Why is sand, sand; and why is glass, glass; While both mainly contain Silicon Dioxide? I know that there is the famous process of heating material and converting them to glass; but what I don't know is, what is the chemical process of the creation of glass? Is it crystallization?

How could it result in so much visual differences?
What is the description of such a radical process?

Is it crystallization?

You are correct. The main difference is that sand is crystalline and glass is not—it is amorphous.

The main component (> 95%) of common yellow sand is quartz (the mineral whose composition is SiO2). Note that not all sand is quartz. There are white sands containing calcite (CaCO3) and black sand (containing various heavy minerals). But the most common sand is indeed quartz sand: SiO2.

Glass, the type you see in your everyday life, on the other hand, is not composed of pure SiO2. It has a bunch of other additives such as Na, K, B, and others. This is done to modify the properties of the glass and make it more suitable for human use. It doesn't matter much though for our discussion.

So if they are made of the same thing, why the difference? The answer is cooling rate. If you cool molten SiO2 slow enough, the atoms have enough time to organize themselves into a crystalline structure. In the case of pure SiO2, this is a network of SiO4 tetrahedra: One silicon atom surrounded by four oxygens. If it cools too fast, then the crystalline structure does not form. It may be completely amorphous, or form into a sub-microscopic array of SiO2 crystals in various structures (CT-opal for example).

What determines the cooling rate? Well, in the case of glass it is a matter of minutes. You've seen glass making: The glass is molten and very quickly it solidifies to a solid. In contrast, most of the quartz sand you're seeing is actually broken fragments of rocks called granite. This type of rock has abundant quartz in it, and it forms deep underground (as in 10s of kilometers) at very slow cooling rates. While a glass maker can take his glass and let it cool in the atmosphere or in water, molten silicate magma ("glass") deep in the Earth is surrounded by rocks that are in the hundreds of degrees. This slow cooling facilitates crystallization of the SiO2 into quartz rather than glass. How slow is this? At least tens of years, more commonly hundreds or even thousands of years. This is much slower than the seconds and minutes in glass making.

Slow crystallization of SiO2 in the Earth is not the only process. There are actually processes that can take molten silicate magma and cool it very rapidly. This commonly happens in volcanic settings: A volcano throws molten magma (again, made mostly out of SiO2) and throws it into the atmosphere where it can cool very rapidly. It's even faster if the magma erupts into water. The result is volcanic glass: sometimes called obsidian.

The difference in the visual appearance is actually not due to the presence or absence of a crystalline structure. It's mostly related to grain size. If you take window glass and shatter it so it's all small grains, it will look just like sand. On the other hand, if you take quartz, and make it big and smooth (just like window glass), it will look like this:

(By Parent Géry (Own work) [Public domain], via Wikimedia Commons)


Determining the reaction order of acetylsalicylic acid to salicylic acid with a spectrophotometer How can I determine the reaction order of the reaction acetylsalicylic acid + water to salicylic acid + acetic acid with use of a spectrophotometer?

I thought of measuring the absorbance spectrum of both substances individually, compare the 2 spectra, and then look at which wavelength the absorbance of acetylsalicylic acid exists whereas the absorbance of salicylic acid at that wavelength is negligible.

Lastly, use that wavelength to calculate concentrations of acetylsalicylic acid at certain time intervals and conclude the reaction order out of the change in concentration vs time. (for example making a graph on Excel for Concentration vs time -> linear = 0 order, exponential = 1st order)

Is this possible or am I thinking too hard/easy? Is there any other other way?

I think your reasoning is sound. However, a factor you may need to take into account is that you have a number of possible related species here, each of which has a different UV absorbance spectrum. You have acetylsalicylic acid, acetylsalicylate anion, salicylic acid, salicylate anion, and salicylate dianion (this one will probably have a markedly different UV spectrum from the others). The relative amounts of these species present would be affected by pH, so it would be important to buffer the reaction solution, to maintain constant pH.


Froth floatation process - Use of stabilizers in froth floatation method How exactly do the stabilizers affect the process of Froth floatation method? I mean how do they stabilize the froth in the process?

