On heating in the Earth's atmosphere, can magnesium react with nitrogen to form magnesium nitride? This question is derived from a question asked in my school test.

What happens when a magnesium ribbon is heated in air?

My first response was the formation of magnesium oxide $(\ce{MgO})$ when oxygen in air reacts with magnesium at a high temperature which can be expressed in the form of a chemical equation like this:
$$\ce{2 Mg(s) + O2(g) ->[\Delta] 2 MgO(s)},$$
but I was wondering if magnesium could react with any other gas in the air to form a compound with that gas and I found out that magnesium does react with nitrogen in the air to form magnesium nitride too:
$$\ce{3 Mg(s) + N2(g) ->[\Delta] Mg3N2(s)}.$$
What determines whether the heated magnesium ribbon will react with the oxygen in the atmosphere or the nitrogen in the atmosphere?
Two possibilities that I can think of are:

composition of the air;
temperature.

I don't think that composition is the answer because on average the atmosphere of Earth has more nitrogen than oxygen, so I think that the answer may be temperature.
I'd also like to know how the factor affects the chemical reaction on an atomic level.

A large pile of grey magnesium powder, when lit in air, produces a smouldering pile which cools down to reveal a crusty white solid of magnesium oxide. However, if you break apart the mound, you can find something quite strange in the middle - a clearly brownish powder that wasn't there before.
Seeing is believing! The author of the video also has a clever idea to identify the brown solid. By adding water and placing some moist pH paper above the puddle, it clearly shows the transfer of some alkaline substance across the gap. This is ammonia gas, $\ce{NH3}$, whose presence is explained by the hydrolysis of magnesium nitride:
$$\ce{Mg3N2(s) + 6H2O(l) -> 3 Mg(OH)2(aq) + 2 NH3(g)}$$
It is important that the pH paper not come in direct contact with the water used to hydrolyze the magnesium oxide, as $\ce{Mg(OH)2}$ is itself also basic, and could also be formed by reaction with either $\ce{MgO}$ or $\ce{Mg}$ directly. Only $\ce{Mg3N2}$ produces a basic gas which forms an alkaline solution in water.
As you can see, magnesium metal does react directly with molecular nitrogen ($\ce{N2}$) when burned in air. However, the reaction is thermodynamically and kinetically less favourable than the reaction with molecular oxygen ($\ce{O2}$). This is almost certainly due to the extreme strength of the bond between nitrogen atoms in molecular $\ce{N2}$, whose bond dissociation energy of $\mathrm{945\ kJ\ mol^{-1}}$ is one of the strongest in all of chemistry, second only to the bond in carbon monoxide. For comparison, the bond dissociation energy of molecular $\ce{O2}$ is drastically lower, at $\mathrm{498\ kJ\ mol^{-1}}$.
So why did the Chem13 magazine article referenced in Aniruddha Deb's answer not find any magnesium nitride? It is likely that 1 g of magnesium metal is far too little for the experiment run under their conditions. It takes a significant amount of "sacrificial" magnesium to completely consume the oxygen in its surroundings. Only once practically all the oxygen is consumed (and while the pile of magnesium is still hot enough from the reaction between magnesium and oxygen) will the remaining magnesium metal react with the nitrogen in air. Alternatively, the reaction would have to be performed in an oxygen-free environment. Magnesium metal is such a strong reductant that many substances can act as an oxidant for it, including pure $\ce{CO2}$ (also shown in the video above) and water (never put out a magnesium fire with water!).


For hydrogen like atom, find velocity from potential energy PE of system is given by $$\mathrm{U(r) = \frac{-Ke²}{r^3}}$$ assume bohr model to be valid Find velocity

For this I used $\mathrm{\frac{-PE}{2} =KE}$ and then used $$\frac{mv^2}{2} =\frac{Ke^2}{2r^3}$$ and $$mvr=\frac{nh}{2\pi}$$

However, I am getting the answer as $$v=\frac{n^3h^3}{16Ke^2\pi^3m^2}$$  but the answer given has a factor of 24 rather than 16. I don't know where I'm going wrong.

The fault in your solution is the part where you assume  $\mathrm{\frac{-PE}{2} =KE}$ this is true only if the force follows inverse square law.

In this particular case we need to find the force first by using

$$F =\frac{-dU}{dr}$$

Which gives us

$$F =\frac{-3Ke^2}{r^4}$$

Now this force is central and provides centripetal force

$$\frac{mv^2}{r} =\frac{3Ke^2}{r^4}$$

This along with
$$mvr=\frac{nh}{2\pi}$$

gives us
$$v=\frac{n^3h^3}{24Ke^2\pi^3m^2}$$


Formation of species in electrolysis I understand the principles of electrolysis of salts in aqueous solutions, but there are two points on which i am unsure.

At the positive electrode, how can you work out which ions will preferentially donate electrons?
If the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?

If so, what decides which ions are formed?

What happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?

For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?

For the last question, I believe the latter would be true because if
  the anions produced at the negative electrode react with the cations
  that are attracted to the electrode, the cations that are attracted to
  the electrode would not be available to accept electrons and hence the
  current would become increasingly restricted until it stops all
  together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic and galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

Why the salts in a salt bridge?
Voltaic and electrolytic cells
Difficulties understanding how salt bridge works
Why is it important to use a salt bridge in a voltaic cell? can a wire be used?

Don’t be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

  Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of $2.2\ \mathrm{A}$ is run through them. A list of standard reduction potentials is as follows.

  $$\begin{alignat}{2}
\ce{O2_{(g)} + 4H+ + 4e- &-> H2O_{(l)}}\quad&&1.23\ \mathrm{V}\\
\ce{F2_{(g)} + 2e- &-> 2F-}\quad&&2.87\ \mathrm{V}\\
\ce{2H2O{(l)} + 2e- &-> H2_{(g)} + 2OH-}\quad&&{-}0.83\ \mathrm{V}\\
\ce{Ni^2+ + 2e- &-> Ni_{(s)}}\quad&&{-}0.25\ \mathrm{V}
\end{alignat}$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^2+}$ and $\ce{F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since fluorine already exists reduced, and it does not like being oxidized. Then you have nickel and water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this equation for the reaction to make sense. That leaves us with:

Equation 1 (this one is flipped) and 4. $\ce{Ni^2+}$ is being reduced while $\ce{H2O}$ is being oxidized.

In general …

Eliminate which reaction cannot take place.
Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)


Why is RO water more corrosive? According to several articles from the American Water Works Association, permeate from RO should be stabilized in terms of pH and hardness to avoid corrosion of the distribution piping.
How does corrosion differ from that of normal water? Would RO water corrode stainless steel faster than normal water if both had the same chloride concentration?

Water passing RO does not have the same chloride concentration.
The raw RO water contains almost no ions, aside of those that passed by leakages of the osmotic membrane.
This disbalances the water natural $\ce{CO2/HCO3-}$ pH buffer, leaving the ionic alkaline buffer part before membrane, so water becomes more acidic with corrosion aggressive $\ce{CO2(aq)}$.


Why isn’t azete aromatic? I’m confused, because it seems like Hückel’s rule is satisfied (two electrons from the nitrogen lone pair and four from the double bonds) and that the molecule should be aromatic. Where am I going wrong here?

The compound is anti-aromatic. While counting the number of π-electrons, you count the electrons which are delocalized over the ring. In this case the nitrogen lone pair is localised and does not participate in resonance. The nitrogen lone pair is in an $\mathrm{sp^2}$ orbital (red) which is orthogonal to the π system (blue):

So, the total number of π-electrons is only four: two from each double bond.

Please read this for a better understanding : http://www.chem.ucla.edu/harding/ec_tutorials/tutorial04.pdf


Extraction of a pyridine tetrazole from the aqueous phase I have synthesised a compound which is basically a pyridine ring connected to a tetrazole. The procedure suggested to precipitate it at pH 1, but it did not precipitate. It then suggested to extract it with ethyl acetate.

The problem is that it is not going to the organic layer but into the aqueous layer. I have tried both at pH 1 and 7 and it still does not move to the organic layer.

Any ideas how to solve this problem?

Concentrate under reduced pressure on the rotovap. Add a miscible co-solvent that does not solubilize the zwitterion at its isoelectric point but dissolves starting materials, like THF.

The pyridine is basic, the terazole is acidic. Catch it on an ion exchange resin column in water, then displace it with a polyvalent counterion.  Possibly down neutral alumina in water to slow it and concentrate it.  You can run a microcolumn in a disposable pipette with a tiny ball of glass wool inside at the taper.

Salt it out as a solid with saturated NaCl or potassium nitrate its isoelectric point. Precipitate it, perhaps as the tetraphenylborate or picrate (careful there).


Neutralizing a sulfuric acid spill From my AP Chem Prep Book:

What should you do if you spill sulfuric acid on the countertop?
A) Neutralize the acid with vinegar.
B) Sprinkle solid NaOH on the spill.
C) Neutralize the acid with NaHCO3.
D) Neutralize the acid with an Epsom salt $(\ce{MgSO4})$ solution.

The answer given was C — stating that it should be $\ce{NaHCO3}$ because it is a weak base solution.
However, I do not understand why the strength of the base is of relevance (given the situation) and how to determine the strength of the base — at least relative to other bases.

First, why other options are not really the options:

A: vinegar, being a weak acid, doesn't neutralize sulfuric acid and only dilutes it;

B: solid sodium hydroxide, a strong base, does neutralize sulfuric acid, but it does so vigorously releasing substantial amount of heat per unit of time:
$$\ce{2 NaOH(s) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l)}$$
Using solid NaOH it is also tricky to guarantee it will cover the spill and there won't be any unreacted acid or excessive hydroxide (which is equally unwanted) left behind as there is little to no visual clue whether the neutralization is complete, unless you test various spots with, let's say, pH paper.

D: Magnesium sulfate solution doesn't react with sulfuric acid and only dilutes it.

Second, sodium bicarbonate solution not only neutralizes the acid

$$\ce{2 NaHCO3(aq) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)},$$

but also can be (and should be!) used in excess to assure complete acid neutralization.
Once the acid is neutralized, you are left with solution of sodium bicarbonate and sodium sulfate which is only slightly basic due to cation hydrolysis.
Another important advantage of using $\ce{NaHCO3}$ is visual control: once the process is complete, gas evolution stops.


Is a 50/50 mix of Acetone and Water inert? We are cleaning a gel out of a fuel tank that is dissolved by Acetone.  We want to neutralize the acetone by adding water to achieve a mix of 50/50 so that (PERHAPS) the acetone will not damage rubber / plastic compounds in the fuel system.

We have an electric fuel pump we are going to use to get the acetone and water out and we need to make sure we don't damage it. THEN there are the two brand new mercruiser marine fuel pumps, fuel filter and carburetor seals we don't want to damage.   We CAN  add more water than a 50/ mix as well.

We know we can't get all the contents of the tank out and so we want to make sure any remaining acetone will not be of a concentration to ruin the fuel system components.  By the way... this is for a famous movie director Roman Coppola.

Acetone is notoriously aggressive towards the plastic, rubber, and other non-metal components of internal combustion engines (and associated parts like fuel pumps). If the concentration of acetone is strong enough to dissolve your jell (assuming that is some organic compound) it is very likely strong enough to attack components of the fuel pump.

As a bit of anecdotal evidence regarding what I said above, note that of the dozen-odd oxygenated additives allowed by the US EPA for the oxygenated fuels program, neither acetone nor any other ketone is on the allowed list. Although this is in part due to the high volatility of acetone, it is also largely due to the fact that acetone would attack engine components as described above.

In short, what you've described is a bad idea for the health of your fuel pumps, even at very low concentrations of acetone.


Homemade reactor for water cooling Background:
Well, it's summer here in Brazil, and a great way to keep cool is drinking a so called Tereré, and, of course, we do need cold water for that.

