What kind of optical activity requires chirality? From "Meso compound" in Wikipedia:

  A meso compound or meso isomer is a non-optically active member of a
  set of stereoisomers, at least two of which are optically active. This
  means that despite containing two or more stereogenic centers, the
  molecule is not chiral.

What is the meaning of "optical activity" in this context? In other words, what kind of photo activity requires chirality?

The meaning of optical activity in this context is the same as usual: the ability to rotate the plane of polarized light. To have it, a molecule must be chiral.

Chirality is the absence of certain type of symmetry. Since most small molecules have that symmetry, many students get the wrong perception that chirality is a property which must be obtained by having some functional group (asymmetric center, etc.), much like having a $\ce{-COOH}$ group makes a compound an acid. This is not quite true. Not only a compound may be chiral without any chiral centers, but it can also be achiral despite having some chiral centers, as are meso compounds.


Why is water "the universal" solvent? This is an old question that our textbook tried to answer but worsened the situation.

Many things are soluble in water. So many, that studying solutions will always require studying aqueous ones. It is true that many non-polars like waxes are not very soluble in water, yet I have never run into a solvent as "good" as water.

But how were we answered when we asked that "why is water a good solvent"? They said since water is polar so the attraction between for example $\ce{O}$ and the positive ions is so much blah blah blah!

So either there are "great" solvents like water out there or there are other things about water that make it the master solvent that are beyond me.

Is there a solvent as "versatile" as water?
Can this kind of solvent be non-polar?
If the answer to the questions above is no, what is special about water?

To directly address where the phrase comes from:

  Water is called the "universal solvent" because it dissolves more substances than any other liquid. -USGS

What are ideal qualities of solvent?

The strength of a solvent can be attributed to the strength of its intermolecular forces like london forces, dipole-dipole forces, ion-induced dipole, and hydrogen bonding. These are forces of attraction and repulsion. Solvation occurs when a molecule is surrounded by the solvent, so when there are strong intermolecular forces, stronger solvation occurs.

The hydrogen bond

Water makes use of the hydrogen bond, a type of intermolecular force experienced when hydrogen is attracted to the electronegative atoms nitrogen, oxygen, or fluorine. Hydrogen bonding is the strongest intermolecular force.

Hydrophobia

Water is a great solvent for hydrophilic molecules but hydrophobic molecules are, by definition, not easily disturbed by water. This makes a clear exception to the "Universal Solvent" idea. I haven't read any literature claiming it to actually be the best solvent, but it does work well in many chemical situations.


Are there any papers linking the bad taste of orange juice after toothpaste to sulfates? Wikipedia states:

  After using toothpaste, orange juice and other juices have an unpleasant taste. This effect is attributed to products of the chemical reaction between stannous fluoride in toothpaste and the acetic acid in the juices. Sodium lauryl sulfate alters taste perception. It can break down phospholipids that inhibit taste receptors for sweetness, giving food a bitter taste. In contrast, apples are known to taste more pleasant after using toothpaste.

The primary reference for this appears to be a comment on the US Department of Energy website.

There are several references to this.

This seems to be an untested hypothesis. My question is Are there any papers linking the bad taste of orange juice after toothpaste to sulfates?

There may be more, but this is one paper examining the effects of membrane-modifying compounds on taste receptors. It suggests that substances like sodium lauryl sulfate and gymnemic acid affect taste response by altering taste receptors' lipid membranes. Though many anionic surfactants have sulfate functional groups, not all compounds with sulfate groups act as surfactants (nor does the sulfate anion on its own). It sounds like this is something that any compound that can perturb the structure of a taste receptor's membrane can do, not something that necessarily applies to all sulfates.


Why is the manufacture of sulfuric acid known as the contact process? I was after some detail as to why the contact process was named as such.

According to this site:

  The Hot gases( $\ce{SO2}$
  ) evolved from burning of sulfur ore comes in
  contact with catalyst bed,  So the name of this process is called
  contact process.


Why is the change in entropy is different for different interpretations? Forgive me if the title is a bit vague. Read on and I'll clarify.

So, today in class, my teacher was explaining entropy. He gave various examples explaining how the degree of randomness changes. One example in particular stuck out to me. Here it goes:

Imagine there is a container with a hypothetical boundary in the middle. Neon gas is on the left of the boundary, and Argon on the right. As long as the boundary exists, the gases can't cross it to intermix.
Now the boundary is removed, and the gases mix. This increases the randomness in the system and thus entropy increases. We get $\Delta S \gt 0$. Thus the process is spontaneous.

My Question:

How can we definitively say that entropy has increased (or that randomness has increased)?

My Reasoning:

So my teacher says that entropy increases. These following cases show that change in randomness depends on your interpretation.

Interpretation 1:

Before mixing (boundary exists):

If we take some volume of gas, say $1 \text{ cm}^3$, at a random position in the container, we might get:

Pure $\ce{Ne}$
Pure $\ce{Ar}$
A mixture of the two gases

The point is that we aren't sure what composition we get, because the distribution of gases isn't homogeneous.

After Mixing (boundary removed):

The gases diffuse and intermix to form a homogeneous mixture.

Now if we take $1 \text{ cm} ^3$ from a random position, we can be $100\%$ sure of the composition of the gases, because the mixture is homogeneous.

This means there is more uniformity now. Or, we can say that there is less randomness. This tells us that $\Delta S \lt 0$.

Interpretation 2:

Using a new approach now.

Before Mixing (boundary exists):

Take any single atom from either side of the boundary at random. My chances of correctly predicting what atom I get (either $\ce{Ne}$ or $\ce{Ar}$) are $50\%.$

After mixing (boundary removed):

I again pick any atom at random from a random point in the container. My chances of correctly predicting what atom I get is again $50\%.$

This tells us that since I still have the same chance of correctly predicting, the randomness of the system has not changed.

This gives us $\Delta S = 0$.

Same physical process, but we get 3 different values of degree of randomness ($\Delta S$) depending on our interpretation. We know that diffusion does take place, so the process is spontaneous, and hence $\Delta S \gt 0$.

My Questions:

How to accurately know whether the degree of randomness increases or not?
How is the exact value of $\Delta S$ measured in actual experiments?

Note:

Assume the gases to be ideal.
Assume the container to be sufficiently large.

From the perspective of Statistical Mechanics:

Multiplicities do depend on the external parameters that specify the energy level of the system.

For ideal gas of $N$ indistinguishable molecules, the multiplicity is given by $$\Omega= \frac{1}{N!}\frac{V^N}{h^{3N}}\,\frac{2\pi^{3N/2}}{(3N/2- 1)!}(\sqrt{2mE})^{3N-1}$$

So, indeed the multiplicity increases when the position space increases.

When the external parameter like volume of the container is increased, the energy levels come closer and then the particles can have more microstates for the same $E$.

Prior to the volume increase, the microstates of the system are $\Omega_{N,E,V_i}\;_;$ when the volume is increased, not will be there the microstates $\Omega_{N,E,V_i}$ still be available after the removal of the constraint, but also, there be addition of microstates. In general, $$\Omega_{V_f}\geq \Omega_{V_i}\;.$$

So, as the volume increases, the multiplicity increases and so does the entropy.

As requested by OP:

From the perspective of Thermodynamics:

Suppose the two dissimilar indistinguishable gases $A$ and $B$ having the same number of particles, energy, volume are partitioned from each other.

Now, when the partition is removed, the energy of the gas $A$ confined to a localised area gets dispersed over a greater volume.

Thus the entropy of gas $A$ increases and that is given by \begin{align}\Delta S_A &= Nk\,\ln \frac{V_f}{V_i}\\ &= Nk\,\ln \frac{2V}{V}\\ &= Nk\,\ln 2\;.\end{align}

Same happens for gas $B$ also.

Therefore, the change in entropy of the system is given by \begin{align}\Delta S_\textrm{system}&= \Delta S_A + \Delta S_B\\ &= 2Nk\ln 2 \gt 0\end{align}

Therefore, entropy increases when the partition is removed.


Why is a theoretical titration curve higher than a practical one? I'm currently in high school and I did two titrations. The first one is a titration of $10.0~\mathrm{mL}$ $0.1~\mathrm{M}$ $\ce{HCl}$ with $x~\mathrm{mL}$ $\ce{NaOH}$. I obtained the following titration curve:

The blue curve is what I measured and the red curve is what the curve would look like in theory, with these data.

The second titration is a titration of $10.0~\mathrm{mL}$ $0.2~\mathrm{M}$ $\ce{CH3COOH}$ with $x~\mathrm{mL}$ $\ce{NaOH}$. I obtained the following titration curve:

Again, my measurements are in blue and the red curve is the theoretical model. I noticed that (looking at the second curve in specific) the measured pH value is most of the time lower than in the theoretical model, and close to the equivalence point, the graph is not as steep. Why is this? Not a valid reason would be that the pH sensor does not measure the right value instantly, because near the equivalence point, only one drop of $\ce{NaOH}$ was added about every two seconds. Do you have a more reasonable explanation?

One thing that that has already been mentioned in the comments is the impact of $\ce{CO2}$ on the equilibrium. If you look at the following figure you can see the pH diagram of carbonic acid.

The equilibrium between $\ce{CO2}$ and $\ce{H2CO3}$ $$\ce{CO2 + H2O <=> H2CO3}$$ depends very much on the pH of your aqeous system. As long as it is sufficiently acidic no carbonic acid can/will react to form hydrogen carbonate or even further to carbonate. But from $\ce{pH}\approx4$ the possibility that hydrogen carbonate will be formed rises. This means for your titration, that you are not only titrating your acid anymore but you are also titrating the dissolved $\ce{CO2}$. The longer you wait for equlibration near the the equlibrium point the more dissolved $\ce{CO2}$ reacts to hydrogen carbonate and distorts your ideal $\ce{CO2}$-free titration curve.

To show you the impact more clearly I created the following figures showing both of your systems with varying carbonic acid concentrations $\left(0.00\ldots0.05~\ce{mol~L^{-1}}\right)$.

At first your system with hydrochloric acid:

And secondly the system containing acetic acid:

There are two remarking differences between both titration curves.

You can determine the equlibrium point of the $\ce{HCl}$-titration (though it's not around pH 7 but at pH 4-5) no matter how many hydrogen carbonate will be formed during your titration.
You have to pay attention during the titration of acetic acid because the inflection points gets distorted above any detectable other inflection point. However it is not that important as it seems because titrations of acetic acid are fast and there is usually not much time to dissolve enough $\ce{CO2}$.

The other points are already mentioned before. Your pH electrode could not be calibrated correctly and your concentrations might not be as precise as you think they are. The titration curve of hydrochloric acid looks like your acid was less concentrated which results in a higher starting pH and in a lower ending pH. The titration of acetic acid indeed looks like the electrode was not properly calibrated because it looks quite good if you just raise the pH values per measuring point and it suffers from dissolved $\ce{CO2}$ which is why the inflection point is shifted to a higher volume.

The mentioned increase of volume that is most times neglected during the calculation of titration curves has no big impact unless the concentration of the base is much lower than that of the acid that has to be titrated.

On the one hand the first four images show that different base concentrations around your ideal concentration don't effect the curve itself very much but only the needed volume to reach the point of inflection.

On the other hand you can see that there is a visible influence on the titration curve if the base is 10 times as diluted as the acid.


Etymology of "click chemistry" According to Wikipedia, the term click chemistry was coined by K. Barry Sharpless in 1998.

What does the word 'click' mean here? I guess it means "join" here but I'm not sure.

