin's reagent completely identical in their oxidizing action? Is there any advantage (or disadvantages) over the other?

In the Wikipedia article of PCC, it is written that:

  PCC offers the advantage of the selective oxidation of alcohols to aldehydes or ketones, whereas many other reagents are less selective.

Further, on the Wikipedia article of Collin's reagent, it is written that:

  This complex is both difficult and dangerous to prepare, as it is very hygroscopic and can inflame during preparation. It is typically used in a sixfold excess in order to complete the reaction. Nowadays, PCC or PDC oxidation have largely supplanted Collins oxidation for these reasons.


How to derive the nuclear spin of 23Na? Is it possible to derive the nuclear spin I=3/2 for $\ce{^23Na}$ from a term scheme or from something else from spectroscopy?

I thought the nucleus spin is empirical (and cannot be calculated from J and F values) or does it?

The nuclear shell model is a useful "first approach" to determining nuclear spin.  It doesn't always work, but it is a relatively simple way to make a first attempt.

Here is a nuclear shell energy diagram.  As you can see it is somewhat analogous to an electron orbital energy diagram.

A set of shells are filled by neutrons and a separate set is filled by protons.  The protons will pair when possible, same applies to the neutrons.  Notice that shells with J=1/2 hold 2 nucleons (a nucleon can be either a proton or a neutron), shells with J=3/2 can hold 4 nucleons, shells with J=5/2 can hold 6 nucleons, and so on.  This is because the shell with (for example) J=5/2 has 3 sub-shells: J=5/2, J=3/2 and J=1/2, each capable of holding 2 nucleons.  Higher J sub-shells are "usually" filled first.

Let's apply this model to a few examples.

$\ce{^2H}$ (deuterium): 1 neutron, 1 proton; the neutron goes into the neutron $\ce{1s_{1/2}}$ shell and is unpaired, spin = 1/2; the proton goes into the proton $\ce{1s_{1/2}}$ shell and is unpaired, spin = 1/2; total nuclear spin = 1/2 + 1/2 = 1

$\ce{^{17}O}$: 9 neutrons, 8 protons; the first 8 neutrons go into the neutron 1s and two 1p shells and are all paired, the 9th neutron goes into the neutron $\ce{1d_{5/2}}$ shell and is unpaired, spin = 5/2; all 8 protons are paired in the 1s and two 1p shells, spin = 0; total nuclear spin = 5/2 + 0 = 5/2

$\ce{^{23}Na}$: 12 neutrons, 11 protons; the first 8 neutrons go into the neutron 1s and two 1p shells and are all paired, the last 4 neutrons go into the neutron $\ce{1d_{5/2}}$ shell and each set of two is paired, spin = 0; the first 8 protons fill the proton 1s and two 1p shells and are all paired, the next 2 protons go into the $\ce{1d_{5/2}}$ J=5/2 sub-shell and are paired, spin = 0, the last proton goes into the $\ce{1d_{5/2}}$ J=3/2 sub-shell and has spin 3/2; total nuclear spin = 0 + 3/2 = 3/2


What is it in the composition of natural fabrics that cause them to retain smoke odors versus synthetics? Just an observation I notice when around smoker friends or when I visit Las Vegas - polyester clothing doesn't seem to stay smelling of smoke for the rest of the day like my cotton shirts do (or also my hair for that matter).

Adsorption is a very complicated phenomenon which involves so many interactions. Smoke has thousands of chemicals too, therefore one cannot pin-point a single cause. It is just like the concept of stains. Some stains on cotton never come off easily, which means that molecular interaction is very strong between the dye in the stain and cotton fiber. As a rough estimate, the -OH groups on cellulose may be providing the hydrogen bonding sites to nitrogen containing aromatic (in a chemical sense) compounds, hence stick for a long time. I am drawing an crude analogy because silica which has plenty of silanols (Si-OH) groups bind amines too strongly.


Why doesn't the expermintal molar solubility match the one calculated by the Ksp value? For $\ce{HgCl2}$ ($M = \pu{271.52 g mol-1}$), the reported $K_\mathrm{sp}$ value is $10^{-15.69}$.

This means that the calculated molar solubility of $\ce{HgCl2}$ in water is $(10^{-15.69}/4)^{1/3}\cdot\pu{271.52 g mol-1} \approx \pu{0.0010 g L-1}$, or equivalently, $\pu{0.010 g cm-3}$.

However, in Pubchem, the reported molar solubility is 6.9 g/100 cc. What is the source of the large deviation between the experimental and the theoretically calculated molar solubility of $\ce{HgCl2}$?

The $\mathrm{p}K_\mathrm{sp}$ value of $15.69$ for $\ce{HgCl2}$ given by your reference must have an error. First of all, $\ce{HgCl2}$ is fairly soluble in water at $\pu{20 ^\circ C}$. Wikipedia lists it as $\pu{65.7 g L-1}$, yet lists solubility of $\ce{Hg2Cl2}$ as $\pu{3.25e-4 g L-1}$. Meantime, I found an University of Arkansas, Little Rock website, which has listed $K_\mathrm{sp}$ value of $\ce{Hg2Cl2}$ as $\pu{1.43e-18}$ ($\mathrm{p}K_\mathrm{sp} = 17.84$).

The molar mass of $\ce{Hg2Cl2}$ is $\pu{472.09 g mol-1}$. Thus, theoretically, the solubility ($s$) of $\ce{Hg2Cl2}$ can be calculated similar way you did:

$$s = \left({\frac{\pu{1.43e-18 mol3 L-3}}{4}}\right)^{\frac13}\times \pu{472.09 g mol-1} = \pu{3.35e-4 g L-1}$$

I think this is close enough. :-)


Why does Mg2+ have the ionic radii bigger than Ca2+ If they were atoms, $\ce{Mg}$ should have a smaller radii. How does this change?

Source: http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Atomic_%26_ionic_radii.svg/551px-Atomic_%26_ionic_radii.svg.png

Ca2+ is bigger than Mg2+


What is the chemical composition of Turnbull's blue? I was studying qualitative analysis from this book and  I found some tests for testing the presence of the cation $\ce{Fe^{++}}$.

There was a test in which we add $\ce{K4[Fe(CN)6]}$ solution to a solution of a ferrous salt which results in the Turnbull blue precipitate. Now according to Wikipedia, the Turnbull blue and Prussian blue are the same compound, ie, ferriferrocyanide $\left(\ce{Fe4[Fe(CN)6]3} \right)$.

But the book says that Turnbull blue is ferroferricyanide $\left(\ce{Fe3[Fe(CN)6]2}\right)$ which kinda makes sense because we are adding ferricyanide to ferrous solution.

I am preparing for an exam in which such questions can be asked SO I just want to settle this once and for all, which is wrong, Wikipedia or the book?

Though your textbook is right, Wikipedia isn't completely wrong as in aqueous solution, $\left(\ce{Fe3[Fe(CN)6]2} \right)$ gets converted and is in equilibrium with $\left(\ce{Fe4[Fe(CN)6]3} \right)$ (a redox change) and hence prussian blue and turnbull's blue are identical to look visually. But in a textual way, the naming is given as your textbook says, i.e. $\left(\ce{Fe4[Fe(CN)6]3} \right)$ is Prussian blue whereas $\left(\ce{Fe3[Fe(CN)6]2} \right)$ is Turnbull's blue.