Bubbles tend to collapse when the liquid that makes up the walls drains away and the walls weaken. The walls can be strengthened by choice of surfactant (and addition rate) and by adding thickeners like water-soluble gums (e.g., xanthan) or polymers or glycerol (also for children's bubbles).

The small particles being floated may act as imperfections that break the walls more easily, so coating the particles with appropriate polymeric dispersants can make the (coated) particles more compatible with the (thickened) wall.

On the other hand, many industrial operations sometimes develop undesirable bubbles/foam which be eliminated by spraying a small amount of low surface tension silicone or non-viscous solvent like isopropanol on the top of the foam. Avoid these if you need to stabilize the foam.


Mechanism for oxidative cleavage of tertiary alcohols by NaOCl I came across this today

  An ester $\ce{C4H8O2}$ on reaction with $\ce{CH3MgCl} $ and hydrolysis gives an alcohol. The alcohol on reaction with $\ce{NaOCl}$ and $\ce{H3O+}$ gives acetic acid. Identify the structure of the ester.

But what I'm facing a problem with is the role of $\ce{NaOCl}$. My teacher said it can cleave and oxidize a tertiary alcohol, say tert-butyl alcohol, to acetic acid.

But I couldn't understand the mechanism. Any help will be appreciated.

Reaction of an ester with a Grignard reagent gives two products: (1) a tertiary alcohol coming from the carbonyl carbon and two equivalents of the Grignard reagent and (2) an alcohol liberated from the ester. I believe it's this liberated alcohol that is reacting with bleach, since a tertiary alcohol is typically resistant to oxidation. However, the liberated alcohol might be primary, secondary, or tertiary depending on the identity of R2.

The information given says that after treatment with bleach, acetic acid is formed. This tells us two things: (1) that liberated alcohol is a primary alcohol, since the oxidation goes all the way to an acid and (2) that liberated alcohol has two carbons, since acetic acid has two carbons. That implies that the liberated alcohol is ethanol. The starting ester has four carbons total, which leaves two carbons for the carboxy fragment. The original ester was ethyl acetate.


Electrolysis of calcium hydroxide in glycerol Calcium hydroxide is soluble in glycerol. Assuming anhydrous conditions, could such a solution be electrolysed to yield calcium metal, or would glycerol OH groups interefere with the process?

Calcium hydroxide is probably slightly soluble in glycerine due to tight chelation of Ca++ ion by the three hydroxyls of (probably 2) glycerine(s). But the hydroxyls will be present too, and on hydrolysis, will yield O2 plus H2O.

But beyond that, imagine you produce some calcium metal: it will be in glycerine, which has hydroxyl groups that will react with the Ca to produce H2. It could be a competition where some conditions allow production of measurable amounts of Ca metal, but it doesn't look promising. PEG 400 (polyethylene glycol with a mw of 400) has been used as a solubility enhancer for Ca(OH)2 (similar to glycerine), but has many fewer hydroxyls, and might do the job better. https://www.mdpi.com › pdf yields a pdf with more information.

Glycerine and PEG seem to be used to increase solubility of Ca(OH)2 in water rather than in the other solvent alone. I suspect increased viscosity is an issue.

Another way to get some calcium would be to use mercury as the cathode; then calcium will deposit and amalgamate (I'm pretty sure. I've done it with sodium and ammonium, but not with calcium.).


Rate-determining step of Claisen condensation If we use a strong base such as LDA or NaH, will the rate-determining step  change from enolate ion formation to nucleophillic acyl substitution step?

Let me get one thing out of the way: the hydride ion is not a strong base. However, once a hydride is protonated to form hydrogen the acid-base equilibrium breaks down which makes it seem like a much stronger base than it is.

Your actual question — whether the rate determining step shifts from enolisation to nucleophilic attack — is harder to answer as it depends primarily on the reaction procedure. It depends on how you perform the Claisen condensation and what you want to condense.

