How to name this cyclic compound with an O in it? How to name this cyclic compound with an O in it?

I was going to name it cyclobutane, but not all of the atoms making up the ring are carbons, one is an oxygen. So, then I decided to name it butane, but that doesn't make sense because there is a ring.

Well, the molecule you draw is tetrahydrofuran, aka THF, but of course the only way you can guess that is if you already know the molecule. It's a very common organic solvent.

If you didn't know its common name, you could try to name it according to IUPAC nomenclature. It's an ether, and it's cyclic, so it can be named as oxacyclopentane. But IUPAC has specific names for cyclic ethers, which in this case is oxolane:

(from this organic chemistry book).


Choosing Parent Chain, Complexity of Side Chains Which parent chain should we choose for this structure? is it A or B?

I followed the rule A-2.6, but it still got me confused.
Both alternative A and B has the same number of side chains (2 side chains) - rule A-2.6(a).
Both alternative A and B has the same number of locants, which are 5 and 6 - A-2.6(b).
Both alternative A and B has the same number of carbon in each locants, which are 4 atoms on number 5 and 9 atoms on number 6 - A-2.6(c).
Both alternative A and B branched side chains are quite similar, how do we decide? - A-2.6(d)

You have already reduced the problem to two possible parent chains:
A: 5-(2-methylpropyl)-6-(3-methyloctan-4-yl)dodecane
or
B: 5-(butan-2-yl)-6-(2-methyloctan-4-yl)dodecane

In order to identify the preferred name, we have to read further in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). In particular, the following potentially relevant rules are actually not helpful in this case:

P-45.2.1 maximum number of substituents cited as prefixes
P-45.2.2 the lower locant set for substituent groups cited as prefixes
P-45.2.3 the lower locant set for substituent groups in order of citation

However, we finally find:

  P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (see P-14.5). Alphabetic letters are considered first in the order that they appear in the name; all Roman letters are considered before any italic letters, unless the latter are used as locants or are a part of a compound or composite locant, for example, ‘N’ and ‘4a’. Then, if still there is a choice, numerical locants are considered in the order of their appearance in the name.

Therefore, the preferred name is 5-(butan-2-yl)-6-(2-methyloctan-4-yl)dodecane since ‘butan-yl’ is earlier alphabetically than ‘methylpropyl’.


Why does a proton prefer to attack a double bond instead of protonating an alcohol in a reaction of unsaturated alcohol with sulphuric acid? A solved problem in Organic Chemistry by Solomons, Fryhle and Snyder asked to write the mechanism for the following transformation in the presence of sulphuric acid ($\ce{H2SO4}$):

The mechanism provided by the book is as follows:

It can be seen that a proton from sulphuric acid attacks the double bond thereby forming a tertiary carbocation as an intermediate. Later the lone pair on the oxygen atom of the hydroxyl group attacks the carbocation to form a cyclic intermediate which after deprotonation gives the required product. I understood the mechanism given in the book.

However, when I attempted the problem with the given reactant and the reagent, instead of attacking the carbon-carbon double bond, I protonated the alcohol. I attained at a much different product by following the mechanism given below:

The reason I chose to protonate the alcohol was I felt that electron density near oxygen is much higher than that in the double bonds. Moreover, I also end up with a conjugated diene, which is resonance stabilized.

Further, the following statement given as an explanation for a different problem supports my method of protonating over attacking the alkene:

  Protonation of an alkene forms a carbocation with an open octet, whereas protonation of methanol does not form an open octet. Therefore, we predict methanol to be a stronger base than alkene. The mechanism starts with the reaction of a stronger base with sulphuric acid.

Now, my question is, why does a proton prefer to attack a double bond rather that protonating an alcohol? Under what conditions, does a proton protonate the alcohol instead of attacking the double bond? Are there any well-defined rules to predict which attack will be preferred over the other in such cases?

Image source: I made them using ChemDoodle.

Both the reaction mechanisms are equally correct. Reaction takes place at numerous sites in most of the organic reactions, many kinds of products are formed, but the thing which actually give the major product is one of the intermediate, which is relatively stable than its peers.

We know, that 3° carbocation is a more stable intermediate than an allyl carbocation (irrespective of the fact that the latter is going to give a conjugated product), so the former will be present in adequate amount in order to give the major product.

Also, the formation of conjugated alkene takes place through multiple intermediates, and therefore it becomes important to care for the stability of each intermediate. I especially doubt the 3rd step, where along with a 1° carbocation, alkene is present, which acts as an EWG, and hence lowering the stability of the intermediate.

So, taking account of these factors, I would likely go with the given solution.

Edit: As per Yusuf Hasan's points on the stability of THF in comments, the given product, being a THF derivative is likely going to solvate the other intermediates and it is itself stable. That's why, THF is an unreactive compound and hence it is used as a solvent in many organic reactions. (source, under the heading "Reactivity of cyclic ethers - Oxacyclopropanes")


Why aren't neutralisation reactions reversible? Let's take
$$\ce{3HCl + Fe(OH)3 -> FeCl3 + 3H2O}$$
This would be a normal one-way neutralisation reaction.

How about hydrolysis of ferric chloride (technically iron(III) chloride)?

$$\ce{FeCl3 + 3H2O <=> 3HCl + Fe(OH)3}$$

As this is reverseable, we should be able to right it with the reactants and products flipped.

$$\ce{3HCl + Fe(OH)3 <=> FeCl3 + 3H2O}$$

Now that is nothing but our original reaction …
Why aren't neutralisation reactions reversible?

Also, does anyone have any good links to chemical equilibrium as in comments?

Neutralization reactions are reversible. In theory, at least, even if not so much in practice, all reactions are reversible.

If you look at things on a molecular scale, there's a principle called microscopic reversibility. In effect, there's no physical reaction mechanism you can create that's a one-way door. Maxwell's demon doesn't exist, and you can't get around that no matter how many steps you chain in whatever order, or which particular reagents you use.

What you can do, though, is set things up so it's exceedingly unlikely that the reaction proceeds in the other direction. That's what's happening with reactions we normally deem "irreversible". There's something about them which makes it difficult, in practice, for products to go back to reactants.

One common feature is the size of the equilibrium constant. If a particular reaction has an equilibrium constant that's really big, the equilibrium for that reaction is going to be vastly tipped to the product side. So much so, that the reactant side of the reaction will not be present in appreciable amounts. (For example, this is why combustion is viewed as irreversible. The reaction is so favorable toward the products that only parts per million or parts per billion of the reactants will be present at equilibrium.)

Another method for promoting "irreversibility" is for the products to partition to another phase. For example, if the reaction forms a gas or a precipitate, that product is removed from the reaction volume, and the concentration of the products drops, pulling the reaction toward the products. Note that even in this multi-stage setup, each step is reversible. That evolved carbon dioxide molecule could re-dissolve in solution, and then react backwards, but it's terribly unlikely for it to do so. You've basically fixed one of the products at a very low concentration, which draws the equilibrium toward the product side.

Precipitates work much the same way, as the activity of the solid is fixed by surface behavior, and does not increase because you increase the amount of solid you add. (The rate of reaction may increase by adding more solid, but the equilibrium of the reaction does not.)

So in your examples, the reason you get hydrolysis or neutralization is due to how you set up the reaction conditions. Have a large excess of HCl? You drive things toward neutralization. Have a way of getting rid of HCl (for example by driving it off as a gas, or reacting it with a base)? You drive things toward hydrolysis.


Polysaccharides from non-cyclic sugars? I have only come across polysaccharides from monosaccharides that has undergone intramolecular cyclization reaction.  I was wondering if it is possible for polysaccharides to form from linear sugars? I think that the answer lies in how the other glycosidic bond is formed, however my knowledge of chemistry is limited so I don't know if bond formation is possible with a linear sugar.

I know a carboxylic acid group can react with a hydroxyl group, and aldehyde can be oxidised to carboxylic acid. So my guess is that it is possible to create a long ester chain from aldoses but not ketoses (as ketone group cannot react and is fully oxidised). Although then I think it would no longer classify as a polysaccharide but a poly(ester)? I'm sure these exist — I have heard of polyester of course! — although I have not come across any others specifically.

In theory and especially synthetically there is no real reason why one should restrict oneself to polysaccharides where all monomers are in cyclic forms. A linear monosaccharide could, for example, act as a bifurcating branching site with two different sugars attached to the aldehyde carbon to build up the required O,O-acetal. Another variant may be to have a methyl group block the second acetal oxygen. In any case, you want to avoid the aldehyde becoming a hemiacetal because these are too reactive and would degrade quickly.

$$\begin{align}\ce{\underset{aldehyde}{R-CH=O}} && \ce{\underset{hemiacetal}{R-CH(OH)(OR)}} && \ce{\underset{acetal}{R-CH(OR)(OR)}}\end{align}$$

You may want to look up acetal and acetalisation in the general organic chemistry textbook of your choice to learn more about why hemiacetals are much less stable than acetals.

In practice, however, linearised monomers in polysaccharides are not interesting. To the best of my knowledge, they do not occur in nature due to the labile hemiacetal intermediate that would be required — but to a much greater extent due to the fact that nature optimised the connection of cyclic monosaccharides into polysaccharides and is fully content with what it has. There is no need to evolve a synthase that is able to shield the highly reactive intermediates from the surroundings and thus it didn’t happen. While the formation of oligosaccharides can also be a spontaneous process given a catalyst, linear monomers in oligosaccharides will not form that way.

For most natural product or bioorganic chemists, it is by far more interesting to synthesise species that are known in nature or logical extensions thereof. While a linearised monosaccharide might qualify as an extension, it is not that logic an extension and thus of much lesser value.


Why does humidifier put measurable PM2.5 in the air? (I know little science so please help me if I'm in the wrong place or if my question needs editing. I could use help with tagging, for example.)

I use two devices that measure PM2.5:

Kaiterra LaserEgg 2+, an air monitor which uses laser-based light scattering technology
Blueair Classic 280i, an air purifier which measures PM2.5 but doesn't say how

Generally, their PM2.5 measurements are within 20% of each other and the readings go up and down together.

Both devices are in a room measuring about 40sqm.

It's winter so the windows are closed and I keep the Blueair Classic on automatic mode, which varies the fan speed depending on the level of PM2.5 it measures. It is rare for the fan speed to increase above the lowest level.

The level of PM2.5 is usually under 10μg/m3 (safe, according to the WHO), as measured by the LaserEgg and the Blueair Classic.

A few days ago we felt that the air was dry (cracked skin, etc.) and our humidity sensors reports humidity around 25%, so we turned on an ultrasonic humidifier (https://www.amazon.co.uk/Humidifier-Improves-Health-Sleep-Focus/dp/B01C5YHFWO). We use tap water in the humidifier.

Within a few minutes the LaserEgg and the Blueair Classic reported a massive increase in PM2.5, to about 400μg/m3 (very unsafe according to the WHO). The Blueair Classic changed its fan speed to maximum but it seemed to have little effect. Perhaps the output of the humidifier is greater than the cleaning capacity of the purifier.