Problem: I was looking for endothermic spontaneous reactions so I can cool the water in a bottle using an inner aluminum bottle in which the endothermic reaction occurs. I already knew that Urea's dissolution is endothermic and spontaneous, but I couldn't find how much endothermic it is, i.e., how many Joules it absorbs per Mol of urea dissolved. Another problem related with urea is that it's dissolution is a slow process (I think, not sure), additionally, the urea I can find is the one use as fertilizer, and this one has lots of impurities that I'll need to take into account when calculating how much I'll need to react.

What I know is that the urea dissolves in the same amount of water in weight, that the reaction time increases as the concentration increases, but have no idea how much energy it absorbs neither the how much time it takes to react, of course it depends on the concentration, granularity, etc.

Question: Considering all of that, are there any other reagent(s) I can buy or synthesize without being a licensed chemist that can be used to cool the water? Or, do you have the information that is missing about urea's dissolution (Joules/Mol absorbed in dissolution, approximated reaction time) so I can figure out if I can use it for this purpose?

Unfortunately, many aqueous endothermic reactions involve some chemicals you may wish to avoid.  Ammonium nitrate is the classic example, but in the US it is becoming more and more unavailable because of its potential use as an explosive.  Another frequently mentioned possibility is reacting hydrated barium hydroxide with ammonium chloride, but $\ce{Ba(OH)2}$ is very caustic and poisonous.

So of what I've read, your best choices might be mixing water with a chloride salt like ammonium or potassium.  I do not know how cold this will get your container. I suspect it will not be as good as ammonium nitrate, which would probably be ideal given the former use of this compound in medicinal cold packs.


Which are stronger intermolecular forces: ionic bonds or hydrogen bonds? Do strong ionically attracted molecules have stronger intermolecular forces than hydrogen bonding forces?

Hydrogen bonds are the strongest intermolecular forces. Stronger than van der Waals' forces, namely (ordained by strength):

Permanent dipole - permanent dipole (Keesom's Force);
Permanent dipole - induced dipole (Debye force);
induced dipole - induced dipole (London's dispersion force);

Strong hydrogen bonds are created by the FONs (fluorine, Oxygen and Nitrogen), which are very electronegatives, what produces a strong attraction between one them and hydrogen (which is less electronegative). As example of this interaction we can cite the water, that's a polar molecule in a "V" shape, take a look at the figure below.

Then, the both hydrogens yields a $\delta^{+}$ that's attracted by the Oxygen (very strong electronegative, the $\delta^{-}$ side) of the another water molecule, yielding the hydrogen bonds, take a look at the figure below.


Normal phase vs reverse phase HPLC What is the difference between normal phase and reverse phase high-performance liquid chromatography?

Normal phase

Normal phase HPLC systems are similar to the flash-column chromatography that you might be familiar with.

A silica stationary phase is eluted with a non-polar solvent such as hexane, or a fairly non-polar solvent mixture such as 2-propanol in hexanes. In normal phase chromatography, only organic solvents are used.

In the normal phase, polar molecules elute slowly, and non-polar (greasy) molecules elute quickly.

Reverse phase

Reverse phase is essentially the opposite of normal-phase.

A non polar stationary phase (often silica in which the free hydroxyl groups are end-capped with something greasy, C18 chains are common but many many variants are possible) is eluted with a polar solvent such as acetonitrile/methanol, or a fairly polar solvent mixture (acetonitrile water mixtures are common, or methanol water mixtures).

In the reverse phase, polar molecules elute quickly, and non-polar (greasy) molecules elute slowly.


detecting/tracking nano particle I am new to nano-sized particles and conducting a very simple experiment using nano particles. In my experiment I have nanomagnetic particles inside water and they are moving due to applied magnetic field. I am looking for a method to track or detect the motion of nano particles. Is there any easy way?

There are a few ways to do nanoparticle tracking, although I'm not sure they will exactly apply to your system.

Dynamic light scattering or DLS has been used to get the distribution of nanoparticle sizes in solution/suspension over time. Generally, the data gives information on the ensemble, not single-particle tracking.

Nanoparticle tracking analysis or NTA also relies on scattered light, but gives frame-to-frame analysis of each particle's trajectories.

There are other related techniques... Another approach is to create an Anti-Brownian ELectrokinetic (ABEL) trap - to monitor the particle or protein dynamics and apply forces to push it back. This technique was developed by the Moerner lab at Stanford, for example Acc. Chem. Res., 2012, 45 (11), pp 1955–1964


Why acidity of p-methoxybenzoic acid is more acidic than p-hydroxybenzoic acid? I know +M effect of $\ce{OCH3}$ is more than $\ce{OH}$, but in my book (ALLEN General organic chemistry), it is given that $\mathrm{p}K_\mathrm{a}$ of para-substituted benzoic acid containing $\ce{OCH3}$ is less than that of para-substituted benzoic acid containing $\ce{OH}$:

$\mathrm{p}K_\mathrm{a}$/Benzoic acid → $4.20$
$$
\begin{array}{lcccc}
\hline
\text{Substituent} & \mathrm{p}K_\mathrm{a} & \textit{o-} & \textit{m-} & \textit{p-} \\
\hline
\cdots \\
\ce{OMe} & - & 4.09 & 4.09 & 4.47 \\
\ce{OH} & - & 2.98 & 4.08 & 4.58 \\
\cdots \\
\hline
\end{array}
$$

I am in doubt whether these data wrong or correct. If correct, what can be an explanation?

Both -$\ce{OCH3}$ and -$\ce{OH}$ groups have exhibited two effects on the aromatic ring: (1) Electron donating resonance or mesomeric effect (+M) and (2) Electron withdrawing inductive effect (-I). For both acids in hand, the electron withdrawing inductive effect (-I) is almost same since both -$\ce{OCH3}$ and -$\ce{OH}$ groups are 4 carbons away from the acid center. However, when -$\ce{OH}$ group is attached to either the para- or ortho-position, it has a more tendency to delocalize its lone pair electrons towards the aromatic ring than that of -$\ce{OCH3}$ group (Order of activating:  -$\ce{O-} \gt$  -$\ce{OH} \gt$  -$\ce{OCH3}$; for additional learning, read articles on Hammett plots).

As a result, the electron density on the carbon para to -$\ce{OH}$ substitution group increases more than that on the carbon para to -$\ce{OCH3}$ substitution group. Thus, the electron density on carbonyl carbon of carboxylic group attached to those para-carbons also increases accordingly, and consequently, the polarity of $\ce{O-H}$ of carboxylic group decreases as a result. That polarity decrease causes the reduced acidity ($\mathrm{p}K_\mathrm{a}$s are $4.58$ and $4.47$, when para-substitutions are $\ce{OH}$ and $\ce{OCH3}$, respectively) compared to benzoic acid ($\mathrm{p}K_\mathrm{a}$ is $4.20$ when para-substitution is $\ce{H}$). The $\mathrm{p}K_\mathrm{a}$s of p-hydroxybenzoic acid is higher than that of p-methoxybenzoic acid here, because as explained above, p-$\ce{OH}$ group caused reduced polarity on $\ce{O-H}$ of carboxylic group than that caused by p-$\ce{OCH3}$ group.


Why do we consider all gasses to be in one phase? Why do we take the phase of all gases 1. If all gases exist in 1 phase then why do we say that some gases are heavier than air, some are light also we collects some gases in lab by air displacement method. Is this all possible if gases are completely miscible.

Just because two fluids are completely miscible does not guarantee that they are uniformly mixed at all times.

For example, if you have a glass half full with sugar water solution, you can gently pour pure water to form a layer on top.  This is the basis for sucrose density gradient centrifugal separation and layered drinks.

So the same can occur for gases, a non-equilibrium state can exist for considerable time where the gases aren't mixed even though they are completely miscible.


Pressure of two flasks connected to each other at different temperatures Two flasks of equal volume connected by a narrow tube (of negligible volume) are at $\pu{27 °C}$ and contain $\pu{0.7 mol}$ of $\ce{H2}$ at $\pu{0.5 atm}$. If one of the flasks is then immersed into a bath kept at $\pu{127 °C}$ while the other remains at $\pu{27 °C}$. What is the final pressure?

Will the pressure of the two flasks which are now kept at different temperatures be the same or will they have different pressures?
In the solution I found of this question, the pressure of the two flasks at different temperatures are taken to be equal. Can someone please explain why this is so?

The easiest way to prove the equality in pressure is by taking a cross-section of the narrow tube and finding the forces acting on it. By extending this, we can prove that the pressures are equal.
Take a cross-section area of the narrow tube. We assume this is at steady state and there is no average motion of particles in the entire system.
Now, since there is no net average motion, this means that there is no pressure gradient across any cross-section. Expanding this cross-section into the tube, we see that the ends of the tube have to be at the same pressure since there is no net average motion of the gas within the two flasks at steady state (which is where the parameters are asked to be measured).
Therefore $$p_\mathrm{left} = p_\mathrm{right}$$


How to balance more complex redox reactions? I'm in an advanced level chemistry course, and so far we've been introduced to balancing redox reactions. The basic ones like
$$\ce{MnO4- + Cl- -> Mn^2+ + Cl2}\tag1$$
are pretty easy for me, but I'm having trouble with equations where it's not apparent what I exactly need to put in my half reactions. Some examples of these kind of equations are:
\begin{align}
\ce{Pb + PbO2 + SO4^2- &-> PbSO4}\tag2\\
\ce{Ag(s) + CN- (aq) + O2(g) &-> [Ag(CN)2]- (aq)}\tag3\\
\ce{Cu + HNO3(aq) &-> Cu(NO3)2(aq) + NO2(g) + H2O(l)}\tag4
\end{align}

I'm not looking for solutions to any of these problems. I'm just looking for ways to correctly construct my half reactions before balancing, adding water and hydrogen ions, and balancing the charges.
Like in the first reaction on the reactants side, the first $\ce{Pb}$ has a charge of $0$ while the next one has a charge of $+4$. The $\ce{Pb}$ on the products side however has a charge of $+2$. It seems like one is being oxidized and the other reduced. And I'm also not sure what to do with the sulfate.

Once you know the method, balancing any redox equation that you will get in AP chem will be quite easy. All you have to do is follow this simple method:

Balance all the elements in the equation except for $\ce{O}$ and $\ce{H}$. For polyatomic ions that are spectator ions, just think as them as one big element.
Check if $\ce{O}$ are balanced. If they aren't, add water molecules to the side, witch doesn't have enough $\ce{O}$.
Check if $\ce{H}$ are balanced. If they aren't, add $\ce{H+}$ ions to the side, witch doesn't have enough $\ce{H}$.
Now that all the elements are balanced, make sure that the charges are balanced. If not balanced, add electrons to the side which is more positive.
Check your equation now to make sure that everything is balanced correctly.

I will do the first equation as an example for you. You should be able to balance rest of the equations by following this procedure.

Balanced elements except for O and H: Here the sulfate ion is a spectator ion, so you can imagine it as one big element: $$\ce{Pb + PbO2 + 2SO4^2- -> 2PbSO4 }$$
Balance oxygen: $$\ce{Pb + PbO2 + 2SO4^2- -> 2PbSO4 + 2H2O }$$
Balance hydrogen: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -> 2PbSO4 + 2H2O }$$
Balance charge: charge is already balanced. Don't need to add electrons.
Check equation to make sure everything is balanced correctly: all the elements and charges are balanced correctly.

So your balanced redox equation for the first example is: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -> 2PbSO4 + 2H2O }$$

Note: If the question asks to provide the redox equation in basic conditions there are two more steps.

Add the same number of $\ce{OH-}$ ions as $\ce{H+}$ to the products and reactants. The $\ce{OH-}$ ions and $\ce{H+}$ will react to form water.
Cancel out the water molecules from both sides (a bit like algebra).

That should give the redox equation in basic conditions.


Can water be a Lewis acid? I asked this question to my teacher a few days ago. In a reaction such as

$$\ce{NH3 + H2O -> NH4+ + OH-},$$

$\ce{NH3}$ is a Lewis base, but is $\ce{H2O}$ a Lewis acid?