Yes, it is detailed here and in numerous other places. It refers to a class of reactions, and does not mean "join."

From the above-cited page on named reactions:

  "Click Chemistry" is a term that was introduced by K. B. Sharpless in 2001 to describe reactions that are high yielding, wide in scope, create only byproducts that can be removed without chromatography, are stereospecific, simple to perform, and can be conducted in easily removable or benign solvents.

Sigma-Aldrich has a primer where they state:

  The traditional process of drug discovery based on natural secondary metabolites has often been slow, costly, and labor-intensive. Even with the advent of combinatorial chemistry and high-throughput screening in the past two decades, the generation of leads is dependent on the reliability of the individual reactions to construct the new molecular framework.
  Click chemistry is a newer approach to the synthesis of drug-like molecules that can accelerate the drug discovery process by utilizing a few practical and reliable reactions. Sharpless and coworkers defined what makes a click reaction as one that is wide in scope and easy to perform, uses only readily available reagents, and is insensitive to oxygen and water. In fact, in several instances water is the ideal reaction solvent, providing the best yields and highest rates. Reaction work-up and purification uses benign solvents and avoids chromatography.

A pretty comprehensive overview is given here, in a manuscript entitled Click Chemistry, a Powerful Tool for Pharmaceutical Sciences. Hein, Liu, and Wang. Pharm Res. 2008 Oct; 25(10): 2216–2230.


Origin of alkenes from Wurtz Reaction According to Wikipedia:

  Also, since the reaction involves free radical species, a side reaction occurs to produce an alkene. This side-reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon.

How would a radical reaction end up producing an alkene in the context of the Wurtz reaction?

What I do see happening is an E2 elimination creating an alkene.

Also, this raises another question - can't Grignards also create alkenes "by accident"?

The best explanation I was able to find is given by Avery A. Morton , John B. Davidson , Barton L. Hakan, J. Am. Chem. Soc., 1942, 64 (10), 2242–2247. If I understand it correctly, this boils down exactly to that you already propose.

  Disproportionation has been so universally accepted as the criterion of a
  free radical that there has been no attempt to show a derivation from other molecular species. Nevertheless, a mechanism for this phenomenon
  by way of an organometallic intermediate appears exceptionally reasonable. If an alkylsodium meets an alkyl halide in such manner that the two alkyl chains are adjacent to each other while sodium halide is being formed, the two alkyl residues will have unlike charges and a proton will be drawn to the alkyl radical having the two unsaturated electrons. The reaction is merely the conventional prototropic change taking place between adjacent portions of two molecules whose inorganic components are effecting, or have just
  completed, a union.

  If this view be accepted, the olefin produced must come from the alkyl halide rather than at random from both participant alkyl radicals. Though conclusions are to some extent obscured by a metal-halogen interchange (see later), all available facts unquestionably point to the correctness of this deduction.

However, there might be another reason, which I cannot back up with literature. This is probably due to the persistence of the radical species; higher concentration of radicals increases the possibility that they react with each other. In other words, one radical abstracts a hydrogen from another radical, hence creating an alkane and an alkene, in a purely radical reaction.
$$\ce{2 R-CH2-CH2. -> R-CH2-CH3 + R-CH=CH2}.$$

But then again radical mechanisms are never simple and it is quite possible that this does not happen at all, yet it would appear to be at least an explanation for the Wikipedia statement.


CO2 Catalytic Converter We all know how CO2, as a greenhouse gas affects the climate and causes global warming. I was thinking of a small device, which could be attached to the exhaust pipe of an automobile. It could convert the CO2 into something having negligible greenhouse effect, such as water. Is this feasible? Thanks!

Unfortunately this is not feasible
Current catalytic converters are designed to remove noxious pollutants from exhausts. Things like carbon monoxide (poisonous) and nitrogen oxides (noxious and irritating).
The trouble with doing this for carbon dioxide is fundamental. Carbon dioxide and water are the ultimate products of combustion of hydrocarbons because they are the thermodynamically deepest products of combustion (ie forming them releases the most energy from combustion). We are not worried about water. But the trouble with being at the bottom of the thermodynamic well is that, to turn the compound into something else, you need to add energy.
There are known chemical reactions that convert carbon dioxide into other, more useful and less greenhouse-contributing products. We can go the whole hog back to methane or, more commonly, methanol. But these reactions are very energy intensive and require fairly special catalysts and conditions. To do them as a way of reducing carbon dioxide emissions would end up using most of the energy from hydrocarbon burning to run the reaction not to move the vehicle. And there are always energy losses in such reactions so we might end up greatly increasing fuel consumption overall to get a small reduction in emission. It is probably far better to make engines and vehicles more efficient.
On an industrial scale the preferred alternative to carbon dioxide conversion is carbon capture and storage (CCS). Separating carbon dioxide is much cheaper than converting it to something else. Even so, CCS plants are in their infancy and none are remotely economical even when subsidised (or at any feasible carbon price). and these reactions require large chemical plants and could never be viable in a car.


What chemical can be used for getting rid of hard water spots on cars? Most water spot removers available on Amazon or automotive stores are just fine abrasives which remove a small amount of your car's clearcoat every time you use them.

I'm looking for a chemical which can dissolve the water spots made out of limestone, chalk and gypsum.

Any acids suggested should not be strong enough to damage clearcoat (diluted), should not be abrasive.
Obviously I will not leave it on my car for very long, just apply, maybe rub around and rinse off with water.

I've not had any success with vinegar.

Maybe the water spots are not limestone, chalk or gypsum, all of which are dissolvable in acid. Maybe they are mud spots, which could contain clays which adhere to the car paint.

If a toilet bowl cleaner based on citric acid (stronger acid than acetic, also a chelator) or Coke (acidic, contains phosphate) do not work quickly (without damage), perhaps an alkaline dispersing agent (like a liquid laundry soap) plus some elbow grease could diminish the visibility of the spots. Dispersing agents for clays include phosphates and anionic polymers like polysilicates (but they are typically quite alkaline - need to check damage potential).

Then, a good waterproofing (waxy) polish could make it easier to resist and remove the spots. Discovering where the spots come from (rain, splash from puddles, lawn sprayer, etc.) could help suggest what materials are present in the spots.


HCl + NaOH: Why H prefers to move to O instead of staying with Cl? The question is - fundamentally why does H "decides" to replace Na? Why is it more preferable than staying with Cl?

Because hydroxide is a stronger base than chloride. Chloride is a much larger ion, and thus its negative charge is spread over more volume, stabilizing it. The negative charge on the oxygen in hydroxide is spread out over less volume, leaving it more reactive. Thus, when a $\ce{H^+}$ ion is available, it will preferentially be retained by the hydroxide rather than the chloride.


What materials reversibly swell up slightly when in a moist environment? I am aware that some "Super Absorbent Polymers" can swell up enormously when they absorb water. However, I am looking for a material which swells up (maybe about 30% to 100%) when in a moist environment, and which will contract again when moved to a less moist environment.

I.E. by measuring the expansion of the material, one could determine the moisture content of the environment the material was in. It would act like a moisture measurement device.

Are there any such materials? Perhaps there are some "Fairly Absorbent Polymers"

A classic hygrometer consists of bundle of a dozen or so human hairs (blonde are supposedly best, because they're typically thinnest, i.e. react fastest to humidity changes), stretched between a metal spring with a pointing device and a fixed point.

A uniaxial effect is much easier to record precisely, and hair is thin (=fast) and far more mechanically reliable than a piece of swollen hydrogel.


What is the oxidation state of nitrogen in ammonium nitrate? The equation :
$\ce{NH4Cl + AgNO3 -> NH4NO3 + AgCl}$

I know that oxidation number of $\ce{N}$ in $\ce{NH4Cl}$ is -3 and in $\ce{AgNO3}$ is +5.
But what I am confused about, is the oxidation number of $\ce{N}$ in the product $\ce{NH4NO3}$. Should I consider $\ce{N}$ as $\ce{N2}$ in this compound or should I get its oxidation number in $\ce{NH4+}$ and in $\ce{NO3-}$ and then get the sum of them?

There are 2 different nitrogen atoms in $\ce{NH4NO3}$, one having oxidation number -3 and one with +5. They are different, having separate oxidation numbers independently of each other. It is perfectly fine to have several atoms of same element having different oxidation state in same compounds, for example in thiosulphate  $\ce{SO3S^{2-}}$ one (central) sulfur atom has oxidation number +4 and another has 0, and $\ce{Fe3O4}$ is actually $\ce{\overset{(II)}{Fe}\overset{(III)}{Fe2}O4}$ and many more.


Human Body and RNA/DNA After research I found that enzymes are responsible for metabolism, specifically catabolism and anabolism. This in turn creates the necessary chemicals in your body. The only question left is: what are enzymes? I asked in Biology and the answer led me to enzymes being created by ribosomes. Ribosomes by RNA. RNA by the four chemicals, adenine, guanine, cytosine, and uracil. Okay so, my question is: what chemical elements are those four chemicals comprised of?

A quick search would tell you what the 4 nucleobases for RNA are made of:

Uracil: 4 Carbon 4 Hydrogen 2 Nitrogen 2 Oxygen

Adenine: 5 Carbon 5 Hydrogen 5 Nitrogen

Guanine: 5 Carbon 5 Hydrogen 5 Nitrogen 1 Oxygen

Cytosine: 4 Carbon 5 Hydrogen 3 Nitrogen 1 Oxygen


Does neopentyl halide undergo SN1? Carbocation formed from neopentyl halide is primary in nature. So it is not stable. Although it undergoes rearrangement reaction to form tertiary one, now it can undergo SN1 easily. So if we say it undergoes directly in SN1 will it be okay?

Neopentyl halides are not good electrophiles in the way that other primary alkyl halides are, primarily due to steric effects but also because of rearrangents and eliminations.

This paper 1 notes in its abstract that "As a rule, the SN2 reactions of neopentyl bromide and tosylate with sodium alkoxides are extremely retarded by steric hindrance, and the yields of alkyl neopentyl ethers are poor because of side reactions such as neopentyl rearrangement, elimination, and decomposition." This is supportive of the view that an SN1/E1 process may be occurring.

This paper 2 confirms rearrangements occur.

I don't think this leads to us being able to say neopentyl bromide undergoes SN1 reactions directly or easily. It is unreactive to nucleophiles; when it does react it does so through mulitple pathways leading to low yields.


Why does 9-fluorenone dissolve in hexane? I know like dissolves like, but I am still unclear.  9-fluorenone has a polar $\ce{C—O}$ bond.

Is it because 9-fluorenone can only form London forces with other 9-fluorenone molecules and hexane can break these forces?

Firstly, as a general rule, organic compounds with more then 6 carbon have higher London forces then polar forces.  This is a very general rule and doesn't work in all circumstances. Thus, this molecule is not as polar.

Secondly, cyclic compounds have much weaker London forces then molecules that have straight chains. Thus, this makes the London forces not so high(relative).

Thirdly, to have something dissolve in hexane it must have weak intermolecular forces, but not weaker then hexane itself, as hexane only has to suspend the solute molecules in solution and prevent the solute from clumping up and falling to the bottom.

So, quite simply, what happens is that the hexane molecules surround the 9-Fluorenone molecules, which diffuse through the solution.


What determines whether a double displacement reaction will occur? In normal displacement reactions, reactivity plays a large role and sometimes the reaction doesn't even happen. So is there anything limiting double displacement reactions? For example $\ce{2KI + Pb(NO3)2 -> 2KNO3 + PbI2}$. If reactivity plays a role in this reaction, why does it occur? Isn't the halogen ion (iodide) more reactive than $\ce{NO3-}$?