We have reproduced this situation several times since - every time the humidifier is turned on.

What could be causing the PM2.5 increase?

Could it be harmful?

P.S. This seems relevant: https://www.researchgate.net/publication/51859262_Indoor_particle_concentrations_associated_with_use_of_tap_water_in_portable_humidifiers

All tap water contains some dissolved solids, and "hard water" contains more, particularly calcium salts. There are also microorganisms found even in treated tap water, and some organisms can build up a biofilm inside the humidifier. Ultrasonic humidifiers are effective at putting these particulates into the air. However, as others stated, you may just be detecting fine water droplets from the humidifier before they evaporate.

To find out if it's water droplets or particulates from dissolved material, do a comparison test, using distilled, filtered, low-particulate water such as High Performance Liquid Chromatography (HPLC) grade water.

Clean out the humidifier thoroughly, rinsing repeatedly with a dilute acid safe for that machine (perhaps vinegar, but check the humidifier manual).
Rinse repeatedly with distilled water and finally with HPLC water.
Refill with pure HPLC water.

If this does not greatly increase the particulates above background, but tap water does, then you've found the source of the issue. If not, then you're probably detecting mostly water mist.

Caveat: Buying HPLC water is not inexpensive. If you can't easily make your own on-site, perhaps just leave the dehumidifier off.


Preparation of Sodium Peroxoborate Write balanced equations for each step, and an overall equation, for the preparation of sodium peroxoborate hexahydrate from the reagents borax, $\ce{NaOH}$, and $\ce{H2O2}$.

Borax was added to a solution of $\ce{NaOH}$, then it was treated with $\ce{H2O2}$, then cooled to form a precipitate. Please help.

(No structures are necessary, simply the chemical formulas will suffice)

Sodium peroxoborate can be prepare using two methods.

Electrolysis of sodium borate with some sodium carbonate
Oxidation of sodium meta borate using $\ce{H2O2}$

Heating borax strongly will give a mixture of sodium meta borate and boric anhydride.

If you can remove boric anhydride from the transparent glass like mixture, you can then add $\ce{H2O2}$

$$\ce{2NaBO2 + 2H2O2 + 6H2O -> Na2[(OH)2B(O - O)2B(OH)2].6H2O}$$


Why does my sodium acetate look like metalic magma? I tried to make sodium acetate.

Ingredients:

Store bought vinegar - 8% acetic acid, water, E150C  (caramel)
Baking soda - $\ce{NaHCO_3}$

Steps:

I first emptied sacks of baking soda to a container
I kept adding vinegar until the resulting fuming was mild
I started heating the solution slowly, I kept adding more vinegar as long as I could see bubbly reaction
I increased heat until all water evaporated. I was left with white-brown chunks of (presumably) sodium acetate and the impurities from vinegar
I kept heating this until it melted into opaque black liquid
I poured the hot molten stuff into ceramic bowl

This is my result:

I might upload better picture tomorrow, when there's more light available.

So, this does not really look as sodium acetate at all. It looks like a volcano. Clearly there's more volcano-ish experiments with vinegar than I thought.

How do I purify this? Is there even any sodium acetate? Could I burn (decompose) it or something?

I wouldn't bother trying to purify it. As Mithoron said in his legendary comment "Caramel made your product a black unholy abomination of science". I would have distilled the vinegar and diluted it (because concentrated acetic acid is a bit dangerous if you don't know what you're doing) and start again. If you want pure product, make sure your starting materials aren't filled with "unholy abomination of science" causing impurities. If you really want to purify it, maybe recrystallise it in hot water? Or a solvent that sodium acetate can become saturated in.


Is there an alternative to the half-reaction method to solving red-ox reactions? The equation for the reaction between a permanganate and iron(II) ions in acidic solution is:

$\text{MnO}_4^- + \text{Fe}^{2+} \longrightarrow \text{Fe}^{3+} + \text{Mn}^{2+}$

My book, as well as a certain equation balancer, shows that the answer is:

$5\text{Fe}^{2+} + \text{MnO}_4^-  8\text H^+ \longrightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text H_2\text O$

They check the charges to ensure that it is indeed balanced. My book says that:

$5(2+) + (1-) + 8(1+) = 17+$

$5(3+) + (2+) + 0 = 17+$

And voila, they must be therefore balanced. However, there must be more than one way to balance such a reaction, like many problems. Is there a different, perhaps a more interesting, method than the half-reaction method for balancing this reaction and others?

There is one other way that I have used, using a linear system to solve this reaction. Although it can be used for other more casual reactions too, such as the combustion of ethanol for example.

As you can see, in this special case (because the reaction is relatively easy to solve already) we see that there must be $H_2O$ molecules on the product side of the equation, because there is no oxygen on the products already, but there is oxygen in the reactants. So start by assuming the products include $H_2O$ with a coefficient of at least 4 (because there is 4 oxygen in permanganate ion).

Also, there is no hydrogen anywhere in the equation, so we can assume that to balance the water (which has at least 4 moles) there must be some hydrogen ions added as reactants to balance the reactants side (it is acidic conditions in this case). As it turns out, since there must be at least 4 moles of water, then there must be at least 8 moles of hydrogen ions to satisfy the convention of having all mole coefficients in the lowest terms.

Now we can start building the linear system:

Since we obviously do not know the coefficients yet, we will assign variables to them. Let a b c d e and f be the molar coefficients of each respective molecule in the following reaction we have established so far:

a$H^+$ + b$MnO_4^-$ + c$Fe^{2+}$ ==> d$Fe^{3+}$ + e$Mn^{2+}$ + f$H_2O$

Now how do we build the linear system for this reaction? We might do it as follows:

Starting with the hydrogen atom, simply because it comes first when reading it from left to right. We write this equation to represent the hydrogen atoms in our reaction:

a = 2f

It is as simple as it looks, there is one mole of hydrogen atoms per mole of hydrogen ions, and there is two moles of hydrogen atoms per mole of water molecules.

We can keep doing this for the rest of the elements in our reaction, until we have three more equations along with the one for hydrogen, because there are three more different elements to account for.

b = e This one is pretty straightforward... This accounts for Manganese atoms.

4b = f Oxygen atoms are now accounted for.

c = d Also straightforward, now iron atoms will be accounted for as well. Notice that c and d are lone variables, they do not show up with the other equations, but at least they must be equivalent

Remember that we established some mathematical axioms when we said that there must be at least four moles of water molecules to satisfy the one mole of permanganate ion. An axiom is just something that we assume to be true, such as f = 4 in this case.

Now look at the equation we established for hydrogen atoms, the first equation a = 2f. Since we are assuming that f = 4 we can therefore safely assume that a = 8 without going against our original axiom. Also if we assume that f = 4 we can also go ahead and solve the equation to account for oxygen atoms 4b = f and solving it gives us b = 1 and the rest is pretty straight forward. If b = 1 and b = e then e = 1.

Given what we have found out so far, lets take a look at our equations again:

a = 8

b = 1

c = d

e = 1

f = 4

So,

8$H^+$ + 1$MnO_4^-$ + c$Fe^{2+}$ ==> d$Fe^{3+}$ + 1$Mn^{2+}$ + 4$H_2O$

So now we can look at the charges to figure out the correct coefficients for iron(II) and iron(III) ions, which should be the same of course (c = d). As it stands currently (ignoring the iron ions), there is a total charge of 7+ on the reactants side, and 2+ on the products side. These should be the same for the equation to be balanced, so we can start by creating another axiom and saying that c > 1. If there are 2 moles iron then our charges would be 11 on the reactants and 8 on the products. Incrementing it to 3 would increment the charge on the reactants side by 2+ while the product's total charge would increment 3 due to the charge of the ions themselves. Eventually things are balanced when the moles for iron reach 5 making both sides have a total charge of 17 and everything is nicely balanced.


Use of basis set in DFT (Density Functional Theory) Basis sets are used to guess the electronic wave functions for Hartree Fock or similar methods, which are quite legitimate since these methods deal with the wave function of each and every electron.

Density functional theory, on other hand, uses the electron density at every point of space for optimization and the calculation of properties. This has led to two doubts which I want to clarify:

Is the basis set used to estimate the initial electron density of the system? If so, how the basis functions of an STO or GTF basis set change in order to exhibit the electron density? If not, what exactly is the application of basis set in DFT calculation?
In HF SCF, the coefficients of basis functions (for basis sets with GTF) along with geometry changes with change in nuclear geometry while optimising the structure. Does optimisation with DFT follow a similar procedure?

P.S.: Please do not suggest me this question, since this is not the answer I am searching for.

This answer only deals with the most common variety of Density Functional Theory, namely, Kohn-Sham DFT. This is what most people mean by "DFT", but, as noted in the comments, things such as orbital-free DFT exist.

Kohn-Sham DFT was created to solve a historical problem of DFT: The electronic kinetic energy term (as used in the Thomas-Fermi model) was not accurate enough. Kohn and Sham proposed to make use of what is typically called a "fictitious, non-interacting reference", that is, a wave-function whose main purpose would be to yield a density to be used in all other (nucleus-electron attraction, Coulomb, exchange, and correlation energy/potential) and whose secondary function is to provide the kinetic energy, which is accurate from wave-function theory. This wave-function works just like a Slater determinant and is called the Kohn-Sham determinant. See this question for pointers on where to read more: Equivalent of Szabo and Ostlund book for DFT.

As often in DFT, practicality showed a way that was justified later. When implementing this idea, one learns that one can basically use a HF code and replace the exchange term by the exchange-correlation (XC) potential (which is projected back on to the AO basis set). The potential needs to evaluated numerically on a grid, a procedure that yields the XC energy for almost all choices of density functionals. Thus one obtains the energy and the KS operator (the equivalent of the Fock operator) and can perform a SCF procedure in a given basis set. For the mathematical details, see e.g. JA Pople, PMW Gill, BG Johnson Chem Phys Lett 199, 557 (1992).

To answer your questions in this context:
1) The initial density depends on the guess, which is a whole different can of worms, but the same is true for HF calculations. (Procedures based on tabulated atomic densities exist, but the initial guess business is a bit of a dark art, and I don't think this is what you meant by your question.) The basis set and its MO-like coefficients (defining the KS determinant) are used on every iteration to yield the density.

2) Naturally, yes. The MO-coefficients of basis sets change on geometry change for a given system, just like in HF. The procedure and workable algorithms are very similar to HF (because it all is).


Is there a difference between Close Cubic Packing and Face Centered Packing? I see some places use close cubic packing (ccp) and face centered packing (fcp) interchangeably. Are they the same thing? Does one provide more information than the other?

The Wikipedia entry for "Cubic crystal systems" says:

...Face-centered cubic (abbreviated cF1 or fcc, and alternatively
called cubic close-packed or ccp)

In another entry it is stated:

...face-centered cubic (fcc) (also called cubic close packed)...

The text of the above references make no statement about one term providing any more or less information than the other.  This does not preclude the possibility that some organization or another may have a preference for one term or the other, but the meaning seems to be the same.