Why does the conversion of SO₂ to SO₃ require high pressure? The conversion of sulfur dioxide to sulfur trioxide in the contact process for the manufacture of sulfuric acid requires the following condition:

Temperature: 450 degrees
Pressure: 1-2 atm, although High pressure would fasten the reaction, creating those pressures is not economical
Vanadium Pentoxide ($\ce{V2O5}$) as catalyst
and excess of oxygen

$\ce{2SO2 + O2 -> 2SO3}$

My textbook says that high pressure favors the rate of the reaction. The reason it states is that the volume of product is less than that of the reactants.

Can somebody explain how the volume of the product being less causes high pressure to be required?

As Philipp stated in the comments it is Le Chatelier's principle in play.

The principle in layman's language states that

  If you try to bring about any physical change in a system the system will try to but not necessarily succeed in cancelling the change.

The reaction you have given is an equilibrium reaction where all the compounds are in gaseous state.

Now if you count the total number of moles on the left-hand side it will be $\ce{2SO2 + O2}$
i.e. $\ce{2 + 1}$ that is $\text{3}$.
And on the right-hand side there are only $\text{2}$ moles of $\ce{SO3}$.

So ultimately if you consider the reaction to proceed in forward direction 3 moles of gas are converting into 2 moles of gas and hence the volume decreases in forward direction.

If you think according to Gas Law(or logically) an increase in pressure generally corresponds to a decrease in volume.

Now use Le Chatelier's principle to say that if I increase the pressure on the the system ,in order the system
to counteract to the change will decrease it's own volume and it has no other way of achieving that but to proceed in the forward direction.

But if the query arises that high pressure should favour any and every reaction take a look at the example below:

$$\ce{PCl5 <=> PCl3 + Cl2}$$

Here the gaseous moles on the left-hand side is $\ce{1}$ and on the right-hand side $\ce{2}$. So increasing pressure makes the reaction proceed in the backwards direction as there are less number of gaseous moles on the left-hand side.

  In fact, in higher level this idea is what is called $\Delta\text{n}_\text{g}$ i.e the difference($\Delta$) of gaseous($_g$) moles($\ce{n}$).

  $$\Delta\text{n}_\text{g} \quad = \text{gaseous moles on right-hand side - gaseous moles on left-hand side}$$

If $\Delta\text{n}_\text{g}$ is positive, then on increasing pressure reaction proceeds backwards and vice-cersa


How do you have pKa for a base? The $\mathrm{p}K_\text{a}$ is the $-\log_{10}$ of the $K_\text{a}$ value for the disassociation of an acid. So how do you get a $\mathrm{p}K_\text{a}$ value for a base? The base produces no hydrogen ions so can’t you only have a $\mathrm{p}K_\text{b}$?

For bases, you can derive the $\mathrm{p}K_\text{a}$ from the $\mathrm{p}K_\text{b}$.

Knowing that

$$\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$$

Let’s examine ammonia.

Ammonia has a $K_\text{b}$ of $1.8 \times 10^{-5}$.

Taking the negative logarithm, we can derive $\mathrm{p}K_\text{b}$:

$$\mathrm{p}K_\text{b} = -\log( 1.8 \times 10^{-5}) = 4.74$$.

Knowing that $\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$:

We can find $\mathrm{p}K_\text{a}$ by manipulating that expression, thus:

$$\mathrm{p}K_\text{a} = 14 - \mathrm{p}K_\text{b} $$

$$\mathrm{p}K_\text{a} = 14 - 4.74 $$

$$\mathrm{p}K_\text{a} = 9.26 $$

Even in very basic solutions, there will be minute quantities of hydronium ions. By taking the anti-logarithm of the $\mathrm{p}K_\text{a}$, you can derive the $K_\text{a}$.


Why do research papers mention experimental details in such detail? The paper Analysis of cyclic pyrolysis products formed from amino acid monomer has the following to say in it's experimental section:

Glycine, phenylalanine, tyrosine, tryptophan, serine, and valine were purchased from Daejung Chemicals & Metals Co. (Korea). Arginine, aspartic acid, glutamic acid, and proline were purchased from Samchun Pure Chemical Co. (Korea). Alanine, asparagine, glu- tamine, histidine, leucine, and threonine were purchased from Junsei Chemical Co. (Japan). Isoleucine and lysine were purchased from Acros Organics (USA), and cysteine was purchased from Merck Co. (Germany). Methionine was purchased from Yakuri Pure Chemicals Co. (Japan). Fumed silica was purchased from Merck Co. (Germany).
...
Pyrolysis–GC/MS was conducted with a CDS Pyroprobe 1500 heated filament pyrolyzer (Chemical Data System, Oxford, USA) coupled to an Agilent 6890 gas chromatograph equipped with a 5973 mass spectrometer of Agilent Technology Inc. (USA)

Why is there so much emphasis on the sources and country of origin of chemicals/instruments in these sections? This behaviour is not restricted to this paper; several other papers go into similar levels of detail regarding their procedures.

Experimental details are very, very important: they are used to ensure that results are reproducible, or at least that is the aim.

Let's talk about this specific example of suppliers and sources. One might naively think that a chemical purchased from supplier X is the same as the chemical from supplier Y. In an ideal world, that would be how it is. Unfortunately, this isn't the case, due to various considerations e.g. method of preparation, packaging, and so on. Therefore, even though the majority of the compound may be the same, there can be different impurities. And unfortunately, these impurities can lead to reactions working or failing. Many experimental chemists will personally know of cases where a reaction doesn't work until a reagent from a different company is used, or a reaction that stops working after buying a different bottle of a reagent.

There are many examples of this in the literature, but I'll choose the first one that came to mind, which is a reasonably recent paper from my old group (open access):

Mekareeya, A.; Walker, P. R.; Couce-Rios, A.; Campbell, C. D.; Steven, A.; Paton, R. S.; Anderson, E. A. Mechanistic Insight into Palladium-Catalyzed Cycloisomerization: A Combined Experimental and Theoretical Study. J. Am. Chem. Soc. 2017, 139 (29), 10104–10114. DOI: 10.1021/jacs.7b05436.

  During these investigations, we also uncovered a critical dependence of the stereochemical outcome of the reaction on the batch of Pd(OAc)2 employed as catalyst. This discovery was made through the chance purchase of Pd(OAc)2 from a different supplier, which led to an unexpected
  ratio of enamide alkene geometries ((Z):(E) = 80:20), rather than the typical ratio of ∼97:3.  Screening of further samples of Pd(OAc)2 gave variable results, the most extreme being a reversal of stereoselectivity to 40:60 in favor of the (E)-isomer.

The exact supplier isn't named, but 1H NMR spectra of various batches are given and the performance in the reaction can be correlated with the exact form of Pd(II) in the catalyst, which ensures that future readers are aware of what to look out for. This is very important for good science, because what's the point of publishing your fancy new method if nobody can replicate it?

Of course, this doesn't always happen. However, it's far better to include the information, just in case it is needed.

As mentioned by Waylander in the comments, commercial samples of samarium diiodide (SmI2) also have a reputation for being inconsistent: see e.g. this report by Szostak, M.; Spain, M.; Procter, D. J. J. Org. Chem. 2012, 77 (7), 3049–3059. DOI: 10.1021/jo300135v.


