How is bonding in the trihydrogen cation and triatomic hydrogen possible? According to wikipedia $\ce{H^+3}$ and $\ce{H3}$ are possible and have been synthesized. It says that $\ce{H^+3}$ is stable too. Can anyone please tell me how is the bonding possible?

Hydrogen has a valency of 1 but it still bonds like Ozone. How is this possible?

Links:

$\ce{H^+3}$ - Trihydrogen Cation
$\ce{H3}$ - Triatomic Hydrogen

Here are the vital parts from the links about $\ce{H3}$ in the question:

  The molecule can only exist in an excited state [...] the electronic state for a trihydrogen cation with an electron delocalized around it is a Rydberg state.

and from second-level link

  Rydberg states of the neutral triatomic hydrogen molecule can be viewed as
  an electron attached to a tightly bound $\ce{H3+}$ core

Basically, at least some states can be viewed as an $\ce{H3+}$ ion working as a nucleus in a bigger hydrogen-like atom. If said 'nucleus' catches the electron, the molecule immediately dissociates. So, let's focus for a bit on $\ce{H3+}$. In this ion a two-electron three-center bond is formed. Such bonding is actually pretty common, even if (almost never) mentioned in school. As long as orbitals of several atoms can overlap effectively and have fitting symmetry, they can form a single highly symmetrical bonding orbital, that can accept two electrons. In this case the bonding order between atoms is fractional, so in $D_\mathrm{3h}$ (triangular) $\ce{H3+}$, each of the three $\ce{H-H}$ bonds has bond order $1/3$. This kind of bonding plays a critical role in the chemistry of boranes (and Group 3 elements in general, maybe except thallium) and some families of compounds of d-block elements.

To obtain more intimate and correct understanding you should invest in molecular orbital theory. Your link for $\ce{H3+}$ contains a comprehensive description of bonding in terms of said theory.


Nitration of aniline (b) Nitration: Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.

Why there is more para-substituted nitroaniline formed than ortho-substituted one even though it can form hydrogen bonding?

In case of nitration in aniline, the Nitric Acid protonates the aniline to form the Anilinium ion.

Now since the nitrogen atom has no lone pair to conjugate, it has no mesomeric effect on the ring, the but as nitrogen is now protonated, it has a high negative inductive effect.

Which simply leads to the fact that the position farthest from the nitrogen substituted carbon will have the highest electron density (as inductive effect decreases as we move away from the atom causing it).

So in this case the best site for attack for the electrophile was para>meta>ortho.

The Hydrogen bonds would stabilise then ortho product once its formed, but the Rate Determining step of an electrophilic substitution on benzene is the sigma complex intermediate wherein the most stable intermediate is formed as major product. The positive charge in sigma complex stays the farthest in para position from the anilinium ion hence(most stable) is the major product.


Regioselectivity of an amination reaction of 2,6-dichloro-3-nitropyridine In the formation of ethyl 4-(6-chloro-3-nitropyridin-2-yl)piperazine-1-carboxylate G, why is the substitution favoured at the 2-position? I would have said that the 6-position substitution is preferred as:

In pyridines, nucleophilic substitution occurs at the 2/6th or 4th position, and as chlorine is an effective leaving group compared to hydride this rules out position 4, leaving 2 and 6 as the most feasible...
The electron withdrawing nature of the nitro group on the ring would favour nucleophilic substitution at the positions ortho or para to itself.
The chlorine at the 2 -position is ortho, and the chlorine at the 6-position is para so both are feasible...
If I didn't know what the product was, I would have said that position 6 is therefore the site of substitution as the amine reagent is quite bulky and the 6-position is less sterically hindered due to the absence of an adjacent nitro group

However, this evidently isn't the case! I found another similar reaction where the same thing occurs:

Is it something to do with the nitrogen in the pyridine ring hindering the electron density transfer to and from the para nitro group? I'm really stuck!

I'd agree that nucleophilic substitution at C-6, leaving the nitro and piperazine substituents para to each other, would lead to the thermodynamic product.

The observed regioselectivity is therefore probably kinetically controlled. Although the nitro group activates both ortho (C-2) and para (C-6) positions towards nucleophilic substitution via resonance, it's worth bearing in mind that the nitro group is also extremely electron-withdrawing via the inductive effect in its own right.

This would make C-2 more electron-deficient and hence prone to nucleophilic attack.

Recall that in an SNAr reaction the nucleophile must attack the π* orbitals of the arene, which are above and below the plane of the ring. In this case, the nitro substituent is also (for the most part) in the plane of the ring, and doesn't lie along the trajectory of nucleophilic attack. So, the nitro substituent doesn't present all that much steric hindrance to the nucleophile.


Tabulated absorption spectra of greenhouse gases? Are there any free sources of tabulated absorption spectra of greenhouse gases from UV/Vis to far infrared (say 70 micrometers)?

This link shows the net absorption over the whole atmosphere (at least as far as I understand it), but I would be interested in the absorption by an infinitesimally small volume/distance. It would suffice to have the spectra for just one temperature, like $\pu{273K}$ or so.

On the NIST site I have found some information, for example IR spectrum of $\ce{H2O}$ vapor, but apart from the limited spectral window the spectrum is pretty useless for making a radiation simulation because concentration info was missing (as said in the disclaimer).

Is there still hope of getting accurate spectra suitable for understanding molecule specific greenhouse effects?

Contributing to the first half of the answer: The MPI-Mainz UV/VIS Spectral Atlas of Gaseous Molecules of Atmospheric Interest (public access here) allows you to search by chemical name, sum formula, CAS registry number, InChi key or author.

The output is a plot, annotated with the literature reference, e.g.

(reference)

It does not cover IR data to cover the second half of your question.  However, why not asking them if they in turn would know about a complementary database and letting the users of ChemSE know about their answer?

(Otherwise, if affilated with a research institution with a department of chemistry, knowing a chemical identifier like the CAS RN could help you to retrieve spectral data in Elsevier's Reaxys [rooted in Beilstein and Gmelin database], too.)


Change of first order rate constant with temperature In the decomposition of $\ce{N2O5}$, the time taken to reduce the concentration of $\ce{N2O5}$ half of the original amount is 12 min at $\pu{50^\circ C}$, while 5 h at $\pu{25^\circ C}$ and 10 days at $\pu{0^\circ C}$.

How can we understand that the given assertion is correct because the temperature is different every time?

First of all, decomposition of $\ce{N2O5}$ is a first order reaction, not a zero order reaction. So, your question headline needs to be corrected.

You can justify the given assertion by simple mathematics. For a first order reaction, we know that, $$t_{1/2} = \frac{\ln \space 2}{k}$$ which is independent of initial concentration. It only depends on the rate constant which will change if you change the temperature.

From Arrhenius equation also, we know the relation,
$$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$

So here we have three rate constants for three different temperature. Thus we will have at least two sets of above equations. Dividing one by another we will have,
$$\frac{ln\left(\frac{k_2}{k_1}\right)}{ln\left(\frac{k_3}{k_2}\right)} = \frac{\left(\frac{1}{T_1} -\frac{1}{T_2}\right)}{\left(\frac{1}{T_2} -\frac{1}{T_3}\right)}$$

Now, from the relation between half life and rate constants we can have

$$\frac{\ln\left(\frac{\left(t_{1/2}\right)_1}{\left(t_{1/2}\right)_2}\right)}{\ln\left(\frac{\left(t_{1/2}\right)_2}{\left(t_{1/2}\right)_3}\right)} = \frac{\left(\frac{1}{T_1} -\frac{1}{T_2}\right)}{\left(\frac{1}{T_2} -\frac{1}{T_3}\right)}$$

Now from the given assertion by putting the values you have to just verify whether both the sides are giving the same value or not. In this case both sides, gives the value ≈ 1.2. So, I guess this assertion is thus correct.


Calculating molar mass from gas data 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

I tried to solve the above question and got 1249.8g. The book's answer sheet lists the answer to be 1247.7 g.

I don't understand how the heck that happened? I'm pretty sure a mole of phosphorous doesn't weigh more than 1.2 kg.

To start your book didn't use significant figures properly. There is no way that the weight can have 3 significant figures in the mass and then conjure 6 significant figures in the molecular weight. To make matters worse the pressure is only given to 1 significant figure. YUCK!

Use PV = nRT to determine the moles of gas present.

$\mathrm{n =}\dfrac{\mathrm{PV}}{\mathrm{RT}} = \dfrac{0.1 \text{ bar} \times 0.03405 \text{ L}}{\text{0.08314 L bar K}^{−1} \text{mol}^{−1} \times 819\text{ K}} = 5.000 \times 10^{-5}\text{ mol}$

$\text{MW} = \dfrac{\text{m}}{\text{mol}} = \dfrac{ 0.0625\text{ g}}{5.000 \times 10^{-5}\text{ mol}} = 1,249.8\text{ g/mol}$

This is of course wrong. It would seem that the pressure should have been 1.00 bar. That would have given an answer of 125 g/mol. Using 31.0 g/mol as the atomic weight of phosphorous would reveal that the phosphorous vapor was $\ce{P_4}$.


Calculating the mass of acceptable carbon monoxide in a room The acceptable concentration of $\ce{CO}$ in the air is $10\:\mathrm{mg/m}^3$. In
  a room that is $19\:\mathrm{m\times4.0\:m\times25\:m}$, what is the acceptable mass in
  kilograms of $\ce{CO}$?

The room has a volume of $1900\:\mathrm{m}^3$. Since there are $10\:\mathrm{mg}$ of $\ce{CO}$ per $\mathrm{m}^3$, there are $19000\:\mathrm{mg}$ of $\ce{CO}$ in this room. That’s $19\:\mathrm{kg}$.

But the book says the answer is $1.9 \times 10^{-2}\:\mathrm{kg}$. How come?

You have to be careful with your conversions - you are missing a step in your conversions.

You have 19000 mg

As 1 mg = 10-3 g  and
1 g = 10-3 kg

So:

19000 mg = 19 g = 1.9 x 10-2 kg


What chemical can cause such a fire? A friend of mine who thinks that she is under influence of black magic met an astrologer. The astrologer asked her to wear a t-shirt for three nights and then bring that to him. The astrologer then cut some lemons and rubbed them throughout the t-shirt. Then he poured some water on her hands and face; and asked her to keep the t-shirt on her palm. After about 15 minutes, the y-shirt caught fire and the astrologer dropped the rest of water to put the fire down.

I'm now wondering what chemical could cause this fire. I know that the water looked muddy. Would you please help me to find out what material was potentially used by the astrologer? I'm suspicious of the acid in the lemons and water which has been used, as well as body fat or sweat perhaps was on the t-shirt too. She was telling me that at first, smoke color was white but then it turned to black after the t-shirt caught fire. The smell of the fire was as bad as rotten meat.

White phosphorus can spontaneously ignite in air, and produces white smoke.

Some oils, like linseed oil, on bunched-up cloth can also spontaneously combust.


Why is the ionic radius of Al(3+) smaller than that of Li+? I was examining the ionic radii of some ions from this site for a school assignment. I noticed a weird anomaly in the ionic radius of $\ce{Li+}$  as compared to that of $\ce{Al}^{3+}$.