(I'm a high school chem student so this may come off as a newbie question, sorry)

The simple answer is whether the reaction leads to the formation of a solid (precipitate) or a gas otherwise there is really no reaction


Qualitative Analysis- Distinguishing between two ions or two salts I know my question is going to be quite below the level of this site but I am 1st grade in university and I can't understand anything.

I am going to give an example from the book: distinguish between $\ce{AgI}$ and $\ce{AgCl}$. Answersheet tells me to add $\ce{NH3}$; only $\ce{AgCl}$ dissolves. I don't even know why $\ce{AgI}$  precipitate and $\ce{AgCl}$ dissolves.

Or there is another question: distinguish between $\ce{Ag+}$ and $\ce{[Ag(NH3)2]+}$. And answer sheet tells to add NaCl and only $\ce{Ag+}$ gives a precipitate. Why doesn't $\ce{[Ag(NH3)2]+}$ also give a ppt.?

How do I know which cations or anions or salts are soluble in which solvent? What should I look when I am trying to distinguish ions or salts? Or how do I know if there is no reaction? I know these questions are too general but I am having a hard time understanding this class.

You can either answer the questions from the book (which I failed to understand) or help me to understand the basis of the subject by answering others. Or both. I would really appreciate any kind of answers coming from you.

Answering this question requires a preliminary discussion of the solubility product constants of three silver halides and the formation constants of three silver complexes. First, note that $\ce{AgCl}$, $\ce{AgBr}$, and $\ce{AgI}$ are all insoluble in water, but insoluble is a relative term in the end. Their respective solubility equilibria and solubility product constants, i.e., equilibrium constants for their dissolution, are:

$$\ce{AgCl(s) <=> Ag^+ (aq) + Cl^- (aq) \quad $K_\mathrm{sp(1)} = \pu{1.8E-10}$              \tag 1}$$

$$\ce{AgBr(s) <=> Ag^+ (aq) + Br^- (aq) \quad $K_\mathrm{sp(2)} = \pu{5.4E-13}$              \tag 2}$$

$$\ce{AgI(s) <=> Ag^+ (aq) + I^- (aq) \quad $K_\mathrm{sp(3)} = \pu{8.3E-17}$              \tag 3}$$

Silver ions also form complexes with ammonia, thiosulfate ion, and cyanide ion. The formation equilibria and associated equilibrium constants are as follows:

$$\ce{Ag^+ (aq) + 2 NH3 (aq) <=> [Ag(NH_3)_2]^+ (aq) \quad $K_\mathrm{f(4)} = \pu{1.6E7}$              \tag 4}$$

$$\ce{Ag^+ (aq) + 2 S2O_3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) \quad $K_\mathrm{f(5)} = \pu{2.0E13}$              \tag 5}$$

$$\ce{Ag^+ (aq) + 2 CN^- (aq) <=> [Ag(CN)2]^- (aq) \quad $K_\mathrm{f(6)} = \pu{1.0E21}$              \tag 6}$$

Now consider the following sequence of aqueous solution additions:

An aqueous $\ce{NaCl}$ solution is added to an aqueous solution of $\ce{AgNO3}$. Then $\ce{AgCl}$ precipitates, as per equilibrium (1).
Next concentrated ammonia is added in excess, i.e., well above the 2 to 1 stoichiometry of equilibrium (4). Then the following equilibrium, obtained by adding equilibria (1) and (4), occurs:

$$\ce{AgCl (s) + 2 NH3 (aq) <=> [Ag(NH3)2]^+ (aq) + Cl^- (aq) \quad $K_\mathrm{sp(1)} K_\mathrm{f(4)}  = \pu{2.9E-3}$              \tag 7}$$

Although the equilibrium constant is less than unity, adding concentrated ammonia in excess (in a hood!) results in all of the $\ce{AgCl}$ dissolving: the equilibrium is driven to the right.

Next $\ce{NaBr}$ is added. This results in precipitation of $\ce{AgBr}$ via the following equilibrium:

$$\ce{[Ag(NH_3)_2]^+ (aq) + Br^- (aq) <=> AgBr (s) + 2 NH3 (aq) \quad $1/(K_\mathrm{sp(2)} K_\mathrm{f(4)})  = \pu{1.2E5}$              \tag 8}$$

This equilibrium is simply the reverse of the addition of equilibria (2) and (4).

Next excess sodium thiosulfate ($\ce{Na2S2O3}$) is added. Then $\ce{AgBr}$ dissolves as per the following equilibrium, obtained by adding equilibria (2) and (5):

$$\ce{AgBr (s) + 2 S2O3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) + Br^- (aq) \quad $K_\mathrm{sp(2)} K_\mathrm{f(5)}  = \pu{10.8}$              \tag 9}$$

Next $\ce{KI}$ is added. This results in precipitation of $\ce{AgI}$ via the following equilibrium:

$$\ce{[Ag(S2O3)2]^3- (aq) + I^- (aq) <=> AgI (s) + 2 S2O3^2- (aq) \quad $1/(K_\mathrm{sp(3)} K_\mathrm{f(5)})  = \pu{6.0E2}$              \tag{10} }$$

This equilibrium is simply the reverse of the addition of equilibria (3) and (5).

Lastly, $\ce{KCN}$ is added: in a hood, with proper safety precautions! Then $\ce{AgI}$ dissolves as per the following equilibrium, obtained by adding equilibria (3) and (6):

$$\ce{AgI (s) + 2 CN^- (aq) <=> [Ag(CN)2]^- (aq) + I^- (aq) \quad $K_\mathrm{sp(3)} K_\mathrm{f(6)}  = \pu{8.3E4}$              \tag{11} }$$

When I did this demo in lectures, I skipped the last step in order to avoid dealing with the hazardous waste issue that cyanide solutions present.

So now, at long last, the OP's questions:

Distinguish between $\ce{AgCl}$ and $\ce{AgI}$ by adding ammonia.

From equilibrium (7), $\ce{AgCl}$ can be dissolved if an excess of concentrated ammonia is added. Trust me: this should be done in a hood! But, comparing equilibria (1) and (7), $\ce{AgI}$ is more than a million times less soluble than $\ce{AgCl}$, so no amount of concentrated ammonia will significantly dissolve $\ce{AgI}$.

The OP's second question involves distinguishing between silver ion and the silver ammonia complex.

The OP's answer sheet claims that adding $\ce{NaCl}$ should result in only silver ion giving a precipitate. But equilibrium (7) shows that sufficiently high chloride ion concentrations should result in $\ce{AgCl}$ precipitation. But silver also forms $\ce{[AgCl2]^-}$, $\ce{[AgCl3]^2-}$, and $\ce{[AgCl4]^3-}$ , when the chloride concentration is high. So adding chloride will not cause AgCl to precipitate, in agreement with the OP's answer sheet. See here as well.

Last thought: I have done this demo in lecture, minus the last cyanide step, and I think the OP should not be expected to know all this stuff, especially if an exam is involved. The entire point of this kind of problem is to show that we can control equilibria to our advantage in, e.g., gold extraction via the cyanide process, and insoluble is a relative term.

Source of all solubility product constants and formation constants:

Daniel C. Harris, Appendix I In Quantitative Chemical Analysis; 7th Ed.; W. H. Freeman & Company: New York, NY, 2007 (ISBN: 0-7167-7041-5; ISBN-13: 9780716770411).


Does the rate of movement of ions in a salt bridge affect the total voltage produced by a Galvanic cell? I was thinking that different salts used in a salt bridge in a voltaic cell may have different rates of flow of ions in them (perhaps ionic radius or some other property affects the rate?)
I was wondering, if some ions in the salt bridge flow through the solution more quickly, will this have any effect on the overall voltage produced by the cell?

Voltage measurement is ideally a zero current process. Therefore the details of the ions in the salt bridge and their movements will not be an issue.

In reality, voltmeters do flow some current to measure the voltages. Yet, these currents are so small that in any self-respecting salt bridge, there will not be an issue. If your salt bridge was broken for whatever reason though, then your measurement will be completely incorrect. Most voltmeters will give you an overload in that case.


Why doesn't curie temperature follow a periodic trend in ferromagnetic materials I was writing up an answer to a question which was asking why the Curie Temperature's of $\ce{Fe, Co, Ni}$ don't follow a periodic trend. The OP voluntarily deleted his question before I could finish and post my answer.

The OP has reasoned as follows: Since Iron, Cobal and Nickel have electron configurations $\ce{[Ar] 3d^6 4s^2, [Ar] 3d^7 4s^2\ \text{and}\ [Ar] 3d^8 4s^2}$, giving them 4 , 3, and 2 unpaired electrons respectively. This according to the OP should suggest a periodicity in Curie Temperature's which isn't observed.

Clearly this isn't the case. For your reference, the Curie temperatures for the three metals are (iron, cobalt, nickel):    $1043\ \mathrm{K},  1400\ \mathrm{K},  627\ \mathrm{K}$

Since this did indicate some ignorance about the origins of magnetism, I decided to to write up a short answer, describing the salient features and at the same time correcting OP's misconception. The answer was kept relatively simple, and was nearly done but OP chose to delete his question.

Not wanting to let a answer I put some effort in go to waste, I am posting it here. Additionally, since the goal of this site is to build a Q&A, it would be great if others could chime in with their views, derivations etcetera.

TL;DR : There's no reason for it to show any periodicity. It doesn't depend in a simple way on the outer electron configuration of atoms. The specifics of crystal structure are important in ferrmoganetic (also in antiferromagnetic, and ferrimagnetic) materials.

This by no means is a complete description of the phenomenon. If you're interested, consult textbooks, the internet etcetera. and come back with more specific questions. The person who originally asked the question seemed like he was a high school student (or lower) so I tried to tailor an explanation that he would understand. I welcome more complete desscriptions

First of all, ferromagnetism is a property that goes beyond the chemistry of a material, and is highly dependant on crystalline structure and microstructure.

Existence of several ferromagnetic metal alloys whose constituents are not themselves ferromagnetic (Heusler alloys), and conversely, the existence non-magnetic alloys, composed almost exclusively of ferromagnetic metals (for example some types of steel) are evidence of this fact.

The OP originally spoke of unpaired electrons, so I am guessing he knew something about electron spin.

Spin, just like electron charge is an intrinsic property, and its behaviour is described using quantum mechanics. Spin and charge on the electron give rise to a magnetic dipole moment (people like to talk about classical charges spinning and generating a magnetic dipole, however, I would like to discourage this analogy).

Due to its quantum nature, the spin of the electron can be in one of only two states; with the magnetic field either pointing "up" or "down" (what I call up and down is somewhat arbitrary; what's important here is that there are two arbitrary, mutually exclusive eigenstates).

The spin of the electrons in atoms is the main source of ferromagnetism, although there is also a contribution from the orbital angular momentum of the electron about the nucleus.

Materials made of atoms with filled electron shells have a total dipole moment of zero, because the electrons all exist in pairs with opposite spin, every electron's magnetic moment is cancelled by the opposite moment of the second electron in the pair.

So of course, a ferromagnetic material needs to have spin unpaired electrons, but that alone isn't sufficient. Hund's rules, posit that the first few electrons in a shell tend to have the same spin, thereby increasing the total dipole moment.

These unpaired dipoles (often called simply "spins" even though they also generally include angular momentum) tend to align in parallel to an external magnetic field, an effect called paramagnetism. Such materials are hence called paramagnetic. Molecules with spin paired electrons are called diamagnetic.