Miscibility of Hexane and Ethanol (Anhydrous vs 96%) In my lab, I mixed 8:2 (mass:mass) of Hexane:Ethanol

(Hexane: 0 ppm of water content, analytical grade, & ethanol: 4% water content, pharmacy grade)

At room temperature, and at -20ºC, there was a clear phase separation.

I repeated the experiment, but using anhydrous ethanol instead (1300 ppm of water content).

Now, at room temperature, and at -20ºC, there was no phase separation.

I added enough water to the mixture to make it equal to the first mixture, and the phase separation came clearly into view.

My question is, is this kind of very specific data of miscibilities available somewhere? Ethanol 96% is what is commonly used, yet pure ethanol behaves differently in this case.

Any link or insight into the importance of such little water is appreciated.

Theory and Application of solubility Parameters are beyond our scope. However, an interesting viewer should have at least read through collection of research by Charles M. Hansen, published during 60s (Ref.1). One of the parameters discussed in this book is Hildebrand solubility parameter ($\delta$), which is a good indication of solubility, according to Wikipedia. The principal utility of Hildebrand solubility parameter is that it provides simple predictions of phase equilibrium, based on a single parameter. These predictions are often useful, particularly for nonpolar and slightly polar systems without hydrogen bonding. According to the tabulated $\delta$ values, ethanol is polar, and has relatively higher value ($\pu {12.92 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$) than that of hexane ($\pu {7.24 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$) (Ref.1). Yet, it is interesting to note that hexane is soluble in absolute ethanol, but is insoluble in methanol ($\delta_{\ce{MeOH}}=\pu {14.28 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$), albeit the difference of relevant Hildebrand solubility parameters (${\Delta\delta}$) is just $\pu {1.36 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$ (from Ref.1 and 2).

It is an interesting aspect of the Hildebrand solvent parameters that the $\delta$ value of a solvent mixture can be determined by averaging the $\delta$ values of the individual solvents by volume. For example, if you mix $V_\mathrm{A}$ of solvent A with $V_\mathrm{B}$ of solvent B, the mixture will have Hildebrand value of $\delta_\mathrm{Mix}$, which would be given by:

$$\delta_\mathrm{Mix} = \delta_\mathrm{A}\left( \frac{V_\mathrm{A}}{ V_\mathrm{A}+ V_\mathrm{B}}\right) + \delta_\mathrm{B}\left( \frac{V_\mathrm{B}}{ V_\mathrm{A}+ V_\mathrm{B}}\right)$$

Let's see what is the $\delta$ of 96% ethanol ($\delta_{\ce{H2O}}=\pu {23.5 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$):
$$ 23.5\left(\frac{4}{100}\right) + 12.92\left(\frac{96}{100}\right) = 13.34 = \delta_{96\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$$

Let's see now, what is the value of $\delta$ of 90% ethanol:
$$ 23.5\left(\frac{10}{100}\right) + 12.92\left(\frac{90}{100}\right) = 14.0 = \delta_{90\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$$

Even though still $\delta_{90\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$, the concept is clear. The value is more than $\delta_{\ce{EtOH}}$ and getting closer by each addition of water droplet and to been not miscible with hexane. Therefore, it is safe to say that these data would gives you some insight to what you are looking for.

Note: For a review of solution systems: See ref.3

References:

C. M. Hansen, In The Three Dimensional Solubility Parameter and Solvent Diffusion Coefficient: Their Importance in Surface Coating Formulation; Danish Technical Press: Copenhagen, Denmark, 1967.
C. M. Hansen, "The Three Dimensional Solubility Parameter Key to Paint Component Affinities: 1. Solvents Plasticizers, Polymers, and Resins," Journal of Paint Technology 1967, 39(505), 104-117.
R. Wiśniewski, E. Śmieszek, E. Kamińska, " Three-dimensional solubility parameters: simple and effective determination of compatibility regions," Progress in Organic Coatings 1995, 26(2-4), 265-274 (https://doi.org/10.1016/0300-9440(96)81583-9).


Does SiCl2 exist and its possible structure It's known that the elements in IVA all have valence +2 and +4. I also know there exists $\ce{GeCl2}$ and $\ce{SnCl2}$. So my question is does $\ce{SiCl2}$, or even $\ce{CCl2}$ exist? If yes, what about their molecular structure?

In addition, what makes +2 valence in $\ce{Ge}, \ce{Sn}$ more stable?

Via Wikipedia, $\ce{CCl2}$, where carbon is in the +2 oxidation state, does exist but it is fleeting due to its high reactivity.

I was able to find a reference on the analogous dichlorosilylene, and I also found some information on difluorosilylene, also known as silicon difluoride, both of which have silicon in the +2 state and react similarly to a carbene.

So yes, both $\ce{CCl2}$ and $\ce{SiCl2}$ exist where the central atom in the +2 oxidation state.

Tin(II) chloride fill its octet by way of either  bridging chloride bonds in the crystal structure or hydration bonds in water (see Wikipedia images). The structure of solid $\ce{GeCl2}$ isn't given, but I would expect something similar. I'll update if I find anything.


Is the t-butyl carbocation more stable than the benzyl carbocation? Various authors have different views regarding stability order of the benzyl and t-butyl carbocations.

$$\ce{PhCH2+ ; (CH3)3C+}$$

In my opinion, resonance effect dominates, so the benzylic carbocation should be more stable. But in the other case, both the inductive and hyperconjugation effects are present, which stabilize the intermediate carbocation.

What should is the correct stability order? Especially when you consider them in SN1 reactions, and what their effect on their rate is.

I am using a very simplistic quantum chemical approach of the following isodesmic reaction:

I have used Gaussian 16 Rev. A.03 and the DF-B97D3/def2-TZVPP level of theory. The summaries of the calculations are included below.

On this level of theory the depicted reaction has an energy change of $\Delta G = \pu{- 37.1 kJ mol-1}.$ Therefore one could assume that the 2-methylpropan-2-ylium cation is more stable than the phenylmethylium cation. These values were estimated at $T = \pu{298.15 K}$ and $p = \pu{1 atm}.$

One certainly can do more calculations, but it is, however, a start.

Calculation summaries

INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               phch2_q1/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -270.5777462860 hartree
zero-point corr.             (ZPE):            +0.114824 hartree/particle
thermal corr.                  (U):            +0.120571 hartree/particle
ther. corr. enthalpy           (H):            +0.121515 hartree/particle
ther. corr. Gibbs en.          (G):            +0.085614 hartree/particle
entropy (total)            (S tot):              +75.561 cal/(mol K)
heat capacity (total)       (Cv t):              +22.578 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               phch3_q0/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -271.4854244270 hartree
zero-point corr.             (ZPE):            +0.125049 hartree/particle
thermal corr.                  (U):            +0.131428 hartree/particle
ther. corr. enthalpy           (H):            +0.132372 hartree/particle
ther. corr. Gibbs en.          (G):            +0.093359 hartree/particle
entropy (total)            (S tot):              +82.110 cal/(mol K)
heat capacity (total)       (Cv t):              +23.831 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               t-c4h10_q0/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -158.4201368210 hartree
zero-point corr.             (ZPE):            +0.128849 hartree/particle
thermal corr.                  (U):            +0.134606 hartree/particle
ther. corr. enthalpy           (H):            +0.135550 hartree/particle
ther. corr. Gibbs en.          (G):            +0.101200 hartree/particle
entropy (total)            (S tot):              +72.297 cal/(mol K)
heat capacity (total)       (Cv t):              +20.521 cal/(mol K)
==== Next file ====
INFO   :  Found route section:
#P B97D3/Def2TZVPP int(ultrafinegrid) freq geom=check guess=read gfinput
gfoldprint iop(6/7=3) symmetry(none)
----
calculation details               : RB97D3               t-c4h9_q1/b97d3tzvpp.freq.log
temperature                    (T):              298.150 K
pressure                       (p):              1.00000 atm
electr. en.                    (E):      -157.5167928240 hartree
zero-point corr.             (ZPE):            +0.114033 hartree/particle
thermal corr.                  (U):            +0.120557 hartree/particle
ther. corr. enthalpy           (H):            +0.121501 hartree/particle
ther. corr. Gibbs en.          (G):            +0.083653 hartree/particle
entropy (total)            (S tot):              +79.659 cal/(mol K)
heat capacity (total)       (Cv t):              +20.965 cal/(mol K)

(Compiled with tools-for-g09.bash, which somewhat surprisingly works for g16.)

Optimised geometries

14
phch2_q1/b97d3tzvpp.freq.xyz
C          0.08663       -0.19347        0.09384
C          0.05691       -0.11475        1.46208
H          1.02702       -0.19311       -0.45239
H         -0.82919       -0.25883       -0.48899
C         -1.20323       -0.11721        2.16356
C         -1.22180       -0.03857        3.53614
C         -0.00341        0.04387        4.24119
C          1.24453        0.04873        3.58467
C          1.28552       -0.02918        2.21254
H         -2.12409       -0.18167        1.59151
H         -2.16067       -0.03932        4.07940
H         -0.02695        0.10566        5.32621
H          2.15902        0.11360        4.16439
H          2.23039       -0.02773        1.67720

15
phch3_q0/b97d3tzvpp.freq.xyz
C          0.00007        0.00000       -0.04694
C          0.00682       -0.00000        1.46143
H          1.01682       -0.00000       -0.44951
H         -0.51968       -0.88199       -0.43936
H         -0.51968        0.88199       -0.43936
C         -1.19805       -0.00000        2.17854
C         -1.20438       -0.00000        3.57251
C          0.00015        0.00000        4.27958
C          1.20614       -0.00000        3.57927
C          1.20606       -0.00000        2.18275
H         -2.14093        0.00000        1.63572
H         -2.14998        0.00000        4.10835
H         -0.00257        0.00000        5.36594
H          2.14940        0.00000        4.11910
H          2.15082        0.00000        1.64397

14
t-c4h10_q0/b97d3tzvpp.freq.xyz
C          0.00369       -0.00664        0.00284
C         -0.01764        0.03020        1.53527
C          1.44139        0.03012       -0.52791
C         -0.74698       -1.23338       -0.52790
H         -1.04360        0.03406        1.92005
H          0.49077       -0.85020        1.94859
H          0.49182        0.92075        1.92008
H          1.46219        0.03417       -1.62347
H          1.97408        0.92055       -0.17569
H          2.00042       -0.85041       -0.18649
H         -0.76068       -1.24961       -1.62346
H         -0.26410       -2.15775       -0.18616
H         -1.78452       -1.24933       -0.17586
H         -0.51485        0.89146       -0.36388

13
t-c4h9_q1/b97d3tzvpp.freq.xyz
C         -0.00184       -0.00056        0.03904
C          0.01681        1.45949        0.05366
H         -0.46367        1.80346       -0.88071
H          1.00698        1.90257        0.15089
H         -0.66770        1.83073        0.83269
C          1.25179       -0.74789        0.02785
H          1.76765       -0.48991       -0.91708
H          1.93393       -0.35706        0.79738
H          1.13840       -1.82916        0.09679
C         -1.27605       -0.71190        0.03180
H         -1.24998       -1.55839       -0.66820
H         -1.35257       -1.20345        1.02233
H         -2.15046       -0.07921       -0.11824

(g09.chk2xyz does not work for g16.)