Freezing point depression and organic molecules Do organic molecules reduce the freezing point of water in the same way that inorganic salts do?  I've recently been swimming in lakes with a high organic content (e.g. mostly tannins.  There's a lake in this area called Tea Lake).  I've got an infra-red thermometer and readings taken from the water at this time of year before total freeze up have been around -1 to -2 degrees C.  Are these reasonable (accepting of course possible inaccuracies and calibration issues with the thermometer)?
Thanks,
Phil

To get a freezing point depression of $\pu{−1.86 °C},$ the solution should be made by adding $\pu{1 mol}$ organic substance to $\pu{1 L}$ water. And $\pu{1 mol}$ per liter is much. For glucose, it is $\pu{180 g}$ per liter. It is nearly a syrup. I doubt your lake has such a high concentration whatever the nature of the solute.
For salts, it is less, because usual salts are made of two ions. For example, $\pu{29 g}$ $\ce{NaCl}$ in $\pu{1 L}$ water is enough to get a freezing point depression of $\pu{−1.86 °C}.$ Anyway, $\pu{29 g}$ salt added to one liter is not negligible. Are you sure of your measurements?


How to name binary (inorganic) compounds given their chemical formula, and vice-versa? I wish to understand how to name binary (inorganic) compounds when given their formula. When I am a given compound like $\ce{B2F2}$ or $\ce{N2O4}$, I have little to no clue on how to write their chemical names. Similarly, I also trip up when trying to convert a given chemical name - say diboron difluoride - to its chemical formula.

Moreover, I have seen that a few times we do not need the numeric prefixes - as in lithium oxide or magnesium chloride - but sometimes we do - as in dinitrogentetroxide or iodine monochloride. So, I also need help understanding when to include numerical prefixes and when to not.

Please limit the answer to simple, binary, ionic and covalent compounds, and exclude hydrocarbons.

Prerequisites

These are the necessary prerequisites for this topic. If you're uncomfortable with any, please first head over to the corresponding links before continuing.

Chemical symbols are a shorthand method of representing an element. Instead of using the full name of an element, we simply refer to it with one or two letters. $\ce{N}$ for nitrogen and $\ce{Na}$ for sodium. More details on the Wikipedia page.
Polyatomic anions/Radicals: anions with more than one element, like nitrate ($\ce{NO3-}$) or sulfate ($\ce{SO4^2-}$). More details on the Wikipedia page.
Oxidation state: is an integer or decimal number assigned to an element in a chemical species. It is a tool that helps us do nomenclature easily. Read a detailed introduction here.
Ionic and covalent compounds: You must understand what ionic and covalent compounds are. You must also know the few elementary examples of each. For example, you should know that $\ce{N2O4}$ would be a covalent compound, while $\ce{NaCl}$ would be ionic. Here's an introduction by LibreTexts if you need a refresher.

Introduction

There are two separate cases here for ionic and covalent compounds.[1]. Some of the logic overlaps, but some does not, and care must be taken to avoid using one's rules for another.

Naming an ionic compound from its formula

Name the cation and then the anion. ("sodium sulfate", but not "sulfate sodium"). (How to identify the cations and anions in an ionic compound? The cation is usually a metal and the more electropositive element, while the anion is a non-metal and more electronegative element)
Keep everything lowercase (details) ("sodium sulfate", but not "Sodium Sulfate" or "sodium Sulfate" or "Sodium sulfate")
Do not apply numeric prefixes to either the cation or the anion. ("sodium sulfate", but not "disodium sulfate" or "sodium monosulfate")
Remember the suffixes for anions. The single atom anions of group 15, 16, and 17 have the suffix ending "-ide". Examples: fluoride, oxide and phosphide. However, polyatomic anions generally do not end in an -ide. See the Appendix for details.
Transition metal cations (and some p-block metals) also need their oxidation-state to be specified in parentheses, as they exhibit variable oxidation states.  Hence, "mercury(II) chloride", but not "mercury chloride". Notice that the oxidation state is denoted in capital Roman numerals.[2]

Let's take an example: $\ce{MgCl2}$. We can identify it's composed of magnesium and chlorine. Thus, the name would be magnesium chloride – in all lowercase and with the cation (metal) preceding the anion (non-metal). There is no need to denote the oxidation state as magnesium isn't a transition metal.

Examples:

$\ce{CsF}$, $\ce{NaNO3}$, $\ce{Al2(SO4)3}$, $\ce{FeCl2}$, $\ce{Pt3(PO4)4}$

Answers: (hover to peek!)

   caesium fluoride, sodium nitrate, aluminium sulfate, iron(II) chloride or ferrous chloride, platinum(IV) phosphate

Writing the formula for an ionic compound from its name

First write the cation and then the anion ($\ce{KBr}$ but not $\ce{BrK}$)
Use parentheses to denote multiple polyatomic anions ($\ce{Mg3(PO4)2}$ but not $\ce{Mg3PO42}$). Avoid them for a single polyatomic anion ($\ce{Na2SO4}$ but not $\ce{Na2(SO4)}$)
Use the trick of cross-transferring oxidation states between the ions to determine the count of each cation and anion involved. This trick works because the formula should denote an electrically neutral species. (you can verify this mathematically) So, to write the formula for aluminium sulfate:

Note that:

In case both oxidation states have a common factor, you should cancel them. Example: magnesium oxide:
.
drop the number 1. Hence, potassium sulfide is written as $\ce{K2S}$ but not $\ce{K2S1}$.

Examples:

potassium phosphate, calcium chloride, magnesium cyanide

Answers: (hover to peek!)

   $\ce{K3PO4, CaCl2, Mg(CN)2}$

Naming a covalent compound from its formula

Name the less electronegative element and then the more electronegative element ("carbon monoxide" but not "oxygen monocarbide")
Apply suffixes to anions in exactly the same way as we did for ionic compounds. ("sulfur hexafluoride" but not "sulfur fluoride") Also, keep the name all lowercase as before.
Use numerical prefixes.[3] Hence "dinitrogen tetroxide" for $\ce{N2O4}$ but not "nitrogen oxide". Here's the first ten suffixes: $$\begin{array}{|c|c|} \hline
\mathrm{prefix} & \mathrm{number}\\ \hline
\mathrm{mon(o)} & 1 \\
\mathrm{di} & 2 \\
\mathrm{tri} & 3 \\
\mathrm{tetr(a)} & 4 \\
\mathrm{pent(a)} & 5 \\
\mathrm{hex(a)} & 6 \\
\mathrm{hept(a)} & 7 \\
\mathrm{oct(a)} & 8 \\
\mathrm{non(a)} & 9 \\
\mathrm{dec(a)} & 10 \\
\hline
\end{array}$$ Note that:

the letter in parentheses is dropped when the anion begins with a vowel ("tetroxide", but not "tetraoxide")
we do not specify the mono- prefix for the first element ("carbon tetrachloride" but not "monocarbon tetrachloride")
we can construct similar numerical prefixes for larger counts by using the Greek system. Example: "undeca" for 11. Look at a larger list on Wikipedia.

Let's take an example: $\ce{P4O6}$. In one molecule of $\ce{P4O6}$, we can see that it has four atoms of phosphorus and six atoms of oxygen. So, its name should be of the form "Xphosphorus Yoxide". The table of numerical prefixes tells us that we should use the prefix "tetra" for nitrogen and "hexa" for oxygen. Since the name "oxygen" begins with a vowel, we shall write "hexoxide" (and not "hexaoxide"). Thus, the correct name for this chemical formula is "tetraphosphorus hexoxide".

Examples:

$\ce{CO2}$, $\ce{P4O10}$, $\ce{N2O}$, $\ce{N2S5}$

Answers: (hover to peek!)

   carbon dioxide, tetraphosphorus decoxide, dinitrogen monoxide, dinitrogen pentasulfide

Writing the formula for a covalent compound from its name

This is much simpler from the previous case, rather it follows an exactly reverse process.