If, for example, you want to perform a hetero Claisen condensation (i.e. couple two different esters), then you would typically make sure to deprotonate quickly and kinetically first before adding the second ester as Zhe implied in a comment. For example, you might use a very strong base such as $\ce{BuLi}$ or $\ce{LDA}$ for complete deprotonation followed by a waiting period before adding another ester. In this case, obviously you are already dealing with a fully deprotonated product and the question becomes moot.

However, you might be interested in a homocoupling product. For that, you might decide to add $\pu{0.5eq}$ of a strong base to a solution of the ester. Here, it depends on whether you add the strong base rapidly, (meaning the acid-base equilibrium is practically immediately established, in turn meaning that now the attack becomes rate determining) or whether you add it slowly and dropwise, in which case you, the experimentor, are rate-limiting deprotonation and causing it to remain the rate-determining step.

Finally, in response to a comment you gave under your question: You cannot use a protic solvent if you intend to use a strong base. Protic solvents are characterised by a relatively high acidity, meaning they will be deprotonated long before the ester ($\mathrm{p}K_\mathrm{a} \approx 25$). This will completely inhibit the reaction.


What angle limits should apply to Graham condensers? I have seen Graham condensers set at an angle (perhaps as low as 30 degrees from horizontal).  I guessed this wouldn't be a problem because siphoning would clear the condensate.  But I recently saw an argument that Graham condensers should only be used at near-vertical angles.  Thinking harder this makes sense to me, because as soon as enough liquid condenses in the first loop to form a trap the condenser can't work on any other vapor.  Absent some vacuum on the receiving end the only way it could clear itself is if there happened to be enough vapor to eventually close a loop and start the siphon effect, or if enough pressure built up on the boiling end to push the liquid through which would be bad!

But I still see diagrams and photos of Graham condensers set at shallow angles.  Here's an example from the first few Google image results:

So when and why, if ever, is it considered a good practice to run a Graham condenser in a non-vertical orientation?

I have no idea why there are so many photos showing Graham condensers in a non-vertical position.  Maybe just because they look so cool and photos tend to be landscape.  The only time that might be safe is if there were a vacuum pulling to the collection side.

Graham condensers should be used vertically, precisely for the reasons you stated: Particularly because otherwise they will bubble, bump, and potentially explode.


Does mass affect rate of reaction? Suppose I am reacting HCl with CuCO3.
Experiment 1: I use 2g powdered CuCO3
Experiment 2: I use 4g powdered CuCO3
All other variables are kept constant.
Will this have any affect on rate of reaction? Will the 4g react faster because of more particles available for collision?

$1$)
Let's admit first that the reaction rate is proportional to the amount of reactants, in the usual way. Whatever the order $1$ or $2$ of the reaction, the rate constant ${k_{1} }$ or ${k_2}$ will not change if the amount of reactant is increased. But the rate of reaction ${r = - dn/dt}$ does increase when the amount of reactant ${n}$ increases, whatever the order.
For $1$st order, the rate is : ${r = - dn/dt = k_1 n}$. So the rate ${r}$ increases by increasing the amount of reactant ${n}$.
For $2$nd order, the rate of rection is : ${r = - dn/dt = k_2 n^2}$. Here too, the rate of reaction increases when the amount of reactant $n$ increases.
$2)$ As the reactant is a solid, the reaction rate may not be proportional to the amount of reactant (in mole). Let's suppose that it is proportional to the surface of the grains. Even so, the surface increases by increasing the amount of powder. As a consequence, the reaction rate does also increase.


Covalent vs Ionic bond in Silicon dioxide I was wondering why silicon dioxide tends to be a covalent bond instead of an ionic bond. Based on lewis structures, the octet rule seems to work for silicon dioxide so shouldn't it also be an ionic bond?

Because silicon and oxygen are nonmetals on the period, hence they form covalent bonds. For ionic bonding, it must be between a metal and a nonmetal.