When ∆Suniv = zero, does the concentration of reactant equal the concentration of product? It is my understanding that when ∆Stot / ∆Suniv = 0 then the reaction does not occur, the position of equilibrium is not shifted in a specific direction. However, does this imply that [reactants] = [products]?

The question 1e part iii asks "What can you understand about the the equilibrium when ∆Stot = 0". The answer to the question is "Kc = 1, or reaction balanced in middle, or no tendency to go in either direction"

To me, this only makes sense if the starting concentrations of the reactants and products are the same. However, at no point is it mentioned that the starting concentrations of reactants/products are equal.

All thermodynamics boils down to looking at $\Delta S_\mathrm{univ}$. $\Delta S_\mathrm{univ}$ is the sum of $\Delta S$ for your system and $\Delta S$ for everything else. We measure $\Delta S$ for the system directly and we measure $\Delta S$ for everything else via $-\frac{\Delta H}{T}$. When these two are equal, notice that $\Delta G = \Delta H - T\Delta S = 0$. Then given $\Delta G = -RT \ln K$, we conclude that $K = 1$.

The reaction favors reactants as much as products, but that specific balance depends on the reaction and its coefficients, i.e., the equilibrium expression could have powers. Consider:
$$\ce{A2 -> 2A}$$

The concentrations of product and reactant are not equal at equilibrium for $K = 1$.

Also, it does not mean that a reaction does not occur. You need to compare the reaction quotient $Q$ against $K$ to determine if there is a reaction or if we are actually at equilibrium.


Heat transfer with surroundings: why considered reversible? I do not understand how the fact that the surroundings are large makes heat transfer with them reversible? I understand that the temperature of the surroundings is not going to change in the limit, but what about processes in which there is a definite temperature gradient with the system, such as a highly exothermic reaction?

I was sure that this question had already been asked somewhere on Stack, but I can't find anything similar – if you know of one, I would really appreciate if you could refer me to it.

The surroundings are typically modelled in thermodynamics as an ideal (reversible) reservoir with infinite heat capacity and negligible temperature gradients (i.e., infinite thermal conductivity).  The assumption is that the temperature gradients are confined to the system, rather than the surroundings, and the entropy generation is confined to the system.  Of course, in real life, this is only an approximation.


Molar heat of combustion with different fuels I'm not sure exactly how to attack this question, I'm sure that you have to use equations such as

$$Q=mc\,\Delta T$$
$$n=\frac{M}{m}$$

But I don't know where to use them. Any help would be appreciated

I just attempted the question, I'm not sure if this is the correct working.

Assuming 1 mol of each substance

Ethyne:
$$\begin{align}n&=\frac{M}{m}\\[6pt]
1&=\frac{26.038\ \mathrm{g/mol}}{m}\\[6pt]
m&=26.038\ \mathrm g\end{align}$$
producing 1630 kJ

Hexane:
Same working as above so:
$$m=86.178\ \mathrm g$$
producing 1300 kJ

Ethane:
$$m=30.07\ \mathrm g$$
producing 1560 kJ

Simplifying each of the mass and energy relationships for each substance gives:

Ethyne: 1 gram for 62.6 kJ

Hexane: 1 gram for 15.09 kJ

Ethane: 1 gram for 51.89 kJ

This would imply that ethyne would be the best fuel, as it uses the least amount of mass and the greatest amount of energy is produced.

Not sure if this is correct. Any tips would be great.

This question is asking which substance has the greatest heat density, or energy released per mass combusted (something like J/g).

$\ce{C2H2}$ has a mass of $26.04\ \mathrm{g/mol}$. Since the $\Delta H$ in its combustion is $1\,630\ \mathrm{kJ/mol}$, you will be looking at a heat density of $\frac{1\,630\ \mathrm{kJ/mol}}{26.04\ \mathrm{g/mol}} = 62.6\ \mathrm{\frac{kJ}{g}}.$

Similarly, for $\ce{C6H14}$, you get $\frac{1\,300\ \mathrm{kJ/mol}}{86.178\ \mathrm{g/mol}} = 15.1\ \mathrm{\frac{kJ}{g}}.$

For $\ce{C2H6}$, you get $\frac{1\,560\ \mathrm{kJ/mol}}{30.07\ \mathrm{g/mol}} = 51.9\ \mathrm{\frac{kJ}{g}}.$

It looks like you were on the right track.


Orbital mixing in square planar D4h complexes Consider a square planar $\ce{ML4}$ complex with $D_\mathrm{4h}$ symmetry. Which orbital should the $\mathrm{a_{1g}}$ SALC of the ligands mix with, $\mathrm{3d}_{z^2}$ or $\mathrm{4s}$? Both of those AOs transform as  $\mathrm{A_{1g}}$.

All orbitals of the same symmetry will interact (mix) to some degree.

So the ligand SALC will interact with both $\mathrm{a_{1g}}$ orbitals on the metal. The degree of interaction will depend on how close in energy the ligand $\mathrm{a_{1g}}$ is to the $\mathrm{4s}$ or $\mathrm{3d}_{z^2}$ orbitals on the metal.

Generally, the ligand orbitals are closer in energy to the $\mathrm{3d}_{z^2}$ orbital than to the $\mathrm{4s}$.


Breaking Down PCB's in lakes I was wondering if there is any natural process which breaks down PCBs (Polychlorinated biphenyl).

This is an extremely harmful contaminant that was pumped into various lakes by chemical plants up until about the 1950s. It is very stable, which is why it was used a coolant in these chemical plants. However this toxic substance has greatly impacted lake systems, as the concentration of the PCB is increased in each trophic level of the lake ecosystem (due to biomagnification). Anyways, I was wondering if there was any method to break down the PCBs.

Since they are a chlorine-containing compound, would I have to use something like a sodium or potassium based compound to safely break them down chemically. What do you guys think? Thanks!

According to this abstract$\mathrm{^{[1]}}$, both aerobic and anaerobic processes play a role in the natural degradation of PCB's.

  Studies have identified two distinct biological processes capable of
  biotransforming polychlorinated biphenyls (PCBs): aerobic oxidative
  processes and anaerobic reductive processes. It is now known that
  these two complementary activities are occurring naturally in the
  environment. Anaerobic PCB dechlorination, responsible for the
  conversion of highly chlorinated PCBs to lightly chlorinated
  ortho-enriched congeners, has been documented extensively in the
  Hudson River and has been observed at many other sites throughout the
  world. The products from this anaerobic process are readily degradable
  by a wide range of aerobic bacteria, and it has now been shown that
  this process is occurring in surficial sediments in the Hudson River.
  The widespread anaerobic dechlorination of PCBs that has been observed
  in many river and marine sediments results in reduction of both the
  potential risk from and potential exposure to PCBs. The reductions in
  potential risk include reduced dioxinlike toxicity and reduced
  carcinogenicity. The reduced PCB exposure realized upon dechlorination
  is manifested by reduced bioaccumulation in the food chain and by the
  increased anaerobic degradability of these products.

1) Aerobic and anaerobic PCB biodegradation in the environment, Environ Health Perspect. 1995 Jun; 103(Suppl 5): 97–99


Can we make glass by heating Martian Soil I was puzzled by a very interesting question. Can we heat up Martian soil and produce glass just the way we do here on Earth? (the basic way)

For info, Martian sand consists of the following -

$\ce{Na2O}$ — $4~\%$
$\ce{MgO}$   — $14~\%$
$\ce{Al2O3}$  — $18~\%$
$\ce{SiO2}$   — $8~\%$
$\ce{P2O5}$  — $2~\%$
$\ce{SO3}$   — $10~\%$
$\ce{Cl}$ — $2~\%$
$\ce{K2O}$  — $2~\%$
$\ce{CaO}$  — $14~\%$
$\ce{TiO2}$   — $2~\%$
$\ce{Cr2O3}$  — $2~\%$
$\ce{MnO}$    — $2~\%$
$\ce{FeO}$ — $4~\%$
$\ce{Ni}$   — $8~\%$
$\ce{Zn}$    — $6~\%$
$\ce{Br}$    — $2~\%$

I can't see anything out of place here, if we heat it up, the silica
should do it at around $1700~\mathrm{^\circ C}$.

If we can figure out ways to make glass up there then it would be easy for Martian colonization if that is ever going to happen.

The problem is sorting. Ergo, unless you bring your separator with you, the best areas for finding naturally sorted materials are in outflows such as found in Candor Chasma. Semi-translucent larger grains of quartz-rich phenocrysts might be found in useful quantities. Fusing these with a solar kiln would produce a brick of about 0.125 meters thick that would both shield from the rain of radiation and pass some of the weak Martian sunlight, easing the circadian rhythm of colonists.
Time to temperature (1700 degrees C) still makes production slow. And the loaders and other equipment need to be transported and successfully landed.

See
A Fused Regolith Structure
Ethan W. Cliffton
Proceedings of Space 90
Engineering, Construction, and Operations in Space
ASCE 1990
Albuquerque
and
Candor Chasma Camp Ethan W. Cliffton
Proceedings of Space 88
Engineering, Construction, and Operations in Space
ASCE 1988
Albuquerque


Why is FeCl3 covalent? There is an electronegativity difference of 1.2 between $\ce{Fe}$ and $\ce{Cl}$. Then why it is covalent?

usually when metal is bonded to a non-metal, they are bonded together by ionic bond but this is not the only condition , the more important condition is the electronegativity that must be greater than 1.7 to be sure that the bond is ionic if not like in your example 1.2 then the bond will be polar covalent bond which means that the electronegativity difference is not enough for the electrons to be lost by iron and gained by chlorine but they are shared and shifted toward the chlorine atom as it has the greater electronegativity.


How does complex ion differ from ligand complex? As I understand, in terms of "colouring", ligand complex cause a transition metal cation centre's d-shell to split, allowing the electrons in that region to absorb specific frequencies of light, and thus resulting in colouring, does the same apply for complex ions such as chromate? Does the oxygen atoms/ions around chromate cause d-shell splitting and thus "colouring" effect?

To quickly pre-answer the question at the end of the body: Yes, the oxygen atoms around the chromium(VI) cause d-orbital splitting. However, chromium(VI) is a $\mathrm{d^0}$ system and thus the colour is not caused by d-electron transitions. Rather, as I highlighted in my permanganate answer, the cause of the colour is a ligand to metal charge transfer transition. For details, see there.

The question in the title basically refers to the difference between a complex such as $\ce{[CoCl4]-}$ and an ion such as $\ce{SO4^2-}$ or $\ce{CrO4^2-}$. These can be considered two extremes of a transition much like the extremes unpolar covalent bond and ionic bond when used to describe bonding.