He said that it isn't. Now I fail to see the reason why. Can someone explain?

Strictly speaking, a Lewis acid is something that forms a bond by accepting an electron pair form another molecule (Lewis base).  Water as such does not do that, rather it is a hydrogen ion from the water that does so.  When ammonia acts as a proton acceptor in water, the Lewis theory calls it a displacement reaction where the acid, $\ce{H+}$, is initially combined with one base ($\ce{OH-}$) and ends up combined with another base ($\ce{NH3}$).


What is the relationship between dipole moment and solubility? I thought a large dipole moment meant the substance was polar, so I'm confused by the following statement (from Wikipedia) that "With a dipole moment of 3.92 D, acetonitrile dissolves a wide range of ionic and nonpolar compounds...." Why would it dissolve many nonpolar compounds?

It's all about the energetics (assuming an isolated system--happy to elaborate in the comments): as acetonitrile has a large dipole moment, it can readily stabilize the individual ions. See this discussion of solvation energy.

The more subtle case is in the case of nonpolar compounds. In this case, the large dipole can create an induced dipole in the nonpolar molecule. Then the dipole-induced dipole attraction would be larger in magnitude than the spontaneous dipole-spontaneous dipole interactions present in nonpolar-nonpolar interactions (which are also still present as small fluctuations in the polar-nonpolar case). See this discussion of how charges/dipoles/multipoles induce dipoles and then subsequently attractively interact with said dipoles.


percent ionization explanation Using a weak acid as an example:
$$\ce{CH_3COOH <=> CH_3COO^- + H^+}$$

The position of this equilibrium shifts to the right with increasing dilution.

I have found textbooks that say this, but do not provide an explanation. Why is there increasing ionization with increasing dilution?

I was wondering if it has to do with the free energy available; since ethanoic acid is a weak acid, energy has to be provided to promote the proton to a higher free energy state (transfer from $\ce{CH_3COOH}$ to $\ce{H_2O}$), therefore, "more" free energy for promotion and less contact with other $\ce{CH_3COOH}$ molecules?

For $\ce{HA <=> A- + H+}$  with a particular $K_a$ value at a particular temperature (Also, remember it usually applies to weak acid as $K_a$ of strong acid is rather meaningless),
Consider its equilibrium expression: $$K_a=\frac{\ce{[A^{-}][H+]}}{\ce{[HA]}}$$ when water (of the same temperature) is added, all three component concentration will decrease with the same extent.

Think about whether the numerator and denominator has a bigger decrease and you will find the answer (temperature is relatively constant). And for your free energy concern, being out of equilibrium necessarily means there is a favorable driving force pushing the reaction in one direction to reach back eqm.

Acquire a physical picture about the initial and equilibrium state and the dynamic state of a reaction in terms of collision theory. Think equilibrium in the form of competing forward and backward reaction and equilibrium constant is actually derived from rate constant.

Another reminder: Temperature and concentration/pressure all affect free energy. At the initial state (when you just add water and no reaction has taken place yet), there is only change in concentration but not temperature. So even if the temperature is changed upon reaction, the percentage of ionization will also be higher in all circumstances.


Determining a molecular formula of the catalyst for polymerisation of butadiene I've got the following question in a workbook which I need to solve:

  A $\pu{50 g}$ sample of an experimental catalyst used in the polymerisation of butadiene is made up of $\pu{11.65 g}$ of $\ce{Co}$ and $\pu{25.7 g}$ of $\ce{Cl}$. If the molar mass of the compound is $\pu{759 g/mol},$ find the molecular formula of the catalyst.

I attempted to first find the percentage by mass of each given element by dividing their masses ($\pu{11.65g}$ and $\pu{25.7 g},$ respectively) by the mass of the entire catalyst $(\pu{50 g}).$ From there, I divided both percentages ($23.3$ and $51.4)$ by the smallest one, and multiplied the results $(1$ and $2.2)$ by $5$ to get two whole numbers. The formula I ended up with was $\ce{Co5Cl11}.$

Now that made sense to me, but somehow the correct answer is $\ce{Co3Mo2Cl11}$. I don't understand how I'm supposed to figure out that $\ce{Mo}$ is required? Could somebody please shed some light on this for me?

Despite the absurd lack of data, this problem attracted my attention, probably because it closer resembles a real-life challenge rather than a textbook problem.
It looks like it's an adaptation of the problem from Schaum’s Outline of Theory and Problems of College Chemistry [1, p. 39]:

3.37 What is the empirical formula of a catalyst that can be used in the polymerization of butadiene if its composition is $23.3\%$ $\ce{Co},$ $25.3\%$ $\ce{Mo},$ and $51.4\%$ $\ce{Cl}.$
Ans. $\ce{Co3Mo2Cl11}$

Here we are dealing with zero knowledge as to what the remaining element(s) are, but with a couple of assumptions we actually can come up with an answer from the textbook, and an answer on top of that with a justification.
Let's dive in and assume we are dealing with an unknown compound with the following formula:
$$\ce{Co_xCl_y}\sum_{i=1}^N \ce{El}_{i,z_i}$$
where we account for $i$ other elements $\ce{El_$i$}$ with the respective coefficients $z_i.$
The first assumption required to advance with the solution is to treat the compound as stoichimetric. e.g. $x, y, z_i \in\mathbb{N}.$
This would immediately allow us to find the exact values for both $x$ and $y$:
$$y : x = \frac{m(\ce{Cl})}{M(\ce{Cl})} : \frac{m(\ce{Co})}{M(\ce{Co})} = \frac{\pu{25.7 g}}{\pu{35.45 g mol-1}} : \frac{\pu{11.65 g}}{\pu{58.93 g mol-1}} = 3.67\tag{1}$$
To satisfy $x, y \in\mathbb{N},$ and taking into account the "hint" decimal part $.67$ (which implies a triple factor to an integer):
$$x = 3, 6, 9, \ldots, 3n~(n\in\mathbb{N})\tag{2}$$
To pinpoint the exact allowed value of $x$ (and $y,$ as $y = 3.67x),$ we can use the total molar mass:
$$M = x × M(\ce{Co}) + 3.67x  × M(\ce{Co}) + \sum_{i=1}^N z_iM_i\tag{3}$$
$$\pu{759 g mol-1} = x × \pu{58.93 g mol-1} + 3.67x × \pu{35.45 g mol-1} + \sum_{i=1}^N z_iM_i\tag{3a}$$
$$\sum_{i=1}^N z_iM_i = (759 - 189x)~\pu{g mol-1}\tag{3b}$$
Since molar mass is a positive number, $759 - 189x > 0$ and the only $x$ to satisfy this criteria would be $x = 3.$
Accordingly, $y = 3.67 × 3 = 11,$ and at this point we are left with the following formula:
$$\ce{Co3Cl11}\sum_{i=1}^N \ce{El}_{i,z_i}$$
and the remaining sum
$$\sum_{i=1}^N z_iM_i = (759 - 189 × 3)~\pu{g mol-1} = \pu{192 g mol-1} \tag{3c}$$
The second assumption is that we are dealing with the common oxidation numbers of the elements, no exotic stuff.
With this in mind, and knowing that cobalt catalyst used in polymerization are vastly only $\ce{Co^{II}}$ and $\ce{Co^{III}}$ species, we can deduce the possible charge of the remaining sum:
$$(\ce{Co3Cl11})^{q-}\left(\sum_{i=1}^N \ce{El}_{i,z_i}\right)^{q+}$$
Since it's likely only $\ce{Cl-},$ the $q$ can adopt only the values defined by $\ce{Co^{II}}:\ce{Co^{III}}$ ratio.
For the total of three cobalt(II,III) atoms $q = 2,3,4,5.$
This will assist us in proposing missing element(s), and one can start to iterate over numbers.
The third assumption is that $i = 1$ and there is only one extra element with single coefficient $z_1$.
The table below summarizes possible outcomes:
$$
\begin{array}{ccccc}
\hline
z_1 & M_1/\pu{g mol-1} & \ce{El_1} & M(\ce{El_1})/\pu{g mol-1} & \text{Formula Example} \\
\hline
1 & 192 & \ce{Ir} & 192.22 & \ce{Co^{II}2Co^{III}Ir^{IV}Cl11} \\
  & & & & \ce{Co^{II}Co^{III}2Ir^{III}Cl11} \\
2 & 96 & \ce{Mo} & 95.95 & \ce{Co^{II}3Mo^{II}Mo^{III}Cl11} \\
  & & & & \ce{Co^{II}2Co^{III}Mo^{II}2Cl11} \\
3 & 64 & \ce{Cu} & 63.55 & \ce{Co^{II}Co^{III}2Cu^I3Cl11} \\
  & & & & \ce{Co^{II}3Cu^{I}Cu^{II}2Cl11} \\
4 & 48 & \ce{Ti} & 47.87 & \ce{Co^{?}3Ti^{?}4Cl11} \\
\hline
\end{array}
$$
Starting with $z = 4,$ there doesn't seem to be an appropriate set of oxidation numbers for the element to comply with the charge balance, so I'd say the candidates for the third elements are only iridium, molybdenum and copper.
Judging from the application (catalysis) as well as from the typical oxidation states, I'd actually propose iridium cobalt chloride $\ce{Co3IrCl11}$ as the answer, yet $\ce{Co3Mo2Cl11}$ would also be possible (AFAIK, molybdenum(II,III) isn't a common composition for halide salts), and $\ce{Co3Cu3Cl11}$ would be a stretch both due to higher deviation from the declared molar mass and presence of copper(I).
References

Rosenberg, J. L.; Epstein, L. M. Schaum’s Outline of Theory and Problems of College Chemistry, 8th ed.; Schaum’s outline series; McGraw-Hill: New York, 1997. ISBN 978-0-07-053709-5.


Why did ammonium chloride fumes turned pink in this synthesis? I was watching the synthesis of tetrasulfur tetranitride performed by Tom on his Youtube channel, Explosion & fire. He followed the basic route: adding ammonia gas to sulfur chlorides ($\ce{SCl2}$ and $\ce{S2Cl2}$). At first, everything was going normal, purple-black precipitate on the flask with white $\ce{NH4Cl}$ fumes emitting as expected and white $\ce{HCl}$ fumes resulting from hydrolysis of sulfur chlorides. But suddenly a problem occurred. The fumes turned pink.  The following is a sequence in chronological order (snapshots from his videos; temperature of ppt. in brackets):

white $\ce{NH4Cl + HCl}$ fumes on top of precipitate ($\pu{T = 25 ^\circ C}$)

fumes turned pink ($\pu{T = 27-28 ^\circ C}$)

fumes entering the condenser from flask turning it red ($\pu{T = 35 ^\circ C}$)

fumes escaping the condenser ($\pu{T > 35 ^\circ C}$)

condenser changed, $\ce{NH4Cl}$ pellets observed above it, fuming doesn't stop ($\pu{T > 35 ^\circ C}$)

Eventually, after so much fuming and snowing, the end-product was achieved (when fumes turned golden brown, the original color of $\ce{S4N4}$). So, why did ammonium chloride fumes turned pink (when it was evolving)? Is it due to the compounds in the precipitate? If so, what species could have caused that color?
I did some research and found out that $\ce{S4N4}$ is thermochromic(changes color according to temperature). From britannica:

Tetrasulfur tetranitride forms thermochromic crystals, which are crystals that change colour with temperature. They are red at temperatures above 100 °C (210 °F), orange at 25 °C (80 °F), and colorless at −190 °C (−310 °F).

There is no mention of $\ce{S4N4}$ turning pink, so I proposed pink to be an intermediate color (orange -> pink -> red) and since it is a volatile compound, it was turning the fumes pink. But if the compound was pink, why did it later turn golden-brown, the original color of $\ce{S4N4}$? The precipitate was a purple-brown-black sludge, so it was definite not the end product. Thus my proposal contradicts.
prepchem says:

Tetrasulfur tetranitride is prepared by dissolving 1 volume of sulfur
dichloride in 8 or 10 volumes of carbon disulfide, cooling, and
passing in dry ammonia gas until the dark brown powdery
precipitate which forms at first has dissolved and an
orange-yellow solution results which contains light-colored flocks of ammonium chloride.