How do Lewis acids interact with 1,4-diketones and amines to form an alkaloid? The following double cyclization was used to generate the ring system of the lycoanitine alkaloids in 2005. Demonstrate your understanding of this C−C bond-forming process by providing a mechanistic rationale (using arrows that represent
  electron movement) for the boxed reaction.

From what I know in my current synthetic organic chemistry class, this question appears to me like a Paal–Knorr reaction, but the product isn't a pyrrole. I don't even know where to start, and I'm not entirely certain how and where the Lewis acid interacts.

Ignore the boxed part, that's for me to do; I'm just curious to know how the carbons attached to the methoxy group on the tertiary amine are able to bond with the C2 carbons on the 1,4-diketone.

This is a variant of the Mannich reaction. The methoxymethyl groups on the amine are a form of captured Mannich intermediates or masked formaldehyde equivalents. The Lewis acid unmasks them by binding to the oxygen of the methoxy group, creating the functional equivalent of H2C=N+. These are reactive electrophiles which react with the active methylenes of the B-ketoester to give the alkaloid.

More on the Mannich reaction here


Verify my expected product of two acid/base reactions and does chosen solvent change reaction? My education in chemistry goes as far as what I research on the internet. I have two different acid/base reactions for which I've tried to predict the resulting product.  Can anyone here verify if I did this right or have an alternative prediction?

Reaction 1: p-toluenesulfonic acid (anhydrous) $\ce{C7H8O3S}$ reacts with potassium hydroxide $\ce{KOH}$. From this I calculated I should expect toluene $\ce{C7H8}$, potassium sulfate $\ce{K2SO4}$ and water $\ce{H2O}$:

$$\ce{C7H8O3S + 2KOH -> C7H8 + K2SO4 + H2O}\tag{1}$$

Reaction 2: p-toluenesulfonic acid (anhydrous) $\ce{C7H8O3S}$ reacts with
sodium bicarbonate $\ce{NaHCO3}$:

$$\ce{C7H8O3S + 2NaHCO3 -> C7H8 + Na2SO4 + 2CO2 + H2O}\tag{2}$$

If correct, then both of these reactions result in toluene. Does anyone think I got it right (or wrong)?

If the p-toluenesulfonic acid is dissolved in hexane, and I wish to neutralize it, neither base is soluble in hexane. If I first dissolved the base in ethanol (anhydrous) which is miscible with hexane (if I'm using the term correctly), will it neutralize the acid in the hexane, and does the fact that hexane and ethanol are involved change the reactions above? Any prediction if the ethanol was 95% instead of anhydrous?

So separating ethanol from hexane is easy if you want to keep the hexane -- just add water and separate. But if I actually have hexane/ethanol/toluene and add water and dispose of the ethanol/water solution, I'm assuming I'm left with a hexane/toluene mix. The toluene has a much higher polarity than hexane, but much lower one than ethanol. Toluene has a boiling point higher than water, and approx. 40 degrees higher than hexane.

Under this circumstance it appears removing toluene from the hexane and solute is not realistic, and would leave me distilling the toluene from my solute at a too high a temperature for the solute ($\pu{111^\circ C}$) whereas the boiling point of hexane is more like $\pu{68^\circ C}$.

If I'm correct, is there any thoughts on how to neutralize the acid in hexane without creating toluene which is not easy to separate from the hexane or am I stuck with toluene as long as I use this particular acid?

You will not get toluene in either reaction, you will get the potassium and sodium salts of p-toluene sulfonic acid for the respective reactions. This is because 1.) the toluene sulfonic acid is not a salt of toluene and sulfuric acid and 2.) aromatic rings (unless special conditions are applied) undergo electrophilic aromatic substitution. i.e. positive groups cause substitution on the ring, which sodium and potassium are not electrophilic ("electron loving") enough to effect such reaction. If you wanted toluene from these reactions you would have to use a dilute mineral acid in water and the $\ce{H+}$ ions from the acid would cause a substitution, removing the sulfonyl group and yielding toluene and sulfuric acid.

If you really need to neutralize this material, just add the mentioned bases as solids and allow them time to react. Bases don't need to be in solution to react with acids.