The ionic radius of $\ce{Li+}$ is about 2/3 as large of that of $\ce{Al}^{3+}$. How is this possible, since the electronic configuration of $\ce{Al}^{3+}$ is $[1s^2, 2s^2 2p^6]$ whereas that of $\ce{Li+}$ is $[1s^2]$? It appears that despite the presence of an extra shell in $\ce{Al}^{3+}$, it has a smaller ionic radius than $\ce{Li+}$.

The answers in this post suggests an exactly opposite outcome. All the answers says that $\ce{Mg^{2+}}$ is above $\ce{Ca^{2+}}$, thus it has a smaller radius. By applying  the exactly same logic i can say $\ce{Li+}$ is above $\ce{Al^{3+}}$ so it must have less ionic radius but that is not the case.

As the below picture suggests,(taken from the same post).

I think it is because of diagonal relationship. As $\ce{Mg^{2+}} = \ce{Li+}$ because of diagonal relationship and $\ce{Mg^{2+}} \gt \ce{Al^{3+}}$ because ionic radius decreases for cations along the period by going left to right.

Am I correct?

The size of the 1s orbital in $\ce{Li+}$ and $\ce{Al^3+}$ is not necessarily the same. In fact, they are quite different because of the much larger effective nuclear charge in $\ce{Al^3+}$. One can easily look up the wavefunction for the 1s orbital and see the radial dependence on $Z_\mathrm{eff}$.

Therefore, merely looking at the electronic configuration cannot tell you anything about the size. If that was the case then one would expect $\ce{Cl-}$ and $\ce{K+}$ to have exactly the same size, which is clearly not true.

  All the answers says that $\ce{Mg^{2+}}$ is above $\ce{Ca^{2+}}$, thus it has a smaller radius. By applying  the exactly same logic i can say $\ce{Li+}$ is above $\ce{Al^{3+}}$ so it must have less ionic radius but that is not the case

You are missing the point here entirely. Mg and Ca are in the same group which is what makes them directly comparable. Li is not "above" Al any more than F is above Na.

  I think it is because of diagonal relationship. As $\ce{Mg^{2+}} = \ce{Li+}$ because of diagonal relationship and $\ce{Mg^{2+}} \gt \ce{Al^{3+}}$ because ionic radius decreases for cations along the period by going left to right.

The diagonal relationship is something that is observed, not a fundamental principle of chemistry. It still has to be rationalised. Saying that it is "because of the diagonal relationship" is akin to saying "electrons move very fast because they have a high velocity".

The diagonal relationship holds because 1) you add an extra shell of electrons going vertically from $\ce{Li+}$ to $\ce{Na+}$, which decreases $Z_\mathrm{eff}$ and increases ionic radius; and 2) you go horizontally from $\ce{Na+}$ to $\ce{Mg^2+}$, which increases $Z_\mathrm{eff}$ (same number of electrons but one more proton) and decreases ionic radius.

Going one more step horizontally to $\ce{Al^3+}$, one can see that $Z_\mathrm{eff}$ should increase again. So your thoughts on the matter are correct, but they just do not get to the crux of the matter.

The origin of the much smaller ionic radius of $\ce{Al^3+}$ is therefore directly attributable to the much larger $Z_\mathrm{eff}$ in $\ce{Al^3+}$ on the 2p valence electrons, compared to that on the 1s valence electrons in $\ce{Li+}$.

Be careful that the ionic radius is not always particularly well-defined and the values can vary from one source to another depending on how they are experimentally obtained.


What does the pressure in the HPLC(High performance liquid chromatography) do? Well I have to do research about the HPLC. It says, that because of the pressure the stationary phase can be made out of smaller particles and the pillars can be shorter. Also the seperation of the contents is supoosed to be superior, compared to other kinds of liquid chromatographies. Can somebody explain that to me?

The pressure in HPLC systems is not an intended feature, but a consequence of how the system works.

There are several adjustable parameters/properties of a HPLC/UPLC system that will influence on the internal pressure inside the chromatographic column. Some of those are:

Solvent flow:  A greater flow of solvents mean a faster chromatophy. On the other hand, faster runs also mean lower efficiency. An increase in flow leads to an increase in pressure.
Column length: A bigger column will improve the separation efficiency, but also increase the duration of the run. Bigger columns leads to a greater pressure.
Particle diameter: The lower the particle, the greater the separation efficiency, but this also leads to a greater pressure.
Particle pore size: This has the same effect as above, but is not the same thing.

Optmizing this parameters and/or improving the technology related to them is what generally drives the research in the HPLC/UPLC development field, besides of course find the appropriate type of column, solvent gradient, etc. You can't simply make a system where the pressure is as high as possible, or some connectiong will eventually burst. On the other hand, it's not interesting to have a chromatographic run that lasts several hours, or doesn't really separate the molecules, specially when dealing with hundreds or thousands of different analytes.

There is much more to the above in several books and thousands of papers, including very detailed descriptions, experiments and calculations that show a much more precise relation between those and other parameters.


discover of the neutron, what happen to 2+ charge James Chadwik found the neutron by shoot α rays to a Beralium foil. α rays are equal to the He²⁺ and he saw that particles without a charge is out from that. He understood the neutrons are those particles without the charge.

But my question is this.

This is the equation that the chadwik found.

Be + He²⁺= C +neutron

In this equation the left side has a 2+charge, but in right there is no any charge. So, What happen to the 2+ charge when this reaction occurs?

Corrected: "James Chadwik identified the neutron by exposing a Beryllium foil to  α-rays (He+2) and showing that the previously discovered emitted collision product were massive neutrally charged particles." The experimental set-up he used (see:http://rspa.royalsocietypublishing.org/content/royprsa/136/830/692.full.pdf) did not determine where any electrons came from, so while electrical charge can be assumed to balance, he couldn't claim any direct knowledge of the source or fate of any electrons in his experiment. For us, today, balancing a nuclear reaction's charge distribution on both sides of the arrow is simply "bookkeeping", it implies facts that were not as well supported in 1932 as they are today. He avoided the temptation to over-interpret his experimental data.


Correct way to draw a carbene What is the correct way to draw a carbene? On most/all material I have seen, both singlet and triplet carbenes are depicted as so:

With 2 "dots" representing the non-bonding electrons on either a singlet or a triplet carbene.

However in a class today my tutor was adamant that a singlet carbene should be drawn with a positive and a negative sign on the carbon. His argument as such was that "you have an empty and a full orbital and so the empty orbital corresponds to the positive sign, the full orbital corresponds to the negative sign".

So my question is - How is a carbene correctly drawn? As above or with a carbon bearing both a positive and a negative sign?

Put simply, there's no precedent for this unusual depiction of a singlet carbene.

Your comment about most carbene depictions, for both singlet and triplet carbenes is true, but a bit lazy. Yes, many chemists, frequently just draw two electrons as a lone pair on the carbon for both. In principle, with a moment's thought, you can often deduce the likely spin state.

As described, by the comment above, the standard for 2D chemical structure depiction is the IUPAC guide Graphical representation standards for chemical structure diagrams

In short, there's a difference in representing a triplet (two unpaired electrons), which should show preferably two separate electron "dots."

For a singlet, the lone pair is indicated, emphasizing the connection between the electron spins.


Does solid aluminum oxide conduct oxide ions at high temperatures? Can alumina behave as a solid electrolyte for oxygen ions? Could a cell be constructed with an oxygen cathode and a molten aluminum anode producing more alumina? Would the resulting alumina deposit on the electrolyte? Would it be crystalline?

Alumina is one of the most abundant compounds on Earth. If it would be also a solid electrolyte for oxygen ions, why in the world would anyone go into the trouble of dealing with the whole $\ce{Y2O3/ZrO2}$ thing? No, alumina doesn't and can't behave like that. Its oxygen ions are held in place pretty tightly.

Some forms of alumina may serve as conductors of $\ce{Na+}$ and similar cations, but that's another story.


Thermal decomposition of ester I have a clear liquid with a faint sweet smell. I know this compound contains an aromatic group and an ester. However, something strange happens when I try to distill it - it forms a white solid. I believe this solid to be the result of thermal decomposition of the ester - the ester reacting with oxygen in the air. I have searched around, but can't find much written about thermal decompositions of esters. What reaction is this? What is the product? Could it be a carboxylic acid?

Well, you are right. Esters decomposes at higher temperatures to form several small carboxylic acids and ketones and at even higher temperature, these carboxylic acids decomposes into carbon monoxide, carbon dioxide, carbon, oxygen and/or water. A paper[1] discusses about the thermal decomposition of ethyl formate, ethyl acetate and ethyl propanate at 1248-1638 K, pressure around 1.5 atm at a chamber filled with argon at a concentration of 2000 ppm.

After the experiment, it was found out that 90% of oxygen has been released as oxygen gas and the pyrolysis mechanism was determined to be a six-centered decomposition producing equimolar ethylene and corresponding acid and at drastic conditions, it was known to have a dehydration reaction releasing water and a decarboxylation reaction to form carbon dioxide.

It was found out that EF gave formic acid, EA gave acetic acid and EP gave methyl ketene as the main products which subsequently decomposed into various smaller products at higher temperatures and the concentration ratios were noted.

Carboxylic acid can form white solid as noted in case of formic acid and acetic acid.

References

https://sutherland.che.utah.edu//USCI2013/PAPERS/1A07-070RK-0061.pdf


Which is the most acidic proton of (3Z)-3-ethylidenecyclopent-1-ene-1-carbaldehyde? At first glance, I thought the most acidic protons would be the ones in red due to their proximity to the carbonyl. However, the answer key has the hydrogens in green as the most acidic. Can someone explain how that is? (I'm guessing resonance, but I need a more concrete explanation.)

If you draw curly arrows for the deprotonations, it becomes clear that only the pathway in green is productive due to the ability to form an extended enolate.

If you deprotonated the green protons, the resulting negative charge is able to be delocalised through the alkene and onto the carbonyl, the product of which is essentially an extended enolate. Due to this stabilisation, the pka of the green protons is relatively low (the resulting conjugate base is stabilised).

Deprotonation of the red proton however doesnt yield a charge which may be delocalised onto the carbonyl ( you would end up with a 5 valent carbon if you draw the arrows ). What you essentially end up with in this case then is an allyl anion (which itself is conjugated to the exocyclic alkene), which although stabilised by resonance isn't quite as favourable as in the green case (the pka of the red protons is higher).


How would one calculate the entropy change for an adiabatic irreversible process? If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic process because:

$\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$

And $W_{\text{reversible}}\neq W_{\text{irreversible}}$

Now, Atkins' Physical Chemistry (Section $3.2$) states that:

  To calculate the difference in entropy between any two states of a system, we
  find a reversible path between them, and integrate the energy supplied as heat at each
  stage of the path divided by the temperature at which heating occurs.

But the problem is that we cannot define a reversible path for an adiabatic irreversible process. If that's the case, how can we calculate the entropy change for an adiabatic irreversible process?

I agree with you over here.

  $\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$

  And $W_{\text{reversible}}\neq W_{\text{irreversible}}$

But this part seems to be wrong.

  If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic

Just because the process is adiabatic irreversible it doesn't mean that the reversible path needs to be adiabatic as well.

The only requisite for calculating the entropy for an irreversible process is simply that the path taken must be reversible.

So until now we have established that the heat transfer does not have to be 0. For an adiabatic process.

Even though $W_{\text{irrev}} \neq W_{\text{rev}}$  the extra parts will be balanced by the heat transfer in the reversible process.