Here, there is no mention of pink. What species could have caused fumes to turn pink in color?
I recommend to watch the synthesis in details on his second channel, Extraction & Ire: Part 1 and Part 2

$\ce{NH3}$ and $\ce{SCl2}$ form $\ce{NH4Cl}$ which evolves as white fumes. $\ce{NH4Cl}$, when passed through $\ce{SCl2}$ (cherry-red liquid), gives the appearance of dense-pink fumes
$$\ce{\underset{(cherry red)}{6SCl2} + 16 NH3 -> S4N4 +2S + \underset{(white)}{12NH4Cl}}$$
That is why the fumes of $\ce{NH4Cl}$ turned pink


Multiple pKa's of Vitamin C Layman here. I was looking at vitamin C serums in skincare and wanted to know more about it. It has been awhile since taking chemistry in college so my understanding is rusty.

I want to better understand this diagram (Source: U of Iowa) that talks about the different forms vitamin C:

All I remember of $\mathrm{p}K_\mathrm{a}$s was that high $\mathrm{p}K_\mathrm{a}$s means the acid is very strong and dissociates all the way while lower $\mathrm{p}K_\mathrm{a}$ means a weaker acid. So I'm a little lost on how to interpret this diagram in terms of the $\mathrm{p}K$ values, ie the $\ce{OH}$ group is deprotonated from $\ce{AsCH2}$ to $\ce{AsCH-}$ but I'm not sure how the $\mathrm{p}K$ value fits in all of this? Layman me wants to say at $\mathrm{pH}=4.1$ is when deprotonation happens, but I have a feeling that isn't right.

You actually have it backwards. A low $\mathrm{p}K_\mathrm{a}$ is a strongly dissociating acid, and a high $\mathrm{p}K_\mathrm{a}$ is a weakly dissociating acid. It would do you well to remember the origin of $\mathrm{p}K_\mathrm{a}$ \eqref{pKa}, as well as the Henderson–Hasselbalch equation \eqref{HHeq}:

$$K_\mathrm{a}=\ce{\frac{[A-][H3O+]}{[HA]}} \Rightarrow \mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})= -\log \left( \ce{\frac{[A-][H3O+]}{[HA]}}\right)\tag1\label{pKa}$$

$${\mathrm  {pH}}={\mathrm  {p}}K_{{\mathrm  {a}}}+\log _{{10}}\left({\frac  {[{\mathrm  {A}}^{-}]}{[{\mathrm  {HA}}]}}\right)\tag2\label{HHeq}$$

From this we can surmise a physical meaning behind what the $\mathrm{p}K_\mathrm{a}$ of an acid is—it is the $\mathrm{pH}$ of the solution when $\ce{[A-]=[HA]}$. This means that for $\mathrm{pH}>\mathrm{p}K_\mathrm{a}$, $\ce{[A-]>[HA]}$. Since ascorbic acid is a diprotic acid, it has two $\mathrm{p}K_\mathrm{a}$ values, one for each deprotonation.


What is physically responsible for different compounds with the same types of atoms? I only have a GCSE in Chemistry but I have discovered a lot of resources myself. One thing I cannot find an answer to is this: What is physically responsible for the different compounds you can get with the same types of atoms.

For example, if you were to mix a bunch of carbon and hydrogen what exact process determines if you get methane, butane, ethyne etc.? Is it the different concentrations of carbon and hydrogen? Then what is the physical explanation? If there is 4 times more hydrogen than carbon might a carbon atom more likely find itself surrounded by four hydrogens by chance and then bond with them to form methane? Similarly if there is equal carbon and hydrogen is it just morel likely that ethyne will form? If this is the case then in general if you mix carbon and hydrogen in different concentrations would you always get different concentrations of methane, butane, ethyne as products just by chance? For example if there is 4 times more hydrogen than carbon then is there still a small chance that at least some ethyne will form?

Okay, I'm sure there's a much more direct, down-to-earth answer that you'd find more helpful, but I'm going for something different. To stop myself from writing a book chapter, there will be some hand-waving and cut out details. With more time and thought it could be unpacked further, but hopefully I'll at least be able to give you a taste.

Right, let's forget chemistry and try to deduce the product for the following simple reaction:

$$\ce{H2 + O2 -> ?}$$

To start off, we establish some basic rules:

Molecules are formed from the combination of integer amounts of atoms
Atoms cannot be formed or destroyed (i.e., the number of atoms is a conserved quantity)
There is a quantity (let's call it potential energy) inherent to each kind of free atom and molecule
The Universe tends towards the lowest possible potential energy

With nothing else to guide us, we are forced to consider every possible product (isomers neglected for simplicity):

$$\ce{H2 + O2 -> H + O + H_2 + HO + O_2 + H_3 + H_2O + HO_2 + O3 +  H4 +...}$$

Now, if the point is to minimize the total potential energy of the system, we can develop a heuristic (which will turn out to be wrong, but entertain the idea for now). First, find the specific atom or molecule among the listed products with the lowest possible energy,$^1$ then make as much of it as allowed by stoichiometry. When no further can be made, look for the next lowest energy molecule, then make as much of that as possible, and iterate the process until no lower energy molecules can be made.

That's nice and all, but without knowing which molecules have lower or higher energy, you don't get anywhere. Here, we either resort to some hefty quantum mechanical calculations, or empirical data. By doing some thermochemical experiments (measuring enthalpies of formation), it turns out that $\ce{H_2O}$ is the molecule with the lowest energy of the list by quite a margin, so it's going to be the main product. Indeed that is what is observed empirically when hydrogen is burnt in oxygen.

So we know that water is the lowest energy product for the reaction, but what happens if we mix $\ce{H2}$ and $\ce{O2}$ with a ratio different from that in water? Say you have two moles of $\ce{H2}$ and two moles of $\ce{O2}$, does that mean you now get $\ce{H2O2}$ or potentially some other products? In fact, this is not the case. The heuristic above still applies to minimize the potential energy of the system, so $\ce{H_2O}$ is preferentially produced. This shouldn't come as too much of a surprise, because the low energy of $\ce{H_2O}$ is an inherent feature of the molecule, and is entirely independent of how much $\ce{H2}$ and $\ce{O2}$ you decide to mix in an experiment.

Then what happens to the excess $\ce{O2}$? Again following the heuristic, it will seek out the lowest energy molecule available. Since there's no more hydrogen left to react, it has a more limited number of possibilities. It could conceivably still react and turn into other molecules (for example, $\ce{O3}$ is perfectly acceptable), but it turns out that of the remaining candidates, $\ce{O2}$ is the molecule with the lowest energy, so the reaction just stops there, and we get a mixture of $\ce{H_2O}$ and left over $\ce{O2}$.

Okay, we've made some progress, but we need to talk about some important complications. In the real world, at any temperature above absolute zero, you can actually retain a small amount of molecules with a higher potential energy. All of those lower energy molecules can randomly gain a bit of thermal energy, and every so often that causes them to react and turn into a higher energy molecule. If the lowest energy molecules are much lower in energy than anything else (as is the case for $\ce{H_2O}$ among the combustion products of $\ce{H2}$ and $\ce{O2}$), then this effect is very small and can be neglected. However, in many cases there isn't such a clear-cut winner, and you end up getting a mixture of different products with similar energies. This is especially true as you get into more complicated chemistry.

Unfortunately, even after all this we're still missing a very big piece of the puzzle. Everything I've mentioned so far is a part of thermodynamics, and that alone cannot always predict the product of a reaction. Allow me to prove it. According to everything I've argued, $\ce{H2}$ and $\ce{O2}$ will always react to form primarily $\ce{H2O}$ since it has the lowest energy, and $\ce{H2O2}$ can never be a significant product. But hydrogen peroxide is made in very large amounts industrially, so how do they do it? A quick search on the internet provides the answer, the anthraquinone process, whose net reaction is simply:

$$\ce{H2 + O2 -> H2O2}$$

What gives? What we've been missing is the concept of kinetics. For many transformations in nature, chemical or otherwise, even if the final state has a lower energy than the initial state, they are often separated by an energy barrier. Even if there can be a net release of energy, the transformation is hindered because to start off it needs a net input of energy.

This is the case even for the reaction $\ce{H2 + O2 -> H2O}$. If you just mix the gasses together at room temperature, they will not react even after a century. Either some initial energy needs to be supplied (a flame, a spark, etc.) or the energy barrier needs to be decreased by using a catalyst (Pt), after which the reaction will proceed in a fraction of a second.

When you consider all possible reactions, and all the energy barriers separating the different molecules, you get the very high-dimensional potential energy surfaces I mentioned in my comment. All our work in Chemistry boils down to finding reliable paths in (or making our own by actively modifying) the crazily complex energy landscapes and exploiting them for our purposes.

$^1$ Technically, the lowest amount of potential energy per amount of atoms the molecule is made of. You can think of it as the lowest potential energy per gram of molecules.

Here's something which might also help. Is there a case where all this potential energy and kinetic stuff doesn't skew the results, and you can tease out the purely statistical effect? Sure, let me present a typical case.

Consider the following transesterification reaction (poorly drawn because I don't currently have access to ChemDraw):

Both sides of the equation have an alcohol and an ester. The alcohols are very similar to each other, and the same can be said for the esters. Because of this, there's barely any difference between the potential energy of the products and the reagents.

Now consider taking a mole of the starting  ester and alcohol. In the beginning, there is only ethanol and no methanol, so statistically you'd expect the reaction to happen in the direction shown. As the reaction progresses, the amount of ethanol available is reduced, disfavouring further generation of products, and the presence of methanol now actually allows the reaction to also go the other way around. By pure statistics, you'd expect the reaction to stop when there's about the same amount of methanol and ethanol, half a mole each, and this is what is observed. In short, simple transesterification reactions tend to have equilibrium constants very close to 1 (e.g. this article).


Identifying the C3, C4, S4, and S6 symmetry operations in the Oh point group The Oh character table has ten symmetry classes. I am trying to identify them, but my stumbling block is in identifying the order of each class.

I will use this diagram as a reference, from Sydney Kettle's Symmetry and Structure: Readable Group Theory for Chemists pg 191 (Wiley, 3rd edition).

The classes (with their respective orders) for the Oh table are:

$$\begin{array}{cccccccccc}E & 8C_3 & 6C'_2 & 6C_4 & 3C_2 & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d}\end{array}$$

$E$ and $i$ are obvious.
$6C'_2$ and $3C_2$ are shown in the diagram.
$8C_3$ is shown in the diagram above; however, it only displays 4 axes. The character table indicates there are 8 equivalent rotations for this symmetry operation. Where do the other 4 axes for the $C_3$ class come from?
$6C_4$ is the same issue. It's only shown as 3 axes. Where are the other 3 $C_4$ axes from?

The improper rotations are given in another diagram, this time on page 194 from the same book.

The mirror and dihedral planes, $\sigma_\mathrm{h}$ and $\sigma_\mathrm{d}$, are both shown to match the order in the character table.
Similar to above for $C_3$ and $C_4$, the $S_4$ and $S_6$ rotations are only half the order supplied in the character table.

In summary, from 10 total classes in the octahedral table, I can identify all the corresponding axes/planes for 6 of those 10 classes. For the remaining 4 classes, $C_3$, $C_4$, $S_4$, and $S_6$, I can only identify half the axes. At first I thought the character table was counting both positive and negative rotations, but then the order for the $C'_2$ and $C_2$ rotations should be doubled if that were the case ($12C'_2$ and $6C_2$).

Can anyone please advise on exactly what the $C_3$, $C_4$, $S_4$, and $S_6$ operations correspond to in an octahedral complex?

At first I thought the character table was counting both positive and negative rotations, but then the order for the $C_2'$ and $C_2$ rotations should be doubled if that were the case ($12C_2'$ and $6C_2$).