We'll just take a look at two examples:

boron triiodide: we see that its formula would be: $\ce{B_$x$I_$y$}$. Now, $x=1$ since boron lacks any numerical prefix, and $y=3$ due to the "tri" prefix. Hence, the formula is $\ce{BI3}$.
dinitrogen pentoxide: we see that its formula would be: $\ce{N_$x$O_$y$}$. Now, $x=2$ and $y=5$ due to the "di" and "pent(a)" prefixes. Hence, the formula is $\ce{N2O5}$.

Examples:

hydrogen bromide, iodine heptafluoride, diboron hexahydride

Answers: (hover to peek!)

   $\ce{HBr}$, $\ce{IF7}$, $\ce{B2H6}$

Naming of hydrogen compounds

Several hydrogen compounds have non-systematic names that have persisted through the years in standard nomenclature and are now well recognized. Note that for alkali metals and alkali earth metal hydrides, the name is simply "metal + hydride", like sodium hydride, on the basis of rules described earlier.

However, covalent hydrides have varying names. There are not much rules around them, but rather "patterns" that the following list would show you:

Group 13: $\ce{BH3}$: borane, $\ce{AlH3}$: alumane, $\ce{GaH3}$: gallane, etc. Pattern: half of the element name suffixed by "-ane".
Group 14: $\ce{CH4}$: methane, $\ce{SiH4}$: silane, $\ce{GeH4}$: germane. Pattern: half of the element name suffixed by "-ane". Methane is a remnant of organic chemistry, and is a part of nomenclature of hydrocarbons.
Group 15: $\ce{NH3}$: ammonia, $\ce{PH3}$: phosphine, $\ce{AsH3}$: arsine, $\ce{SbH3}$: stibine, $\ce{BiH3}$: bismuthine. Pattern: half of the element name suffixed by "-ine". "Ammonia"'s origins, though, are a bit more interesting (read the Wikipedia page).
Group 16: $\ce{H2O}$: water, $\ce{H2S}$: hydrogen sulfide, $\ce{H2Se}$: hydrogen selenide, $\ce{H2Te}$: hydrogen telluride. Note that while "oxidane", "sulfane", "selane", etc. are acceptable, in the case of oxidane, that name is only used when denoting water derivatives. Read Do chemists refer to water as "dihydrogen monoxide"? and Does water have a chemical name?.
Group 17: they are simply named as "hydrogen halide". Example: "hydrogen bromide" $\ce{(HBr)}$. It is not named "hydrogen monobromide", as you may believe from earlier examples, and that it because hydrogen can only form a single halide with the halogens (see Note 3 below).

The curious students would also notice that – when writing chemical formulae – for groups 13, 14, 15, the hydrogen atom is written after the group atom $\ce{(NH3)}$, while in groups 16 and 17, it is written before the group atom $\ce{(H2O)}$. See this answer to understand why it is so.

Notes:

It is true that no compound is 100% ionic or 100% covalent. The above rules for "ionic compounds", thus, apply to compounds with dominant ionic character. Similarly, rules for "covalent compounds" apply to compounds with dominant covalent character.
Another way to write such a name is to use an "-ic" suffix for a higher oxidation state and "-ous" suffix for a lower oxidation state. As in "mercuric chloride" for "mercury (II) chloride". But, according to the IUPAC Recommendations 2005, it is no longer acceptable.
Why are numerical prefixes necessary? Note that in covalent compounds, a given anion can be present in more than one count. For example, the $\ce{F}$ in $\ce{BrF_x}$ can be present as $\ce{BrF}$, $\ce{BrF3}$, or even $\ce{BrF5}$. Hence, it is necessary to indicate if it is a monofluoride, a trifluoride, or a pentafluoride.
Several covalent compounds and ionic compounds have their own unique, common/trivial names. These names do not follow any systematic rules and are based on the source (location/discoverer) or properties of that compound. Learning these common names and their formulae comes by experience over time. Examples include: diborane $\ce{(B2H6)}$, Epsom salt $\ce{(MgSO4)}$, oil of vitriol (sulfuric acid), etc. Read more on the Wikipedia page.

Appendix: polyatomic anion names

Apart from the regular -ide suffix for monoatomic anions, there are some more suffixes for the polyatomic anions.

most pseudohalides retain the "-ide" suffix. (except cyanate derivatives)
the other two suffixes are "-ate" and "-ite". "-ate" is used when the central atom is in a higher oxidation state, whereas "-ite" is used for the lower oxidation state.
the "per" prefix is used when the central  halogen atom is in its highest oxidation state, while "hypo" is used when it is in its lowest oxidation state.

A few examples would make these rules clearer:

Sulfate $(\ce{SO4^2-})$ and sulfite $(\ce{SO3^2-})$: both are sulfur and oxygen anions, but in sulfate, sulfur has higher oxidation state as compared to that in sulfite. Similar case applies to nitrate, nitrite and phosphate, phosphite.
Hypochlorite $\ce{(ClO^-)}$, chlorite $\ce{(ClO2^-)}$, chlorate $\ce{(ClO3^-)}$, perchlorate $\ce{(ClO4^-)}$: have chlorine in increasing oxidation numbers. Similar series exist for other halogens (except fluorine).

Of course, there are many more polyatomic anions in chemistry, but these ions are relevant to our discussion at an elementary level.


Which of the following compounds do not undego electrophilic aromatic substitution? Which of the following does not undergo electrophilic aromatic substitution?

Answer is option (D). But I am not able to understand why. All of them have lone pairs. I can notice only two differences. Functional group in Option (d) has the largest size and lowest electronegativity.

I think that may be because of a large difference in size of Selenium and Carbon, pie bond formation in between them will be a bit difficult, but that bond is not playing any role here except stabilizing the compound.

Low electronegativity seems like a good thing to me as it does not deactivates the compound.
So what am I missing here?

Wikipedia disagrees.  Selenophene does undergo EAS, and apparently faster than thiophene.

None of the proposed answers to the original question is correct.


Why is there a z² label for d orbitals, but no x² and y² labels with corresponding shapes? I know that the $\mathrm{d}_{z^2}$ orbitals are named as a result from the conversion of the angular component of the hydrogen-like wave function in terms of Cartesian coordinates, but how come we lack analogous orbitals along other axes, such as $\mathrm{d}_{x^2}$ or $\mathrm{d}_{y^2}$? My naive assumption would be due to the fact that $z$ axis is considered the principle axis in regards to chemical bonds, but I don't feel that this is the best answer, nor can I seem to come across anything in textbooks or articles online.

TL;DR The $\mathrm{d}_{z^2}$ orbital is a result of solving the Schrödinger equation for the hydrogen atom in the most mathematically convenient way.