Ferromagnetism involves something extra. In a few substances the dipoles tend to align spontaneously, giving rise to a spontaneous magnetisation, even when there is no applied field. This is ferromagnetism.

Now let's develop some specifics. For instance, how and why do spins align? Additionally, why can we find iron in both magnetised and unmagnetised states, shouldn't it always be magnetised?

When two nearby atoms have unpaired electrons, whether the electron spins are parallel or antiparallel affects whether the electrons can share the same orbit as a result of the quantum mechanical effect called the exchange interaction. The exchange interaction is related to the Pauli exclusion principle, which says that two electrons with the same spin cannot also have the same "position".

When the orbitals of the unpaired outer valence electrons from adjacent atoms overlap, electrostatic repulsions are reduced when spins are parallel to each other. In simple terms, the electrons, which repel one another, can move "further apart" by aligning their spins, so the spins of these electrons tend to line up. This difference in energy is called the exchange energy. In ferromagnetic materials, the exchange interaction is much stronger than the competing dipole-dipole interaction. For example, in iron (Fe) the exchange force is about 1000 times stronger than the dipole interaction.

Therefore, below the Curie temperature virtually all of the dipoles in a ferromagnetic material will be aligned. (we will get to Curie temperature in a minute)

So why are iron and other ferromagnets ever found in an "unmagnetised" state ?

The reason for this is that a bulk piece of ferromagnetic material is divided into tiny regions called magnetic domains. Ferromagnetic materials spontaneously divide into magnetic domains because the exchange interaction is a short-range force, and over long distances of many atoms the tendency of the magnetic dipoles to reduce their energy by orienting in opposite directions wins.

If all the dipoles in a piece of ferromagnetic material are aligned parallel, the space surrounding it contains a lot of magnetostatic energy, which the material wishes to  reduce, and it does so energy by splitting into many domains pointing in different directions, so the magnetic field is confined to small local fields in the material, reducing the volume of the field.

This means that within each domain, the spins are aligned, but (if the bulk material is in its lowest energy configuration, i.e. unmagnetised), the spins of separate domains point in different directions and their magnetic fields cancel out, so the object has no net large scale magnetic field.

The domains are separated by thin domain walls a number of molecules thick, in which the direction of magnetisation of the dipoles rotates smoothly from one domain's direction to the other

Although the exchange interaction keeps spins aligned, it does not align them in a particular direction. Without magnetic anisotropy, the spins in a magnet randomly change direction in response to thermal fluctuations and the magnet is super-paramagnetic. There are several kinds of magnetic anisotropy, however, let's not go into that here

If one now places such a material in a strong enough external magnetic field, the domains will reorient so more of the dipoles are aligned with the external field. Moreover, the domains will remain aligned when the external field is removed, creating a magnetic field of their own extending into the space around the material, thus creating a "permanent" magnet.

The domains do not spontaneously return to the original minimum energy configuration when the field is removed because the domain walls tend to become fixed in place by the crystal structure, and defects in the bulk solid. Thus the magnetisation, and the resulting magnetic field, is baked into to the crystal structure of the material, making it very difficult to demagnetise (it is metastable in this sense).

As the temperature increases, thermal motion, competes with the ferromagnetic tendency for dipoles to align, and when the temperature rises beyond a certain point, (the Curie temperature), the system can no longer maintain a spontaneous magnetisation, as reorientation takes place.

This is an example of a second order phase transition

Thus the material looses its ferromagnetism; unpaired electrons are still present, and thus it is still paramagnetic.

Now clearly, you can see that ferromagnetic properties don't depend in a simple way on unpaired spins, and there is no reason for the curie temperature to follow a periodic trend the way you expect it to.


Total number of stereoisomers of 2,3,5,6-tetrachlorooctane Total number of stereoisomers for:

I tried the following combinations:
RRRR, SSSS, RRRS, SSSR, RRSS, RSRS, SRRS, RSSR, RSRR, SRSS, RRSR, SSRS, SSRR, SRSR.
So, total number of stereoisomers I got is 12, but my teacher says it's 16. I don't know how to figure out what other four isomers are.

If you have studied combinatorics, and if you know the fundamental principal of counting, then this is basically making a four letter word with two letters- R and S. Each place has two choices, and you have four spaces to fill, so you get 2×2×2×2= 16 optical isomers for this molecule. Since the terminal alkyl groups are different, so there is no chance of a meso-compound. And as Ivan rightly stated in the comment, to list the isomers, make all possible combinations of R and S(you have missed a few as pointed out by porphyrin in the comments)


Can carbon dating tell how long a species would survive? Carbon ($\ce{^{14}C}$) is used to determine how long a particular species has lived or how old it is. Can we use the same isotope to derive how long a species may survive? (My understanding is still quite at infant level on half-life concept)

Is it prohibited or is it impossible? If not possible, why?

No, it can't be used for that. Read the mechanism of radiocarbon dating.

  Radiocarbon, often abbreviated as $\ce{^{14} C}$, is constantly being created in the atmosphere by the interaction of cosmic rays with atmospheric nitrogen. The resulting radiocarbon combines with atmospheric oxygen to form radioactive carbon dioxide. This is then incorporated into plants by photosynthesis, and animals acquire $\ce{^{14}C}$ by eating the plants. When the animal or plant dies, it stops exchanging carbon with its environment, and from that point the amount of $\ce{^{14} C}$ it contains begins to reduce as the $\ce{^{14}C}$ undergoes radioactive decay. Measuring the amount of $\ce{^{14}C}$ in a sample from a dead plant or animal such as piece of old wood or a fragment of bone provides information that can be used to calculate when the animal or plant died. The oldest dates that can be reliably measured by radiocarbon dating are around 50,000 years ago, though special preparation methods occasionally permit dating of older samples.


Why does oxygen in OH have an oxidation state of 2-? I am watching videos on Khan Academy in order to go over redox reactions (a really weak area for me.) In Sal's video, "Oxidation state trends in periodic table" at 7:10, he says:

  If you had to pretend this was an ionic bond, then maybe this hydrogen
  would fully lose an electron, so it would get an oxidation state of
  plus 1. It would be oxidized by the oxygen.

  And the oxygen actually has fully gained one electron. And if we're
  forced to think about this is an ionic bond, we'd say it fully gains
  two electrons. So we'd have an oxidation state of negative 2.

I'm having a hard time understanding how he is able to deduce that oxygen has an oxidation state of 2- given the above reasoning. He does add that the states add up to the overall charge which is helpful, but I think I need a little more clarity than that. My reasoning for oxygen being 2- is that "it usually is because my textbook says it usually is," which is probably an even worse explanation.

Could anyone clarify this for me?

First step: draw the Lewis structure of $\ce{OH-}$ with all lone pairs. You should be getting three lone pairs on oxygen and one $\ce{O-H}$ bond.

Second step: ‘[cleave] [all bonds] heterolytically in such a way that the electronegative partner gets both electrons.’[1] You should now have eight electrons around the oxygen and zero on hydrogen.

Third step: count the electrons on both (or all) atoms and check what the difference is to their elemental valence electron count. Oxygen formally has eight electrons on it which is two more than the six it would in its elemental state. The two extraneous electrons give an oxidation state of $\mathrm{-II}$. Likewise, hydrogen has lost its electron for an oxidation state of $\mathrm{+I}$.

Control step: add up the oxidation states of all atoms. If you did everything correctly, they will add up to the entire charge of the ion. Here, this works out:

$$(-2) + (+1) = -1$$

[1]: https://chemistry.stackexchange.com/a/66242


How do we get g and E values for various levels to calculate electronic partition function When we calculate electronic partition function using the formula:
$$q_\mathrm{elec}=\sum^\infty_{n=1}g_ne^{-E_n/k_BT}$$
How can I get the $g_n$ value and $E_n$ values for $n=0,1,2,3...$ levels?
I need to calculate the $q_\mathrm{elec}$ for various species like $\ce{H, H2, OH, H2O, O2, O, O+, H+}$.

$E_n$ represents the energy of the $n$th electronic state relative to the ground state (i.e. the ground-state electronic energy $E_0 = 0$). Often, electronic states are very high in energy and the excited-state energies $E_1, E_2, \cdots$ are far greater than $k_\mathrm{B}T$, such that $\mathrm{e}^{-E_n/k_\mathrm{B}T} \approx 0$.
Under these conditions, the sum can be collapsed to just a ground-state term
$$q_\mathrm{elec} = \sum_n g_n \mathrm{e}^{-E_n/k_\mathrm{B}T} \approx g_0 \mathrm{e}^{-E_0/k_\mathrm{B}T} = g_0$$
where the latter equality is because $E_0 = 0$. $g_0$ is the degeneracy of the electronic ground state: to find this out you will need to draw some MO diagrams and/or construct the term symbols.
If the ground state is nondegenerate (e.g. $\ce{H2O}$) the whole thing is just equal to 1. Otherwise, for an atom with ground-state term symbol $^{2S+1}L_J$, the degeneracy is $2J + 1$. Similar considerations can be applied to molecules.
However, you will have to watch out for cases where there are low-lying energy states where $E_n$ is on the order of $k_\mathrm{B}T$. I am not aware of an easy way to predict this, though – you will have to know some typical examples, failing which you will just have to look it up, as porphyrin suggests. The NIST WebBook is a good starting place for this.


Writing Mass Balance Equations I was curious if there is a specific reason why when writing mass balance reactions we always leave a $\ce{H+}$ on the left side of the equation as the professor did not explain so, or is it just convention?

For example, write the mass balance equation of $\ce{H3PO4}$:

$$\ce{H2O <=> H+ +OH-}$$
$$\ce{H3PO4 <=> H+ +H_2PO4^{-}(aq)}$$
$$\ce{H2PO4- <=>H+ + HPO4^{2-}(aq)}$$
$$\ce{HPO4^2- <=> H+ + PO4^{3-}(aq)}$$

There is no such convention explicitly telling what comes out first.
Here, for a traditional formula of phosphoric(V) acid $\ce{H3PO4}$ writing a proton first serves only a didactic purpose.
It's visually easier for students to keep a track on dissociation as the order of the elements both in formula and among the products is preserved.

On the other hand, shall one use a coordination formula of phosphoric(V) acid $\ce{[PO(OH)3]}$, it probably would make more sense to use a reversed order and put $\ce{H+}$ at the end:

$$
\begin{align}
\ce{[PO(OH)3] &<=> [PO2(OH)2]- + H+}\\
\ce{[PO2(OH)2]- &<=> [PO3(OH)]^2- + H+}\\
\ce{[PO3(OH)]^2- &<=> [PO4]^{3-} + H+}
\end{align}
$$

One more minor thing.
Equations are usually aligned about arrows, and a tabular array of products and reactants emerges when the same repeating compound ($\ce{H+}$) is shown first.
Note how easier it is to grasp the information when more compounds are aligned:

$$
\begin{align}
&\color{green}{\text{aligned}} & &\color{red}{\text{misaligned}}\\
\ce{H3PO4 &<=> H+ + H_2PO4^{-}(aq)} &\quad \ce{H3PO4 &<=> H_2PO4^{-}(aq) + H+} \\
\ce{H2PO4- &<=>H+ + HPO4^{2-}(aq)} &\quad \ce{H2PO4- &<=> HPO4^{2-}(aq) + H+} \\
\ce{HPO4^2- &<=> H+ + PO4^{3-}(aq)} &\quad \ce{HPO4^2- &<=> PO4^{3-}(aq) + H+}
\end{align}
$$


Can alkyl fluorides be formed by the Hunsdiecker reaction? We know that the Hunsdiecker reaction is

$$\ce{RCOOAg + X2 ->[CCl4] RX + AgX + CO2}.$$

This reaction is successful with chlorine and bromine. Also, we know that if $\ce{X}$ is iodine, we will get an ester.