All coordination complexes are nothing more than dative bonds, which in turn are nothing more than a special subset of weak covalent bonds. From a qualitative bonding point of view, there is really no difference between either compounds: For both you get orbital mixing, a bonding and an antibonding set of orbitals and an overall stabilisation of the system. And if you totally ignore energy differences and the like, you will find that each of the three compounds written up there have more or less the same MO scheme.

The difference from a theoretic point of view is the bond strength and the energy difference between the central atom’s and the ligands’ orbitals. For a complex ion, the orbital energies of the unbonded atoms are very similar to each other resulting in a large energy difference upon forming a bond and therefore a large difference between bonding and antibonding orbitals. For typical ligand (or coordination) complexes, the ligands have much lower orbital energies than the central metal and thus the energy gained in the bonding process is comparatively small. On this scale between $\ce{[CoCl4]-}$ and $\ce{SO4^2-}$, chromate(VI) would be somewhere just to the sulphate side of the axis’ centre.

What does this difference mean from a practical point of view? Well, one typical reaction of coordination complexes is the ligand exchange reaction: Add a better ligand into the solution and more often than not it will exchange with a worse ligand. Sometimes this results in a different geometry. $\ce{[CoCl4]-}$ can be synthesised by preparing an aquaeous cobalt(III) solution which contains $\ce{[Co(H2O)6]^3+}$ ions and adding excess chloride ions — the complex will form spontaneously. The reverse reaction is possible by adding silver nitrate which will precipitate the chloride ions, reforming the hexaaqua complex. Ligand exchange is simple.

The sulphate ion, as the extreme opposite example, has such strong $\ce{S-O}$ bonds that they will not ‘simply dissociate by offering a better ligand.’ (In fact, I don’t know if a better ligand even exists.) It is impossible to displace the oxygens to attach other ligands — and most notably, there is no good oxophilic compound known that could simply remove (i.e. precipitate) those oxygens. Note, however, that sulphuryl chloride $\ce{SO2Cl2}$ will readily react with water to form sulphuric acid (diprotonated sulphate) and hydrogen chloride, so one type of ‘ligand exchange’ is still possible.

Chromate(VI), as written earlier, is closer to sulphate than to the cobaltate complex. It will — much like permanganate — not take part in displacement reactions, the metal–oxygen bonds are pretty stable. Again note that an analogous chloride complex exists with chlorochromate $\ce{CrClO3-}$.

Note that not always will the ligand exchange I highlighted above occur easily. Ammonia should form a more stable complex with iron(III) than water. However, when adding ammonia to an iron(III) solution, iron hydroxides will precipitate rather than an ammin complex being formed. So the practical point of view is a really rough approximation that only works sometimes.


Why is the pairing energy not considered when calculating the crystal field splitting energy? Consider the low-spin octahedral crystal field splitting diagram for a d4 system.

When we calculate the crystal field splitting energy we simply observe the absorbed color of light and calculate the frequency of a photon required to promote an electron from the t2g to the eg orbitals. Why is the spin pairing energy not considered (PE) in these calculations? If we had a simple d3 system, I would understand why the energy of the photon absorbed is equal to the crystal field splitting energy since we are not dealing with any paired electrons; however, in the case of the d4 system, doesn't "removing" one of the paired electrons actually release energy? Why do we say the energy of the photon absorbed is equal to Δo and not Δo - PE?

It is actually the two single-paired electrons are more likely to be promoted than the two paired electrons.
Promoting the electrons always cost energy whether the electrons in question are paired or not. In the classical examples of hybridization (where they might actually promote a paired electron), this energy cost is compensated by the formation of stable covalent bonds, but in this case, a complex, no bonds formed here.
So why the single electrons are promoted before the paired ones?
The spin rule in the coordination complexes tell that the allowed d-d transitions must not involve the change in the total spin of the electrons.
In your case of a low-spin d4, promoting the the paired electrons will leave the d-orbitals with single electrons with opposite spin, which is highly unfavorable. The changing of the spin of the opposite-spin electron is forbidden per the spin rule since it will change the total spin of the system.


How do we identify the molecular formula from the empirical formula of a completely unknown substance? Suppose you are given a sample of new (undiscovered) hydrocarbon with 600/7% $\ce{C}$ and 100/7% $\ce{H}$. (From combustion or any other method) We can find the empirical formula to be CH. Then how the we figure out whether it will be $\ce{C2H2}$ or $\ce{C4H4}$ or even something else?

Now we have identified the different compounds and can identify it through various reactions, but what would we do if we had something entirety new? How did chemists initially differentiate between these compounds?

If the empirical formula is known, then any method that will determine the molecular weight will do. Of merely historical interest is the use of freezing-point depression, nowadays, one would use mass spectrometry (electron-spray ionisation will leave most molecules intact).

This assumes that it is a reasonably small molecule, but only for those will the empirical formula be known before the molecular one.


Why does iron have an abnormally high ionization energy? Along a period the ionization energy should increase because the atomic number is increasing, but there is negligible increase in shielding. However, $\mathrm{IE}_\ce{Mn} < \mathrm{IE}_\ce{Fe} > \mathrm{IE}_\ce{Co} > \mathrm{IE}_\ce{Cu}$, so why does iron have such a high ionization energy?

I've considered that iron has a $\mathrm{4s^2 3d^6}$ configuration that goes to a $\mathrm{4s^1 3d^6}$ configuration upon ionization, while $\ce{Co}$ has a $\mathrm{4s^2 3d^7}$ that goes to $\mathrm{4s^0 3d^8}$ configuration, $\ce{Ni}$ has a $\mathrm{4s^2 3d^8}$ that goes to a $\mathrm{4s^0 3d^9}$ configuration and $\ce{Cu}$ has a $\mathrm{4s^2 3d^9}$ that goes to a $\mathrm{4s^0 3d^{10}}$ configuration (Lang, J. Chem. Ed., 2003).  In the Long paper, the authors explain the increase of IE at iron is due to "electronic structure" and I would like to know exactly what about the electronic structure it is that causes iron to have a high IE than expected based on a general periodic trend.

I was thinking that adding an electron to the $\mathrm{3d}$ orbitals might be energetically favorable enough to lower the ionization energy of $\ce{Ni}$, $\ce{Co}$ and $\ce{Cu}$, so rather than iron having a high ionization energy the surround metals just have a low ionization energy because the ion is stabilized by having another $\mathrm{d}$-electron added? Does that sound reasonable?

Another thing I was thinking was that iron is the first transition metal to have a paired $\mathrm{d}$-electron, could that effect the shielding and increase the ionization energy?

"I mean "abnormal" in that it defines a general trend" makes no sense to me at all. Did you mean "defies"?  Take a look at the chart here: https://en.wikipedia.org/wiki/Ionization_energy.
By my count, for the s-block elements the "trend" always holds.  For the p-block elements, it holds in 5/6ths of the cases. And for the d-block, I can't tell from the chart, but claiming it is increasing for them seems agross exaggeration to me; I'd say they might have a very slight tendency to increase, but they're so close to being flat-lined, that it seems to be not a useful claim. If you count the exceptions to the rule, and you expect your "rules" to hold at least 90% to 95% of the time (this isn't the Social Sciences, after all), then it is a just plain lousy rule. Nothing unusual about that: I think the Periodic Table is best when used to "predict" Atomic Number (they increase moving right and down, without exception), and so untrustworthy with the other periodic trends as to not be something in my "mental toolbox". Use it at your own peril. (The at. no. "trend" was a joke, of course)


How do you solve a net ionic equation without a balanced equation? In class, my professor went over a method for finding the net ionic equation without determining the balanced equation (e.g., if finding the balanced equation was too difficult for us students). It involved writing the ions of each of the solutions being mixed and determining which ions would be spectator ions.

For example, finding the net ionic equation for the following mixture of solutions:

$\ce{NaC_2H_3O_3 + BaCl_2 +NH_4NO_3 + K_2CrO_4}$

Unfortunately, I don't quite grasp the process. How does one determine which ions and compounds will remain in the net ionic equation?

This sort of question is really all about solubility. What you're really asking when you're trying to get a net ionic equation is What combinations of the ions are insoluble? Net ionic equations involve only ions that end up as precipitate on the right side of the equation. If you do enough of these problems, you get a feel for the general solubility trends of common ions, but there should be a table in your textbook or have a look here.

For this specific example, salts with alkali metal and ammonium cations are soluble. Nitrate and acetate anion salts are soluble. Therefore, all of those end up as spectator ions and we're left with just $\ce{Ba^2+}$ and $\ce{CrO_4^2-}$. Chromate salts other than those of alkali metal cations or ammonia are insoluble (rule 8 in the link), so the net ionic equation is:

$$\ce{Ba^2+_{(aq)} + CrO_4^{2-}_{(aq)} -> BaCrO_{4(s)}}$$

In this case, it's already balanced, but that may not necessarily be true.


Why does sulfonation of aniline occur para rather than ortho? According to a textbook by K. L. Chugh:

  Sulfonation [of aniline] is a reversible reaction. The p-isomer, being more stable, does not get desulfonated easily. Hence, p-sulfanilic acid is the major product.

I am curious as to why the ortho isomer (2-aminobenzenesulfonic acid) is not more thermodynamically stable. There is a possibility of intramolecular hydrogen bonding (in a six-membered ring):

Why is the para isomer (4-aminobenzenesulfonic acid) favoured?

The sulfonyl group is kind of bulky due to the size of the sulfur, so steric hindrance is probably making it energetically disfavored for the group to be placed ortho. It's unlikely you can see this easily from a model or a drawing. If you have access to computational resources, you might be able to see the issue.


Why is there oxidation in the anode? In a battery, I don't understand what makes the element in the anode separate from the anode to become a positive charge in the electrolyte and leave some electrons inside the anode.

Also, why does the molecule separated from the anode leave electrons on the anode?

We define anode to be the electrode where oxidation takes place.

Oxidation signifies loss of electrons and Reduction signifies gain of electrons.

For example, in this electrochemical cell,

Copper has more reduction potential than Zinc. So it will be easy for copper to reduce (or gain electrons) and if copper is reducing (or gaining electrons) then there must a source of these electrons. These electrons are supplied by Zinc (which gets oxidized).

At anode, $Zinc(s)$ in losing electrons and converting into $Zn^{+2}(aq)$ ions and at cathode, $Cu^{+2}(aq)$ ions are gaining these electrons lost by $Zinc(s)$ and converting into $Cu(s)$.

These reactions occur because they are spontaneous $(\Delta G =-ve)$. Spontaeous reactions occur in nature by themselves without any external driving force. A system aims for least energy $(\Delta H = -ve)$ and maximum entropy $(\Delta S = +ve)$ and its the combination of both these factors that result in spontaneous processes. $(\Delta G = \Delta H - T\Delta S)$.