For example let's take adiabatic expansion.

$$W_{\text{irrev}} > W_{\text{rev}}$$

Numerically speaking in an absolutely hypothetical situation.

\begin{align}
\ce{ ∆U &-> 10~J}\\
\ce{ W_{irrev} &-> 10 ~J}\\
\ce{ W_{rev} &-> 5~J}\\
\end{align}

So the possible proposed reversible path will automatically have a transfer of heat into the gas . Whose value will be 5J.
 .


Using group theory to make molecular orbital diagrams Because of Orthocresols' meta post I ask this question about application of group theory to molecular orbital theory.

In my undergraduate lecturers, we've seen many examples of making qualitative MO Diagrams by inspection of what orbitals could line up to form bonding / anti bonding orbitals.

This works fine for simple things, but I know it's possible to use symmetry to also work out mathematically which orbitals can mix with which other orbitals by using group theory.

Perhaps using example of water could somebody show how this could be achieved?

Group theory really just formalises the process that you're going through when you construct molecular orbital diagrams by inspection (though as the molecules get bigger it gets significantly harder to intuitively guess at what will mix and in what combinations).

In order to construct an MO diagram for water, we'll take a stepwise approach:

first, determine the way in which the hydrogen atoms can combine (in phase or out of phase, as per the standard dihydrogen MO diagram) which will give us a set of symmetry adapted combinations
second, we mix these symmetry adapted orbitals with the atomic orbitals on the oxygen, bearing in mind the 'rules' of the game

only orbitals of similar energy are able to combine
only orbitals of correct geometry/orientation are able to combine (i.e. head on or side to side)
only orbital of correct symmetry are able to combine. This is where group theory becomes useful

1. Symmetry of the central oxygen's orbitals

The symmetry of the oxygen orbitals (2s and 2p, the 1s isn't relevant to the bonding) can easily be read off from a character table. In this case, water has C2v symmetry, the character table for which is given below:

$$\begin{array}{c|cccc|cc} \hline
C_\mathrm{2v} & E & C_2 & \sigma_\mathrm{v}(xz) & \sigma_\mathrm{v}'(yz) & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & z & x^2, y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_1} & 1 & -1 & 1 & -1 & x, R_y & xz \\
\mathrm{B_2} & 1 & -1 & -1 & 1 & y, R_x & yz \\ \hline
\end{array}$$

To summarise the information obtained:

the s-orbitals have a1 symmetry
the px orbital has b1 symmetry
the py orbital has b2 symmetry
the pz orbital has a1 symmetry

(as @Orthocresol noted in the comments, there is no convention on axis in the C2v point group,meaning x and y can switch depending on what book you get the character table from. The one given is the same as the commonly used set from Atkins' Physical chemistry, but overall it doesn't really matter as long as you're consistent)

It's worth noting that you can derive these symmetries above by considering how the atomic orbitals moves under the various elements within that point group, this is shown below for the hydrogens, and not really needed to be repeated here.

2. Symmetry of the hydrogen group orbitals

To construct the group orbitals for hydrogen (which will go on to be mixed with the oxygen atomic orbitals), we simply have to consider the possible combinations and then work out the symmetries of those combinations.

For the case of two hydrogens interacting, only two cases are possible: in phase, and out of phase.

If we consider these two cases pictorially, we can then determine what happens to them under the various symmetry elements within the C2v group. If the orbitals stay the same, we assign a character of 1, if they switch, we assign -1.

Once we have the full set of characters for both possible combinations, we can assign symmetries, again, using the character table for the relevant point group:

the in-phase combination has a1 symmetry
the out of phase combination has b2 symmetry

3. Mixing

Mixing is the part of this process that you're likely more comfortable with, except now rather than considering whether interaction (on symmetry grounds) is likely by inspection, we can do it by mixing orbitals with the same symmetry labels:

The diagram above is the one that you would likely have arrived at without any consideration of group theory (you can actually derive a more complex diagram in which the molecular orbitals also mix), some key points to note:

the oxygens orbitals have all (qualitatively) been placed lower than those of hydrogen. Electronegative atoms have more contracted orbitals and are lower in energy
two of the p-orbitals end up non-bonding, these are the lone pairs on oxygen
the s-orbital on oxygen interacts with the in-phase combination of hydrogen, this is the lowest energy combination rather than using the p-orbital (also with a1 symmetry)

Further reading

The general approach taken here was a mix between that taken in Group Theory for Chemists by Molloy, and Cottons' Chemical Applications of Group Theory. Most (if not all) books on group theory will provide examples of how to construct MO diagrams based on formal symmetry arguments, and many approaches exist.

I've followed a more qualitative method, relying on the character table to gain the required information, but there is a more formal approach in which irreducible representations are used... possibly someone else will provide an answer with this more formal method (the amount of MathJax required put me off doing this, especially given the ease at which the information can be accessed and the fact that given you're asking the question, you're probably not at the point at which this has been formally taught).


Vapour absorption in 'normal conditions' When is described: "In normal conditions, viscose absorbs 11 to 14% of water vapour. In liquid water, they swell and can absorb 80 to 120% of water." (Clothing Technology)

What do they mean with 'normal conditions'? In open air?

Normal or standard conditions are usually ment pressure of 101.325 kPa and temperature of 298.15 K or 293.15 K. Please see the link for reference.


Why are the densities of europium and ytterbium anomalously low? Why do europium and ytterbium have lower densities than expected in comparison to other lanthanides? I know it has something to do with the fact that they have half-full and full 4f subshells in the +2 oxidation state, but how exactly does it relate to density? And how does it affect the melting point?

Expanding upon @Ivan's comment: Eu and Yb can access the +2 oxidation state instead of +3, due to the +2 ions having a relatively stable half-filled or filled f subshell.
That includes the embedded ions in the structure of the metal. Eu and Yb have $\ce{M^2+}$ instead of $\ce{M^3+}$ ions in the metal. The extra electron and reduced effective nuclear charge on the outer subshells make those $\ce{M^2+}$ ions larger, therefore these metals have lower densities than the surrounding lanthanides. The lower ion charge also means fewer electrons binding the ions together, making Eu and Yb easier to melt.
The difference in ion charge can also show up in other ways. Eu and Yb, for instance, resemble heavy alkaline earth metals more than other lanthanides by dissolving in liquid ammonia.  Europium is also known to appear in some rocks alongside alkaline earth metals rather than alongside other lanthanides (see Wikipedia):

In anaerobic, and particularly geothermal conditions, the divalent form is sufficiently stable that it tends to be incorporated into minerals of calcium and the other alkaline earths. This ion-exchange process is the basis of the "negative europium anomaly", the low europium content in many lanthanide minerals such as monazite, relative to the chondritic abundance.


Does the current of an electrolytic cell increase if the surface area of the Cathode is increased I am currently in high school and will be conducting an experiment looking at the effect of surface area on electroplating. I will be changing the surface area of the Cathode during my experiment but am unsure whether the amp and voltage would be controlled or dependant variables.

Just remember two major rules in electrolytic electrochemistry:

If you fix voltage, current cannot be controlled. This is called constant potential electrolysis
If you fix the current, voltage cannot be controlled. This is called constant current electrolysis

Now your electrolysis cell has resistance:

(i) It will depend on electrolytes

(ii) Surface area of the electrodes

(iii) Distance between the electrodes

So let us know your results and findings. Plot a graph systematically by choosing condition 1 or 2.


Acid hydrolysis of ethyl acetate For my chemistry IA, I am planning to determine the reaction rate and activation energy of the acid catalyzed hydrolysis of ethyl acetate. Where can I find literature values for these, so i can compare my results and evaluate my erros?

Check out this 1953 article from ACS on this exact reaction: https://pubs.acs.org/doi/abs/10.1021/ja01103a521

Does a variety of systems and calculates rate constants for comparison. I'd then recommend using similar terms they use to locate further resources on particular molecule hydrolysis, and even Google Scholar's (or Web of Science, Scopus, JSTOR, etc...) "cited by" option to work your way forward in time to papers that deal with this type of work. Using search terms from any articles you come across in databases (ACS, ScienceDirect, Scifinder, etc...) should get you going with plenty of benchmarking material as well.


Why are there only two irreducible representations in the C3 group? There are three elements in a $C_3$ group and there are three classes in this group cause it is an Abelian group.

\begin{array}{c|ccc|cc}
\hline
C_3 & E & C_3 & C_3^2 & & \varepsilon = \exp(2\pi\mathrm{i}/3) \\ \hline
\mathrm{A} & 1 & 1 & 1 & z, R_z & x^2+y^2, z^2 \\
\mathrm{E} & \left\{ \begin{aligned}1 \\ 1\end{aligned} \right. & \begin{aligned}\varepsilon \\ \varepsilon^* \end{aligned} & \left. \begin{aligned}\varepsilon^* \\ \varepsilon \end{aligned}\right\} & (x, y), (R_x, R_y) & (x^2-y^2,xy), (xz, yz) \\ \hline
\end{array}

According to group representation theory, the number of classes of a group should be the same as the number of irreducible representations of the group, which means there should be 3 irreducible representations for $C_3$ group.

In a character table, this means there should be 3 rows.

Further more, according to $$\sum_{i}l_{j}^2= h$$
in which $l_{j}$ stands for the dimension of the $j-th$ irreducible representation, and $h$ stands for the order of the group.

All of the three irreducible representation of $C_3$ group should be of 1 dimension.

But in the link above, for $C_3$ group, there are only two irreducible representations, and one of the irreducible representation is of 2 dimension, so it seems it violates the basic results of group representation theory.

So, what I have misunderstood?

You got everything right. It is just that two of these representations would use complex numbers ($\varepsilon=e^{2\pi i\over3}$ and $\varepsilon^2$), and we don't like that, so we stick them together into one representation of dimension 2, which makes the imaginary part vanish. Afterwards we treat this representation as "kinda irreducible". In fact it can be reduced, but we just don't want to go that way.

The same applies to other cyclic groups.


Equilibrium between barium carbonate, and barium oxide and carbon dioxide The USNCO 2017 Question 33 is as follows:

Barium carbonate, $\ce{BaCO3}$, is stable at ambient temperatures, but decomposes to barium oxide and carbon dioxide at higher temperatures.
$$\ce{BaCO3(s) <--> BaO(s) + CO2(g)}$$
At a certain temperature, this system in in equilibrium in a closed system and contains appreciable amounts of all three compound. Which changes will lead to an increase in the pressure of $\ce{CO2}$ present at equilibrium?
I. Adding more $\ce{BaCO3(s)}$
II. Increasing the volume of the container
(A) I only
(B) II only
(C) Both I and II
(D) Neither I nor II

What I understand so far:

Concentrations of reactants and products at equilibrium are affected by:

Increasing/decreasing concentrations
Change in pressure/volume
Change in temperature

Temperature remains unchanged, so that factor can be ruled out

(I) does not change concentration, as $\ce{BaCO3(s)}$ is considered to have a constant concentration, and it could be assumed that change in volume due to the addition of $\ce{BaCO3(s)}$ is negligible. Hence, (A) nor (C) are correct

(II) If $\ce{CO2}$ was in a closed system itself, increasing the volume of its container will lead to a decrease of pressure and decrease of concentration.