Can we form alkyl fluorides using this reaction? If not, why?

The classic reaction of silver salt of fatty acid and halogen to produce alkyl halide was first studied by Alexander Boronin in 1861. But the reaction was generalized by Heinz Hunsdiecker due to which the reaction is named Hunsdiecker reaction. Hunsdiecker used bromine to make methyl bromide. Chlorine was then used to make alkyl chloride. When iodine was used, then ester was formed. This reaction was called Simmonini reaction. Fluorine was not yet discovered then and hence alkyl fluoride was not made through this reaction (fluorine was discovered in 1886 by Henry Moissan).

But there are some speculation discussed among the chemist community whether Boronin actually used some fluorine containing reagent to make alkyl fluoride. It is discussed in this forum that Boronin might have used "acidic potassium fluoride" or $\ce{KHF2}$/potassium fluoride($\ce{KF}$) to make benzoyl fluoride in 1862 but there is no report or scientific paper on this reaction making it a speculated information. (Note: fluorine was discovered by electrolyzing $\ce{KHF2}$).

Fluorine containing compounds were made through Hunsdiecker reaction though proper alkyl fluoride synthesis through this reaction is yet to be reported:

Perfluoroalkyl iodides were made through this reaction:
$$\ce{R_fCO2Ag + I2 ->[\Delta] R_fI + CO2 + AgI}$$
This compound is used to make numerous fuctionalized fluorocarbons and
first Grignard reaction of this class.
Perfluoroalkyl diiodides has also been prepared by heating perfluorocarboxylic acid in presence of iodine. It has been used to make diacids having ether linkage.
Fluorinated ester of type $\ce{RFCO2CF2RF′}$ has been was obtained from reaction of silver 3,6,9-trioxa-F-undecanoate and iodine with a diluent at 130° C. Substitution products, $\ce{RFCOY}$ (Y = nucleophile) and pyrolysis products, $\ce{RFCOF}$ and $\ce{RF′COF}$ were isolated and characterized. It appears probable that the acyl hypoiodite and the iodo-F-alkane reacted by an $\ce{S_Ei -}$ type process to give the completely fluorinated ester.
Fluoro-functionalized graphene oxide  is made via the Hunsdiecker reaction. It is made by treating of carboxylated graphene oxide with selectfluor at 90 °C for 10 h under an atmosphere of nitrogen, using silver nitrate as a catalyst.

References:

Organofluorine Chemistry: Principles and Commercial Applications
edited by R.E. Banks, B.E. Smart, J.C. Tatlow - page 18
Well-Architectured Fluoropolymers: Synthesis, Properties and Applications
By Bruno Ameduri, Bernard Boutevin - page 108
Brace, N. O. Some approaches to the synthesis of fluorinated alcohols and esters. I. Completely fluorinated esters from the hunsdiecker reaction of silver F-alkanoates with iodine. J. Fluorine Chem. 1981, 18 (4), 515–524. DOI: 10.1016/S0022-1139(00)82668-8.
Xing, R.; Li, Y.; Yu, H. Preparation of fluoro-functionalized graphene oxide via the Hunsdiecker reaction. Chem. Commun. 2016, 52 (2), 390–393. DOI: 10.1039/c5cc08252h.


Why does electrolysis take so much time? What makes electrolysis or electroplating take so long ? What stage takes longer when ions are attracted to electrodes or when they gain or lose electrons at electrode.
And if we are talking about an electrolyte composed of positive ions and electrons , which is faster, electrons entering the circuit at positive electrode or gaining electrons by positive ions at negative electrode?

It depends: The rate of electrolysis is limited either by the electric source or by the ion concentration. I am inexperienced with electrolysis but I imagine that quality may increase as the the process slows, much like crystals grow with more imperfections when the rate of crystallization is fast.

There is no such thing as an electrolyte solution composed of positive ions and electrons.

  "Activity coefficients of single ions cannot be measured
  experimentally because an electrolyte solution must contain both
  positively charged ions and negatively charged ions." -Debye–Hückel
  theory

Electroplating systems tend to use an electrode source that replenishes the solvent ions plated onto the target electrode.


Reducing power of ammoniated electrons I came across a question asking which of the following conversion/reductions can be accomplished using ammoniated electrons.

$\ce{O2}$ to $\ce{O2^{2-}}$
$\ce{K2[Ni(CN)4]}$ to $\ce{K4[Ni(CN)4]}$
Aromatic ring
Non-terminal alkyne

I have been able to find on Wikipedia that $\ce{O2}$ is reduced to $\ce{O^{2-}}$. So I believe 1 should be incorrect. I am unsure of 2 and could not find any literature regarding the same. About 3, I know about Birch Reduction. About 4 I don't know.

The answer is 1, 2, 3, 4.

The reducing power of alkali metal such as $\ce{Na}$ and $\ce{K}$ in liquid $\ce{NH3}$ is duee to the solvated electrons ($e^-$), which can reduced certain compounds into unusual oxidation states. One such example is reduction of oxygen ($\ce{O2}$). The the solvated electrons can be reduced $\ce{O2}$ to $\ce{O2^{.-}}$ (superoxide ion) first and then to $\ce{O2^2-}$ (peroxide ion) (Ref.1). However, a solution of $\ce{Na}$ in liquid $\ce{NH3}$ is treated with oxygen, the peroxide ion is generally formed (Ref.2). When a solution of $\ce{Li}$ in liquid $\ce{NH3}$ is rapidly oxidized with $\ce{O2}$ at $\pu{-78 ^\circ C}$, a bright lemon-yellow solution is formed (which is believed to be a solution of $\ce{LiO2}$). When the solution is warmed towards $\pu{-33 ^\circ C}$, the yellow color faded and white suspension is formed (which is believed to be solid $\ce{Li2O2}$). The color change is not reversible (pp. 131-132, Ref.2).

Another example is conversion of $\ce{Ni^{II}}$ to $\ce{Ni^{I}}$ and $\ce{Ni^{II}}$ to $\ce{Ni^{0}}$. Potassium tetracyanonickelate(II) ($\ce{K2[Ni(CN)4]}$) has been reduced by $\ce{K}$ in liquid $\ce{NH3}$ at $\pu{-33 ^\circ C}$ to the red potassium tetracyanonickelate(I) $(\ce{K3[Ni(CN)4]})$ (Ref.3), which is slowly reduced further to yellow potassium tetracyanonickelate(0) $(\ce{K4[Ni(CN)4]})$ at $\pu{0 ^\circ C}$ (Ref.4).

It is known fact that solutions of alkali metal in liquid $\ce{NH3}$ reduce several organic compounds including alkynes to trans-alkenes and aromatics to 1,4-cyclodienes (Birch reduction; e.g., reduction of benzene to cyclohexa-1,4-diene).

Therefore, correct answer to the question is all of (1)-(4).

References:

William H. Schechter, Harry H. Sisler, Jacob Kleinberg, “The Absorption of Oxygen by Sodium in Liquid Ammonia: Evidence for the Existence of Sodium Superoxide,” J. Am. Chem. Soc. 1948, 70(1), 267-269 (https://doi.org/10.1021/ja01181a083).
Nils-Gösta Vannerberg, “Chapter 3: Peroxides, Superoxides, and Ozonides of the metals of groups  Ia, IIa, and IIb,” In Progress in Inorganic Chemistry; Volume 8; F. Albert Cotton, Ed.; John Wiley & Sons, Inc.: Easton, PA, 1962, pp. 125-198 (ISBN:978-04-7017-6733).
John W. Eastes, Wayland M. Burgess, “A Study of the Products Obtained by the Reducing Action of Metals upon Salts in Liquid Ammonia Solutions. VII. The Reduction of Complex Nickel Cyanides: Mono-valent Nickel,” J. Am. Chem. Soc. 1942, 64(5), 1187-1189 (https://doi.org/10.1021/ja01257a053).
George W. Watt, James L. Hall, Gregory R. Choppin, Philip S. Gentile, “Mechanism of the Reduction of Potassium Tetracyanonickelate(II) and Potassium Hexacyanocobaltate(III) with Potassium in Liquid Ammonia,” J. Am. Chem. Soc. 1954, 76(2), 373-375 (https://doi.org/10.1021/ja01631a016).
Additional reference: V. Gutmann, "Chapter III: Coordination Chemistry in Proton-containing Donor Solvents," In Coordination Chemistry in Non-Aqueous Solutions; Springer-Verlag: Wien, Austria, 1968, pp.35-58 (ISBN-13: 978-3-7091-8196-6).


Electron configuration of iron(I) cation What is the electron configuration of $\ce{Fe+}$ cation?
\begin{align}
\ce{Fe+} &\!:~ [\ce{Ar}]\mathrm{(3d)^6(4s)^1}\label{chm:1}\tag{1}\\
\ce{Fe+} &\!:~ [\ce{Ar}]\mathrm{(3d)^7(4s)^0}\label{chm:2}\tag{2}
\end{align}
What I've been told is that \eqref{chm:2} is correct since the $\mathrm{4s}$ electron jumps to $\mathrm{3d}.$
Which representation is correct?

Isolated in a vacuum, in the absence of external fields, the first configuration is correct - $[\ce{Ar}]\mathrm{(3d)^6(4s)^1}$ is the ground electronic state of the iron(I) cation $\ce{Fe^{+}}$. More specifically, the ground state also has the term symbol $\mathrm{^6D_{9/2}}$.
NIST's Atomic Spectra Database has compiled the ground state electronic configurations of almost all possible ions of almost every known element (including such exotic things as $\ce{U^{58+}}$), obtained from a combination of experimental data and theoretical calculations. For example, search for iron, and you will be presented a table with all ground state electronic configurations, from the neutral atom $\ce{Fe^{0}}$ to the hydrogen-like $\ce{Fe^{25+}}$. You can find a description of the output data here.
I'm sure you'd like a reason for why $\ce{Fe^{+}}$ has the first configuration you suggest and not the second, but there is almost certainly no explanation which is both simple and correct. Unfortunately the predictive power of the aufbau principle is somewhat overstated, even when taking into consideration qualitative corrections. Depending on your notions, there are arguably several dozen "unexpected" electronic configurations, across all possible ions of all elements. The worst offenders are the heavier elements (say, the sixth period and onwards), which are rarely (if ever) covered in early chemistry classes.


carbon dioxide and its behaviour in acid base balance In human body acid base balance, why and how does the dissolved carbon dioxide increases the hydrogen ion concentration with increasing in its partial pressure..? At the same time bicarbonate ion is staying in same concentration while hydrogen ion are increasing?

The disolved carbon dioxide increases the acid content of the human body. This happens when a carbon dioxide molecule bumps into a water molecule. Carbon dioxide makes water break into hydrogen and hydroxide ion. Hydroxide merges with the carbon dioxide molecule creating bicarbonate ion. Sometimes the hydrogen ion can become part of increasing acidity. Sometimes the hydrogen ion combines with the bicarbonate molecule and make carbonic acid. Carbonic acid could turn into carbonate ion and two hydrogen ions. This is what makes the human body acid base balance. The amount of bicarbonate is staying the same because the bicarbonate that does get converted into carbonic acid or carbonate ion an equal amount of bicarbonate is created from water and carbon dioxide. This just results in a huge amount of hydrogen ions.