Synthesis of a lactone I'm solving some synthesis problems for practice and I came across this one:

I thought of going for compound A to B(aldehyde) with this procedure:

$\ce{t-BuOK}$
$\ce{BH3,THF,NaOH,H2O2}$
$\ce{PCC,CH2Cl2}$

For the second path, in order to get to a lactone, I don't know what to do. I see that the carbonyl bond is moved to another carbon. I've learnt a way, in order to get an ester, is to add an alcohol $\ce{ROH}$ to an acyl chloride. But I'm so confused.

Here is a way to make the lactone:

To your aldehyde in dry THF at 0 °C, add the commercially available 3-butenylmagnesium bromide. Work up and collect the resulting secondary alcohol.
React the alcohol with dihydropyran in dry $\ce{CH2Cl2}$ to form the THP ether.
React this THP ether with $\ce{BH3.THF}$ as you did before, but do an oxidative work up with PDC to get the carboxylic acid according to Brown et al. [1], or do the basic peroxide as before to get the aldehyde, then oxidise that.
Treat the product acid with PTSA in MeOH to remove the pyran ether, the lactone will form spontaneously.

Your choice of alcohol protecting group will be driven by the oxidation conditions you use for the borane. THPO is known to be stable to PDC/$\ce{CrO3}$ in pyridine.

Reference

Brown, H. C.; Kulkarni, S. V.; Khanna, V. V.; Patil, V. D.; Racherla, U. S. Organoboranes for Synthesis. 14. Convenient Procedures for the Direct Oxidation of Organoboranes from Terminal Alkenes to Carboxylic Acids. J. Org. Chem. 1992, 57 (23), 6173–6177. DOI: 10.1021/jo00049a024.


Significant figures addition vs multiplication I was finding the molecular weight of $\ce{Br2}$ the other day and noticed that if I add the weight of $\ce{Br}$ to itself, $79.90+79.90=159.80$, that answer for the weight has $5$ significant figures. But if I just multiply the weight by $2$, $79.90 \times 2 = 159.8$, the weight has $4$ significant figures. If adding something to itself is the same as multiplying it by $2$ mathematically, which accuracy is more justifiable for my calculations in cases like this?

Your calculations are in accordance with the rules of the so-called significance arithmetic even though they yield different results.

When multiplying numerical values, according to the rules of significance arithmetic, the result is rounded to the number of significant digits in the factor with the least significant figures. When you multiply the value of the molar mass $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ by 2, the value of the molar mass $M_{\ce{Br}}$ has four significant digits, whereas the factor 2 is assumed to be exact. Accordingly, the result is rounded to four significant digits:

$$M_{\ce{Br2}}=2M_{\ce{Br}}=2\times79.90\ \mathrm{g/mol}=159.8\ \mathrm{g/mol}\tag1$$

When adding numerical values, according to the rules of significance arithmetic, the result is rounded to the position of the least significant digit in the most uncertain of the numbers being summed. When you add the molar mass $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ to itself, the value of the molar mass $M_{\ce{Br}}$ is significant to the hundredth place. Accordingly, the result is rounded to the hundredth place:

$$M_{\ce{Br2}}=M_{\ce{Br}}+M_{\ce{Br}}=79.90\ \mathrm{g/mol}+79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag2$$

However, this concept of significant digits is just an oversimplified form of the concept of uncertainty as a quantifiable attribute. In your question, you are testing the limits of the applicability of this simplification, which is certainly a good idea.

When a number is given without any further information, it is generally interpreted so that the last digit is rounded. Thus, the molar mass of $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ is generally assumed to represent a value between $79.895\ \mathrm{g/mol}$ and $79.905\ \mathrm{g/mol}$, or $M_{\ce{Br}}=\left(79.900\pm0.005\right)\ \mathrm{g/mol}$.

Therefore, the assumed uncertainty of the molar mass $M_{\ce{Br}}$ is $u(M_{\ce{Br}})=0.005\ \mathrm{g/mol}$. (Note that, normally, the uncertainty corresponds to the estimated standard deviation; in this case, however, the uncertainty just indicates an interval corresponding to a high level of confidence.)

When you multiply the value of the molar mass $M_{\ce{Br}}$ by 2, you also have to multiply its uncertainty by 2 (the factor 2 is assumed to be exact):

$$M_{\ce{Br2}}=2M_{\ce{Br}}=2\times79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag{3a}$$

$$u(M_{\ce{Br2}})=u(2M_{\ce{Br}})=2u(M_{\ce{Br}})=2\times0.005\ \mathrm{g/mol}=0.010\ \mathrm{g/mol}\tag{3b}$$

When you add the value of the molar mass $M_{\ce{Br}}$ to itself, you also have to add the uncertainty to itself:

$$M_{\ce{Br2}}=M_{\ce{Br}}+M_{\ce{Br}}=79.90\ \mathrm{g/mol}+79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag{4a}$$

$$\begin{align}u(M_{\ce{Br2}})&=u(M_{\ce{Br}}+M_{\ce{Br}})=u(M_{\ce{Br}})+u(M_{\ce{Br}})=0.005\ \mathrm{g/mol}+0.005\ \mathrm{g/mol}\\&=0.010\ \mathrm{g/mol}\tag{4b}\end{align}$$

(Note that such a simple addition of uncertainties is not correct when the errors of the added values are uncorrelated, for example when calculating $M_{\ce{HBr}}=M_{\ce{H}}+M_{\ce{Br}}$.)

Both calculations yield the same results for the molar mass $M_{\ce{Br2}}$ ($\text{(3a)}$ and $\text{(4a)}$) as well for its uncertainty ($\text{(3b)}$ and $\text{(4b)}$). Thus, the molar mass of $M_{\ce{Br2}}$ is estimated to represent a value between $159.79\ \mathrm{g/mol}$ and $159.81\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.80\pm0.01\right)\ \mathrm{g/mol}$$

By way of comparison, the result of the multiplication according to the rules of significance arithmetic $\text{(1)}$, $M_{\ce{Br2}}=159.8\ \mathrm{g/mol}$, represents a value between $159.75\ \mathrm{g/mol}$ and $159.85\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.80\pm0.05\right)\ \mathrm{g/mol}$$ Thus, this approach overestimates the uncertainty of the result.

The result of the addition according to the rules of significance arithmetic $\text{(2)}$, $M_{\ce{Br2}}=159.80\ \mathrm{g/mol}$, represents a value between $159.795\ \mathrm{g/mol}$ and $159.805\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.800\pm0.005\right)\ \mathrm{g/mol}$$ Thus, this approach underestimates the uncertainty of the result.


LiOH hydrolysis of methyl 2,2-dimethoxyacetate not giving product? I am forming 2,2-dimethoxyacetic acid from its methyl ester, and below is the experimental protocol I am following (from Tetrahedron 2016, 72 (3), 420–430):

  2,2-Dimethoxy Acetic Acid Formation

  Add lithium hydroxide monohydrate ($\pu{89.5 mmol}$, $\pu{3.75 g}$, $\pu{1.2 equiv}$) to a solution of methyl dimethoxyacetate ($\pu{74.6 mmol}$, $\pu{10 g}$, $\pu{1 equiv}$) in dioxane ($\pu{45 mL}$) and water ($\pu{45 mL}$) at $\pu{0 ^{\circ}C}$.
  Stir the mixture for $\pu{1 h}$ at $\pu{0 ^{\circ}C}$ and one additional hour at room temperature.
  Add sodium hydroxide ($\pu{1M}$ aq soln, $\pu{75 mL}$) to the mixture.
  Extract the mixture with diethyl ether ($\pu{375 mL}$).
  Acidify the recovered aqueous layer with hydrogen chloride ($\pu{6N}$ aq soln) until a pH of 1-2.
  Extract the organics with diethyl ether ($\pu{375 mL}$).
  Dry the combined organic layers with magnesium sulfate and filter.
  Evaporate the solvent slowly under reduced pressure to obtain 2,2-dimethoxyacetic acid.

I repeated this reaction many times but every time I check the $\mathrm{^{13}C}$-NMR, I only see a single peak at $\delta \: \pu{67.12 ppm}$. I believe this corresponds to C-OMe. By theory (and by literature) I should see 3 peaks.
$\mathrm{^{1}H}$-NMR does show 6:1 ratio of singlets at $\delta \: \pu{3.704 ppm}$ (6H) and $\delta \: \pu{3.458 ppm}$ (1H) but it is not well aligned with literature; By theory, a single hydrogen atom adjacent to $\ce{C=O}$ group should be more deshielded than $\delta \: \pu{3.458 ppm}$ (i.e. be placed at higher ppm). I thought through what I have done and I still can't figure out which step has gone wrong. Any thoughts and ideas? Would appreciate any advice.

If your reaction has worked and you get the expected product, you should see following NMR data as the sought literature reported:

$\mathrm{^{1}H}$-NMR: $\delta$ 3.44 (s, $\ce{6H}$), 4.85 (s, $\ce{1H}$), and 10.32 (broad s, $\ce{1H}$).
$\mathrm{^{13}C}$-NMR: $\delta$ 54.1 ($\ce{2CH3}$), 98.6 ($\ce{CH}$), and 170.7 ($\ce{C_q}$).

Instead, you got two peaks, $\delta$ 3.704 (s, $\ce{6H}$) and $\delta$ 3.458 (s, $\ce{1H}$), in your $\mathrm{^{1}H}$-NMR. And, you got only one peak at $\delta$ 67.12 in your $\mathrm{^{13}C}$-NMR. All this data telling me is your reaction has not worked as expected. What you have is 1,4-dioxane during the ether extract, $\mathrm{^{1}H}$-NMR of which gives only a singlet at $\delta$ 3.71 when $\ce{CDCl3}$ is the solvent. It also gives one peak at $\delta$ 67.14 in $\mathrm{^{13}C}$-NMR under same solvent (Ref.1).
I believe your compound may still be in the aqueous phase. You may need to saturate the aqueous phase with salt ($\ce{NaCl}$) before extract the product to organic phase. Note that acetic acid is very soluble in water. Therefore, I expect 2,2-dimethoxyacetic acid to be highly soluble in water as well (a $\ce{C4(H2O)4}$ compound like a monosaccharide).
References:

H. E. Gottlieb, V. Kotlyar, A. Nudelman, “NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities,” J. Org. Chem. 1997, 62(21), 7512–7515 (DOI: 10.1021/jo971176v).


What is the order of the reaction with respect to NO? The following reaction mechanism has been proposed for the reaction of $\ce{NO}$ with $\ce{Br2}$ to form $\ce{NOBr}$
  $$\begin{align}
\ce{NO(g) + Br2(g)&<=>NOBr2(g)}\\
\ce{NOBr2(g) + NO(g) &-> 2NOBr(g)}
\end{align}$$
  If the second step is the rate determining step, what is the order of the reaction with respect to $\ce{NO}$?

Since the 2nd step is the rate determining, shouldn't the order with respect to $\ce{NO}$ be 1? According to my book though, the correct answer is 2. What am I missing here?