Why are rare earth metals and platinum group metals are often found clustered together in ores Rare earth and platinum group metals are often found clustered together in the earth's crust. Mining for platinum, for instance, also yields Rhodium and Ruthenium belonging to the same group. Likewise, rare earth elements such as Neodymium, Europium and Samarium also cooccur together in the same ore so much so that they are difficult to chemically separate.
It could be reasoned that it's the result of nucleogenesis where elements are formed consecutively based on their atomic number. While it might explain the first row and the second row of each group, where each metal is only one atomic number apart, it doesn't explain why metals from both rows are found together which are much further apart.
Alternatively, the similar chemistry of each group could explain the clustering. The two groups are the only group with this property. It fails to explain, however, how these metals found each other in a molten soup of heterogeneous elements. There may be some geological factors in the clustering, but it's unclear.
Why are the two groups of elements found clustered together?

The factors that generate mineral concentrations are complex and often only partly known
Introduction: geology is complicated
The one thing we can be very certain about is is that the distribution of minerals in the earth's crust has very little to do with the primordial origins of the component elements (that is where they came from in the early solar system and how they were originally generated). Most "heavy" elements are originally formed in the cores of supernovae and not in either the big bang or in normal stars.
The distribution of elements in the earth is mostly unrelated to the cosmic origins of elements because the earth's crust is not static but is frequently churned up by a variety of processes on a geological timescale. If we go back far enough in the history of the planet, everything was molten and this allowed some of the denser components to separate out before the surface cooled enough to be solid. The led to the core being mostly metallic (and consisting of mostly iron and nickel). Higher layers contain less dense minerals containing a lot of silicate minerals. At the top there is a thin layer, the crust, which is where we find useful minerals and it is even more concentrated in silicate minerals and even less dense.
But those early processes are mostly irrelevant to what we see in the crust. The crust is subject to a variety of processes that churn up the content, many of which concentrate specific components.
On a very large scale we have plate tectonics where large arts of the crust are both made and recycled on a geological timescale. To simplify greatly, new continental rock at emerges at one place (eg the mid atlantic ridge) and is consumed by subduction at other plate boundaries (eg the Andean belt in South America).
On a smaller scale (though often related to plate tectonic boundaries) volcanism takes molten rock from relatively deep in the crust and spews it out to the surface, bringing minerals of new compositions to the surface and altering the surrounding minerals by heat and pressure.
Also the surface is subject to weather which causes erosion (leading to both chemical and physical separation of the mineral content of rocks) and the uncovering of rock layers originally formed much deeper in the crust by the cooling of liquid mantle. Another product of erosion is sedimentary rocks where the things being eroded reform into new types of rock.
Plus life itself leads to the creation of some rocks. Some creatures collect carbonate minerals to make their protective bodies, for example, and if these concentrate when they dies, they may deposit layers which over time become new rock types (eg limestone).
To make things even more complicated these processes may interact. Volcanic heat or pressure caused by burial or major stresses may cause major changes to other types of rock, altering their mineral content in the process. Limestone may be recrystallised into marble; plant deposits may be transformed into coal or oil. And other separations are caused by related processes. Metamorphism is often associated with fluid flow which, depending on the composition of the fluid, may move specific minerals around, extracting specific components from some minerals and recrystallising them in voids left by the stresses associated with the metamorphic processes.
In short, things are pretty complicated in the crust and the dynamics will churn things up a lot over geological time. The one thing we can be sure of is that what we see now is not  primarily dominated by the primordial origin of the elements.
There are three major processes that concentrate specific ores
I'm going to greatly simplify some of the things that matter here but there are basically three important processes that concentrate things. Not that geologists can typically reach a consensus on what specifically happened with many deposits.
The three processes are:

separation due to erosion
separation due to differential crystallisation of liquid rocks
separation caused by metamorphic processes