You were very close. The missing symmetry operations are indeed ‘positive and negative’ but only those that lead to non-identical transformations. In a $C_2$ axis, it does not matter whether you turn clockwise or counter-clockwise. Consider the following mock-up of the central planar square of an octahedron with the ligands labelled A, B, C and D:

 A
BMD                      (1)
 C

If you apply $C_2$, you will arrive at the same orientation (2), regardless of which direction you rotate in:

 C
DMB                      (2)
 A

However, when rotating by $C_4$ clockwise and counterclockwise actually lead to two different results (3) and (4):

 B
CMA                      (3)
 D

 D
AMC                      (4)
 B

These two are not identical transformations. In fact, they have distinct symbols; depending on your direction of rotation one is $C_4$ and the other is $C_4^3$. You can also interpret them as rotation by $90^\circ$ and rotation by $3 \times 90^\circ = 270^\circ$. $C_2$ and $C_4^2$ correspond to the same transformation as well.

The same case can be made for $C_3$, the reverse rotation is $C_3^2$.

The $S_n$ cases are not quite as simple. Well, actually $S_4$ is ($S_4$ and $S_4^3$) but that’s a coincidence. $S_6$ shows this more clearly. Remember that $$S_6 = C_6 \oplus \sigma_\mathrm{h}$$

Therefore, performing $S_6$ twice is equivalent to $C_3$:

$$S_6 \oplus S_6 = S_6^2 = C_6 \oplus C_6 \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} = C_3 \oplus E = C_3$$

Thus, $S_6$ applied twice and $S_6^5$ applied twice give $C_3$ and $C_3^2$, respectively, so $S_6^2$ and $S_6^4$ are already taken care of. A last case needs to be made for $S_6^3$: This is identical to $i$ as can be seen below:

$$S_6^3 = S_6 + S_6 + S_6 = C_6 \oplus C_6 \oplus C_6 \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} = C_2 \oplus \sigma_\mathrm{h} = i$$

Thus, while you could imagine up to six different $S_6$ transformations, only two of those ($S_6$ and $S_6^5$) give new transformations not covered elsewhere.


What is the correct structure of sodium thiosulfate? Which structure of sodium thiosulfate is correct? One book says the second one (the one with co-ordinate bonds) and the other one says the first one?

According to me the structure 2 is correct as in that structure the octet of the middle S atom is complete whereas in the first one there's an excess.

Both of your Lewis structures are wrong.

In both of the shown structures, the solid line from sodium to oxygen/ sulfur implies that there is a discrete bond between these atoms. This is not the case.

The sodium ions and the thiosulfate ions form ion pairs (or ion triples) and are attracted to each other via diffuse electrostatic interactions. Usually that is what we consider ionic bonding.

In aqueous solution the ions will be very mobile. Sodium ions may coordinate to one or more thiosulfate ions, and therein to more than one site of the thiosulfate ion. At the same time thiosulfate ions may coordinate to more than just two sodium ion, or fewer. In such solutions, the solutes also coordinate to the solvent.
Ignoring any external field, the oxygen within the thiosulfate ion are equal.

The Lewis structure best describing the thiosulfate molecular ion is having a +2 formal charge at the central sulfur, and a -1 charge on each of the ligand atoms oxygen/ sulfur. (See below for details.)
One possible (and more accurate) Lewis structure is therefore:

(Note that the dashed lines do not represent bonds, but weakly coordinating interactions, but not necessarily all of them.)

More common are probably the following Lewis structure representations:

In solid state coordination of the ions depend on various factors, especially how much water is incorporated into the crystal structure. In any of these cases, it would be wrong to draw any bonds between the (molecular) ions. I'll include a few links to publications at the end of the post.

Additional thoughts of your input structures

  Which structure of sodium thiosulfate is correct? [...]

You noted correctly that in the first structure the central sulfur needs to "expand the octet". This would only be possible if d orbitals of said sulfur are involved. It has been proven many times that the contribution of d-orbitals in these hyper-coordinated species is minimal (usually below 1%).

The second structure is technically, or strictly speaking, not a Lewis structure, because dative bonds are not part of this description. However, if we allowed them as an extension, which has been proposed multiple times, then the arrow would need to point from the oxygen towards the central sulfur, as in those occurrences the more electronegative element would assume the higher electron density.

Disclaimer: The following section is a bit more advanced, and might not be appropriate to the casual reader, as another user pointed out that it might obfuscate the most important points about the structure of the anion.

Gas phase approach to determining the most likely Lewis structure

If we assume this molecule to be an ion-pair and calculate it in the gas phase, then we can draw a couple of conclusions, how a more accurate representation in the Lewis framework would look like.
I have performed a quick calculation on the DF-BP86/def2-SVP level of theory with Gaussian 09 and performed a natural bond order analysis (NBO 6.0) on it. The following picture shows the optimised geometry, alongside some inter-atomic distances.

As you can see, the sodium ions are coordinated to two oxygen, or one oxygen and a sulfur, respectively. The natural atomic charges reproduce well what I have initially stated, and reproduce well the Lewis structure I have suggested.

\begin{array}{rrr}
\text{Atom}& \text{No}& \text{Natural Charge}\\\hline
  \ce{S}  & 1 &   1.98 \\
  \ce{S}  & 2 &  -0.73 \\
  \ce{O}  & 3 &  -1.01 \\
  \ce{O}  & 4 &  -1.01 \\
  \ce{O}  & 5 &  -1.01 \\
  \ce{Na} & 6 &   0.91 \\
  \ce{Na} & 7 &   0.86 \\\hline
\end{array}

In terms of the NBO analysis, we can back this up, as it indicates no lone pair at the central sulfur (S1), three lone pairs at the oxygen (O3, O4, O5) and the ligand sulfur (S2), and four single bonds. There are vacant acceptor orbitals at the sodium ions (Na7, Na8). Here is an abridged version of the analysis:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
(skipping core orbitals)
  24. (1.98855) LP ( 1) S  2            s( 90.65%)p 0.10(  9.34%)d 0.00(  0.01%)
  25. (1.88728) LP ( 2) S  2            s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
  26. (1.84883) LP ( 3) S  2            s(  1.80%)p54.53( 98.09%)d 0.06(  0.11%)
  27. (1.97170) LP ( 1) O  3            s( 77.50%)p 0.29( 22.49%)d 0.00(  0.01%)
  28. (1.83558) LP ( 2) O  3            s(  0.67%)p99.99( 99.25%)d 0.11(  0.07%)
  29. (1.82232) LP ( 3) O  3            s(  0.03%)p99.99( 99.89%)d 2.42(  0.08%)
  30. (1.97170) LP ( 1) O  4            s( 77.50%)p 0.29( 22.49%)d 0.00(  0.01%)
  31. (1.83558) LP ( 2) O  4            s(  0.67%)p99.99( 99.25%)d 0.11(  0.07%)
  32. (1.82232) LP ( 3) O  4            s(  0.03%)p99.99( 99.89%)d 2.42(  0.08%)
  33. (1.96648) LP ( 1) O  5            s( 77.67%)p 0.29( 22.33%)d 0.00(  0.01%)
  34. (1.83753) LP ( 2) O  5            s(  0.02%)p99.99( 99.91%)d 4.78(  0.07%)
  35. (1.82147) LP ( 3) O  5            s(  0.00%)p 1.00( 99.93%)d 0.00(  0.07%)
  36. (1.95903) BD ( 1) S  1- S  2
               ( 56.47%)   0.7515* S  1 s( 25.58%)p 2.86( 73.06%)d 0.05(  1.36%)
               ( 43.53%)   0.6598* S  2 s(  7.70%)p11.86( 91.35%)d 0.12(  0.95%)
  37. (1.97694) BD ( 1) S  1- O  3
               ( 34.82%)   0.5901* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 65.18%)   0.8073* O  3 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  38. (1.97694) BD ( 1) S  1- O  4
               ( 34.82%)   0.5901* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 65.18%)   0.8073* O  4 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  39. (1.97758) BD ( 1) S  1- O  5
               ( 34.64%)   0.5886* S  1 s( 25.28%)p 2.91( 73.66%)d 0.04(  1.06%)
               ( 65.36%)   0.8085* O  5 s( 22.34%)p 3.47( 77.48%)d 0.01(  0.18%)
 ---------------- non-Lewis ----------------------------------------------------
  40. (0.05958) LV ( 1)Na  6            s( 95.91%)p 0.04(  3.91%)d 0.00(  0.18%)
  41. (0.10019) LV ( 1)Na  7            s( 96.97%)p 0.03(  2.93%)d 0.00(  0.10%)
  42. (0.25853) BD*( 1) S  1- S  2
               ( 43.53%)   0.6598* S  1 s( 25.58%)p 2.86( 73.06%)d 0.05(  1.36%)
               ( 56.47%)  -0.7515* S  2 s(  7.70%)p11.86( 91.35%)d 0.12(  0.95%)
  43. (0.24584) BD*( 1) S  1- O  3
               ( 65.18%)   0.8073* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 34.82%)  -0.5901* O  3 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  44. (0.24584) BD*( 1) S  1- O  4
               ( 65.18%)   0.8073* S  1 s( 24.89%)p 2.97( 74.04%)d 0.04(  1.07%)
               ( 34.82%)  -0.5901* O  4 s( 21.84%)p 3.57( 77.98%)d 0.01(  0.19%)
  45. (0.24610) BD*( 1) S  1- O  5
               ( 65.36%)   0.8085* S  1 s( 25.28%)p 2.91( 73.66%)d 0.04(  1.06%)
               ( 34.64%)  -0.5886* O  5 s( 22.34%)p 3.47( 77.48%)d 0.01(  0.18%)

Key: LP = lone pair, BD = bond, LV = lone valency, BD* = "antibond"

Alternative coordinations

There are at least two other structures of the stoichiometry $\ce{Na2S2O3}$,  essentially with the same bonding motif, but different coordinations of the sodium atoms. The first is about $\pu{2.4 kJ mol-1}$ more stable than the previously shown.

The second is about $\pu{3.2 kJ mol-1}$ higher in energy.

These are electronic energies only, but they show that the sodium atoms are not discretely bound and may interchange positions quite freely.
Given that disodium thiosulfate probably only exists in a crystal lattice which cannot be described by Lewis structures, or in solution where we have to worry about other effects and equilibria, it is obvious, that there cannot be a single Lewis structure for this compound. It simply does not exist as a discrete molecule.

Literature (selection) of crystal structures

The crystal structure of anhydrous sodium thiosulphate. Sandor, E.; Csordas, L.; Acta Cryst. 1961, 14, 237-243. DOI: 10.1107/S0365110X61000802 Crystallography online: Crystal Structure of Na2 (S2 O3)
On Polymorphism of $\ce{Na2S2O3}$. von Benda, H.; von Benda, K.; Z. Naturforsch. B 1979, 34, 957-968. DOI:
10.1515/znb-1979-0716 Alternative direct link (pdf) Crystallography online: Crystal Structure of Na2 (S2 O3), Crystal Structure of Na2 (S2 O3), Crystal Structure of Na2 (S2 O3)
Rerefinement of sodium thiosulfate pentahydrate. Prasad, S. M.; Rani, A.; Acta Cryst. E, 2001, 57, i67-i69. DOI: 10.1107/S1600536801011175 Crystallography online: Crystal Structure of Sodium thiosulfate pentahydrate
Structure of a new sodium thiosulfate hydrate. Hesse, W.; Leutner, B.; Böhn, K.-H.; Walker, N. P. C.; Acta Cryst. C 1993, 49, 363-365. DOI: 10.1107/S0108270192009399 Crystallography online: Crystal Structure of sodium thiosulphate 2/3-hydrate
The crystal structure of sodium thiosulphate pentahydrate. Taylor, P. G.; Beevers, C. A.; Acta Cryst. 1952, 5, 341-344. DOI: 10.1107/S0365110X52001027


Quantitative Analysis of Urea in aq. solution What would be a good way to assay Urea in aq. solution? The test doesn't need to be super accurate. 1% accuracy is fine.