However, according to Le Châtelier’s Principle, the system should react by favouring the forward reaction, leading me to believe (D) is the correct answer

How do I judge whether or not the increase in volume of the container can be balanced by the increased rate of the forward reaction, such that the pressure of $\ce{CO2}$ remains unchanged at the new equilibrium?

Your conclusion is correct.

In the expression for equilibrium, we consider the activities for the solids to be 1. This ultimately leads to:$$K_p=p_{\ce{CO2}}$$

[Note that $K_P=f_{\ce{CO2}}$, where $f_{\ce{CO2}}$ is the fugacity of $\ce{CO2}$. Under "ideal" conditions, $f_{\ce{CO2}}=p_{\ce{CO2}}$, where $p_{\ce{CO_2}}$ is the partial pressure due to $\ce{CO2}$.]

In other words, the equilibrium partial pressure of $\ce{CO2}$ will be constant. One may compare this to the aqueous tension in case of water VLE.

Increasing the volume would cause more $\ce{BaCO3}$ to dissociate, producing more $\ce{CO2}$. The ideal gas law tells us that: $$n_{\ce{CO2}}\propto p_{\ce{CO2}}\cdot V$$

Your equilibrium pressure due to $\ce{CO2}$ does not change.


What is the difference between exponential factor and orientation factor in case of Arrhenius equation? As per the book (Nivaldo J. Tro)

The Frequency Factor: The number of approaches to the activation barrier per unit time.

The Exponential factor: Number between 0 and 1 that represents the fraction of molecules that have enough energy to make it over the activation barrier on a given approach. The exponential factor is the fraction of approaches that are actually successful and result in the product.

Collision Frequency: It is the number of collisions that occurs per unit time.

Orientation Factor: Usually between 0 and 1, which represents the fraction of collisions with an orientation that allows the reaction to occur

I am unable to distinguish the between the exponential factor and orientation factor. Aren't they saying the same thing? How do they differ from each other?

Another doubt is what does the number of approaches mean? i.e. number of approaches taken by a single reactant per sec, or all reactant per sec.

The orientation factor is really an empirical add-on fix as way of trying to make the Arrhenius equation seem more reasonable. Most collisions between molecules do not lead to reaction, which is why the activation energy is there, but even when a molecule has obtained enough energy to react ( by random collisions in solution or vapour)  we may know from its structure that it can only react from one side or at one end, then an orientation factor is added.


Why does the distance between molecules increase when the temperature is raised? I have learnt that when we heat ice-like substances it changes to water and when I asked my teacher she said  that the distance between molecules increases.

When I thought about it a bit more a question arose: Why does the distance between the molecule increase as we raise the temperature?

The answer "the distance between molecules increase" is incomplete if not plain wrong.

Temperature is an effect of energy present. Basically, it's an effect of little movements and vibrations of molecules and atoms due to their energy.

In an crystal, the energy of the molecules is so low, that they don't vibrate and move enough to break the structure. The more energy you put into the system, the more the molecules move. At one point, the movement is too much to keep the molecules in place, the crystal structure breaks apart, the ice melts.

A liquid (NOT WATER, IT IS A SPECIAL CASE) has lower density than the crystal because the molecules are moving around a lot and "need more space".


What is the structure of 92 m/z fragment for aspirin Aspirin has the following structure:

What would the fragment structure be for m/z = 92?

Even though $\ce{C6H4O}$ is a molecular structure with mass 92, I don't think this will be formed since two bonds needs to be broken. Also the 92 peak is a major peak and the probability of two bonds breaking to lead to major peak is unlikely. The mass spectrum is as follows

This spectrum is recorded using electron impact. That means that the molecule is positively charged by losing one electron.
Since that is a very unstable structure, the molecule starts fragmenting losing neutral fragments.
More processes occur, but these are the ones that we see in the spectrum.
The peaks are the positively charged fragments.

A fragment with m/z of 92, 93 or 94 is typical for phenolic compounds.
The aromatic $\ce{C-O}$ bond is very strong and therefore first all the rest is lost.
In this case the fragmentations are:

Loss of the acetyl group: $180-42=138$
Loss of water from the acid group: $138-18=120$
Loss of $\ce{CO}$ from the acid group: $120-28=92$
Loss of $\ce{CO}$ from the phenolic group: $92-28=64$

These are all typical fragmentations.
The height of the peaks indicate the relative stability of the positively charged fragments.


Including H⁺ from the dissociation of the acid in a titration pH calculation? Let's say you are doing a standard titration problem, for example:

  Find the pH after adding $10~\mathrm{mL}$ of $0.3~\mathrm{M}$ $\ce{NaOH}$ in the
  titration of $25~\mathrm{mL}$ of $0.3~\mathrm{M}$ $\ce{HF}$ with $0.3~\mathrm{M}$ $\ce{NaOH}$.
  The $K_a$ of $\ce{HF}$ value is $6.6 \cdot 10^{-4}$

Most sources start by considering the reaction between $\ce{HF}$ and $\ce{OH^-}$, finding the pre-equilibrium concentrations of $\ce{HF}$ and $\ce{F^-}$ and then using Henderson-Hasselbalch or the equilibrium expression directly to find $\ce{[H^+]}$ (as seen here).

However, don't you have to consider the dissociation of $\ce{HF}$ first,

$$\ce{HF <=> H^+ + F^-}$$

and then consider the reaction between those produced $\ce{H^+}$ and the added $\ce{OH^-}$? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?

There are numerous approximations made when pH calculations like these are made.

In a problem like this the approximations that would usually be made are

pH = -log[H+] (as opposed to pH being -log (H+ activity) which is the true definition).
NaOH is an infinitely strong base, with NaOH completely dissociated. (really, [Na+][OH-]/[NaOH] is about 4)
The dissciation of water  can be ignored (really (Kw =[H+][OH-]= 10^-14).
There exists a constant Ka for HF that is [H+][F-]/[HF], (whereas a true equilibrium constant would be a function of activities, rather than concentrations, of these species, and of water.  The Ka expressed as concentrations is dependent upon ionic strength for example in accordance with the Davies equation approximation).
Particular to this example, it is being assumed that the only fluorine species are HF and F- (really there is also HF2- and another equilbrium constant for it).

So you can take away one or more of these approximating assumptions to try to more accurately calculate pH, but it makes the problem very difficult.

  However, don't you have to consider the dissociation of HF first, and then consider the reaction between those produced  H+  and the added OH-? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?

At the end of the titration (or at any other point during the titration along as the solution is homogeneously mixed) the system will be in equilibrium.  The correct approach is to consider all the species that are present and all the equilbrium constants amoung the species.

Here, the species would be H2O, H+, OH-, HF, F-, Na+, and NaOH (and if I want to be picky HF2-).

The equilbrium constants are Kw, Ka of HF, Kb(NaOH), (and Ka of HF2-).

You need the same number of independent equations as unknowns to solve the problem.

[F-] + [HF] = (0.1/0.35)0.3M

[Na+] + [NaOH] = (0.1/0.35)0.3M

[H+] + [Na+] = [F-] + [OH-]

[H20] = 55M (or, if that is too much of assumption [H2O] + [OH-] + [NaOH] = constant by conservation of oxygen atoms)

plus the 3 equilibrium constant equations gives you 7 equations and 7 unknowns.   That would be a comprehensive consideration of the problem dropping assumptions number 2 and 3.


Could we reach a pH of 4.9 by blowing through straws? According to Wikipedia's article on Breathing:

  The permanent gases in gas we exhale are 4% to 5% by volume more carbon dioxide and 4% to 5% by volume less oxygen than was inhaled.

If I started with a clean beaker of pure water and measured the pH carefully and quickly, I might get a value close to 7.0. If I left it out in the air for a while, far from concentrated sources of pollution, it would equilibrate with the local CO2 concentration which would be 300 to 500 ppm depending on local variables and natural sources (people, plants), and that would lead to a (carefully) measured pH below 6, and probably somewhere around 5.8 or 5.7. Is that about right so far?

Now suppose that I then took turns with lab mates and constantly blew bubbles into the water through long straws that contained some charcoal and other filters to absorb organics from my breath, leaving only the added 4 to 5% CO2 from our respiration.

If the average area of all the bubbles present was about the same or greater than the average area of the water exposed to the air, would I be roughly correct in using a number like 0.02 or 0.03 atmospheres for the CO2 partial pressure in the plot as displayed below, and getting a pH of about 4.9?

Is this what would happen, or are there significant effects that I've neglected?

related: answer to Formation of carbonic acid from breath?

Screenshots from Effect of Dissolved CO2 on the pH of Water, Byck, Harold T., Science  19 Feb 1932, Vol. 75, Issue 1938, pp. 224, DOI: 10.1126/science.75.1938.224.

See here for partial pressure of $\ce{CO2}$ from exhaled breath that I used (28 mmHg = 0.037 atm). I am assuming Henry's law is applicable to calculate:

${7.4*10^{-4} M} =  [\ce{CO2}_{(aq)}]$

Used Henry's Law Constant of $2*10^{-3} \frac{M}{atm} $

See here for equilibrium definition and equilibrium constant value ($K=4.45*10^{-7}{M}$) I used:
$$K = \frac{[H^+]^2}{[\ce{CO2}_{(aq)}]}$$
I then solved for pH from the above equation after rearranging to solve for $[H^+]$, and got a pH of 5.2; so I don't think we can achieve pH of 4.9 from only breathing air into the beaker.


Conditions for free radical bromination using NBS and peroxides? In the free radical bromination of 1-methylcyclohexene using NBS and peroxides a very important consideration to get the appropriate product is:

A. The wavelength of light used for initiation
B. The polarity of the solvent
C. Temperature
D. None of the above

You would need a high enough temperature / low enough wavelength to form the radicals but he said nothing about polarity of the solvent, and there can only be one answer.

From practical experience, I can tell you that the answer is "All of the above".

First, your peroxide initiator needs some energy to undergo homolytic cleavage. This can either be achieved through UV photolysis (A) or heating (B).

$$\ce{RO-OR <=>[h\nu][\mathrm{or\ }\Delta] RO. + .OR}$$

The amount of energy needed will correlate to the strength of the O-O bond. The energy available from UV irradiation is dependent on wavelength because $E=hc/ \lambda$. The amount of energy available from heat is dependent on temperature.

I've initiated NBS reactions both ways, but I prefer heat since I can more easily heat a larger vessel.

  What about solvent polarity?

Solvent is the tricky point, and depending where you are in your studies, you may not have covered the other things that NBS can do. Under the right circumstances, NBS is also an electrophilic source of bromine, especially in polar protic solvents.

Here are two examples from my personal experience using the same substrate with different conditions:

Using methanol as a solvent without a radical initiator, I got electrophilic bromination of the arene.

Using 1,2-dichloroethane as a solvent with heat and a radical initiator, I got radical bromination of the benzyl position and electrophilic bromination of the arene.

Using 1,2-dichloroethane as a solvent with UV irradiation and a radical initiator, I got radical bromination of the benzyl position.


Do certain cumulated dienes present geometric isomerism? Let's take, for example: 2,3-pentadiene.

Can't this compound present geometrical isomers, like 2-butene?

Properly speaking, these sorts of allenes/cumulated dienes do not exhibit geometric isomerism in the way that alkenes can. Rather, when each of the two respective carbons bear two different substituents, the molecule is chiral and will exhibit enantiomerism. More specifically, these asymmetric allenes exhibit axial chirality. This is consequent to the two consecutive $\pi$-bonds being mutually perpendicular, which results in the molecule lacking any mirror plane of symmetry.