Shape of PEG in water According to the chemical formula $\ce{H−(O−CH2−CH2)_n−OH}$ of polyethylene glycol, or PEG, I assume that this organic compound has a long chain structure when it is dry and in powder form. (This is also what guys in the chemistry field say.)
But when PEG is dissolved in water, would it form some structure like a cluster or would it remain a long chain?
Can we somehow calculate the diameter of a dissolved PEG molecule when $n$ is given? Or the dissolved PEG molecules band randomly so there are no calculations for an average diameter statistically at all?

PEG is the term usually reserved for shorter chains, while PEO is  used to refer to longer chain molecules. When polyethylene glycols are dissolved in water, they increase the viscosity of the solution. While it might be easy to model short-chain polymers in dilute solution as some kind of folded chain approaching a spherical volume, as the chains get longer and the solution becomes more concentrated (>1%), there is significant entanglement.

The picture above is for a long-chain polyacrylate, which can be adjusted to be very hydrophilic by adding alkali to generate anions all along the chain. The images show various shapes in water: some fairly tight, some looser (spread out), some spread out but very much entangled. At some point in increasing chain length or increasing concentration, it becomes difficult to project the image of a single molecule of PEO into useful predictions of solution properties. Measurements of dilute or shorter chain molecules would include viscosity; as the solution becomes more concentrated or the chain longer, the property includes some elasticity, so we have viscoelasticity, and at very high chain lengths, we have pituosity (more commonly known as stringiness or ropiness). A video shows this effect dramatically:
https://www.neatorama.com/2017/10/09/Fun-with-Polyethylene-Glycol/.
Since PEGs and PEOs are so soluble and compatible with water, the first molecules to be added just fill all the volume of water; you could calculate how long a PEO molecule would be if completely stretched out. Stirring can extend the molecule, showing effects due to viscoelasticity and pituosity.


Comparing C=O bond length in CO2 and CH2O Why is the C=O bond in $\ce{CO2}$ shorter than that in $\ce{CH2O}$? How can I use hybridisation theory to explain this?

Drawing the Lewis structure for CO$_2$ and CH$_2$O shows that the C atom in CO$_2$ is sp hybridized and therefore forms a double bond (which is composed of a $\sigma$ and $\pi$ bond) with either O by overlapping one of its sp orbital with one of the sp$^2$ orbitals of O (since O is sp$^2$ hybridized) to form the $\sigma$ bond and one of its p orbitals with the p orbital of O to form the $\pi$ bond.

In CH$_2$O, the C atom is sp$^2$ hybridized, meaning the C$=$O bond is formed by an overlapping sp$^2$ orbital from C and an sp$^2$ orbital from O (since O is sp$^2$ hybridized) for the $\sigma$ bond and an overlapping p orbital from C and a p orbital from O for the $\pi$ bond.

So, the difference in the C$=$O bonds in the two compounds is in the constituent orbitals of their respective $\sigma$ bonds. The $\sigma$ bond of the C$=$O bond is shorter in CO$_2$ because it involves the overlap of an sp orbital with an sp$^2$ orbital. An sp orbital has more "s character" than an sp$^2$ orbital (you can think of an sp orbital as 50% s character, 50% p character and an sp$^2$ orbital as 33% s character, 67% p character), therefore the electron density in an sp orbital is on average closer to the nucleus of the C atom than in an sp$^2$ orbital, leading to a shorter $\sigma$ bond length. Because the $\pi$ bonds are formed from identical (up to orientation) p orbitals in each compound, the shorter $\sigma$ bond in CO$_2$ gives it a shorter C$=$O bond length than CH$_2$O.


Does water freely absorb atmospheric CO2? So, there has been a fairly groundbreaking new invention that can convert CO2 dissolved in water to Ethanol1. At face value, it's cheap, easy, and relatively energy-lightweight. All it requires is the catalytic surface and CO2 to be dissolved in water.

I'm just curious, this invention looks like the holy grail of Green news, but not being a chemist, and doubting that the reporters are (all) chemists, I have to ask "Where's the catch", and the best I can find is how quickly CO2 dissolves in water, so, is this step of the above process fairly easy, or would it require additional research?

Note: This question is about how quickly atmospheric CO2 dissolves in water. Comments on the discovery would probably be better served as a comment then an answer.

Er, the quick, simple technical answer is "Yes, water will absorb $\ce{CO2}$". However, there is a "but" coming. Actually, two "buts".

First, plain water will not be able to absorb much $\ce{CO2}$. Henry's Constant for $\ce{CO2}$ in $\ce{H2O}$ is $K_H=29.76 \mathrm{atm\over mol/L}$. Considering the partial pressure of $\ce{CO2}$ in air is roughly $3.5\times10^{-4}~\mathrm{atm}$, one liter of water can absorb, at best $1.18\times10^{-4}~\mathrm{mol}~\ce{CO_2}$ which is only about $0.1$% of a gram of $\ce{CO2}$. Not very much. (Note, I'm ignoring the conversion to carbonate, $\ce{CO3^{2-}}$, and bicarbonate, $\ce{HCO3^{-}}$, because in otherwise neutral water, they don't exist in appreciable quantities (see: Carbonic Acid)

Second, that's the thermodynamic best you can do. However, you have to contend with the kinetics and diffusion limitations as well. Basically, it takes time for the $\ce{CO2}$ to go from the air to the water, and then to get well mixed within the water.

Now, there's a couple things you can do to help $\ce{CO2}$ absorption. First, instead of using neutral water, use and alkaline solution. For example, if you add $\ce{NaOH}$ to your water, the equilibrium will shift toward the carbonate ion, $\ce{CO3^{2-}}$:

$$\ce{2NaOH + CO2->Na2CO3 + H2O}$$

This allows orders of magnitude more $\ce{CO2}$ to be absorbed than would possible, basically, you become limited by the amount of sodium hydroxide you have and the kinetics.

Speaking of kinetics, how can we help that out? Packed bed towers! Imagine having a tall column filled with beads. Blow your air up from the bottom, while allowing the alkaline solution to flow from the top. This will maximize contact between the two phases and enhance the transfer of $\ce{CO2}$ from the gas to liquid.

And that is basically how scrubbers work in chemical plants, and one method of carbon sequestration. However, the very low concentration of $\ce{CO2}$ in air ($\sim\!400~\mathrm{ppm}$) makes pulling appreciable amounts of $\ce{CO2}$ from the air infeasible and an inefficient use of resources.


Bromination of acetanilide Why does acetanilide gives exclusively para isomer. I know that -I of nitrogen must decrease the yield of ortho product, but still it should be made in accountable amounts. Where am I going wrong.
I am inserting the question and solution screenshots for reference. The question is from IIT JEE 2016 exam, chemistry section of paper 1

Question

Solution

Turning the amino group into an amide serves two purposes:

We are introducing a sterically bulky group. We expect the amide bond to be as complanar as possible with respect to the benzene ring to maximise electronic interactions. That should block one of the two ortho positions sterically.
We are turning an electron-rich, $+M$ aromatic system into an electron-poor, $-I$ one.

Especially the second transformation will greatly reduce the reaction rate. Furthermore, a bromine atom is mildly deactivating, too, due to its $-I$ effect (the weak $+M$ effect is neglegible and only responsible for the ortho/para directing abilities). Thus, it is very likely that we can selectively stop the reaction after monobromination.

The question still arises why we only obtain the para-product. Obviously the amide behaves much like a bromide and displays a weak $+M$ effect in spite of the amide resonance. Yet we can imagine the the proximity to the bulky amide group inhibits substitution at both ortho-protons sufficiently to allow isolation of the para-product.


What does "percentage by mass" mean in the context of a compound dissolved in a solution? Copper can be recovered from low-grade ores by 'leaching' the ore with dilute $\ce{H2SO4}$, which converts the copper compounds in the ore into $\ce{CuSO4(aq)}$. In a $\pu{50 cm^3}$ leach solution, the concentration of copper ions is $\pu{7.8 x 10^-3 mol/dm^3}$. Find the percentage by mass of copper in the leach solution.

Isn't it impossible to find the percentage by mass in this case? We don't know the total mass of the leach solution, and it could be anything. Is the question wrong, or is the definition of percentage by mass different than what I'm using?

Also, how does $\ce{H2SO4}$ react with copper compounds to form $\ce{CuSO4}$? I thought copper couldn't reduce hydrogen ions.

is the definition of percentage by mass different than what I'm using?

Basically, the questioner wants you to calculate mass by volume percentage contrary to your assumption that the question asks for a mass by mass percentage. The existence of a lot of terms referring to the same mass percentage can be a common confusion.

Nevertheless, as a rule of thumb, you'd say, I advise you to go for the calculation of mass by volume percentage unless the questioner explicitly mentions or you've been given enough data.

  Find the percentage by mass of copper in the leach solution.

Now that you already have the concentration of $\ce{Cu^2+}$ ions with you, you just need to calculate the number of moles, then the mass and finally divide it with the volume of the mixture.

  Also, how does $\ce{H2SO4}$ react with copper compounds to form $\ce{CuSO4}$? I thought copper couldn't reduce hydrogen ions.

Oxidic sources of copper are well known for their amenability to sulfuric acid leaching, for eg. azurite, malachite, tenorite, cuprite, chryosocolla, and brochantite dissolve easily in dilute sulfuric acid. Note that too dilute concentration is refrained from since we don't want hydrous ferric oxides to precipitate (they can adversely effect copper extraction).

And as a side-note, if you remove copper sulfides and iron sulfides, the $\ce{H2SO4}$ requirement would be about five times the weight of the dissolved copper (concentrated $\ce{H2SO4}$ is a good oxidizer).

Interestingly, this question was asked in University of Cambridge international examinations October/November 2010, your question is perhaps a simplification of this:


Group 15... Lewis Acid, Lewis Base or Not At All? I am a long-time reader of this forum, but this is my first time post. I am a chemistry undergrad with a question about Lewis acidity or basicity of Group 15 compounds.

I know Lewis acids act as an electron pair acceptor and a Lewis base as an electron pair donor. Then sometimes a dative bond is formed between the two, forming a Lewis adduct (as per classic $\ce{NH3}$ and $\ce{BF3}$ reaction). But this is where my knowledge becomes a little sketchy, and hence my question:

When considering group 15, is it possible for the elements of this group to form compounds which are either Lewis acids or bases?

Firstly, the requirements for Lewis acids of group 15, is it empty d-orbitals or an incomplete octet which give a compound the ability to accept two electrons?
Therefore, would this make $\ce{PF5}$ an acid (d-orbitals)? Would $\ce{AsF5}$ be classed as an acid? (If so, is this due to $\ce{As}$'s ability to violate the octet rule coupled with the fact $\ce{As}$ is very positive in this compound).

Then are the requirements for something to be a Lewis base purely the presence of two non bonding electrons? Would this then make compounds such as $\ce{AsF3}$ and $\ce{PF3}$ bases?

I am a little lost, would like to know if the above is along the right lines. Do things like electronegativity and atom size play a part?

I think this is a base?

I think this is an acid?

For anyone else with a similar question above... I have been researching this for the last day or two, I have found Group 15 should be spilt into two groups - either nitrogen or heavier G15 elements when considering basicity or acidity.

Nitrogen - Trihalides

NF3 < NCl3 < NBr3 < NI3 (with NI3 as the most basic due to increasing relative electronegativity of the nitrogen). (These compounds are generally only weak bases.)