To properly understand this, it is necessary to go back to the fundamentals. The complex d-orbitals are obtained by solution of the Schrödinger equation. In general, these d-orbitals are made up of a radial part $R(r)$ and an angular part $Y(\theta,\phi)$. The radial part is not relevant to the question, because its value does not depend on the direction in which you look, so it cannot explain why there is a $\mathrm{d}_{z^2}$, but not a $\mathrm{d}_{x^2}$. So, it is instructive to look at the angular component (the spherical harmonics), which are tabulated below. The normalisation factors are ignored here:

$$\begin{align}
\mathrm{d}_{+2} &\propto R(r) \cdot \sin^2\theta \cdot \mathrm{e}^{\mathrm 2i \phi} \\
\mathrm{d}_{+1} &\propto R(r) \cdot (-\sin\theta \cos\theta) \cdot \mathrm{e}^{\mathrm i \phi} \\
\mathrm{d}_0 &\propto R(r) \cdot (3\cos^2\theta - 1) \\
\mathrm{d}_{-1} &\propto R(r) \cdot \sin\theta \cos\theta \cdot \mathrm{e}^{-\mathrm i \phi} \\
\mathrm{d}_{-2} &\propto R(r) \cdot \sin^2\theta \cdot \mathrm{e}^{\mathrm -2i \phi}
\end{align}$$

These are combined in an appropriate manner to generate the real d-orbitals. It turns out that the $\mathrm{d}_{z^2}$ orbital is simply the same as the original $\mathrm{d}_0$ orbital, which is already real (the radial component $R(r)$ is real). Since $\cos\theta = z/r$ and $r^2 = x^2 + y^2 + z^2$ (spherical coordinates), the angular component can be rewritten as

$$\begin{align}
3\cos^2\theta - 1 &= 3\left(\frac{z}{r}\right)^2 - 1 \\
&= \frac{3z^2 - r^2}{r^2} \\
&= \frac{2z^2 - x^2 - y^2}{r^2}
\end{align}$$

and conventionally we denote this as $\mathrm{d}_{z^2}$. There is no analogous $\mathrm{d}_{x^2}$ orbital because the conventional way of deriving real d-orbitals does not yield any orbital with an angular form of $(2x^2 - y^2 - z^2)/r^2$.

But that still does not properly answer the question. To do so, we should look more closely at how the complex d-orbitals are obtained. The Schrödinger equation, $\hat{H}\psi = E\psi$, only prescribes that we need to find eigenstates of $\hat{H}$, but does not say how. It turns out that the easiest way to accomplish this is to look for solutions which are simultaneous eigenstates of $\hat{H}$ (energy), $\hat{L}^2$ (total orbital angular momentum), and $\hat{L}_z$ (projection of orbital angular momentum onto the z-axis). The eigenvalues of the solutions under these operators are directly tied to the quantum numbers $n$, $l$, and $m_l$, respectively.

Fundamentally, the $\mathrm{d}_{0}$ orbital (i.e. the $\mathrm{d}_{z^2}$ orbital) arises because we have chosen it to be an eigenstate of the operator $\hat{L}_z$. If we were to solve the Schrödinger equation in a different way, by searching for simultaneous eigenstates of $\hat{H}$, $\hat{L}^2$, and $\hat{L}_x$ (instead of $\hat{L}_z$), we would indeed get a $\mathrm{d}_{x^2}$ orbital. This is an equally valid solution of the Schrödinger equation (as it is an eigenstate of $\hat{H}$), but it is simply not the conventional solution.

So, why choose $\hat{L}_z$ and not $\hat{L}_x$ when solving the Schrödinger equation? The answer is simply mathematical convenience: the Schrödinger equation is most easily solved in spherical coordinates, and in spherical coordinates, $\hat{L}_z$ has a particularly nice form of being equal to $-\mathrm{i}\hbar(\partial/\partial \phi)$. The eigenstates of this are none other than $\mathrm{e}^{\mathrm{i}m\phi}$, which is exactly what we observe in the complex d-orbitals at the beginning of this post ($m$ is, of course, quantised, so there are only five valid solutions with $l = 2$, i.e. d-orbitals).

But why does $\hat{L}_z$ have a nice form and not $\hat{L}_x$? Ultimately, this mathematical convenience is a result of the definition of spherical coordinates. Because $\theta$ is chosen to be the angle made by the vector with the z-axis, the z-axis has a special status in spherical coordinates:

$$\begin{align}
z &= r\cos\theta \\
x &= r\sin\theta\cos\phi \\
y &= r\sin\theta\sin\phi
\end{align}$$

So, at the end of the day, it is a completely arbitrary convention. And it should be, too, because no axis should have an inherently special status in a spherically symmetric system like the hydrogen atom. As mentioned previously, you could obtain a $\mathrm{d}_{x^2}$ orbital by solving the Schrödinger equation in a different way. You could even choose to redefine the axes in such a way which gave you a $\mathrm{d}_{x^2}$ orbital (it's quite easy; simply rename $(x,y,z)$ to $(y,z,x)$).

But there is no way you can have both at the same time. Because $\hat{L}_x$ and $\hat{L}_z$ do not commute with one another, it is not possible to "solve" the Schrödinger equation in a way which gives you both $\mathrm{d}_{x^2}$ and $\mathrm{d}_{z^2}$ orbitals at the same time.


Knowing about Sensor Film Kit I am very new to this site and this is my first ever question. I want to know what is this black material in this sensor film kit. I am creating a weight sensor by using this very sensor kit. Now I am short of this black film. I want to purchase it now locally but do not know the composition of this material.
This material behaves as a resistance between two aluminium foil, just to inform.
If any one can help me in this.

It's called a force sensing resistor. It's basically a film of small conductive particles mixed into a non-conductive polymer matrix on top of a conductive substrate. The more force on the film, the more conductive particles can touch the layer underneath, reducing the resistance. Sensitronics owns the original patent for the technology and their site shows how they're constructed, though you'll probably have to dig up the patent to find out exactly what's in the force sensing layer.


Why do SN2 reactions of alkyl halides proceed differently with KCN and AgCN? Alkyl halides react with $\ce{KCN}$ to form alkyl cyanides as the main product, whereas the use of $\ce{AgCN}$ leads to isocyanides as the chief product. Why does this happen?

The cyanide anion is an example of an ambident nucleophile, these are nucleophiles which, usually due to delocalisation, are able to attack an electrophile from two or more atoms. In the case of the cyanide anion reaction from carbon leads to nitriles, whilst reaction from the nitrogen leads to isocyanides.

To give the full IUPAC definition from the Gold Book:

  A description applied to a chemical species whose molecular entities each possess two alternative and strongly interacting distinguishable reactive centres, to either of which a bond may be made in a reaction: the centres must be connected in such a way that reaction at either site stops or greatly retards subsequent attack at the second site. The term is most commonly applied to conjugated nucleophiles. [...]

To answer your actual question about why $\ce{KCN}$ forms nitriles, whilst $\ce{AgCN}$ forms isocyanides, March's Organic Chemistry (7th ed., p 449) gives a pretty concise answer in which we consider the extent to which the cyanide anion is associated with the counter-cation in solution.

  All negatively charged nucleophiles must of course have a positive counterion. If this ion is $\ce{Ag+}$ (or some other ion that specifically helps in removing the leaving group), rather than the more usual $\ce{Na+}$ or $\ce{K+}$, then the transition state is more $\mathrm{S_N1}$ like. Therefore the use of $\ce{Ag+}$ promotes attack at the more electronegative atom. For example, alkyl halides treated with $\ce{NaCN}$ generally give mostly $\ce{RCN}$, but the use of $\ce{AgCN}$ increases the yield of isocyanides ($\ce{RNC}$).

                           [See footnote]

This, of course, is not a complete picture, as solvent effects greatly change the reactivity. In protic solvents (MeOH, to give a possible example) the most electronegative atoms (nitrogen in this case) will be solvated to a greater extent via H-bondng. In aprotic solvents (THF/DMF, to give possible examples) the anion isn't solvated quite so much (on either atom of the nucleophile), but the cation is somewhat more solvated, leaving the nucleophile more able to attack (via the most electronegative atom as this is the most nucleophilic in the absence of other factors).