The rate of that second reaction will indeed be first order with respect to $\ce{NO}$:

$$
R = R_2 = k_2[\ce{NO}][\ce{NOBr2}].
$$

The problem is, the overall order of the reaction is determined from the rate expression involving only the reactants.  The above rate expression is written in terms of $\ce{NO}$ (reactant) and $\ce{NOBr2}$ (intermediate), not in terms only of reactants.

To convert the reaction rate expression to one involving only reactants, exploit the fast equilibrium of the first reaction,

$$
K_1 = \frac{[\ce{NOBr2}]}{[\ce{NO}][\ce{Br2}]},
$$

to find an expression for $[\ce{NOBr2}]$ in terms only of $K_1$ and the reactants:

$$
[\ce{NOBr2}] = K_1[\ce{NO}][\ce{Br2}].
$$

The expression for $R = R_2$ can thus be rewritten as

$$
R=R_2 = k_2[\ce{NO}]\left(K_1[\ce{NO}][\ce{Br2}]\right) = k_2K_1[\ce{NO}]^2[\ce{Br2}],
$$

which is indeed 2$^{\mathrm{nd}}$ order in $\ce{NO}$.


Can a mixture of water miscible solvent and water be separated by fractional freezing on a dry ice bath? If a mixture of water with acetone, methanol, ethanol, or other water miscible solvent, is cooled below 0 degrees, would the water freeze and aggregate out as ice, separating out of the mixture. Or would the frozen 'water particles' be dissolved in the solvent, similar to how solid organics dissolve in solvents?

I suppose it really depends on the concept of solubility vs miscibility. I've been researching this for a while now and haven't been able to find a clear definite answer.

The short answer is

  Yes

This is exactly how acetic acid is purified: Acetic acid freezing distilation

In this case, acetic acid (freezing point $16\ ^\circ \text{C}$) freezes first, leaving a water-enriched fluid behind. Conceivably you could remove water from methanol or acetone in the same way (except that water would freeze first). In practice, fractional distillation can be done more economically on a large scale. Fractional distillation requires 1 step (with lots of theoretical plates), while you need to repeat fractional freezing multiple times to get to high purity. Acetic acid is purified by fractional freezing perhaps because it may azeotrope with water.


What is the relationship between entropy and heat capacity? The molar heat capacity of hydrogen gas and deuterium gas are nearly the same, $\pu{28.8 J K-1 mol-1}$ and $\pu{29.2 J K-1 mol-1}$, respectively, but the absolute entropy of deuterium ($\pu{145.0 J K-1 mol-1}$) is significantly larger than that of hydrogen ($\pu{130.7 J K-1 mol-1}$).
I have seen the equation that associates the heat capacity to entropy, and I understand the equipartition theorem and different modes of motion contribute to the heat capacities. Since the heat capacities are fairly the same, what other factors can affect the entropy in these species?

The entropy depends on the mass. The translational part is given by the Sakur-Tetrode eqn., the rotational part depends on the moment of inertia, and the vibrational part on the vibrational frequency. In each case the partition function $z$ has to be calculated then the entropy via $\displaystyle S=R\ln(z)+RT\left(\frac{\partial z}{\partial T} \right)_V -R\ln(N)+R$ (for a perfect gas). The physical reason is that for the heavier molecule there are more energy levels up to a given energy that can be occupied and so more ways of populating these levels and so more entropy.

The heat capacity is the slope of internal energy $U$ with temperature (at constant volume) $\displaystyle C_V=\left({\partial U}{\partial T}\right)_V$ and the internal energy is related to the partition function as $\displaystyle \frac{U}{T}=RT\left(\frac{\partial z}{\partial T}\right)_V$ and so also to the entropy. The classical thermodynamic relationship is $\displaystyle S=S_0+\int_0^T \frac{C_p}{T}dT$ where $C_p$ is a function of temperature.


Does desalination work? Does salt vaporise? If I boil salt-water and leave it there until half of the water vaporises, will it be twice as salty as in the beginning? Or exactly the same? Or somewhere in between?

This is a classic science experiment. Where saltwater (using $\ce{NaCl}$) is heated, as it does, the water evaporates and most, if not all, of the salt remains - increasing the concentration of salt.

Evaporating half the water would approximately double the concentration of $\ce{NaCl}$.

Eventually, the salt concentration becomes supersaturated and crystallises. The reason is due to:

  As the water evaporates less and less water molecules are present to keep the salt particles apart. The salt therefore recrystalises and can be collected.

Source

This process is also used in salt evaporation ponds, where

  seawater or brine is fed into large ponds and water is drawn out through natural evaporation which allows the salt to be subsequently harvested.

As Curt F mentioned in the comments, different salts and mixtures of salts can have different outcomes in terms of vaporisation.


What is "stuffy"/"stale" air composed of? When a room is not ventilated for a long time (days if people are present, weeks if not), humans perceive a change in air quality commonly described as "stale" or "stuffy" air.

What is going on here chemically? How is "stale" air different from normal air?

Note that the difference cannot be the mere presence of biological exhalations from the inhabitants, as the phenomenon definitely also occurs when nobody lives in the building at all (though it takes a lot longer in that case).

It may not be a complete answer but I can think of a few distinguishable features that may or may not be chemical reasons:

chemicals created by bacteria and organisms that are present in everyday households (mold, light activating some production from bacteria and micro-organisms, $NH_3, S^{2-}$ related compounds (they smell bad), etc).
Compound this with the acknowledged lack of circulation and you have funky smelling air.
high humidity: how one can "feel" "stale, muggy" air.
high carbon dioxide concentration can make the air feel heavier (from interior design sources).


Colour due to transmission and reflection It makes sense to me that when looking through a sample (observer | sample | light), it should appear as the opposite of the light absorbed, but it does not make sense to me to expect the same when not looking through it, just standing by the same side as the light source (observer | light | sample, or light | observer | sample).
In the first cenario, the observer sees light emitted - light absorbed (the transmitted, that barely interacts with the sample). In the second, the observer sees light reflected (to my understanding, light emitted from the de-excitations of excitations caused by the light source).
However, when I see a solution of $Cu^{2+}$, it looks the same when observed in both settings. Thereby implying that transmitted colour = reflected colour. Why is that?
Somewhat related question but without a satisfactory answer: Transmission, absorption, and reflection of light

Since your experimental observation with copper salts negates your original hypothesis, it implies that the way you are trying to explain it is wrong. The main culprit and the source of all these problems is the color wheel which is taught in schools. Misconceptions persist for long. When you are looking at the copper solutions in two different settings, it is misleading to think that copper solution is reflecting blue light back to you. It is not.
In each case, copper(II) solution is absorbing a small portion of the red light from the visible spectrum and it appears to you like a pure blue solution.
Indeed it is our brain which has been created in such a way that when the visible spectrum has a certain red portion missing, it perceives the remaining spectrum as "blue".
Hint: Water in the ocean also appears blue? Water also very very weakly absorbs the red portion of the visible spectrum. You just need tons of water to perceive this effect.
There is a beautiful book by the name of The Physics and Chemistry of Color. The same author wrote an article "The fifteen causes of color: The physics and chemistry of color." It is certainly worth consulting. Article-behind paywall

The fifteen causes of color derived from a variety of physical and chemical mechanisms are summarized in five Croups in this article. Vibrations and simple excitations explain the colors of incandescence (e.g., flames), gas excitations (neon tube, aurora), and vibrations and rotations (blue ice and water). Ligand‐field‐effect colors are seen in transition‐metal compounds (turquoise, chrome‐oxide green) and impurities (ruby, emerald). Molecular orbitals explain the colors of organic compounds (indigo, chlorophyll) and charge‐transfer compounds (blue sapphire, lapis lazuli). Energy bands are involved in the colors of metals and alloys (gold, brass), of semiconductors (cadmium yellow, vermillion), doped semiconductors (blue and yellow diamond), and color centers (amethyst, topaz). Geometrical and physical optics are involved in the colors derived from dispersive refraction (rainbow, green flash), scattering(blue sky, blue eyes, red sunset), interference (soap bubbles, iridescent beetles), and diffraction (the corona aureole, opal).

Now you can imagine that all is not that simple.


Why are atoms with eight electrons in the outer shell extremely stable? Atoms that have eight electrons in their outer shell are extremely stable. It can't be because both the $s$ and the $p$ orbitals are full, because then an atom with 13 or 18 valence electrons would be extremely stable. ($d$ has 10, and 5 is also stable).

Why is it that atoms with eight electrons in the outer shell are extremely stable?

First, this isn't quite true.  It is true for the first row of the periodic chart (from lithium to neon). It is almost true for the second row (from sodium to argon.  But there are exceptions there.  Beyond that it really isn't true at all for the elements beyond the first two columns.

The reason for the increased stability for the first two rows lies in quantum mechanics.  Classically we can note that there are no $d$ electrons there.  Another ways of looking at it from a classical point of view is that the early elements are too small to allow too many other atoms or groups of atoms around them.  That tends to go away as you go down the periodic chart and the atoms get "fatter".   A typical example is chloroplatinic acid which has six chlorines around it.

Most transition metals also can have more than four groups around them as well.

I suspect that this isn't an exceptionally useful explanation.  As I said, the answer really lies in quantum mechanics.  In looking up "molecular orbital theory",  one reference can be found here.


Why is the force constant for double and triple bonds approximately double and triple of that of a single bonds? I am new to spectroscopy, and I am referring to Donald Pavia's Introduction to Spectroscopy. I came across this statement in the book, under the chapter Infrared Spectroscopy:

  $K$ is a constant that varies from one bond to another. As a first approximation, the force constants for triple bonds are three times those of single bonds, whereas the force constants for double bonds are twice of those of single bonds.

As far as I know, double bonds are not considered as two single bonds. Double bonds are rather one $\sigma$ bond, and one $\pi$ bond. Neither the bond energies of double bonds are twice the energy of corresponding single bonds, nor is the bond length of double bonds half of the single ones. Then how is the force constant doubles as the bond order doubles?

Force constant is not the same as bond strength. Remember that an approximate measure of bond energy from the bond as a simple harmonic oscillator would be $$E=\frac{1}{2}kx^{2}$$ where $x$ is the bond length. In this case, you would also need to account for differing lengths of single, double, and triple bonds to recover the bond energy trend.

Note that this is not how a quantum mechanical harmonic oscillator is normally described. There, you would have quantized energy levels of $(n+\frac{1}{2})\hbar\omega$, where $\omega = \sqrt{\frac{k}{m}}$. But since the higher order wave functions "concentrate" at the classical turning points, you might still be able to use the classical harmonic oscillator as a model for what is going on, with strong caveats.

For reference, consult: https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator


The effect of concentration on the rate of a homogenous reaction I have to do an experiment where I will find out what the influence of the concentration on the rate of a homogenous reaction is.
We are asked to create 2 solutions, respectively A and B:
$$\ce{KI + Na2S2O3 + Starch + water}$$
and
$$\ce{H2O2 + H2SO4 + water}$$
As long as A and B aren't added together, the reaction cannot start.
Now what I'm supposed to do is make 10 different A and B solutions and add these together and measure the time it takes for the solution to turn blue (Iodine + Starch as an indicator).