Generally, an even number of successive double bonds in a cumulene produces a potential axis of chirality (again, assuming each carbon at the end of the double-bond system bears two different substituents), while an odd number results in the substituents at the ends of the double-bond system being coplanar, which permits geometric isomerism akin to that seen in alkenes.


What's the exact difference between (-)-BTX and (+)-BTX I quite liked this article, which describes the synthesis of (-)-batrachotoxin as well as a steroisomer, (+)-batrachotoxin. I would like to know what the precise difference is between the two: are they precise mirror images of each other, or is there something more subtle going on? Is there something specific about the (±) notation (as opposed to S and R, for example) with relevance to the difference between the two?

I had a look online but the original paper is thin on structures that I can understand, and just going on google buries me under a slew of (-)-BTX results.

From a cursory glance, batrachotoxin has 8 stereocenters. No single Cahn-Ingold-Prelog R or S designation is going to be enough. As a result, we simply identify which enantiomer we're talking about by using the standard optical rotation of the substance. The (+) and (-) forms are enantiomers. They rotate plane polarized light in equal and opposite directions and are non-superimposable mirror images of each other.


Comparison of boiling points I was comparing boiling points of $\ce{CHCl_3}$ and $\ce{CH_3Cl}$. According to me B.P. of $\ce{CHCl_3}$ should be higher due to it's higher molecular mass than $\ce{CH_3Cl}$, but the answer is opposite ,ie. B.P. of $\ce{CH_3Cl}$ $\gt$ B.P. of $\ce{CHCl_3}$.Why?

Molecular mass has NO pertinence to BP, water and ammonia.  Look up the BP of perfluorocyclohexane, hexafluorobenzene, and the hydrocarbons.  If I want to volatize iron, ferrocene.  If I want to volatize iron more, $\ce{Fe(hfac)_3}$ where the ligand is the bidentate enolate anion of 1,1,1,5,5,5-hexafluoro-2,4-pentanedione.  Big MW.  What are the MW and BP of $\ce{UF_6}$?  $\ce{-Si(CH3)_3}$ also confers volatility.

Boiling points are more about polarizability, intermolecular van der Waals and such. Bigger molecules are overall electronically floppy.  Benzene, toluene, xylenes – nice BP steps, but not because of MW. Look up $\ce{SF_6}$ bp.  $\ce{I(CF3)_7}$ boils near 0 °C.  Calculate that MW, and its vapor’s mass/liter at STP compared to dry air at 29 g/liter.


Why Calcium Oxalate is insoluble in acetic acid? Calcium oxalate is insoluble in acetic acid, but not it's carbonate, oxide and hydroxide.

From internet, it is even considered as a test to distinguish calcium oxalate from it's carbonate and oxide. The reason from the resources which i had searched, only gave factual reasons such as

High Lattice Energy and
Low solvation energy of $Ca(C_2O_4)$

But intuitively the "short sized" compounds like $CaO$ and $Ca(OH)_2$ must have better packing and more lattice energy, then why much bigger calcium oxalate have higher lattice energy?

Also, i am clueless about how to compare solvation energy of different compounds. I think it has to do with polarity of the solute and the solvent. Any help will be appreciated.

Edit:
I had omitted the relative basicity of oxide, hydroxide and carbonate of calcium but i am more interested in knowing how to compare the lattice energy and solubility of the given compounds.

Calcium oxide, hydroxide and carbonate react with acetic acid to form calcium acetate. Calcium oxalate does not react.

Edit: In real life, you'll of course will end up with a little oxalic acid and calcium acetate, as calcium oxalate is very slightly soluble and then you'll mix all available ions.


For alkali and lanthanide metals, is there a better liquid barrier to oxidation than mineral oil? Most often mineral oil is used to store, protect reactive metals from oxidation. By far the best choice is to hermetically seal the metal in pure argon or some other inert gas, since, although mineral oil tends to slow down the oxidation process, it's still soluble to oxygen and slowly but surely the metal will begin to oxidize.

While argon is a sure thing it requires special equipment, and once sealed you have to break the container to access the metal.

My question - among liquids is there a better choice than mineral oil? A liquid that will not react with the metals, and provide a better barrier to migration of oxygen?

Not really as all liquids will, to some extent, dissolve a small amount of oxygen or water. The only liquid that comes to mind that will prevent either is mercury, but you probably don't want your stuff contaminated with mercury. If you really wanted to prevent water contamination, adding molecular sieves might help, but other than that, no not really.


How semi-permanent hair dye stays in place? Semi-permanent hair dye washes off after several months. What mechanism(s) keep the dye in the hair? (I assume it's not primary chemical bonds, otherwise it would be permanent)

According to this: http://questions.sci-toys.com/node/72, hair dyes work by creating pockets within the hair. The colored dye molecules then get stuck in these pockets - requiring multiple washes to escape. Short term hair dyes use larger dye molecules so that they cannot get stuck in the hair. On the other hand, longer term dyes actually react with you hair to stay in there for an extended duration of time. For the most part, all of these hair-dye interactions are the result of intermolecular forces.

Semi permanent hair dyes tend to be small compounds that get stuck in your hair. This means that they are not permanently bound to your hair, but they will require a lot of washes to escape. I hope this helps!


Why does 1,2-dichloro-4-nitrobenzene undergo SNAr para and not meta to the nitro group? In the reaction of 1,2-dichloro-4-nitrobenzene with sodium ethoxide, why does ethoxide end up substituting the chlorine para to the nitro group rather than the chlorine meta to the nitro group?

I think that C–2 should be the most electron-poor carbon (and hence prone to nucleophilic attack) since it experiences a combined electron withdrawal from the nitro and chloro groups flanking it. Shouldn't the nucleophilic ethoxide anion attack there instead?

You can obtain an answer by looking at the electronic structure.  $\ce{NO2}$ is a meta director for electrophilic aromatic substitution (EAS).  The question asks about nucleophilic aromatic substitution (NAS). Aromatic substitutions involve a free pair of electrons resonating around a structure. Electrophiles may add at any resonance site where the pair of electrons are. Nucleophiles may inversely add to any site where electrons are not present shown in the picture below. Therefore if a functional group is meta directing for EAS it will be ortho/para directing for NAS and vice versa.


Is there a general way to predict products of a chemical reaction Is there a general way to predict products of a chemical reaction, given only the atomic masses of masses of all elements involved? for example:

$\ce{CuFeS2 + O2->}$
$\ce{K2CrO4 + H2SO4->}$
$\ce{FeOCr2O3 + K2CO3 + O2->}$
$\ce{K2Cr2O7 + H2SO4->}$

If there's no method to do so, are we supposed to memorize all chemical reactions?

PS: In case you are writing the mechanism of chemical reactions, please bear in mind that products are not known beforehand.

Often times it is possible due to previous data. Trends appear and you can make deductions from such trends. Reactions may not happen at all. There may be deviations from the trends as well. The deviations may be for unknown reasons or there may be unproven theories that are included in textbooks. Inorganic chemistry reactions are not as clear cut as organic chemistry reactions. You don't have to memorize all the reactions. Know the trends, conduct the experiment, and if your prediction fails, try to think hard about the conditions in which the experiment took place. Sometimes the explanation may lay in the container that you used for the reaction, or it may be stoichiometry.

As in all sciences, you need experiments.


Electrode potentials at interfaces? My questions relates to the fundamental concept of electrochemistry, more specifically the electrode potentials.

1) First, why is there a potential difference at the interface of two phases? Considering the simplest case of a metal rod dipped in the solution of its ions, why should there be a potential difference. Out of many ways, one of the ways of understanding this provided by this site here,is in terms of equilibrium between the metal atoms and ions. But then, why is the presence of the other phase (ions for the metal piece) required to effect this equilibrium? For example, for the equilibria of ammonia and nitrogen/hydrogen, even if we take pure ammonia, the equilibrium is set up according to the surrounding conditions. Next,then can this equilibrium be described in terms of the equation $\Delta G=-RT\ln(K)$. Lastly, if this is an equilibrium, then there must also be an equilibrium between the metal and the hydrogen or the hydroxyl ions in solution as these might also take up the electrons from the metals?

2) When we measure the difference in electrode potentials of two different half cells, we are just measuring the potential difference between the two metal electrodes. How can we get an idea of the potential difference of the metal piece and its solution as the absolute value of potential of the two solutions might also differ. That is, if the two solutions are at different electrostatic potentials ( though not measurable), the potential difference between the two electrodes will not be the difference of their electrode potentials(which are the potential difference between the electrode and the electrolyte in each case). So, in order for the potential difference of the electrodes be equal to the difference in electrode potentials of the two half-cells, the solutions should be at the same potential?

There are a bunch of of sub parts to these questions; let me see if I can tease some of them out.
Why does a potential difference occur at an interface?
The easiest way to answer this question is through thermodynamics.  If we consider the situation of a reactive metal rod (e.g. magnesium) being exposed to water, the reaction $$\ce{Mg -> Mg^{2+} + 2 e-}$$ is favorable.  In other words, the $\Delta G$ for this reaction will be less than zero.  I've done nothing more than summarize the website you reference in your question, which does a nice job at describing what happens when a piece of metal comes in contact with water, so I don't feel the need to repeat that work.  We then get to your next question:
Why is a second phase required to generate a potential difference?
If we have a universe that consists just of magnesium, then there would be no place for the above reaction to occur.  In this trivialized case, there is no driving force to make magnesium give up its electrons.  That doesn't mean it would not happen, only that any dynamic equilibrium that is established would highly favor the pure metal.  The meat of question one, then, is (may be?):
Why doesn't a potential difference between a metal and gas occur?
Actually it does.  There's an interesting paper (which is behind a paywall if you are not at a university that subscribes to this journal) that describes the measurement of potentials across metal/gas interfaces.  The challenge in studying metal/gas potentials is the ability of controlling this potential externally.  With the metal/liquid scenario, an electrochemist (or budding electrochemist) can add another electrode to the solution and apply a potential between the electrodes.  It's much harder to do the same thing with a gas.  (Well, technically it's *not harder to connect two electrodes to one another and hold them in air, but the redox work functions, resistance and capacitance issues that arise in gas-phase electrochemistry make this a really challenging feat.)
The paper I reference above uses chemisorbed species that have active IR or Raman vibration modes to measure the surface potential using spectroscopy and something called Stark tuning (not the best reference, but it's also not behind a paywall) which states that the vibrational frequency of an harmonic oscillator is proportional to an applied electric field.  Not many metal/gas interfaces have been explored using this method, but those that have display surface potentials fairly close to 0 vs. SCE or 5 eV vs. vacuum.
What about the rest of the stuff in solution?
Back to the metal/liquid scenario, the last part of your first question asks about other ions in solution.  They most definitely play a role in defining the potential of a metal in solution.  In order to accurately predict electrode potentials, one must know concentrations and possible reactions (including states of matter) for all species present.
What about question 2?
I'm afraid question 2 might be based on the false premise that we can measure the potential of an electrode in the absence of anything else (liquid, gas, ions or molecules).  When electrochemists refer to a 'half cell', they have in mind both the reduced (in this case, metal electrode) and oxidized (metal ions) forms of a redox couple contained within the same system.  (Think about a beaker with a copper wire placed in it and the beaker is filled with a solution containing 0.1 M $CuCl_2$.  That half-cell or system has a given potential based upon (primarily) the standard reduction potential of $Cu^{2+}$ to $Cu$ and the concentration of $CuCl_2$.  If I have misread your second question, please edit and I will update my answer as needed.