Footnote: As an aside at this point, it's important to realise that these are not fully covalent bonds (in $\ce{AgCN}$ and $\ce{KCN}$), despite having some level of covalent character. We're somewhat bastardising the Winstein model of nucleophilic substitution in which we define $\mathrm{S_N1}$/$\mathrm{S_N2}$ based upon the separation between the cation and the anion at the point when substitution takes place, with the two mechanisms being extremes of a continuum:


Diagonal relationships in the Periodic Table I have often wondered about diagonal relationships between elements on the periodic table, and the most often cited explanations revolve around charge-density considerations.

But other than that, what other factors could possibly contribute to this phenomenon?

EDIT: What I mean by "diagonal relationships" is that there are certain similarities in chemical properties that have been observed in diagonally adjacent neighbors in the 2nd and 3rd period example: Li-Mg, Be-Al, and B-Si.

Link to a wikipedia article (not much to be found here though)

The explanation that I personally have most often encountered is that the  charge-to volume ratios (charge density), say for Li-Mg cations is roughly the same, and hence could account for some of the similarities in behaviors.

What I am looking for is what can be some other contributing factors (if any) to this phenomenon. I am not expecting very concrete answers because I believe this phenomenon is not very well understood.

What I remember from studying C-P 'diagonal relationships' (for multiple bonds e.g. carbene vs phosphinidene, alkyne vs phosphaalkyne) is that similar electronegativity also played a role, also by affecting the valence orbitals.

However, perhaps it is better viewed from a different vantage point:
the main group elements behave rather similarly from 3rd row onwards, but there is a distinct difference between the 2nd and 3rd row elements and the explanation is pretty straightforward:
The 2s and 2p orbitals are roughly similar in 'size' since the 2p orbitals are the first p-orbitals and are pretty compact. Therefore 2s and 2p orbitals mix happily.
Going to the third row, the 3p orbitals need to be orthogonal to the 2p orbitals and as a consequence they are much more diffuse than the 3s orbitals, resulting in less efficient s-p hybridization. This also manifests itself in the so-called 'inert pair' effect whereby e.g. Si will resist hybridization in favor of a lone pair. There is some beautiful work by Knutzelnigg from the 80s (?) describing these effects.
This 2nd-3rd row discrepancy is somewhat counterbalanced if you step to right, increasing Z and thereby contracting the orbitals giving rise to this apparent diagonal relationship.

Similar arguments make that for the 3d TMs the hybridization is more efficient for 4s-3d than for the 4d/5d TMs. Consequently, 3d TMs tend to behave differently from their heavier sisters, which in turn are closer together (although for the 5d's the relativistic effects start to kick in more heavily...).


Azathioprine CIF data file rendering problem in Accelrys I have been working on a CIF data file based on experimental data [1] (free full text isn't available, but the page 2 with the experiment details can be found in the free preview provided by Springer).

Unfortunately, I don't know an awful lot about crystallography parameters (although I have been consulting the CIF core dictionary). The CIF file I linked to was created using EnCIFer & VESTA (I mostly used VESTA to get a handle for the text formatting of CIF files). The non-coordinate data mentioned in this paper includes:

\begin{array}{ll}
\hline
Z & 4 \\
\text{Formula} & \ce{C9H7N7O2S} \\
M_r & 277.3 \\
\text{Coordinate system} & \text{monoclinic} \\
\beta & 105.99^\circ \\
a & \pu{4.488(2) Å} \\
b & \pu{31.886(4) Å} \\
c & \pu{8.067(2) Å} \\
V & \pu{1149.5 Å^3} \\
D_m & \pu{1.59 mg mm^{-3}}\\
D_c & \pu{1.60 mg mm^{-3}} \\
\lambda & \pu{0.7107 Å} \\
\text{Space group} & P2_1/c \\
\text{Crystal dimensions} & 0.7\times 0.1 \times 0.7\, \pu{mm} \\
\hline
\end{array}

The problem is that Accelrys DS Visualizer so far renders my CIF file as:

Notice how some of the hydrogen atoms are disconnected from the main chunk of the azathioprine molecule.

Acharya, K. R. Proc. Indian Acad. Sci. (Chem. Sci.) 1984, 93 (2), 183–187 DOI: 10.1007/BF02842141.

The coordinates of the hydrogen atoms of the methyl group in question are there (H16a, H16b, H16c).

The problem is a rendering problem of the Accelrys visualizer.

Other tools don't have a problem with this CIF file. The image below is a medium-quality PovRay rendering generated through PLATON.


Isolating barium(II), copper(II) and zinc(II) from aqueous solution Q15 Each metallic ion was separated from an aqueous solution containing $\ce{Ba^2+},$ $\ce{Cu^2+},$ and $\ce{Zn^2+}$ by the procedure shown in the following figure. From ①–⑥ in the table below choose the correct combination of ions that predominates in the precipitates a, b and the filtrate c.

For the precipitate a, I know the answer is $\ce{Ba^2+}$ ion because it has a highly negative standard reduction potential than the other ions, so it tends to displace the sulfuric acid to form insoluble barium sulfate.
But I don't know which one can displace $\ce{NaOH}.$ $\ce{Na+}$ ion reduction potential is a lot more negative than both $\ce{Cu^2+}$ and $\ce{Zn^2+}$ ions. I suspect there is something in 'added in excess' part, but I don't know how can it be any different.

Once $\ce{Ba^2+}$ is eliminated as poorly soluble barium(II) sulfate, there is a solution of soluble copper(II) and zinc(II) sulfates.
Considering this has been done by careful/gradual addition of sulfuric acid, and the fact that both remaining sulfates undergo hydrolysis, the solution is going to be slightly acidic — this is one reason for sodium hydroxide "added in excess".

Another reason, as both Ivan and Waylander pointed out in the comments, is the difference between reactivity of copper(II) and zinc(II) in alkali medium.
Long story short, you are probably supposed to know the specific properties of the common transition metals cations as the reduction potential isn't really helpful here.

Copper(II)

Copper(II) with $\ce{NaOH}$ first forms colloidal copper(II) hydroxide; the gel is then solidifies to light-blue amorphous substance upon subsequent addition of hydroxide:

$$\ce{CuSO4(aq) + 2 NaOH(aq) -> Cu(OH)2(s) + Na2SO4(aq)}$$

$\ce{Cu(OH)2(s)}$ appears to be precipitate b.
Sodium tetrahydroxocuprate(II) $\ce{Na2[Cu(OH)4]}$ doesn't form here: it's a very moisture-sensitive and in the air quickly turns dark-brown:

$$\ce{Na2[Cu(OH)4](aq) -> CuO(s) + H2O(l) + 2 Na(OH)(aq)}$$

Zinc(II)

Zinc(II), on the other hand, with $\ce{NaOH}$ first forms white precipitate of zinc(II) hydroxide:

$$\ce{\ce{ZnSO4(aq) + 2 NaOH(aq) -> Zn(OH)2(s) + Na2SO4(aq)}}$$

which is then dissolved again forming colorless soluble sodium terahydroxozincate(II) (filtrate c):

$$\ce{Zn(OH)2(s) + 2 NaOH(aq) -> Na2[Zn(OH)4](aq)}$$

Excess of $\ce{NaOH}$ is also necessary to keep $\ce{Na2[Zn(OH)4]}$ in solution: upon diluting (lowering pH, adding water/acid), terahydroxozincate(II) decomposes to zinc(II) hydroxide.