But it needs to be relatively fast & robust to carry out.

NaHCO3 is a co-dissolved impurity so the test needs to tolerate NaHCO3 interference.

In clinical chemistry, urea is quantitativly determined as follows:

Hydrolyze urea with urease to ammonia and carbon dioxide
react the ammonia with 2-oxo-glutarate in the presence of glutamate dehydrogenase and $\ce{NADH}$: $\ce{NADH}$ is converted to $\ce{NAD+}$
monitor the decay of the UV absorption of $\ce{NADH}$ at $\lambda$= 340 and 410 nm


Why do Allylic Halides prefer SN2 reaction over SN1? As explained by my instructor,

  Allylic Halides form highly stable Transition States. They are stable because they are:

  $sp^2$ hybridized
  possess a conjugated system of pi bond and empty p orbital

However, if we consider the $S_N1$ mechanism, a carbocation intermediate is stabilized by resonance. Why is the formation of the transition state more favorable than the formation of the resonance stabilized intermediate?

You and your instructor are both correct, it is your premise that's wrong.

Your instructor's explanation about the allyl double bond stabilizing the $\mathrm{S_{N}2}$ transition state is correct. The following figure shows the stabilizing interaction.

image source

If we remove the attacking nucleophile and the leaving group and place a positive charge in the system, then we have the $\mathrm{S_{N}1}$ allyl intermediate.  As you point out, this intermediate is resonance stabilized, and therefore, the transition state leading to it is also stabilized.

Your premise,

  Why do Allylic Halides prefer SN2 reaction over Sn1?

is not true.  There are many cases where allylic halides react preferentially by an $\mathrm{S_{N}1}$ process.

Since both the allylic $\mathrm{S_{N}1}$ and $\mathrm{S_{N}2}$ reactions are stabilized, there is a delicate balance between the two pathways.  We can shift from one mechanism to the other by changing reaction conditions.

For example, allyl chloride reacts by an $\mathrm{S_{N}1}$ mechanism to produce allyl alcohol when we place it in a 50:50 mix of $\ce{H2O:EtOH}$ at 45°C.  If instead we run the reaction in the presence of ethoxide ion in ethanol at 45°C, the corresponding ethyl ether is formed by an $\mathrm{S_{N}2}$ process.


Get the molar mass of an unknown liquid in a dissolution A dissolution has $10$g of some unknown liquid and also $90$g of
  water. The freezing point of this dissolution is $-3.33^\text{o}$C.
  What is the molar mass of the unknown liquid in g/mol?

Alright. First of all, the formula to calculate the freezing point of a dissolution is

$$T_f - \Delta_f$$

Where

$$\Delta_f = \text{m} \cdot K_f$$

In our case, it would be

$$-3.33 = 0 - (\text{m}\cdot 1.86)$$

The molality m is

$$\frac{\text{moles of unknown liquid}}{\text{kg of water}} = \frac{\text{moles of unknown liquid}}{0.09}$$

We have to solve the equation:

$$-3.33=-\frac{\text{moles of unknown liquid}}{0.09}\cdot 1.86$$

Which yields

$$0.16 = \text{moles of unknown liquid}$$

However, I need the molar mass of this unknown liquid. How do I get it?

As long as you work with sufficently low concentrations, you do not need the molecular mass of your unkown.  Indeed, this was (is?) your initial question.

Freezeing point depression (salting the footwalk to melt ice in the winter), elevation of the boiling point (adding salt to cooking water that increases it's boiling temperature, too) and osmotic pressure are colligative properties.  What you need is to look up the crysoscopic constant, for water (same page) of $1.853 \mathrm{~K·kg/mol}$.

On the other side, if you already state the equation of $0.16$ moles corresponding to $10\mathrm{~g}$.

Despite advent of mass spectroscopy, such a determination may be still helpful today.


Is there a way to dehydrate hydrated aluminum chloride for use as a Lewis acid? I heard that one can use thionyl chloride to dehydrate it, but I'm not really sure if it would be available for amateur chemists since it's listed on the UN Convention for (the Elimination of) Chemical Weapons.

And no, I'm NOT thinking of getting cryolite either.

I think obtaining anhydrous aluminium(III) chloride from hexahydrate $\ce{AlCl6 · 6 H2O}$ isn't worth it, both economically and time-wise.
Thionyl chloride might work, but, as you said, it's an unpleasant reactant to work with, which is also likely more expensive than dry $\ce{AlCl3}$ (or is on par, depending on the purity of aluminium trichloride).
A common way to prepare water-free $\ce{AlCl3}$ in the lab is to heat pure aluminium metal with the excess of dry $\ce{HCl}$ gas at $\pu{150 °C}.$
The product is then sublimated (above $\pu{180 °C}$) and collected in the condenser (air-cooled).
The quirk here is you have to make sure pure aluminium is used, not an alloy; but this is usually evident from the color of the product, which should be white/colorless.
If you already have $\ce{AlCl3}$ labelled as anhydrous, but you suspect it was slightly moisturized, it can be regenerated via sublimation [1, p. 636]:

Aluminium chloride (anhydrous) […] Sublime it several times in an all-glass system under nitrogen at 30-50 mm pressure.  It has also been sublimed in a stream of dry $\ce{HCl}$ and has been subjected to a preliminary sublimation through a section of granular aluminium metal.

An extensive description is provided in Experimental part in [2, p. 1230]):

Aluminum Chloride.—The aluminum chloride (Baker Chemical Company) was purified by subliming three times down a straight Pyrex tube (with constrictions) which contained a small amount of nitrogen gas (50 mm.). In the first sublimation, the aluminum chloride mas sublimed through a section of granular aluminum metal, which was supported with plugs of glass wool. After the final sublimation and before sealing the tube off, aluminum chloride mas sublimed back down the tube until the remaining material formed a loose plug.
In order to remove the aluminum chloride from the tube, the tube is scratched and then the scratch touched with a hot rod in such a manner that the tube cracks but does not open. The tube may then be weighed, the tube tapped slightly to open. and after removing the aluminum chloride, the empty tube weighed. Using these techniques, it is possible to prepare very pure samples of aluminum chloride and weigh and transfer them without bringing them into contact with air.

Remember to store anhydrous $\ce{AlCl3}$ in an air-tight vessel and perform all operations with it in a fume hood.
Also, keep in mind that you can purchase $\ce{AlCl3}$ dissolved in organic solvents such as nitrobenzene, THF, or even $\ce{AlCl3}$-functionalized silica gel.
References

Armarego, W. L. Purification of Laboratory Chemicals; Elsevier: Boston, MA, 2017. ISBN 978-0-12-805457-4.
Jensen, F. R. Substitution of Polynuclear Aromatic Compounds. I. The Friedel-Crafts Benzoylation of Naphthalene. J. Am. Chem. Soc. 1957, 79 (5), 1226–1231. DOI: 10.1021/ja01562a051.


Assignment of the IR spectrum of 2,4-pentanedione (acetylacetone) The IR spectrum is shown here:

My assignments so far:

The 1728 and 1708 stretches correspond to $\ce{C=O}$ stretches in the keto form - the symmetric/asymmetric stretches respectively.
The 3003 stretches are typical $\ce{C-H}$ stretches.
The 1606 stretch corresponds to the $\ce{C=C}$ in the enol form. However, hidden under this peak there is also the $\ce{C=O}$ stretch in the enol form, which is lowered by conjugation to the $\ce{C=C}$ and the $\ce{O}$ of the $\ce{-OH}$ respectively.

I have two questions about the spectrum, however:

Why is there no $\ce{-OH}$ peak? Even if the enol is hydrogen bonded to itself, there should still be an $\ce{-OH}$ peak in the $\pu{3400 cm^{-1}}$ range?
Does the fact that the carbonyl peak in the enol form is hydrogen bonded change its shift? Assuming the only contribution to the lowering of the $\ce{C=O}$ stretching frequency is conjugation to the one $\ce{C=C}$ bond, this would lower the peak to ~$\pu{1655 cm^{-1}}$: Surely this would be seen as a peak in its own right? It seems to be shifted even lower to be fully hidden by the 1606 peak.

OH str for beta diketones is tabulated from 3200 to 2400 1/cm. The shift is explained by resonance of the hydrogen bonded structure.

Its absorption features is small or unseen as for

commonly the ketonic form is highly dominant; i.e. we won't expect to see OH str in the acetaldehyde spectrum, even
owing to the above mentioned resonance,  the dipole change associated to symmetric OH str is null, and it is minimal for the AS one.

The symmetry of 2,4-pentanedione makes the last point even more important.

Concerning the second part. The fact that the O of the carbonyl of the enoform is H bonded does further shift down the related CO str. Moreover, in addition to resonance with the C=C, note that again there will be resonance with the CO in beta as above, and this further shift the frequency to lower wn.


Dehydration of amide with PCl5 Can $\ce{PCl5}$, phosphorus pentachloride, react with an amide?

I saw a reaction where the corresponding cyanide was formed in this process. What other reagents could do the same?

There's a couple of newer methods, using

zinc + N-Methyl-N-(trimethylsilyl)trifluoroacetamide (MSTFA)
copper + N-Methyl-N-(trimethylsilyl)trifluoroacetamide (MSTFA)
diethyl chlorophosphate $\ce{(EtO)2POCl}$
$\ce{PdCl2}$ in $\ce{CH3CN/H2O}$
$\ce{(COCl)2}$ with DMSO at -78 °C (Swern-type conditions)

Thionyl chloride, $\ce{SOCl2}$ is almost a classic here.

Under these conditions, the side products of the amide dehydration are $\ce{SO2}$ and $\ce{HCl}$.

$\ce{R-CONH2 + SOCl2 -> R-C#N + SO2 ^ + 2 HCl ^}$

On the first view, $\ce{PCl5}$ looks like leave quite a mess. Thinking in the workup, that's probably the last thing I would try.


Why is magnesium ascorbate, which is a conjugate base of a weak acid, acidic? Why are magnesium ascorbate and potassium acetate, which are salts of strong bases with weak acids, acidic?
I discovered with litmus paper that magnesium ascorbate is acidic and found out from https://foodb.ca/compounds/FDB015417 that potassium acetate should also be acidic.
How can this be? Shouldn’t conjugate bases of weak acids, ascorbate and acetate in this case, be basic and the strong dissociated spectator ions, $\ce{K+}$ and $\ce{Mg^2+}$, neutral?

$\mathrm{p}K_\mathrm{a}$ relation in the answer of Kav

Wikipedia says pKa's are 4.10 (first), 11.6 (second)

gives the hint a monobasic salt with alkali metals should be very mildly alkalic, as pKa2 is farther from pH 7.
$pH \approx \frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}}{2} = \frac{4.10+11.6}{2}=7.85$
OTOH, it will probably form a complex with Mg, affecting acidity of the 2nd hydrogen. Like
$$\ce{2 H2A(aq) + Mg^2+(aq) -> [MgA2]^2-(aq) + 4 H+(aq)}$$
respectively
$$\ce{Mg(HA)2(s) ->[H2O] [MgA2]^2-(aq) + 2 H+(aq)}$$
See also citrate or EDTA complexes for this effect. AFAIK, before suitable metalochrome indicators came, metal content with excess of Na2H2-EDTA had been sometimes determined by back titration using NaOH.


Hottest region of Bunsen flame Since region 1 is closer to the source, I presume it to be the hottest as complete combustion takes place there. Also, this is the part where the gas mixture(responsible for flame) reacts with oxygen first(as soon as valve is open). So, the reaction should take place much quicker and more heat should be released. But in my book, the answer is given as region 2. Where am I wrong?

I know there exists as question similar to this in our community : Bunsen burner and hottest part but the answers don’t answer my question.

The question itself is poorly written as given.

The diagram should have "regions" and "points."

So what seems to be labeled "Region 2" and "Region 4" I would name as "Point 2" and  "Point 3" respectively.