Crystallization from a solution I was wondering if there is any other way to crystallize materials from a solution other than using supersaturation?

Thanks,

No. Per definition you need to have at least local supersaturation. You can archive it by different means (cooling, evaporating of solvent, mixing with other solvent...) but in the end you need supersaturation. See: http://xray.chem.ufl.edu/growing%20tips.htm or http://web.mit.edu/x-ray/cystallize.html

Of course you can use techniques like sublimation but your question was about solution (btw in case of sublimation you also have supersaturation, only difference is that your solvent is gas).


As mentioned, you don't want a direct answer so I am just going to ask you some question to think about, that hopefully help in getting you to the answer. Please reply in the comments whenever something is unclear to you!

You're expecting first order in both reactants, so if you would write down the reaction rate equation, what does that look like?

With the equation you should have one unknown only: $k$ (or two that make up $k$: $k_0$ and one other, which?, if you are interested in the temperature effect as well).

Now calculate $k$ for each reaction separately. Are they the same? Do you expect them to be? If they're not, should you average them?


SI Units for Isothermal Incompressibility and Specific heat capacity at constant volume I am looking for the SI units of isothermal incompressibility and specific heat capacity at constant volume.

What I found was that isothermal compression was measured in " $\mathrm{m^2}\ \mathrm{N^{-1}}$" so I'm guessing isothermal incompressibility would be the same?

I also found that specific heat capacity was " $\mathrm{J}\ \mathrm{K^{-1}}$"so is this the same units for constant volume also?

The units of a partial derivative $\displaystyle\left(\frac{\partial Y}{\partial X}\right)_Z$ are the units of $Y$ divided by the units of $X$.  The units of $Z$ don't matter.

Isothermal compressibility is $-\displaystyle\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$, so the SI units are reciprocal pressure, $\rm Pa^{-1}$ or $\rm m^2\ N^{-1}$.
Heat capacity at constant volume is $ \displaystyle\left(\frac{\partial E}{\partial T}\right)_V$, so its SI units are energy/temperature or $\rm J\ K^{-1}$. When you say specific heat capacity, that means per unit mass, so the specific heat capacity at constant volume is $\rm J\ kg^{-1} K^{-1}$.


How does Nickel harden steel? Nickel :

Does not harden austenitic stainless steel
Ease the tempering of martensitic stainless steel
Harden maraging steel by forming precipitates

I also read people saying that it harden ferritic steel. Is it true ?

The chemicals reasons behind all of this are quite unclear for me. Could you explain how it works?

Thanks

I have the following comments about nickel steels:

Tensile strength is different than hardness. I assume you are specifically asking about hardness, not strength.
Hardness has two different aspects: resisting wear and machinability. Different steels will exhibit different degrees of the two aspects. Ideally, you want a steel that is hard wearing, yet highly machinable.
There are different alloys involving nickel, notably the nickel alloys and chromium-nickel alloys. These two types of steel are different.
Different alloys have very different hardening characteristics. Even very minor changes to the amount of carbon or nickel in an alloy can have dramatic effects on its hardening characteristics.
The main effect of nickel when present in a ferritic or pearlite matrix is to increase the fineness of the grain structure.
Nickel slows the growth in size of ferritic/pearlite grains. This makes it easier to obtain a consistent carburization on the surface of a workpiece. For this reason, small amounts of nickel are valuable in alloys which will be case hardened.
Nickel preferentially dissolves into the ferritic crystals of certain steel phases, therefore it can potentially increase the hardness of a steel without harming its ductility.


Gibbs free energy=potential free energy? (Transition state theory) Online I found the plot Gibbs free energy vs. reaction coordinate:

https://en.wikipedia.org/wiki/Transition_state_theory

and on many books I found the plot potential energy vs. reaction coordinate (e.g. page 237 Atkins, Physical Chemistry for the Life Sciences, 2nd Ed).

In the first case activation energy is defined as $\Delta G$, whereas in the second case is defined as difference of potential energy.

Thus is the Gibbs free energy a potential energy?

Thank you very much.

Yes both mean exactly the same thing.

By definition gibbs free energy is the total amount of non pv work that can be extracted from the system. Non Pressure-Volume work refers to the work which does not allow change in both volume and pressure of the system. An example of non pv work is electrical work which is done by cells and batteries in creating a potential difference about its ends.

Now pv work is related to kinetic energy of the molecules of the system. We can see this by two ways. First is that pressure and volume both are dependent on movement of molecules. Hence increasing pressure at constant volume is the same thing as increasing kinetic energy of the gas due to which there are more collisions of higher magnitude. Similarly increase in volume at constant pressure also means increase in kinetic energy of system. Second is that pressure and volume are both related and dependent on Temperature by gas equation where temperature of a system is directly related to kinetic energy of the gaseous system.

So whenever pv work is done by or on system changes take place only in kinetic energy of the molecules of system and not in the potential energy of system.

Hence whenever non pv work is done by or on system, changes in only potential energy of the system takes place. So Gibbs free energy is same as potential energy of system.


derivation of enthalpy internal energy relationship A question asks us to derive a relationship between the enthalpy change and the change in internal energy for a reaction.
Now, I know the relationship is:
$\Delta H = \Delta U + V \Delta p + p \Delta V$
But i thought enthalpy was conveniently defined this way ( not derived) so that enthalpy change equals heat change at constant pressure. So which

First a brief note on the mathematics of Legendre Transforms

Consider a function in two independent variables: $ f(x,y)$

We right the differential of this function as follows:

$df(x,y) = \Big( \frac{\partial f}{\partial x}\Big)_ydx + \Big( \frac{\partial f}{\partial y}\Big)_xdy$

Now, I shall define $u := \Big( \frac{\partial f}{\partial x}\Big) $ and $w := \Big( \frac{\partial f}{\partial y}\Big)$

so, $df = udx + wdy$ [equation 1]

Here, u and x (and similarly w and y) are called conjugate variables. (i.e $ux$ and $wy$ will have the same units as $f$)

[Think of it this way: In thermodynamics, PV, TS have the same units as H, U,G,A]

Now, our goal is transform our original function $f(x,y)$ to a function $g(x,w)$. We wish to replace y, with it's conjugate variable and the function $f $ and $g$ will have the same units.

so, we take the differential of the product $wy$ : $d(wy) = ydw + wdy$

We subtract this from [equation 1] and get $d(f-wy) = udx - ydw $
and can deduce that our Legendre Transformed function $ g(x,w) := f-wy$

Now, we can easily apply this method to Internal Energy (U) to obtain Enthalpy (H)

We have $U(S,V)$ and $dU(S,V) = \Big( \frac{\partial U}{\partial S}\Big)_VdS + \Big( \frac{\partial U}{\partial V}\Big)_SdV $

Now, here we seek $H(S,P) \equiv g$

a)  $f \equiv U$

b) $S \equiv x$ is the variable we hold fixed

c) $V \equiv y$ is the variable we switch

d)  $\Big( \frac{\partial U}{\partial S}\Big)_V := T \equiv u $(the conjugate of the variable we are holding in place

e) $\Big( \frac{\partial U}{\partial V}\Big)_S := -P \equiv w$ (the conjugate of the variable we change

$ g =f-wy $

so, $H(S,P) = U-(V)(-P)$

Or, $H(S,P) = U + PV$

From here, you can proceed to write down: $dH = dU + PdV + Vdp$

and for macroscopic changes: $\Delta H = \Delta U + P\Delta V + V\Delta P $


Why does Oxygen act as Nucleophile over here? We were taught in class that when a carbon atom and an oxygen atom both have a negative charge in a resonating structure, then carbon atom attacks the nucleophile rather than oxygen atom because it acts as a better nucleophile.

In the question I have attached, in the first step the hydroxide ion abstracts the hydrogen from the oxygen of o-cresol. The negative charge so formed is delocalised onto ortho and para position.

There are 4 possibilities:

Oxygen atom attacks $\ce{CH3I}$.
The carbon atom attached to $\ce{CH3}$ attacks.
The carbon atom adjacent to oxygen (not attached to $\ce{CH3}$) attacks.
Carbon at para position attacks.

Now, based on the statement that Carbon atom should attack, Option 1 should be ruled out.
Option 2 also is unfavourable because the carbon atom attached to $\ce{CH3}$ would have less negative charge due to inductive destabilisation by $\ce{CH3}$.
Option 3 is unfavourable because the product formed would be sterically crowded.
Option 4 seems correct.

But the answer given to the above question is Option 2.

I confirmed from my teacher and he says that the statement that Carbon atom is a better nucleophile than oxygen is true but here attack from oxygen would take place (he doesn't explain why).

Also, in the well known Riemer Tiemann Reaction and Kolbe Reaction of Phenol, the attack takes place from Carbon side.

Question- where is my reasoning wrong? According to my reasoning, answer should be Option 1. Why isn't it?

You may be seeing the impact of the aromatic ring.  Putting a negative charge on carbon in your structure disrupts the aromatic coupling.  So carbanion contributions are reduced and the oxyanion contribution, where the oxygen is nucleophilic, is favored.


What impact does heating or cooling have on the oxidation reduction potential of water? There are alkalinizing water ionizers on the market that lower the ORP of water. Does heating or cooling that water have an impact on its ORP?

I see two questions here.  The one that is stated (temperature effects on ORP) and a hidden one about water ionizers.

First, the temperature effect on ORP.  The thermodynamics of oxidation and reduction reactions is governed by the Nernst equation:
$$E(ORP) = E_o - \frac{2.3RT}{nF}\times log\frac{[reductant] }{[oxidant]}$$

In this equation, R and F are the gas law and Faraday's constants, respectively, $E_o$ is the standard reduction potential of the system of interest, T is temperature, n is the number of electrons transferred in the reaction and the concentrations of the reductant and oxidant are found in the log term.  The coefficient $\frac{2.3RT}{nF}$ in front of the log term can be simplified to $\frac{0.00198\times T}{n}$ and when we perform dimensional analysis on the term we find that it will provide units in Volts.  If we are looking for the largest effect that temperature would have on ORP then we want to use the smallest value of n, which would be 1 electron, further simplifying the coefficient to approximately $0.002T$.  T has to be in Kelvin so if we take our two extreme values as the freezing (273 K) and boiling (373 K) points of water, we find that the difference in ORP between these two extremes (all else being the same) would be about 200 millivolts.

Now to the hidden question about water ionizers.  There's a lot of hype about these systems with little scientific evidence to back up the claims.  There is at least one website useful for exploring the science behind water ionizers that is (I think) accessible to the general public.  A number of claims made by the manufacturers of these products, such as ion streams and water clusters simply don't stand up to scientific experimentation, and consumers need to be wary of the health benefit claims.


If the electron of a hydrogen atom transitions from energy level n=5 to n=3, which level corresponds to n1 and which level corresponds to n2 which level corresponds to n1 and which level corresponds to n2 in the Rydberg equation?