Note: chemical reactions are adapted from [1].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Acidic nature of boric acid when its concentration high Following statement was given for boric acid $(\ce{H3BO3}):$

  At low concentrations $(\leq\pu{0.02 M})$ essentially $\ce{B(OH)3}$ and $\ce{B(OH)4-}$ are present, but at higher concentration the acidity increases and pH studies are consistent with the formation of polymeric species such as $$\ce{B(OH)3 <=> H+ + [B3O3(OH)4]- + H2O}$$

Please explain the meaning of the italicised statement i.e. pH studies are consistent with the formation of polymeric species?

In the equation $\ce{H2O + H3BO3 -> H+ + B(OH)4-}$, one negative charge on the anion is shared among 4 electronegative oxygens. Yes, one positive charge on the proton is also shared with some water molecules, so the orthoborate ion probably shares some negative charge with some waters, but let's keep it simple.

Boric acid is known to complex with sugars to generate higher acidity. If boric acid complexes with itself, as in $$\ce{3 H3BO3 -> H+ + [B3O3(OH)4]- + 2 H2O}$$, the one negative charge on the anion can be considered to be distributed among 7 electronegative oxygens, or maybe still distributed mostly on the OH groups, but since they will be separated more, the stability of the ion will be greater, and the tendency to lose a proton will be greater: greater acidity. The electronegativity of boron is less than that of hydrogen (2.02 vs 2.2, electronegativity, Wikipedia), so when the atoms in the ions are counted, the first ionization equation has one hydrogen and one boron as the metallic atoms getting the positive charge, while the second equation has one hydrogen and 3 borons donating electrons - more sharing of the positive charge here, tho the hydrogen still gets the positive charge.

But this structure may be only a suggestion. Well-known polyborate anions include the triborate(1-), tetraborate(2-) and pentaborate(1-) anions. The condensation reaction for the formation of tetraborate(2-) is as follows Wikipedia

$$\ce{2 B(OH)3 + 2 [B(OH)4]− ⇌ [B4O5(OH)4]2- + 5 H2O}$$

The italicized text mentions an observation of pH that is consistent with the complexation of boric acid with itself. It doesn't prove the existence of the triborate ion only, and perhaps is only a way to keep the text shorter without ignoring complexation completely. Then along comes a student who looks into the sentence and sees that something was omitted or not explained thoroughly. (Keep asking questions!)


How Does The Mordant, Ferric Acetate, Interact With Tannins When Ebonizing Wood? Ferric Acetate (II) and (III) are "mordants," which intensify wood staining in direct proportion to the tannin concentration in the wood. Some wood stainers add tea to increase the tannin content and thereby make the stain darker. How does Ferric Acetate act as a mordant in this case of ebonizing (web archive) wood?

A mordant is a substance used to set dyes on fabrics or tissue
  sections by forming a coordination complex with the dye which then
  attaches to the fabric or tissue (source).

Tannin is a biomolecule consisting of many carbohydrate and polyphenol residues with plenty of hydroxyl and carboxyl groups. Tannic acid is a common type of tannin:

$\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ form colored complexes with these oxygen-containing functional groups. These iron-tannin complexes change the color of the wood. The color intensity increases with tannin content of the wood and can be increased by the addition of tannin-rich tea. Structures with a varying coordination sphere of iron are formed, some of which are shown here (sources: 1 and 2):


If the same atoms (cells) that now compose my body get arranged in N years away, am I that person? Probably a very stupid question, even may deserve to be downvoted, but I have always wondered: Hypothetically, let's say somehow tomorrow I die in an accident. Now lets say that after a long time (e.g. centillion of years) for one reason or another the cells that made up my body -right before I died- are part of a new body arranged exactly the same doing the exact same functions. My questions are:

1) It is possible? -Possible in the sense that atoms that are cremated tomorrow or ate by worms, can they go back to their previous chemical state?-

2) Am I the same person? Not philosophically, I mean are there in my brain the same synapses? or are other variables to take into account? (I guess this one is more about neurochemistry than chemistry)

Let's not discuss the likelihood of happening, at least not much, because time is infinite one can argue that it may be bound to happen. But time being infinite or not and related questions may be more of an cosmology issue.

Let me start from the bottom up: If we assume the earth as a closed system, i.e. exchange of energy with the surrounding space is possible, but exchange of matter impossible, than naturally our body contains atoms previously present in other form of matter.  Already our breakfast is matter we incorporate, subject to chemical reaction (digestion), and eventually excrete in one form, or an other.  Hence it is likely that we contain some atoms that once constituted a mammoth, or a dinosaur, or by your tomato in the back yard.

Digestion often splits the larger molecules present in our food into smaller ones.  This eases transport of matter across membranes, for example.  Think of proteins de-constructed into amino acids; our organism uses the smaller to build new proteins in need.

Beside chemical degradation as in aforementioned example of the proteins, digestion includes physical degradation, too -- chewing food, for example increases the surface area the chemicals responsible for digestion may access.  By  both means of action, physical and chemical degradation, entire cells face only a tiny chance to remain intact "as such".  It is more likely that they are dismantled, are split into their constituents, eventually -- as molecules -- harvested in the ways briefly mentioned earlier.

To sum up: so far there is no evidence a synapse of yours will be recycled, maintaining the original form and function, to find its way back as synapse in an other brain, similar to a spare tire.  Atoms of yours however, as easily they entered your body, eventually will leave you; at their pace.


How does alum improve the settling of particles to produce 'clean' water? How does the addition of alum — potassium aluminium sulfate or $\ce{KAl(SO4)2}$ — help settle the dust particles in water?

There was such a statement in my Year 9 textbook, and so I was wondering how exactly alum does it.

Edit: I have posted an answer below based on what I've found, but I'd love it if some else could verify and fill in the missing gaps in it.

OK, so after a bit of researching this is what I've found.

First of all,why does the dust not settle in the water in the first place? Well, dust forms a kind of a suspension.

The particles are all similar and ionized, so therefore they have a similar charge (positive, I think) causing them  repel each other and stay separate.
And because they're so small, the gravitational force acting on them isn't very strong either.
Also, the dust particles, previously dispersed in the air, have a lot of kinetic energy already.

All these 3 reasons coupled cause the particles to remain suspended for so long.

So what does adding alum do? Alum has a negative charge and tends to disperse in water very fast and very well (Why so? Because it's a colloid?).

This causes it to join up with all of the offending particles and neutralize them. Now that the particles don't have any repelling charges, they tend to clump together into 'flocs'.

The increased size as well as the lack of repelling charges cause the alum particles to settle down at the bottom or rise up and float in the water.

Why do the particles clump together and settle? Apparently even this isn't as simple as it seems.

After the particles are neutralized, they clump together because of something called the London Dispersion Force which are part of the Van der Waals forces. Basically, weak intermolecular force arising from quantum-induced instantaneous polarization multipoles in molecules causes even nonpolar particles to attract each other due to the correlated movements of the electrons in interacting molecules. More info can be found inside.

As to why they then settle, I think it can be sufficiently explained by gravity.

Oh well, that's all. All for explaining something even the ancients knew.


What is the significance of enthalpy? I an getting confused between enthalpy and internal energy. My textbook says that change in enthalpy is the sum of change in internal energy and the amount of work done in expansion or compression but isn't it the same as the amount of heat lost or given. So why do we need enthalpy?