In my research I have found the greater the electronegativity of the lone pair atom in comparison to its ligands (i.e. the halides in this instance), the more free the lone pair is to participate in a dative bond to form a Lewis adduct with a Lewis acid. The more free the lone pair is the stronger the base is, hence NI3 is the strongest base out of the trihalides.
This is why NH3 is such a great base as the lone pair on N is not inductively pulled towards the ligands because H is relatively less electronegative. Whereas in NF3 the lone pair is so drawn away from nitrogen this compound is sometimes not even classed as a base, hence it is the weakest of the trihalides.
The reason nitrogen is on its own is because it has max coordination number 4.

P, As and Sb - Trihalides or Pentahalides

I have found halides, either tri or penta, are generally classed as Lewis acids.
PX3, AsX3, SbX3 (X - F, Cl, Br or I)

First the trihalides, these act as mild Lewis acids towards trialkylamines and halides. This is even though the P, As or Sb will have a lone pair. As before,  this is partly due to electronegativity reducing the availability of the lone pair.

My textbook points out PF3 is probably a little bit unusual in this group because it has empty d-orbitals (making it a weak sigma donor yet a strong pi acceptor.)

The pentahalides of heavier group 15 elements are pretty strong acids
PX5, AsX5, SbX5 (X - F, Cl, Br or I)

The trihaldies of nitrogen are explained nicely here -
https://courses.lumenlearning.com/introchem/chapter/trihalides-boron-halogen-compounds/

Also a clever link about NCl3 -
http://www.chm.bris.ac.uk/motm/ncl3/ncl3h.htm

The heavier G15 information and pictyre came from a textbook Inorganic Chemistry (Weller, Overton, Rouke, Armstrong)


Is there a finite number to the possible state of electron in an atom? Okay, I know you may say, there are infinite numbers of possible state of an electron between its ground state and ionic state, since energy levels get continuous as electron gets farther away from the nuclear. But that's not what I mean.

Here is my concern: Any linear combination of eigenstates is also a state for an electron in an atom. For example, an electron in 3d orbital (n=3, l=2) is thought to have only five possible states: dxy, dxz, dyz, dx2-y2, dz2, with literally no m eigenvalue. Yet, what about there linear combinations? Aren't they also possible states of an electron in 3d?

Thanks for any help!

For degenerate eigenstates any (normalized) linear combination of them will be again an eigenstate of same energy. This in fact means that there are infinitely many of them.

So where does the 5 (in case of $d$-orbitals) come from, when we actually have infinitely many? It means we can select at most 5 of them, which are orthogonal (or more generally speaking: linear independent). We are free to choose which 5, and thus there are (infinitely) many set of 5 $d$-orbitals.

Here are 2 possible sets (again for the example of $d$-orbitals) that are typically used:

The general solutions are $Y_l^m(\theta,\phi)=f_l^m(\theta) g_m(\phi)$ for $l=2$ and $m=\{-2,-1,0,1,2\}$. Where we have
\begin{align}
f_l^m(\theta) &= (-1)^m \sqrt{\frac{(2l+1)(l-m)!}{4\pi (l+m)!}} P_l^m(\cos \theta) \\
 g_m(\phi) &= e^{i m \phi}
\end{align}
and $P_l^m(\cos \theta)$ are the associated Legendre polynomials. As those are complex functions due to $g_m(\phi) = e^{i m \phi}$ (except for $m=0$), those are hard to plot or imagine.
The real linear combinations from the above:
\begin{align}
    d_{z^2} &= Y_2^0(\theta,\phi) \\
    d_{x^2-y^2} &= \frac{1}{\sqrt2}\left[Y_2^{+2}(\theta,\phi) + Y_2^{-2}(\theta,\phi)\right] \\
    d_{xy} &= \frac{1}{\sqrt2 i}\left[Y_2^{+2}(\theta,\phi) - Y_2^{-2}(\theta,\phi)\right] \\
    d_{xz} &= \frac{1}{\sqrt2}\left[Y_2^{+1}(\theta,\phi) + Y_2^{-1}(\theta,\phi)\right]\\
    d_{yz} &= \frac{1}{\sqrt2 i}\left[Y_2^{+1}(\theta,\phi) - Y_2^{-1}(\theta,\phi)\right]\\
\end{align}


Is there a point at which Ethanol (E10) fuel becomes harmful to gas tanks or engines if not used? This is not the typical chemistry question on this website, but I think it's an important practical question.

When I got a gas string trimmer, the woman who worked at the gardening shop told me that E10 (fuel commonly sold in the USA - 10% Ethanol, 90% gas) left sitting in a gas tank for months would be harmful. So I wonder, is this true? And if so, what about for cars and other engines (boats, tractors, etc).

This effects my life because I was considering at some point getting a plug-in hybrid car because during most weeks I would never have to use or buy fuel. However, could I be damaging this car by leaving the same fuel inside the tank for months?

There are various claims about ethanol either helping or hurting things, but I have not seen any supported by science. What is the chemical basis of these claims? A layman translation is appreciated.

Well first off, pure ethanol is hygroscopic; it attracts water, to the point that it will pull it out of the air. Ethanol and gasoline will mix, but ethanol, gasoline and water will not; the ethanol-water mixture will come out of solution and settle on the bottom of your tank. Add a little oxygen to the mix, and you get rust. However, the more common side effect of this is more immediate; turn the car on, and the fuel pump will draw the water from the bottom of the tank into the engine, where it will promptly kill it, and require a costly dry-out process.

Down here in the south, where it's hot and humid most of the year, you hear a news story or two about this every year, usually when a gas station didn't properly purge its storage tank on a regular basis. The regular addition of thousands of gallons of fresh ethanol fuel "recharges" this hygroscopic process, causing a rather large pool of water to form at the bottom of the tank which the station is supposed to drain regularly. If they don't, and you fill up when the store's tank is just a little too low, or just a little too soon after the tanker's refilled it, the pump will draw a water-gas emulsion, or even straight water, into your tank, and the water will settle to the bottom.

The usual fix for long-term storage of ethanol fuel is to add a stabilizer, which will contain a high concentration of isopropanol. The main thing that will do is form an azeotropic mixture with water, meaning that its components will evaporate at the same rate, so when the isopropanol evaporates, it takes the water with it. This same mixture, in sufficient concentration, will also burn, so while it's not the greatest thing for your engine it can at least be flushed out the normal way.

Second, gasoline, as you probably well know, isn't a pure substance; it was originally what was left over after the refinery had extracted the "useful" heavier compounds (kerosene, diesel fuel, lubricating oil, paraffin waxes) out of crude oil by fractional distillation. Nowadays it's much more carefully purpose-made by thermal decomposition of heavier alkanes, but it's still an amalgam of relatively lightweight carbon compounds (usually between 6-12 carbons, with the dividing line for "octane rating" drawn between heptane and octane), plus additives and pollutants. It contains three components of interest; sulfur (a naturally occurring pollutant and difficult to remove completely; "summer blend" gas in the U.S. is more expensive because they have to get rid of more of it to prevent smog and acid rain), oxygen (modern fuels are oxygenated to make them more clean-burning), and ethanol (added to make the fuel burn cleaner and be more sustainable to produce than neat gasoline).

In the presence of water (pulled in by the ethanol) and oxygen, sulfur naturally forms sulfuric acid by a variation of the "contact process"; it oxidizes easily to sulfur dioxide, then more slowly to sulfur trioxide, which when dissolved in water becomes sulfuric acid. This reaction, specifically the formation of sulfur trioxide, is sped up in industrial processes by heat and by contact with catalyst metals such as vanadium oxide (hence "contact process"), but it will also happen on its own without any special treatment (i.e. acid rain).

This sulfuric acid will do a lot of corrosion damage, reacting with the tank's sheet metal producing hydrates of iron sulfate. When burned, those sulfates decompose into iron oxides and sulfur dioxide again, which do more damage to your exhaust system on the way out (rust is a catalyst for more rust, and we've discussed sulfur dioxide's contribution to the party in detail). In addition, sulfuric acid is even more hygroscopic than ethanol; at concentrations in water as low as 10%, it'll attract more water. Lastly, acids and alcohols form esters, in our case ethyl sulfate, and esters and alcohols form ethers, here diethyl ether (with the sulfuric acid and water as a byproduct). These are reversible and ultimately circular reactions, which will form an equilibrium of roughly equal concentrations of the intermediates. A dehydration reaction to produce ethylene is also possible in very high acid-to-water concentrations of sulfuric acid (possible when filling up in the winter, when it's dry and the gas has more sulfur). Ethylene will form a number of intermediate products with just about anything in that gas tank, some better for your engine than others (ethylbenzene is an antiknock compound; methyl ethyl ketone is a solvent and a pollutant byproduct of combustion engines).


Oxidative chlorination mechanism (sulfide to sulfonyl chloride) I cannot find much guidance in the literature. Any ideas as to what the mechanism of this reaction might be?

As Ben Norris already pointed out, these processes are typical a bit complex. It is however conceivable that the direct oxidation of the thioether 1 is among the first steps involved.

While the oxidative cleavage of PMB (= 4-methoxybenzyl) ethers, often applied as protecting groups, is assumed to take place in the methoxy-substituted arene, here the electron transfer mostly likely proceeds from the sulfur atom, leading to the sulfur-centred radical cation 2.

In the presence of a nucleophile, this species is likely to break down en route to the radical 3.

Whether a disulfide, formed upon dimerization of 3, will take part in the further course of the reaction is beyond these musings.


Here are four observations of the reactions of four metals Here are four observations of the reactions of four metals.

Metal F and G react slowly with water
Metal H displaces metal G from its chloride salt solution
Metal E does not react with dilute acids
Metal H does NOT displace metal F from its sulphate salt solution.

What is their order of reactivity, starting from the most reactive metal to the least reactive metal?

F>H>G>E;  H>G>F>E;  H>E>F>G;  G>H>F>E;  None of these

I'm really unsure how to order them     I would say F>H>G>E but thats half guessing.
Can someone give me a clue to arrange them right?

Before starting of with my reasoning , I must tell you that your answer ( whether guessed or logically reasoned ) looks correct according to me .

I am assuming that you are familiar with the Metal Reactivity Series ( if not look here )

We shall take use of a well presented reactivity series :

Now to crack the answer clue by clue :

1.

  Metal F and G react slowly with water.

Hence F and G are somewhere at the top.

2.

  Metal H displaces metal G from its chloride salt solution

Clearly H>G.

3.

  Metal E does not react with dilute acids

E will come somewhere at the bottom half so F,G>E (from 1) and we can further say H>G>E (from 2)

4.

  Metal H does NOT displace metal F from its sulphate salt solution

This gives F>H.

Combine last two clues' result to get F>H>G>E .


How does bismuth's lone pair affect its properties as a metal and alloy? Bismuth is really the only metal to exhibit a lone pair in its outermost shell, this means that it and its salts should be a Lewis base. As bismuth can act as a Lewis base, how does this affect its properties as a metal specifically in regards to alloying? For instance, if I have a 1:1 mole ratio of aluminum and bismuth in an alloy, will the aluminum and bismuth form dative bonds?

If you look at the chemistry of lead(II) and the work by Tony F. Hill (and others) on platinum group metals you will see that bismuth is clearly not the only metal to have lone pairs.
What Tony Hill did was rather clever, normally metal-ligand coordination is all about the donation of lone pairs to the metal and sometimes the retrodonation of electron density onto pi acidic ligands such as carbon monoxide and cyanide.
What Tony did was to get a borane to bind to a lone pair on a low valent metal such as ruthenium, osmium or rhodium. Here is a diagram of an example of one of his compounds.