Saturated vs Supersaturated The way I was told is that a saturated solution contains the maximum amount of solute and a supersaturated solution contains more solute than a solution can hold at a particular temperature. But I was also told that if you add more solute than it can hold, that it will not dissolve and will settle at the bottom as a solid. So how is that supersaturated? Since it didn't dissolve, isn't it still just saturated? It was also mentioned that a supersaturated solution needs to be cooled, but if the temperature dropped, wouldn't the solute just come out of solution and settle at the bottom (like sugar and cold coffee) and remain saturated?

A saturated solution has the maximal amount of solute at equilibrium.

If equilibrium is not required, solutions can have more solute.  For example, when a bottle of soda is first opened, it is supersaturated in CO2, but it is not at equilibrium and CO2 effervesces.


Reduction of conjugated double bond by sodium borohydride Some of my teachers were debating about whether a double bond in conjugation with a ketone is reduced or not by $\ce{NaBH4}.$ One of my teachers presented the evidence from Clayden which had an example where a cyclic ketone was reduced alongside the conjugated double bond.

However, there was no example for a straight chain compound. Also, the book said that steric hindrance may affect the reduction process and only favour the reduction of ketone.

I want to know that is there any authentic source which has an example of reduction of a conjugated double bond and ketone via $\ce{NaBH4}.$

Scifinder shows lots of reactions such as you describe. Often (but not always) copper is used to help favour the 1,4-addition (alkene reduction) process.

The simplest example I can see is the reduction of 3-penten-2-one[[2]] - when treated with NaBH4 the majority (89%) is reduced at the ketone only, while 11% of the doubly-reduced product is formed:

Obviously different substrates would give different product ratios, but certainly this is a plausible (and precedented) reaction.


Does the aluminium chloride assay (Ordonez et al.2006) for total flavonoid content work on coloured extracts? It is said that after applying aluminium chloride assay, the extracts will turn yellow due to formation of Al-Flavonoid complexes. So I was wondering if the assay still works if the extracts are already yellow.
For example, the yellow extracts are applied aluminium chloride assay and determined for total Flavonoid content using colorimetric method.
Then, would the result show the TFC (total flavonoid content in those yellow extracts?

Depending on the intensity of the yellow color (specifically, the degree of absorption of light at the wavelength of your measurement) you might be able to subtract out the yellow from the extraction in order to differentiate between that and the color arising from the aluminium chloride assay.

In other words, just measure the absorbance of your material before the addition of your reagents, and subtract that value from your final absorbance reading.

Alternatively, since this is a matrix interference issue, you might do well using the method of standard additions. This is described in this Wikipedia article:

  The method of standard addition is a type of quantitative analysis approach often used in analytical chemistry whereby the standard is added directly to the aliquots of analyzed sample. This method is used in situations where sample matrix also contributes to the analytical signal, a situation known as the matrix effect, thus making it impossible to compare the analytical signal between sample and standard using the traditional calibration curve approach.

This article gives a reasonable outline of the procedure that should be sufficient should you decide to try this method.


Which dicarboxylic acid has the most acidic hydrogen? Which of the following acids (maleic, fumaric, succinic, or malonic) has the most acidic hydrogen?

I think that malonic acid should have the most acidic hydrogen due to the presence of an active methylene group.
The extraction of hydrogen atom from malonic acid would lead to high resonance stabilization, which is not the case in other acids.
However the answer key says that maleic acid has the most acidic hydrogen. Can someone please explain it to me why?

The $\mathrm pK_\mathrm a$ of the $\ce{CH2}$ group in a dicarbonyl compound is roughly $10$, whereas the $\mathrm pK_\mathrm a$ of a carboxylic acid is roughly $5$. For example, dimethyl malonate has a $\mathrm pK_\mathrm a$ of $13$ whereas acetic acid has a $\mathrm pK_\mathrm a$ of $4.7$. So, the C–H bond acidity is not likely to be under consideration in any of your compounds.

Let's draw up some $\mathrm pK_\mathrm a$ data (only first ionisations, all data from Wikipedia):

$$\begin{array}{cc} \hline
\text{Carboxylic acid} & \mathrm pK_\mathrm a \\ \hline
\text{Maleic} & 1.9  \\
\text{Fumaric} & 3.03 \\
\text{Succinic} & 4.2 \\
\text{Malonic} & 2.83 \\ \hline
\end{array}$$

Clearly, all of these relate to ionisation of the $\ce{CO2H}$ group; if the acidity of dimethyl malonate is anything to go by, the $\ce{CH2}$ group in malonic acid is less acidic than the $\ce{CO2H}$ group by a factor of $10^{10}$.

What remains is to analyse this trend. Why is maleic acid the most acidic? The answer is – at least partly – intramolecular hydrogen bonding. When the first carboxylic acid group is deprotonated, the resultant monoanion is stabilised as such:

You are probably familiar with the mantra that a more stable conjugate base implies a more acidic compound. This is precisely what happens here.

Obviously, for fumaric acid, this is not possible because the double bond holds the two carboxyl groups apart. One could, however, argue that similar stabilisation could be derived for malonic and succinic acids. Why is it not a factor, then? It has to do with conformation: for the malonate or succinate anions to enjoy the benefits of intramolecular hydrogen bonding, they have to twist themselves into one specific conformation where the carboxyl groups are close to each other. This results in a loss of configurational entropy (the molecule is not as "free" as it would like to be to explore different conformations). For maleic acid, though, this is not a problem because the (Z)-double bond plonks those two carboxyl groups next to each other, and they don't have a choice about it. This bears some similarity to the concept of preorganisation, where two or more groups are "organised" in a fashion which alters their properties.

That's not the only reason why maleic acid is especially acidic, however. Part of the reason is also because an $\mathrm{sp^2}$ carbon is more electronegative than an $\mathrm{sp^3}$ carbon. An $\mathrm{sp^2}$ orbital contains 33% s-character, compared to an $\mathrm{sp^3}$ orbital which has 25% s-character. s-Orbitals are closer to the nuclei than p-orbitals, and therefore, an electron in an $\mathrm{sp^2}$ orbital experiences a greater effective nuclear charge from carbon; this translates into a greater electronegativity. (For similar reasons, alkenes are more acidic than alkanes.)

So, relative to the succinate anion, both the maleate and fumarate anions are stabilised by electron withdrawal via the inductive effect. This explains why fumaric acid is more acidic than succinic acid; the intramolecular hydrogen bonding explains why maleic acid is more acidic than fumaric acid.

Lastly, of course, malonic acid is more acidic than succinic acid because the electron-withdrawing $\ce{CO2H}$ substituent is fewer bonds away. In fact, succinic acid is barely more acidic than acetic acid, which doesn't have an electron-withdrawing substituent on it.


Reduction of nitro compound using protection group I've tried researching things like cyclisation, nitro reduction and all I've found is little to nothing. I know that for the 1st arrow I need to protect the carbonyl and the 2nd arrow is the carboxylic acid but I'm not sure how to go about with the reaction or mechanism. Can you someone guide me along please?

For the first arrow you are correct that you need to protect the ketone before reducing the nitro group. For this I would use a dimethyl ketal. There are many sets of conditions for this; trimethyl orthoformate in refluxing MeOH with catalytic PTSA is a good bet. This will also give the methyl ester of the carboxylic acid but that is of no consequence as it will be reacted in the next step. Your next step is then the nitro group reduction. I see no reason why you should not use hydrogenation, just don't have any acid present as that may remove the ketal and you still need that for step 3. The aniline thus produced will cyclise easily, just heat it in neutral solvent if it does not spontaneously cyclise. Step 3 is reduction of the amide, LiAlH4 should do this. Finally you deprotect the ketone with a mild acid hydrolysis.

For the second arrow reduction of the nitro group by hydrogenation will give indole-2-carboxylic acid by spontaneous cyclisation as formation of the aromatic [5] is favoured over the non-aromatic [6]. This has been reduced to your target indoline-2-carboxylic acid by sodium/liquid ammonia  https://patents.google.com/patent/US4485241


What is a level 4 carcinogen? My friend spilt a "level 4 carcinogen" on their lap.  What is it and how dangerous is it?

They had washed if off after.

There are several different ways to rate carcinogens.  According to its MSDS, Pyridine belongs to Group 3, A3:  Known to cause cancer in animals, inadequate evidence in humans.

If it was just contact with the skin, your friend should wash it off with soap and water immediately for 15 minutes (this is again from the MSDS).  Short term exposure can cause "irritation (possibly severe), allergic reactions, sensitivity to light."

So, if any irritation occurs, it is probably not serious, but I would recommend seeing a doctor anyways.  I doubt there is anything to worry about, but cancer is not something to mess around with.


Chemical Volcano- Experiment In a science exhibition in 2011 , I and my  friend made a volcano that used vinegar and baking soda for the 'effects'. There was another group that made volcano too and they had brought their own supplies. But before the exhibition started, the members of other group came to us and asked us if we could share our baking soda because their baking soda wasn't 'working'. When I asked them why it wasn't working they told me that there wasn't 'salt' in it. (We were only 8-9 years old). When I tried using it it was forming bubbles but very slowly. Eventually our supplies ran out before the exhibition was over. One of my friends suggested that we should use chalk powder instead of baking soda(I don't know how he figured it out at age 9). So we ground up chalks and tried using it. In this case , it was just like the baking soda that didn't work , it was producing bubbles ,but very slowly.
What could be the issue in the two cases? Was it a matter of purity?
We had used synthetic vinegar used for food purposes(5-10%)(There was no issue with the vinegar because it worked well with our own baking soda). The chalks were ordinary chalks used in classrooms. Regarding the other group's Baking soda I do not have any other information cause it happened 10 years earlier.(And I'm no longer in contact with them)
Was reaction less fast in case 2(Chalk powder) because of the low solubility product of Calcium carbonate? But Solubility product is a thermodynamic factor and not a kinetic factor right ?

Baking soda is sodium bicarbonate NaHCO$_3$. It is stable in air indefinitely. Baking powder is a mix of sodium bicarbonate and a powdered acid, typically an aluminum or calcium hydrogen phosphate, which, in the absence of moisture, is stable for a long time (maybe a year? maybe more). After exposure to moisture, it will release CO$_2$ and become inactivated.
The third item, "chalk", has many definitions. Blackboard chalk is not just a stick of CaCO$_3$. A stick of CaCO$_3$ could be marble, or limestone - oh - no, you would like it to be nice and powdery but sticking together. Yep, it can be done: glue. Nah, too scratchy. Plaster of Paris! Mix with water, add some CaCO$_3$ chalk, maybe some white clay... You get the idea. Blackboard chalk is not just CaCO$_3$. Not even that, but CaCO$_3$ reacts very fast with hydrochloric acid (but that's too hazardous for grade school experiments), and its reaction with vinegar (5% CH$_3$CO$_2$H) is slower initially, and slows down markedly as the reaction progresses because the acetate ion produced inhibits the ionization of the remaining (weak acid) CH$_3$CO$_2$H.
The key lesson is that demonstrations are very temperamental, and depend on doing the same thing with the same materials many times, to see how the "same" experiment can vary so much. You may even begin to doubt that any experiment can ever be duplicated - but it can, if you keep everything constant.


Does Henry's law depend on curvature of gas/liquid interface? Could you tell me Henry's law depends on curvature of gas/liquid interface or not? It would be better with references.

No, but not because the vapour pressure of mixtures doesn't depend on curvature (it does), but because Henry is a narrowly-defined law that doesn't contemplate surface curvature.