So the innermost conical area would be "Region 1", the next conical area would be "Region 2" and the third conical area would be "Region 3."

Region 1 is where the mostly unburned gas and oxygen mixture are pushing above the lip of the Bunsen Burner. Region 1 exists because the gas coming out of the tube is cool. If the air port is open and the gas flow is too low then the gas will start to burn down the tube and you'll get a "strike back" where the flame is either (1) blown out or (2) burns at the jet. If the gas flow is too great you can blow the burning region off contact with the upper tube. If you increase the gas flow even more then you can in fact blow the flame out.

Region 2 would be a reducing region within the flame. This region is hot and burns the fuel and oxygen coming out of the tube.

Region 3 would be a oxidizing region of the flame. Here oxygen from "outside air" (oxygen which didn't come up the tube) is migrating into the flame to burn the excess fuel which is not bunt in region 2.

You'd use the reducing and oxidizing regions when doing bead tests for identification.

Point 2 would be the hottest part of the flame as shown by the composite image of a paperclip in the flame from the YouTube video.


Is entropy change of the system for reversible process same as entropy change of the system for irreversible change? Is entropy change of the system for reversible process same as entropy change of the system for irreversible change? Is Q for reversible change equal to Q for irreversible change? Or the change in entropy of the surrounding is same in any case because surrounding is bigger than system, so the change in surrounding is small?
And finally, does internal energy of the system depend on entropy of the system and how?

I registered just so I could ask this question. I would be very grateful for answer.

Being a state function, entropy is independent of path, so the entropy change of the system is indeed the same for both a reversible and an irreversible process. The transferred heat, however, is not, and does depend on process.

You've probably heard mentioned that the entropy change through a reversible process is less than that through an irreversible process. At first glance this proposition seems suspicious, because we've just said that as a state function entropy is independent of path$-$the resolution is that the entropy change of the system remains the same regardless of process, but the entropy change of the environment is indeed different, such that the total entropy change is greater for an irreversible process than for a reversible process. [Interpreted differently: the environment is in a different state after the system undergoes a reversible process than after it undergoes an irreversible process.]

The internal energy does depend on the system's entropy$-$by the first law we may write $$\mathrm{d}U = T\,\mathrm{d}S + \mathbf{f\cdot\mathrm{d}X}$$ (where $\mathbf{f\cdot\mathrm{d}X}$ represents a generalized work term; commonly this is $-p\,\mathrm{d}V$), so we see that $U$ is a natural function of $S$ and $\mathbf{X}$; $U = U(S,\mathbf{X})$. To give a functional form for $U$ as a function of $S$ and $\mathbf{X}$, we'll need in general an equation of state, an equation for energy, and a type of process$-$let's restrict ourselves to the case of an ideal gas, for which the two equations are $$pV = nRT \quad \text{and} \quad U = \frac{3}{2}nRT$$ (assuming also a monatomic gas), and work out how $U$ varies over an isobaric process. We want to evaluate $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$; let's start by working out $-p\,\mathrm{d}V$: $$-p\,\mathrm{d}V = -p\,\mathrm{d}\left(\frac{nRT}{p}\right) = -nR\,\mathrm{d}T = -\frac{2}{3}\,\mathrm{d}U.$$ Now plug that into our starting equation, substitute, and rearrange... $$\frac{5}{3}\,\mathrm{d}U = T\,\mathrm{d}S = \frac{2U}{3nR}\,\mathrm{d}S \quad \Longleftrightarrow \quad \frac{5}{2}nR\,\frac{\mathrm{d}U}{U} = \mathrm{d}S,$$ which integrates to $$\Delta S = \frac{5}{2}nR\ln\left(\frac{U_2}{U_1}\right) = \frac{5}{2}nR\ln\left(\frac{T_2}{T_1}\right).$$ This is no different from working out $q_\text{rev}$ for such a process and then using $\Delta S = q_\text{rev}/T$; the role of energy is simply more explicit in this approach.


Electrochemistry - reduction potentials - an IB Chemistry Question I have re-read the Electrochemistry chapter from my textbook just now to remind my self of the exact formulae. I still don't understand how to approach this question, however - how can we work it out? I can either give you the mark scheme answers straight away and we can work backwards, or I can wait for an answer and then tell you if that is the correct solution. What ever is the norm on this forum! Thank you.

First put your redox potentials on a horizontal line by order of increasing standard redox potentials. You will see that it starts on the left-hand side by $\ce{Zn^{2+}/Zn}$ at $\pu{-0.76 V}$. Then, going to the right, you find $\ce{V^{III}/V^{II}}$ at $\pu{-0.26 V}$. Now you find $\ce{Pb^{2+}/Pb}$, and so on.
Now try to learn how to use this scale. The reduced species of any couple can react with any oxidized species of any couple situated to its right. For example, $\ce{Zn}$ is the reduced species of the couple $\ce{Zn^{2+}/Zn}$. So $\ce{Zn}$  can reduce all oxidized species, and for example all oxidized forms of Vanadium ions. But $\ce{Zn^{2+}}$ cannot be reduced, because there is no reduced species at the left hand side of $\pu{- 0.76 V}$.
On the other end of the scale, you find $\ce{Cl2/Cl-}$ at $\pu{+1.36 V}$. Hopefully you admit that $\ce{Cl2}$ can oxidize all reduced species, and $\ce{Cl-}$ cannot reduce anything.
Now let's consider $\ce{Pb^{2+}/Pb}$, at $\pu{-0.13 V}$. $\ce{Pb}$ can reduce all oxidized species situated on the positive side of the axis, for example $\ce{VO^{2+}}$ which is at $\pu{+0.34 V}$. But it cannot reduce $\ce{V^{3+}}$ to $\ce{V^{2+}}$, because $\ce{V^{3+}/V^{2+}}$ is on tis left-hand side. And this is exactly your first question.
Now think a little bit about how to solve the second question. It is exactly the same approach.


What is the acid dissociation constant of cacodylic acid? What is the $\mathrm pK_\mathrm a$ of cacodylic acid? I cann't seem to find one answer on the internet. I have read 1.67, 6.37, and 4.8 in my $\mathrm pK_\mathrm a$ table. What is the reason for this? Also, the problem I just did had it as 0.34.

I found this article (TW. Shin, K. Kim, IJ. Lee, J. Solution Chem. 1997, 26, 379. doi: 10.1007/BF02767677) with precise analysis of the pKa of cacodylic acid - it says that in 25 °C it's estimated to be 6.27. Values lower than 6 are wrong, but found often. Sometimes it's said to be a diprotic (which it isn't) with first pKa 1.57 and second 6.27 - pubchem says 1.57 citing Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 76th ed. Boca Raton, FL: CRC Press Inc., 1995-1996., p. 8-45 - it might be one of the sources of problem.


What does (sol) in a chemical equation mean? I'm reading Shrivers Inorganic Chemistry book and some of the equations have a (sol) state on them. I'm sure the book has already defined this but I cannot find where.
Examples:

\begin{align}
  \ce{[NH3OH]Cl (sol) + NaOBu &-> NH2OH (sol) + NaCl (s) + BuOH (l)}\\
  \ce{SbF5 (l) + HF (l) &-> H2F+ (sol) + SbF6- (sol)}
\end{align}

The state (sol) or (solv.) stands for "solvated". Since you are dissolving $\ce{HF}$ in liquid $\ce{SbF5}$, there is no water in that system (which would cause hydrolysis). So it is not correct to use (aq), since it is a nonaqueous system.

Here are some other examples:

Autoprotolysis of liquid ammonia:
$$\ce{2NH3 (l) <=> NH4+ (sol) + NH2- (sol)}$$
Autoprotolysis of nitrosyl chloride:
$$\ce{NOCl (l) <=>  NO+ (sol) + Cl- (sol)}$$
Dissolving sodium metal in liquid ammonia creates solvated electrons:
$$\ce{Na (s) <=>  Na+ (sol) + e- (sol)}$$


How would one make a non-toxic pen out of osmium? Osmium is the densest element in the periodic table. However, it is toxic in its pure form. Is there a way to create a standard pen out of osmium but coat it in a dense material that isn't toxic to have the heaviest pen?

How would this pen be made?

It would be a cap pen (like the Uniball pen pictured below) with no wasted space for plastic mechanical parts.

Osmium alloys with many metals, so it could be covered by an alloy. Osmiridium, an alloy with iridium, is even harder than pure osmium.

It would seem possible to dip the osmium pen blank in an iridium or osmiridium bath (at rather high temperature) to produce a long-lasting casing, rather like dipped chocolate cordials.

If you accept this answer, I'd be glad to receive an award of one of these pens. ;-)


Why is the pure tungsten (W) melting point higher than that of its oxides (either WO3 or WO2)? Why is the pure tungsten ($\ce{W}$) melting point way higher than that of its oxides $\ce{WO3}$ or $\ce{WO2}$? And why does pure $\ce{Al}$ react differently? I.e. the oxides have a higher melting point than pure $\ce{Al}$.

As far as I know,

$\ce{WO2}$ and $\ce{WO3}$ have a covalent characteristic, and thus lower melting point than the metal.

while, $\ce{Al2O3}$ have more ionic characteristic, and thus higher melting point than the metal.

$\ce{W}$ is a d-block metal from the $\ce{3^{rd}}$ row which make it possible to have covalent materials rather than ionic materials.

this article might prove this point
http://pubs.acs.org/doi/abs/10.1021/jp963724z?journalCode=jpcbfk


What are GAMESS and Gaussian? I've been seeing a lot of GAMESS and Gaussian questions around here, and most of them seem very hard to answer. I extrapolated that they might be some sort of Chemistry programs, but what do they do exactly?

Instead of looking it up, I feel like most of you would be more knowledgeable in this.

There exist a physical theory called quantum mechanics which describes the Universe at atomic and subatomic length scales. Chemical systems, such as, for instance, molecules, are one types of systems one encounter at this level, and as such, they can in principle be completely described by quantum mechanics. What is meant by a "complete description" is that all the properties of a system (say, a molecule) can be theoretically predicted by the application of the principles of quantum mechanics.

However, the direct application of the principles of quantum mechanics to chemical systems is practically impossible in the absolute majority of cases due to the complexity of the resulting description. To simplify matters a series of approximation has to be introduced. But even the approximate description is usually too complex to be handled by a human being analytically (on a piece of paper), and thus, numerical methods has to be developed to shift this burden to our digital mates (computers). All these questions (approximations plus numerical treatments) are addressed by a branch of chemistry known as quantum chemistry.

GAMESS[1] and Gaussian are indeed programs; programs that implement different methods of quantum chemistry. And just for your information there are a lot of other such programs. These programs allows us to calculate the properties of molecules, to predict their behaviour in different environments, etc. And all this can be done theoretically using computers without actually synthesising a molecule in question and then doing some real experiments on it. To understand what these programs do exactly, you have to spent a lot of years by learning some advanced mathematics, quantum physics, quantum chemistry, programming, numerical methods, ... At this point I can only tell you that they are doing some magic. The quantum magic.

[1] There are in fact 3 different programs: PC GAMESS (now Firefly), GAMESS (UK), and GAMESS (US).


Which double bond will react with m-CPBA to form an epoxide? The solution to this reaction sequence given was,

What I don't understand is the last step. How do we know which double bond will react to form an epoxide?
My guess was that the double bond adjacent to the methyl group should react as it has a more double bond character than the other one which is in resonance with the lone pair of oxygen.

Any help would be appreciated.

When two or more double bond are present then the more electron rich double bond will epoxidise. In above example the double bond containing the methyl group  has  electron donating  inductive  effect while the double bond containing the methoxy group  will have an electron donating resonance effect. The resonance effect is dominant over the inductive effect. Therefore the double bond containing the methoxy group is more electron rich. Therefore epoxidation occur on it.