The synthesis of this rhodium compound can occur from $\ce{[RhCl2(Ph)(PPh3)2]}$, this rhodium complex can react with $\ce{Na[HB(mt)3]}$ (mt = methimazolyl) to form an intermidate $\ce{(mt)B(H)(mt)2Rh(Ph)Cl(PPh3)2}$ which then decomposes to lose a molecule of benzene to form $\ce{B(mt)3RhCl(PPh3)}$. (Chem. Comm., 2005, page 221).
I am sure that tin has a stereochemcially active lone pair, if we consider the strcuture of $\ce{Cp2Sn}$, this is not quite like ferrocene. Instead of the two Cp rings being parallel. They are not. So I think that bismuth is not the only metal to have lone pairs on it. Even the bis Cp* complex of lead(II) (bis(h5-Pentamethyl-cyclopentadienyl)-lead) is bent away from the ferrocene type strcutre suggesting that the lead has a stereochemically active lone pair.


Is center of symmetry only valid for symmetric substituents? The given compound is optically active. Initially I had reasoned that the compound has center of symmetry( two xylene groups across each other). Why doesn't it have a center of symmetry?

Source: MS Chouhan, Advanced Problems In Organic Chemistry, Page 116 (Chapter Isomerism Level 2 q6a)

You assume too much, starting with the words "The given compound". There is no given compound. The picture corresponds to quite a few different compounds, and here is why.

As I told you before on a remotely similar question, the biphenyl fragment totally can't be planar. The xylene substituents must be rotated so their ortho methyl group points either up or down relative to the plane of the central ring. Since there are two of them, we have three options: "both up", "both down", and "one up, one down". The first two compounds are indeed chiral, and enantiomers of each other. The third one is centrosymmetric, just as you've reasoned.


CO poisoning - What I've been taught is a hoax? In elementary school, I happened to ask my teacher:

  What is $\ce{CO}$ they talk about these days?

I was then taught that Carbon monoxide "competes" with the oxygen in our blood (we were  taught that oxygen gets to our cells via blood at that time) and replaces it and thus, doesn't let enough O get to cells and this will be fatal in extreme cases.

I didn't do any research about it after that because I later made the implication that $\ce{CO}$ is able to replace O and react with Hb instead; but

I recently became doubtful about my previous lessons and decided to do a research to see what really was going on:

  Carbon monoxide mainly causes adverse effects in humans by combining with hemoglobin to form carboxyhemoglobin (HbCO) in the blood. This prevents hemoglobin from releasing oxygen in tissues, effectively reducing the oxygen-carrying capacity of the blood, leading to hypoxia. Additionally, myoglobin and mitochondrial cytochrome oxidase are thought to be adversely affected. Carboxyhemoglobin can revert to hemoglobin, but the recovery takes time because the HbCO complex is fairly stable. Wikipedia, CO poisoning

Wikipedia states that the problems are:

$\ce{CO}$ is able to combine with Hb to form HbCO.
This compound is fairly stable, so we'll have a hard time removing $\ce{CO}$ from hemoglobin.

Other sources, such as How stuff works, Answers.com and CDC.gov have given either only more explicit explanations of wikipedia, or the explanations weren't "chemical-biology" enough.

I haven't seen a single reference in which the first assumption was declined, or agreed with. Is what I'd learnt non-scientific?

Each hemoglobin molecule has 4 subunits (4 hemes each having one iron atom). Each time a molecule (such as oxygen or CO) binds to one subunit, it changes the binding properties of the other subunits. This is referred to as positive "cooperative binding" in the case of hemoglobin, because binding one molecule makes the other sites bind more easily.

Initially, considering hemoglobin with no CO or oxygen bound, CO competes with oxygen for the same binding sites (ligating one of the iron atoms).  CO has a binding constant 300 times lower than oxygen, meaning a factor of 300 lower CO concentration will result in an equal percent of bound protein.

Considering CO has reached a level such that 2 out of the 4 sites can have CO bound, the remaining 2 sites would still provide enough oxygen carrying capacity for a person to live if the sites were function as normal; however, the 2 remaining sites then bind oxygen too tightly and don't release it.

Other iron-containing proteins such as myoglobin can also bind CO and are also inhibited.

See Pharmacokinetics and Mechanisms of Action of Carbon Monoxide for more infomation.


For homogeneous equilibrium, why are liquids and solids included in the equilibrium constant (when they aren't in heterogeneous equilibria)? In a heterogeneous reaction (where the states are varied) we do not include liquids and solids in the equilibrium equation because their concentrations do not change.

E.g. Chemguide.co.uk

However, when it is a homogenous equation we DO include solids and liquids.

E.g. Chemguide.co.uk

1) Why do we include liquids and solids in the Kc equation in homogeneous equilibrium equations? I get that if we didn't there would be nothing on the right hand side of Kc=.... , but why is it physically different to the case where there are varied phases?

2) What if there was an equation that involved only liquids and solids  (if this is possible)? As this is a heterogeneous equation would we still not include solids and liquids? If so how would you write the Kc equation where there is nothing but solids and liquids?

Chemguide is simplified for A-levels and therefore in this case is strictly speaking incorrect.

The equilibrium constant $K$ is defined as a product of activities. I described this in a previous answer here.

The crux of the matter is that the activity of a pure solid or pure liquid is equal to 1, which means that it can be omitted from the expression for $K$ without affecting the value.

In your first reaction

$$\ce{H2O(g) + C(s) <=> H2(g) + CO(g)}$$

the chunks of carbon in the reaction are necessarily pure because they don't mix with the gases.

In the second reaction (yes, I am lazy, please feel free to edit for me)

$$\ce{EtOAc(l) + H2O(l) <=> AcOH(l) + EtOH(l)}$$

none of the liquids are pure, hence their activities deviate from unity.

Just as a final example, in the dissociation of a weak acid

$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)}$$

water is omitted from the expression for $K_\mathrm{a}$ because water, as the solvent, is in large excess over $\ce{HA}$ and is therefore effectively "pure".


When MgCl₂ · nH₂O is heated to the boiling point of H₂O When magnesium chloride hydrate is heated, does $\ce{Mg(OH)2}$ form or what? I'm curious, because Wikipedia says that it decomposes to release $\ce{HCl}$ gas.

Wikipedia pretty much has it all already.
There are numerous magnesium chloride hydrates $\ce{MgCl2 · x H2O}$ $(x = 1, 2, 4, 6, 8, 12),$ however in the temperature range from 0 °C to 115 °C the most stable one is hexahydrate $\ce{MgCl2 · 6 H2O}.$
Whether $\ce{MgCl2 · 6 H2O}$ decomposes to hydroxo-salt or only dehydrates depends on the conditions.
When heating takes place in atmosphere of $\ce{HCl},$ anhydrous $\ce{MgCl2}$ is formed:
$$\ce{MgCl2 · 6 H2O (l) ->[>\pu{115 °C}][HCl(g)] MgCl2(s) + H2O(g)}$$
Otherwise, magnesium hydroxochloride is formed [1, p. 522] and $\ce{HCl}$ gas is released:

The product obtained is always the hexahydrate, $\ce{MgCl2 · 6 H2O}.$ It is dehydrated to anhydrous magnesium chloride by spray drying and heating with dry hydrogen chloride gas. In the absence of $\ce{HCl},$ heating hexahydrate yields the basic salt, $\ce{Mg(OH)Cl}:$
$$\ce{MgCl2 · 6 H2O -> Mg(OH)Cl + HCl + 5 H2O}$$

References

Patnaik, P. Handbook of Inorganic Chemicals; McGraw-Hill handbooks; McGraw-Hill: New York, 2003. ISBN 978-0-07-049439-8.


Why objects usually don't exibit fluorescence/phosphorescence? We learn in school that the color we perceive is determined by the light that is reflected back at us. Not all wavelengths of light reflect back at our eyes, some wavelengths are absorbed.

I'm confused about this absorption. From my understanding, molecules can absorb light typically rotationally, vibrationally, and electronically. rotationally and vibrationally are associated with the microwave and infrared region. Electronic excitations can be in the visible, but I always had the impression that the excited electrons would collapse down to the ground state and emit light. Thus, I don't think I know any mechanism which visible light can be absorbed without emitting light, albeit, longer wavelength light typically.

You are correct in most of what you write. Absorption in the visible (and uv) parts of the spectrum are of far larger energy that IR and microwave and the lowest electronic transition involves promoting an electron from the highest populated orbital into the lowest unoccupied one. Once the excited state is formed it can emit via fluorescence and phosphorescence, which are both at a slightly lower energy that the light absorbed and this is because higher energy vibrational levels can be populated in the ground state. (Look for picture of a Jablonski diagram).

However, fluorescence is emitted into all directions ($4\pi$ radians due to molecular motion between absorption and emission) and only a small part of this could enter the eyes but also and most importantly there are non-radiative processes that competitively remove the energy from the excited state ultimately populating high vibrational levels in the ground state from which energy is rapidly lost into the surroundings. These non-radiative events mean that for most types of molecules the fluorescence yield is small, which means that only a few % of the excited state's energy ends up as photons. The non radiative processes are called intersystem crossing between singlet and triplet excited state and internal conversion between states of the same spin state. The processes causing these (besides and chemical reaction) are anharmonicity in molecule vibrations and particularly heavy atoms or unpaired electrons as part of the molecule itself or in nearby molecules such as in a solvent. Some molecules may also have symmetry forbidden optical transitions in emission but not in absorption: carotenes are of this type.


Reaction mechanism for the synthesis of 9,10-phenanthrenequinone In my lab course I had to synthesize 9,10-phenanthrenequinone (phenanthrene-9,10-dione) from phenanthrene and chromium(VI) oxide. The synthesis was made as explained in Organic Syntheses, Coll. Vol. 4, p.757 (1963)

The reactants are: phenanthrene (1.00 eq), chromium(VI) oxide (7.50 eq), sulfuric acid (15.0 eq), water (83.3 eq)

I now have to explain the reaction mechanism, but unfortunately I don't find it in any of my textbooks, neither on the internet.

Here is my assumption:

The chromium(VI) oxide reacts with water and forms chromic acid.
The chromic acid can oxidize an alcohol to a ketone (Jones-Oxidation)

So what misses is the alcohol. Somehow I have to get the 9,10-phenanthrenediol, which could react as alcohol in the Jones-Oxidation.

How could this possibly work. Remember, I don't have $\ce{OsO4}$, which could easily dihydroxilate an alkene.

EDIT 1: As an answer to the proposed mechanism by @Zhe:

Is this ok in your eyes?

EDIT 2: An other idea for the first step (dihydroxilation)

A person of the scientific staff of my university told me, that the $\ce{CrO3}$ should react in a very similar way as $\ce{OsO4}$ does. As I'm using 7.50 eq of chromium(VI) oxide, there is enough for the dihydroxilation and the Jones-Oxidation. He said that a radical mechanism is rather unlikely, because I'm working with water as solvent and water is an inhibitor for radical reactions. This is the mechanism I came up with, regarding these new information.

What do you think about this possibility?

Here's what I was thinking.

  This probably happens via a single electron oxidation of that double bond to a radical cation as a first step. The cation can trap water and then we can oxidize the radical to a cation. Eventually, you get to a diol which you can oxidize to a diketone.

I ommited the steps for chromium oxidation from the alcohol to the ketone. Sounds like you are already familiar with those.