Now, my problem is that the following questions are also asked:
"Write down the different reactions that take place"
This seems dubious to me but might not be.
e.g. Does dissolving into ions also count as a reaction?
Does the Starch (for which I have no given chemical formula) react in solution A?
I've also read online that adding $\ce{H2O2}$ and $\ce{H2SO4}$ creates a "Piranha" solution ($\ce{H3O+ + HSO4- + O}$ or $\ce{H2SO5 + H2O}$, I've found both online but I don't know which is correct here.)
In conclusion I just don't know which reactions I should give (Everything that happens in A and B isolated from eachother or everything that happens when I add A and B together? Either way, I'm also not sure what the reactions would be because, for instance, I don't have any certainty on what the chemical composition of the Starch is.)
Is the reaction a multistep reaction or a single step reaction? Why? If it's a multistep reaction, write down the partial reactions
Assuming that adding B to A is "The" reaction, how do I find out if there are multiple steps or not? How do I find these steps?
Given that this question is asked already makes me think that it is in fact a multi-step reaction.

I've also read online that adding $\ce{H2O2}$ and $\ce{H2SO4}$ creates a "Piranha" solution [...]

This is true - in principle. In reality, Piranha solution is made from concentrated sulfuric acid and 30% hydrogen peroxide.
In the case of your experminent, you will most likely use diluted sulfuric acid and aqueous solutions of hydrogen peroxide with a much lower concentration. Consequently, the Piranha effect will not play a role in your experiment, neither is the oxidizing power of sulfuric acid relevant here.
On a side side note, treating a sample with conc. $\ce{H2SO4}$ is a method for the qualitative determination of iodide: Upon heating, one can observe the violet vapour of iodine.
In your case, you need the sulfuric acid to keep the pH value low. Under these conditions, $\ce{H2O2}$ acts as an oxidant:
$$\ce{H2O2 + 2 H+ + 2 e- -> 2 H2O}$$

Does dissolving into ions also count as a reaction?

No.

Does the Starch (for which I have no given chemical formula) react in solution A?

Starch is a mixture of two natural polymers (amylopectin and amylose), both made from D-glucose. Amylopectin doesn't play a role here, but it doesn't disturb either and it's just cheaper to let you work with the natural mixture.
It's the helical amylose that is relevant here - and it acts as a reagent.
What actually happens in the reaction is the following:
$$\ce{2 I- -> I2 + 2 e-}$$
$$\ce{I2 + I- -> I3-}$$
The linear triodide ion forms a strongly coloured inclusion complex with the amylose in the starch.


Why is SiO2 a solid while CO2 is a gas? I was under the impression that chemistry almost exclusively involves valence electrons because there isn't enough energy to strip off electrons located closer to the nucleus.

If that is true, and elements in the same period have similar properties because they have the same number of valence electrons, then why is $\ce{SiO2}$ a solid and $\ce{CO2}$ a gas? Surely a difference in mass by a factor of 2.5 can't be that big a deal.

Is it because of a van der Waals force difference, because silicon has extra electrons which result in compounds being formed from it being more symmetrical than those formed from carbon?

The reason why carbon dioxide is a gas and silicon dioxide is a solid is because their chemical structures are different.

Carbon dioxide is a linear structure with two double bonds between carbon and oxygen. It is a small molecule and non-polar with only weak bonds between the molecules. Hence it is a gas.

Silicon dioxide is not formed of small molecules. It consists of an infinite array of silicons where each silicon is bonded to four separate oxygens (and each oxygen is shared between two silicons). This creates a strong refractory solid (glass and sand are mostly silicon dioxide aka silica). So the same apparent overall formula doesn't describe the actual structure of the compounds at all. But the structures explain the difference in behaviour.

Of course this doesn't explain why silicon prefers to bond with four oxygens when carbon prefers just two. This is not completely simple and results from the relative bond strengths of carbon-oxygen bonds, carbon-oxygen double bonds and the equivalent bonds for silicon and oxygen. The simple version is that silicon oxygen bonds are strong relative to their double-bond equivalents whereas carbon-oxygen double bonds are strong relative to their single bond equivalents. Or, more precisely, if we could make a carbon-oxygen network solid with the equivalent structure to silica, it would tend to fall apart into carbon dioxide. If we could make silicon dioxide molecules, they would react with the release of energy to give silica.

Deeper explanations would need to look at why the relative strengths of double and single bonds turn out that way, but that would get into molecular quantum mechanics and would not be much more useful as an explanation.

The simplest explanation is the fact that the structures are different.


What is the dissociation percentage of a polybasic organic acid at a certain pH? For my Master's Thesis I am working with the dicarboxylic acid muconic acid $(\ce{H2MA},$ $\mathrm{p}K_\mathrm{a1} = 2.9,$ $\mathrm{p}K_\mathrm{a2} = 4.0).$ It thus dissociates into $\ce{HMA-}$ and $\ce{MA^2-},$ respectively.
I am aware of how to calculate the $\mathrm{pH}$ and the subsequent dissociation percentage when dissolving a weak acid in water. But I am wondering how you can calculate it the other way around. Say I dissolve $\pu{1 g}$ of the acid in water and adjust the $\mathrm{pH}$ to 4 with a strong acid such as $\ce{H2SO4},$ how do I know what the concentration is of each species $(\ce{H2MA},$ $\ce{HMA-}$ and $\ce{MA^2-})$ at that given $\mathrm{pH}?$
In general, how can I calculate the dissociation percentage of a weak polybasic organic acid when the pH is given?

Reused my answer to how-to-numerically-model-a-phosphoric-acid-titration-curve, edited for diprotic cases:
First, calculate the common denominator $CD$:
$$a_1 = [\ce{H+}]^2$$
$$a_2 = [\ce{H+}] \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$
$$a_3 = K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$
$$CD = a_1 + a_2 + a_3 $$
.. and then reuse it and its additive terms:
$$[\ce{H2A}]=c_0 \cdot \frac {a_1}{CD}= c_0 \cdot\frac{[\ce{H+}]^2 }{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$[\ce{HA-}]=c_0 \cdot \frac {a_2}{CD}= c_0 \cdot\frac{ [\ce{H+}] \cdot K_\mathrm{a1} }{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$[\ce{A^2-}]=c_0 \cdot \frac {a_3}{CD}=c_0 \cdot \frac{K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$c_0$ is the total concentration of respective particular acid forms.
For the general degree of "N-proticity", the expression terms span from $\ce{[H+]}^N$ to $K_1 \cdot ... \cdot K_\mathrm{aN}$


Autoprotolysis Constant of Water My chemistry textbook states that $[\ce{H3O+}][\ce{OH-}] = K_w$ in all aqueous solutions, which we know of as $K_w$. However, I was wondering why this was true. The textbook states that it is true because $K_w$ is an equilibrium constant, but why are we allowed to assume that the other ions take no part in this reaction?

To clarify further, consider the following reaction
$$\ce{HCl + H2O (l) <=> H3O+ + OH- + Cl- }$$
Of course this isn't balanced, but why doesn't the hydrogen ion on the left affect the equilibrium constant?

Your textbook is correct, the $K_\mathrm{w}$ expression is as follows:
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]=1.0\times10^{-14}$$

Since for every $\ce{OH-}$ there must be one $\ce{H+}$ ($\ce{H+}$ in water is synonomous with $\ce{H3O+}$) the mol ratio is 1:1, if moles are equal the molarities are equal. I'm going to substitute $x$ in for both
$$K_\mathrm{w}=[x][x]=[x]^2=1.0\times10^{-14}; x= 1.0\times10^{-7}.$$

So the molarity of $[\ce{OH-}]$ and $[\ce{H3O+}]$ are both $1.0\times10^{-7}$ That's a really really low amount, small enough to ignore.

However, a lot of reactions (especially in Organic Chemistry) you'll see a negative ion, for example, $\ce{Cl-}$ pulling the $\ce{H+}$ off of a water. So when $[\ce{OH-}]$ or $[\ce{H3O+}]$ are needed, water can be torn apart to make it.


What is the significance of the 2 and 4 positions in 2,4-DNP? (2,4-Dinitrophenyl)hydrazine, a popular laboratory reagent is used to identify carbonyl compounds by forming 2,4-dinitrophenylhydrazones, which are often red or yellow colored precipitates.

Why is the specific 2 and 4 positions that cause precipitate formation with carbonyls, and make it a useful reagent? Can (2,6-dinitrophenyl)hydrazine work this way?

Probably any DNP could work if you could get it.  But, because of the directing  character of substitutents in electrophilic substitution you inherently favor the 2,4 isomer:

the hydrazine group (or more accurately the chlorine atom which is converted to a hydrazine group) is ortho/para directing.
the nitro groups are deactivating and meta directing.

So you get the nitro groups meta to each other and ortho/para to the hydrazine group.  Therefore 2,4 DNP.


Solution, hardener and metal required to cast key duplicates I just watched this YouTube video for a DIY key duplication system and was trying to figure out what all the material components could be:

The "mold solution" (parts 1 and 2)
The "hardener"
The "low melting point ingots"

The mold gel has to be resilient enough to withstand molten metal (even if the melting point is low), but I can't figure out what materials would fit the bill. Any ideas?

From the video, the material looks like some kind of silicone. There aren't many other materials that have that translucent elastomeric quality. If that's the case, the larger of the two mould solutions will be mostly some kind of silane precursor such as SiOH. The second part will contain the catalyst for the polymerization. Depending on the exact chemistry used it could be perhaps tin-based. My guess for the "hardener" is that it's simply more catalyst. Two part silicones with longer working lives are more common than very rapid cure ones so adding more/stronger catalyst as the third part may be just to make it cure quickly. Many silicones can withstand temperatures of 200+ °C.

As for the low melting point alloy, there are a number of things it could be. Given that the application is keymaking and people touch keys with their hands, I hope it's safe to assume that the alloy doesn't contain lead or cadmium. The video says its melting point is around 90 °C. The two closest common alloys without toxic metals are 58% Bi, 42% Sn @ 132 °C and 66.3% In and 33.7% Bi @ 72 °C, from the wikipedia page on fusible alloys, so it's likely some alloy with similar composition to one of these. Bismuth and tin are far cheaper than indium, so it wouldn't surprise me if the alloy was some mixture of those two elements.


Why is mercury a "fancy proton"? I remember that one of my old organic chemistry profs referred to mercury as being a "fancy proton". I think it was in the context of oxymercuration reduction.

What might have been his rationale for offering this comparison?

It is a large, soft, Lewis acid. One can use protic acid and water to promote the same reaction, but the product distribution can sometimes be different.
 In the oxymercuration reaction Hg can form a pi complex with alkenes or alkynes, and these "bridged" cations will not undergo rearrangments to the extent that those produced by protic acids might.
 Bottom line - they're both essentially performing the same function in this reaction - activating the pi bond towards attack - but the oxymercuration is milder.