Why is hydrated beryllium ion acidic? The beryllium ion, $\ce{Be^{2+}}$, forms the aquo complex $\ce{[Be(H2O)4]^{2+}}$. According to LibreTexts, this complex is acidic in solution: $$\ce{[Be(H2O)4]^{2+} + H2O -> [Be(H2O)3OH]+ + H3O+}$$
It is not clear to me why it should be acidic. I read here that "the hydrolysis happens because the $\ce{Be−O}$ bond is very strong and so in the hydrated ion this weakens the $\ce{O−H}$ bonds hence there is a tendency to lose protons."
I have two questions:

What is meant by a "very strong" $\ce{Be-O}$ bond? Is it stronger than those in other metal aquo complexes? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size?
Why does a strong $\ce{Be-O}$ make the $\ce{O-H}$ bond weaker? I know that oxygen forms a coordinate bond with metal cation so the electrons in the $\ce{O-H}$ bonds would get shifted towards oxygen. But as far as I can tell this only makes the $\ce{O-H}$ bonds more polar. Why does a polar bond have to be a weaker one?

What is meant by a "very strong" Be−O bond? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size?

You got that absolutely right.
$\ce{Be\bond{-}O}$ is a strong bond because of the small size of Be. Smaller cation size means a stronger pull on the $\ce{O}$ electrons, thus reducing the bond length and hence increasing the bond strength.

[...]the electrons in the O−H bonds would get shifted towards oxygen. But as far as I can tell this only makes the $\ce{O\bond{−}H}$ bonds more polar.

You struck at the right point. Let's look at the original question.
I'm sure you must know that a stronger acid has a higher tendency to release $\ce{H+}$ ions. This means that the electron pair should be drawn away from $\ce{H}$ atom in the $\ce{O\bond{−}H}$ bond. Thus, a polarized bond makes it easier for the H+ ion to 'escape' from the molecule.

Why does a polar bond have to be a weaker one?

It is not appropriate to answer this question in definitive terms. The strength of a bond is defined by its bond length, not its polarity. To address the confusion you're having, look back at the context of the question. $$\ce{[Be(H2O)4]^{2+} + H2O -> [Be(H2O)3OH]+ + H3O+}$$
We're talking about a compound with polar O-H bonds dissolved in water. Water has a high dielectric constant, and can easily break ionic/polar bonds since it liberates a high amount of solvation energy. This is because of its high dielectric constant.
So, are polar bonds weaker than non-polar bonds in water? Sure, I can see your line of thinking. But, always remember that ionic bonds can easily be broken by water, but that doesn't mean that ionic bonds are weaker than non-polar bonds.


Reaction of water and methane Methane is a non-polar compound, i.e. it shows no bipolar movement but even after that it reacts with water which is a polar compound and polar and non-polar compounds do not react. Why does this reaction take place, then:

Q: The reaction of methane and water is a way to prepare hydrogen according to
$$\ce{CH4 (g) + H2O -> CO2 (g) + H2 (g)}$$
...

Polar and non-polar liquids do not mix, but that doesn't tell you anything about their ability to react, which depends on the thermodynamics and kinetics of the reaction. The reaction shown is in the gaseous state.

The reaction of methane gas with gaseous water, which you show, is a common reaction called Methane reformation. Methane gas is heated with steam (gaseous water) to produce hydrogen (and $\ce{CO2}$).

As shown in that link, the reaction takes place in two steps:
$$\begin{align}\ce{CH4_{(g)} + H2O_{(g)} &->  CO_{(g)}  + H2_{(g)}} &&\text{(Endothermic)}\\
\ce{CO_{(g)} + H2O_{(g)} &-> CO2_{(g)} + H2_{(g)}} && \text{(Exothermic)}\end{align}$$

It appears it probably requires some heating to get this reaction to proceed (the first step is endothermic), I'll edit when I have a chance to look up the thermodynamic properties and see if we can glean what drives the reaction, or if it requires catalysis to avoid the energy barrier.

Side-note: by "bipolar movement" I think you mean "dipole moment"

Edit: According to this source, the first step of the reaction happens at ~3–20 atmospheres of pressure. The second step uses a catalyst to make the reaction more energetically favorable.

The thermodynamic variables for the individual species (at $25~\mathrm{^\circ C}$) in the net reaction are:

$$
\begin{array}{c|cc}
 & \Delta S^\circ \left( \frac{\mathrm{kJ}}{\mathrm{K\cdot mol}} \right) & \Delta H_\mathrm{f}^\circ \left( \frac{\mathrm{kJ}}{\mathrm{mol}} \right) \\
\hline
\ce{CH4_{(g)}} & 0.1862 & -74.81 \\
\ce{H2O_{(g)}} & 0.1887 & -241.8 \\
\ce{CO2_{(g)}} & 0.2136 & -393.5 \\
\ce{H2_{(g)}} & 0.1306 & 0 \\
\end{array}
$$

The balanced net reaction is:

$$ \ce{CH4_{(g)} + 2 H2O_{(g)} -> CO2_{(g)} + 4 H2_{(g)}}$$

Therefore:

$$
\begin{array}{l|ccc}
 & \Delta S^\circ \left( \frac{\mathrm{kJ}}{\mathrm{K\cdot mol}} \right) & \Delta H^\circ \left( \frac{\mathrm{kJ}}{\mathrm{mol}} \right)  \\
\hline
\text{Reactants} & 0.5636 & -558.41 \\
\text{Products} & 0.7576 & -393.5 \\
\hline
\text{Net Reaction} & 0.2136 & 164.9 \\
\end{array}
$$

Right away, it's apparent that the reaction is entropically favorable ($S_{\text{rxn}} > 0$), but enthalpically unfavorable ($H_{\text{rxn}} > 0$) at room temperature. To find out if the reaction is spontaneous, consider:

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

The reaction is spontaneous when $\Delta G < 0$. Because $\Delta S$ is so close to $0$ ($0.194~\mathrm{kJ/(K\,mol)}$), and $\Delta H$ is so high ($164.91~\mathrm{kJ/mol}$), we should expect already that this reaction is non-spontaneous at room temperature. This is consistent with the fact that methane doesn't readily decompose in moist air, as any cow pasture will tell you.  But we can check:

$$\Delta G = 164.91\ \frac{\mathrm{kJ}}{\mathrm{mol}}-298.15~\mathrm{K}\cdot0.194~\frac{\mathrm{kJ}}{\mathrm{K\cdot mol}}=107.07\ \frac{\mathrm{kJ}}{\mathrm{mol}}$$

Quite non-spontaneous. However, since the overall reaction is entropically favorable, we can use $\Delta S$ to overpower the $\Delta H$, by jacking up $T$. Since the industrial reaction takes place between $700$ and $1000~\mathrm{^\circ C}$, let's plug in $700~\mathrm{^\circ C}\ (973.15~\mathrm{K})$ and check that the reaction is indeed spontaneous:

$$\Delta G = 164.91\ \frac{\mathrm{kJ}}{\mathrm{mol}}-973.15~\mathrm{K}\cdot0.194~\frac{\mathrm{kJ}}{\mathrm{K\cdot mol}}=-23.85 \frac{\mathrm{kJ}}{\mathrm{mol}}$$

Voilà! A spontaneous reaction!

Do note, however, that this calculation:

ignores any temperature-dependence of the thermodynamic variables;
ignores the role of pressure.

Tldr: This reaction is not energetically favorable at room temperature. However, the reaction is entropically favorable. If you heat it up enough (i.e. $\gtrapprox 700~\mathrm{^\circ C}$), the reaction will proceed spontaneously.

Source for thermodynamic values: Oxtoby, Principles of Modern Chemistry Seventh Edition


Cubic crystals anisotropy My textbook states

  "Cubic crystals contain four three-fold axes, converting the ellipsoid to a sphere."

What indeed is the ellipsoid and sphere in the lattice? I know the meaning of three fold axes but unable to figure out what and how ellipsoid is converted to sphere.

From the limited excerpt of your book (title? author?) it is likely it just deals with the interaction of linearly polarized light with condensed matter.  If there is near- and far range order (as regular consecution of building elements), the solid is crystalline.  Among the crystaline states, there are (Bravais) lattice systems, that differ in symmetry, i.e. because they possess different symmetry elements, some are of higher symmetry (like cubic) than others (like monoclinic).  And vectorial physical properties of the material are coherent with the crystalline symmetry met.

(image source).

By means of an index ellipsoid, you may map the orientational dependance of physical properties, in other words, anisotropy.  In the case of mineral- and more general in polarized light microscopy, such an ellipsoid may be used as representation to map refractive indices along three directions orthogonal to each other. Now

either the three axes of the indicatrix differ from each other: optical biaxial system, characteristic Bravais lattice of lower symmetry: orthorhombic, monoclinic and triclinic.
two of the three axes of the indicatrix are identic, the third is different: optical mono- or uniaxial system, characteristic for hexagonal, trigonal and tetragonal Bravais lattice.
all three axes of the optical indicatrix are identic, in other words, the material is optical isotropic: this is characteristic for the cubic Bravais lattice.  (And: for the state of randomness, the glasses).

So in short: it is the number, the degree, and the relative orientation of the symmetry elements of the cubic Bravais latice that renders the more general case of an ellipse into the special case of a sphere.  And this means, the question should be "cubic crystal's isotropy".

Further information may be found for example here.


Why when deriving the Clapeyron equation can you assume that dH = dq, as surely the pressure is not constant? Part of the Clapeyron equation involves replacing $\Delta S$ with $\Delta H/T$. I can see the step from $\Delta S$ to $\Delta q/T$ but I'm not sure why you can replace $\Delta q$ with $\Delta H$. From my understanding $\mathrm dH = \mathrm dq_V +V\mathrm dp$ when assuming no work other than $pV$ work is done. I don't understand how you can substitute $\Delta H$ for $\Delta q$ without assuming that the pressure is constant?

At any given temperature, the change from a liquid to a vapor occurs at constant pressure.  What we are trying to find with the Clapeyron equation is how this constant pressure changes as the given temperature changes.

At equilibrium at a given temperature and constant pressure, we have $$G_{vapor}=G_{liquid}$$ and $$S_{vapor}-S_{liquid}=\frac{(H_{vapor}-H_{liquid})}{T}$$

Now, if we want to find out how temperature and pressure change in tandem along the equilibrium line, we can write:$$dG_{vapor}=-S_{vapor}dT+V_{vapor}dP$$and
$$dG_{liquid}=-S_{liquid}dT+V_{liquid}dP$$Since, along the equilibrium line, we must have $dG_{vapor}=dG_{liquid}$, we can write:$$-S_{vapor}dT+V_{vapor}dP=-S_{liquid}dT+V_{liquid}dP$$or equivalently:$$\frac{dP}{dT}=\frac{(S_{vapor}-S_{liquid})}{(V_{vapor}-V_{liquid})}=\frac{(H_{vapor}-H_{liquid})}{T(V_{vapor}-V_{liquid})}$$This is the Clapeyron equation.


Can I solve the partial derivative (dG/dH)s ? Does it make sense? It is possible to express   ,   in terms of Cp, Cv, V, P, kt etc. ?

I know that G and H are both derived from the Legandre transformation of U, so they are the same equivalent representation of internal energy but expressed like G=G(T,P) and H=H(S,P).

So I'm wondering if this partial derivative makes sense...

$$dH=TdS+VdP$$
At constant S, $$dH=VdP$$
$$dG=-SdT+VdP$$
If S is constant, then $$\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP=0$$But,$$\left(\frac{\partial S}{\partial T}\right)_P=\frac{C_p}{T}$$and$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$So, $$dT=\frac{T}{C_p}\left(\frac{\partial V}{\partial T}\right)_PdP$$and$$dG=\left[V-S\frac{T}{C_p}\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$Dividing by dH then gives:
$$\left(\frac{\partial G}{\partial H}\right)_S=1-\frac{\alpha T}{C_p}S=1+\frac{(G-H)\alpha}{C_p}=1+\frac{\alpha T}{C_p}\left(\frac{\partial G}{\partial T}\right)_P$$