The internal energy $U$ is the sum of all the energies stored in the chemical bonds. If a chemical reaction happens, the bonds are losing or getting energy. Heat is getting in or out of the system, and this heat $\Delta Q$ can be measured. $\Delta U$ =$\Delta Q$, if the volume is constant.  But if the transformation is carried out in contact with the atmosphere, at constant pressure, the volume may change. A fraction of this energy is converted into work $\Delta w$ to repel the atmosphere, because the volume of the system may change. Usually this volume change is small (except if a gas is produced or consumed). And this volume change may be difficult to know with precision. But it must be taken into account to calculate $\Delta U$. So if this small amount of work is neglected, a correction must be added to the internal energy $U$, that depends on pressure and volume. Neglecting this volume correction gives a sort of “apparent internal energy”, which is called enthalpy $H$, with $H = U + PV$. And then, in all transformations made at ordinary pressure, the following expression is valid : $\Delta H$ = $\Delta Q_p $.
Enthalpy is a sort of "apparent internal energy". There is a similar difference between the weight and the "apparent weight", which is measured in air. The weight of an object should be measured in a vacuum. The weight, as measured with a dynamometer in the usual atmosphere, is a little bit smaller than in a vacuum, because of Archimedes.


Can cis and trans isomers have widely different flash points? Sigma seems to have two products listed for sale as "Cinnamic Aldehyde". One compound clearly seems the trans isomer but the other one seems like an unspecified mixture. (They even have two different CAS nos.: 14371-10-9 vs 104-55-2 )

What surprises me is that the corresponding SDS lists the Flash Point of one as 71°C and the other as 125°C.

Is this really possible or is this just an error? The boiling points listed seem fairly close: 250°C vs 248°C.

P.S. Even if one accounts for methodological differences in Flash Point Determination e.g. closed cup vs open cup those are typically at worst 10°C apart. Here the flash points seem a good 55°C apart.

What gives?

It is not totally surprising giving that they are stereoisomers. Think about maleic and fumaric acid, which due to their very different properties were given different names.
Moreover, but perhaps you are aware of, there is not in general a relationship between boiling point and flash point. In the first case factors which depends on the actual configuration - as molecular symmetry, polarity - should be considered for how they affect intermolecular forces. For the flash point, their influence to the intrinsic stability of the molecule and to the overall shape and accessibility for oxygen, should be considered instead.

For the less experienced ones: in few words volatility is required to reach the concentration (in air) at which the vapour can be ignite at a given temperature (flash point). More a liquid is volatile, easily that concentration is reached. But this does not imply a lower flash point, which depends on the thermodynamics and the kinetics of the reaction between the vapour molecules and oxygen in air.


Can the RDS be different in different direction in a multi-step reversible reaction? The following mechanism is given in Peter Sykes when they are "talking about de-esterification" of an ester where the alcohol part is bulky-

Here the second step is given as RDS. According to me, in the reversed reaction(esterification), this step will remain RDS and so the reaction is bimolecular. Someone argues that RDS will be the formation of carbocation then and so the reaction is unimolecular.

After discussion with OP in chat, the issue in logic can be rectified as follows. Simply put:

$\ce{A<=>B<=>[slow]C <=> D}$ doesn't imply that $\ce{B <=> C}$ is the RDS in both $\ce{A -> D}$ and $\ce{D -> A}$.

Rather than this, we need to consider $\ce{A->D}$ and $\ce{D-> A}$ as two separate reactions. Doing so for the two reactions given in the question (acid catalysed hydrolysis ($\ce{A ->D}$) and  esterification ($\ce{D -> A}$)), we see the mechanisms are as follows:
Acid catalyzed hydrolysis

Here the RDS is the formation of the carbocation $\ce{^+CMe3}$.
Esterification of an acid and a bulky alcohol

Here the RDS is the formation of carbocation but it happens at a different step and not the same step.
Therefore the two reactions have different RDS steps as they are different reactions


What is the correct name for 2,4-diethyl-4-ethoxyhexane? A classmate asks you to draw the structure of 2,4-diethyl-4-ethoxyhexane, which he cannot find in a chemical reference manual. Draw the structure of 2,4-diethyl-4-ethoxyhexane.

  (Question continued...) The reason your classmate cannot find the compound in a chemical reference manual is that 2,4-diethyl-4-ethoxyhexane is the incorrect name for the compound. What is the correct name?

Here are my steps:

I count the longest carbon chain = 7 carbons = -heptane
I look at all the substituents: 1 ethoxy group, 1 ethyl group and 1 methyl group
I number the carbon chain starting from the end closer to the ethoxy group (alphabetical order)
I add all the substituents and the carbon chain together

In this case, this is the name I come up with: 3-ethoxy-3-ethyl-5-methylheptane, which is incorrect. Can somebody tell me where I went wrong, please?

In short, your answer is correct. The steps that you have taken are correct, so it's likely the online homework tool has a key mistake.


Is it possible for the kinetic energy integral to be negative? Is it possible for the kinetic-energy integral, Tij, to be negative? I was messing around with some HF code and found that the integral became negative on some off-diagonal terms. (This could also be a bug with the code.)

If the kinetic energy integral is negative, what would the physical meaning be?

Generally, the off-diagonal elements of matrices, especially in physical systems, are interpreted as the coupling between whatever the $i$ and $j$ elements correspond to. So, in this case, a negative element of $T_{ij}$ corresponds to a negative kinetic coupling between atomic orbital basis functions. In plain language, this means that basis functions $i$ and $j$ tend to mutually lower the kinetic energy of an electron placed in one of those orbitals. I am not sure it really makes sense to give much of an interpretation to this, however, because what one is really interested in is diagonalizing the Fock matrix, of which the kinetic energy is only one part.

Also, it is always possible to choose a basis where the kinetic energy matrix is diagonal, but again, this would not get you anywhere as in solving the HF problem, you will diagonalize the Fock Matrix which will surely put you in a basis in which the kinetic energy matrix is non-diagonal. What might be more interesting would be to take the actual molecular orbitals which are part of the solutions of the Roothan-Hall equations and re-compute the kinetic energy matrix over these orbitals. Looking at the coupling between these orbitals may be more easily interpretable for e.g. aromatic $\pi$-systems.


Is it safe to store muriatic acid, hydrogen peroxide and acetone? I am recently getting into PCB etching. So I use 10 M HCL solution, 3% hydrogen peroxide mixed in 1:1 ratio for my etchent. Currently, I am storing my HCL and hydrogen peroxide in my shed which can vary in temperature from -10 degrees Celsius to 25 degrees Celsius. The shed is outside my house. I also store my etchent next to HCL and hydrogen peroxide in the shed as well. Will it be safe to store all of these three chemicals inside a shed?

I do have metal tools in there but I don't care if it rusts. I am just afraid of hydrogen gas that might form because of HCL fumes. Also, I store my acetone indoor inside a plastic container because I heard that its flash point is -20 degrees Celsius. I don't want it to spontaneously catch on fire if I did store it in the shed. So are these storing method adequate or do I need more precautions.

You don't need to take any special precautions when storing these chemicals together compared to how you would store them individually. They should all preferably be stored in their original, properly labeled containers with properly fitting lids.

Regarding your concern for hydrogen gas forming from the HCl fumes, that is simply not a reaction that will take place from only the compounds you've listed, or in combination with anything else you would typically find in a utility or storage shed. You will want to avoid breathing HCl fumes while working with it of course, but as far as a storage concern even this will not be an issue for you.

I commend your extreme caution here in general, although your fears of the acetone igniting from sitting in a hot shed are unwarranted. The flash point of a compound is the lowest temperature at which it can be ignited with an ignition source, like a spark. This is commonly confused with the autoignition temperature, which is the lowest temperature at which a compound can spontaneously combust without an ignition source. So, while you really don't want your acetone near a flame or spark source at any temperature, it's autoignition temperature is $\pu{465^oC}$ and so you don't have to worry about it igniting just from sitting in a hot shed.