Henry's law is, after all, a barely better description of vapour pressure in mixtures than Raoult, and that's because it introduces an experimental value (the Henry constant) for each solute/solvent pair. The only reason why it's (rarely) used is that it's extremely easy to operate with. And pretty much the only reason why it's taught is because it's a simple, mathematically trivial treatment that moves away from ideality; it's sort of the minimum, simplest step away from ideality, which is convenient for a classroom.

But if you attempt to combine Henry with the effect of surface curvature, say by combining it with Kelvin's equation, you run into a serious problem. The equilibrium difference of pressure between a liquid, $p^L$, and its vapour, $p^V$, when the liquid has a curvature of radius $r$ is

$$ p^L = p^V + 2\sigma/r$$

where $\sigma$ is the surface tension of the liquid. And that's when you run into a problem - surface tension of a liquid mixture also depends on the composition, so the entire point of Henry's law (that concentration of a species in solution is proportional to its partial pressure with a proportionality constant $H^{cp}$) is lost. Even if you accept the simplest case (that surface tension behaves ideally with composition following Gibb's isotherm) you are stuck with a $H^{cp}$ that is itself a function of $c$, and so the mathematical model has definitely moved away from Henry.

By which point you are better off using other, more realistic models for your non-ideal mixture. Note that even then, you are stuck with iteratively computing the crossed effects of surface tension on vapour pressure and vice versa, as the relationship can't be solved analytically as far as I know.


How to calculate the solubility of lead (II) carbonate in water, given that bicarbonate will also form? Calculate the solubility of lead (II) carbonate, $\ce{PbCO3}$, ($K_{\mathrm{sp}} = 1.5 \times 10^{-15}$) in water, given that carbonate is a stronger base than water, so will also form bicarbonate. When bicarbonate acts as an acid: \begin{align}
\ce{HCO3- &<--> H+ + CO3^2-} & K_{\mathrm{a}} &= 4.7 \times 10^{-11}.
\end{align}

I think the equation is, in light of airhuff's suggestion, $$\ce{PbCO3 -> Pb^2+ + CO3^2-}.$$
What I'm most confused about is how the bicarbonate and $K_\mathrm{a}$ comes into play in this question. It makes a traditional $K_\mathrm{sp}$ question a little trickier.

Possible answer?:
\begin{align}
\ce{PbCO3 &<--> Pb^2+ + CO3^2-}
&\implies K_1 &= K_\mathrm{sp} = 1.5\times10^{-15}\\
\ce{CO3^2- + H2O &<--> HCO3- + OH-} &\implies K_2\\
\ce{PbCO3 + H2O &<--> Pb^2+ + HCO3- + OH-} &\implies K_3\\
\ce{HCO3- &<--> H+ + CO3^2-} &\implies K_4 &= K_\mathrm{a} = 4.7\times10^{-11}\\[3ex]
K_3 &= K_1 \cdot K_2\\
K_3 &= \frac{K_\mathrm{w}}{K_4}\\
K_2 &= \frac{1\times10^{-14}}{4.7\times10^{-11}}\\ &= 2.1\times10^{-9}\\
K_3 &= (1.5\times10^{-15})\cdot(2.1\times10^{-4})\\ &= 3.15\times10^{-19}\\
S_{\ce{PbCO3}} &= [\ce{Pb^2+}][\ce{HCO3-}]\\ &=[\ce{OH-}]\\
K_3 &= [\ce{Pb^2+}][\ce{HCO3-}][\ce{OH-}]\\ &= S^3\\
S &= \sqrt[3]{K_3} (3.15\times10^{-19})\\
  &= 6.8\times10^{-7}~\mathrm{M}\\
\end{align}

Does the way I solved it look okay?

List known reactions and equilibrium constants

\begin{align}
\ce{PbCO3 &<--> Pb^2+ + CO3^2-} \quad \quad \quad \quad \quad K_\mathrm{sp} = 1.5\times10^{-15}\\
\ce{HCO3- &<--> H+ + CO3^2-} \quad \quad \quad \quad \quad \quad K_\mathrm{a2} = 4.7\times10^{-11}\\
\ce{CO3^2- + H2O &<--> HCO3- + OH-} \quad \quad \quad \quad \quad K_\mathrm{b2} = \dfrac{K_\mathrm{w}}{K_\mathrm{a2}} = 2.13 \times 10^{-4}\\
\end{align}

Start off by just evaluating quick and dirty

Assume $\ce{[Pb^{2+}] = [CO3^{2-}]}$, $\ce{[Pb^{2+}] = \sqrt{1.5\times10^{-15}} = 0.387 \times10^{-7}}$

OK, the assumption is no good and this is going to be fairly nasty...

The pH is close to the autoionization of water
$\ce{CO3^{2-}}$ is a strong base so $\ce{HCO3^{-}}$ can't be ignored.

GIVEN:

Starting off with pure water so $\ce{[H+] = [OH-] } = 1\times 10^{-7}$ before $\ce{PbCO3}$ is dissolved.
$\ce{[CO3^{2-}] = 0}$ and $\ce{[HCO3^-] = 0}$ before $\ce{PbCO3}$ is dissolved.

WORK

Know that

The final pH > 7 due to the formation of $\ce{HCO3^{-}}$
$\ce{[Pb^{2+}] = [CO3^{2-}] +[HCO3^-]}$
$\ce{[OH^-]_{final} - [OH^-]_{water} = [HCO3^-] }$ or $\ce{[OH^-]_{final} = 1.00\times 10^{-7} + [HCO3^-] }$

For the ratio $ \dfrac{\ce{[CO3^{2-}]}}{\ce{[HCO3^-]}}$ we can use the expression of the $K_\mathrm{a2}$ equilibrium to find that:

At pH 7 ratio = 0.00047
At pH 8 ratio = 0.0047
At pH 9 ratio = 0.047

So it would seem that $\ce{[HCO3^{2-}] >> [CO3^-]}$ is a reasonable assumption, which also means that $\ce{[Pb^{2+}] \approx [HCO3^-]}$.

Now we have the two equilibrium equations

$K_\mathrm{sp} = 1.5\times10^{-15} = \ce{[Pb^{2+}][CO3^{2-}]}$

$K_\mathrm{b2} = \dfrac{K_\mathrm{w}}{K_\mathrm{a2}} = \dfrac{\ce{[HCO3^-][OH-]}}{\ce{[CO3^{2-}]}}$

which we can multiply together to get

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} = \ce{[Pb^{2+}][HCO3^{-}][OH^-]}$

APPROXIMATE CUBIC SOLUTION

if we let $x = \ce{[Pb^{2+}]}$ then since $x = \ce{[Pb^{2+}] \approx [HCO3^-]}$ and $\ce{[OH-] = [HCO3^-] + 1.00\times10^{-7}}$ we have

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} =(x)(x)(x + 1\times10^{-7}) =x^3 + (1\times10^{-7})x^2$

Now if we multiply both sides by $10^{21}$ we are trying to solve:

$319.149 = x^3 + x^2$

neglecting the square term we get $x=\sqrt[3]{319.149} = 6.834$, or $\ce{[Pb^{2+}] = 6.834\times10^{-7}}$.

But looking at our cubic equation again

$319.149 = x^3 + x^2 \text{ =?= } 6.834^3 + 6.834^2 = 365.821 $

which is too far off since we need values to about 1%. Solving the cubic by iteration we get $x = 6.516$

    Using 2 guesses then interpolation
    6.834           365.876
    6.000           252.000
    6.492           315.758
    6.517           319.257
    6.516           319.116
    Check
    $319.149 = x^3 + x^2 \text{ =?= } 6.516^3 + 6.516^2 = 319.116 $

which gives:

$\begin{align}
\ce{[Pb^{2+}]} &= 6.516 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.516 \times10^{-7}\\
\ce{[OH^{-}]} &= 7.516 \times10^{-7}\\
\end{align}$

We can calculate

$\ce{[CO3^{2-}]} = \dfrac{K_\mathrm{sp}}{\ce{[Pb^{2+}]}} = 0.023\times10^{-7} $

$\ce{[H+]} = \dfrac{K_\mathrm{w}}{\ce{[OH-]}} = 0.133\times10^{-7}$

CHECK:

$\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^{-}]}} = 4.695\times10^{-11}$

So to the correct number of significant figures we have

$\begin{align}
\ce{[Pb^{2+}]} &= 6.5 \times10^{-7}\\
\ce{[CO3^{2-}]} &= 0.023 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.5 \times10^{-7}\\
\ce{[H^{+}]} &= 0.13 \times10^{-7}\text{,}\quad pH = 7.88\\
\ce{[OH^{-}]} &= 7.5 \times10^{-7}
\end{align}$

EXACT FORMULIC SOLUTION

Starting with the formula

$\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} = \ce{[Pb^{2+}][HCO3^{-}][OH^-]}$

We let $x = \ce{[Pb^{2+}]}$ as before.

$\ce{[HCO3^-] = [Pb^{2+}] - [CO3^{2-}] =} x - \dfrac{K_\mathrm{sp}}{x} = \dfrac{x^2 - K_\mathrm{sp}}{x}$

$\ce{[OH-] = [HCO3^-]} + \sqrt{K_\mathrm{w}} = \dfrac{x^2 - K_\mathrm{sp}}{x} + \sqrt{K_\mathrm{w}} = \dfrac{x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}}{x}$

$\begin{align}
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}} &= (x)(\dfrac{x^2 - K_\mathrm{sp}}{x})(\dfrac{x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}}{x})\\
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}}x &=(x^2 - K_\mathrm{sp})(x^2 +(\sqrt{K_\mathrm{w}})x - K_\mathrm{sp}) \\
\dfrac{K_\mathrm{w}K_\mathrm{sp}}{K_\mathrm{a2}}x &= x^4 + (\sqrt{K_\mathrm{w}})x^3 - 2K_\mathrm{sp}x^2 - K_\mathrm{sp}\sqrt{K_\mathrm{w}}x + K^2_\mathrm{sp}\\
0 &= x^4 + (\sqrt{K_\mathrm{w}})x^3 - 2K_\mathrm{sp}x^2 - \dfrac{K_\mathrm{w}K_\mathrm{sp}+ K_\mathrm{sp}K_\mathrm{a2}\sqrt{K_\mathrm{w}}}{K_\mathrm{a2}}x + K^2_\mathrm{sp}\\
\end{align}$

$x = 6.531198\times10^{-07}$, and other values same as Excel also...

EXACT SOLUTION WITH EXCEL

Solved first by using an excel spreadsheet (with too many significant figures, but extra digits help show that the calculations are consistent)...*

$\begin{align}
\ce{[Pb^{2+}]} &= 6.531199 \times10^{-7}\\
\ce{[CO3^{2-}]} &= 0.022967 \times10^{-7}\\
\ce{[HCO3^{-}]} &= 6.508231 \times10^{-7}\\
\ce{[H^{+}]} &= 0.133187 \times10^{-7}\text{,}\quad pH = 7.8755376\\
\ce{[OH^{-}]} &= 7.508230 \times10^{-7}
\end{align}$

CHECKS:

$\ce{[H^{+}][OH-] = 0.9999986\times 10^{-14}}$

$\ce{[Pb^{2+}][CO3^{2-}]} = 1.500020\times10^{-15} $

$\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^{-}]}} = 4.700057\times10^{-11}$