Some major mineral deposits of economic importance (such as the Klondyke deposits in Canada) are caused by the first type of process. Some major platinum and related mineral deposits are from similar processes. What is basically happening is that the metals or minerals containing the metals are concentrated by flowing water because they are denser than the bulk of other minerals from the source rock. This is the same process that prospectors practice when panning for gold in rivers (fast flowing water tends to wash away the less dense clay minerals and leave the denser specs of gold). Over a geological time scale this process sometimes leads to very significant concentration of "heavy" minerals. The association depends on the density of the specific minerals and their specific presence in the rock being eroded. But, if the eroding rock contains many dense minerals, then the process can concentrate them all in the alluvial deposit.
But why would some rocks originally contain more than their fair share of specific minerals? One reason is the second geological process that leads to selective concentration of some minerals. This is that, as liquid rocks in the mantle cool, different minerals will crystallise from the mix at different times. a visible manifestation of this process can be seen in many of the polished granites used to decorate kitchen tops or the floors and walls of buildings. Granites consist of three key minerals: feldspars, quartz and mica each with very different mineral contents. The rock is usually formed deep in the crust when a large body of liquid rock cools. But the feldspars crystallise first, giving the large colourful crystals that make the polished surfaces so attractive. Sometimes the large crystals even show patterns of flow in the liquid source rock. The important general point is that the composition of the liquid changes as crystals form and this may concentrate some components. But this process of selective crystallisation is very general and explains why some particular concentrations exist.
The largest concentration of platinum group minerals is in a formation in southern africa called the Merensky "reef". The key concentrations of minerals appear to have been caused by separation as the rock crystallised (according to this book):

“Present opinion is that the Complex was intruded from a magma that was undergoing some differentiation, but was intruded in discontinuous phases. There is an overall trend in the mineralogy and chemistry of the basic rocks that is normal, and the layering of individual rock units is thought to be due to settling out of crystals according to density modified by connection currents flowing in the magma.”

The third process is also important for many valuable minerals. Metamorphic processes, for example, often involve heat and pressure but also hydrostatic processes. The pressure may crack existing rock leaving voids which are filled with high pressure liquid water containing various minerals that can selectively solubilise the contents of the rock and, later, deposit it in the voids. Once chemical property that can selectively separate gold and platinum group metals occurs when the water contains a lot of sulfur, selenium or tellurium and is a reducing environment (eg sufides rather than sulfates). Many precious metals will go into solution in such environments only to be deposited later from the solution as the state changes. Many gold deposits are like this as are some platinum group deposits.
Rare earth (lanthanide-containing) minerals are thought to be concentrated by the same sorts of processes but are, despite rare earths being pretty common in the crust (far more common that precious metals), much less common as economically viable deposits. This may be because the specific chemical process that concentrate them in the first place are less common. This is partially because they have large atomic radii and high charge and that means they don't fit well into the commonest silicate types of mineral. The key minerals they do occur in are carbonate and phosphate based. And these are often thought to occur near locations where continental crust is being consumed at a plate boundary where metamorphic processes may increase the presence of phosphate and carbonate in the resulting rock. See this presentation for some ideas. But they can also be concentrated by sedimentary processes where solutions eroded from source rocks are selectively absorbed into clay minerals elsewhere.
In general the reason why rare earths are found together is that they are chemically similar so there are few natural processes that will separate them (heck, the industrial processes are very hard and expensive).
Summary
There are many processes, both physical and chemical, that can cause concentrations of particular minerals in the earth's crust. Some of these depend on chemistry but others are simple physical processes. But geological chemistry is pretty messy and complicated and, even now, geologists can't always agree what particular process created a deposit. But there are broad types of process that probably contribute some of which depend on the similar chemistry of particular groups of elements. Rare earths are very chemically similar, for example, and precious metals have some chemical similarities (especially solubilisation in reducing environments with sulfur and other group 16 elements) that enable geological processes to concentrate them in some locations.
Some material here is from The Atlas of Economic Mineral Deposits which, though fairly old (1979) is worth a read to get a feel for how complex this subject is.


Chemoselectivity in reduction of 3′-nitroacetophenone When 3′-nitroacetophenone is treated with sodium borohydride, only the carbonyl group is reduced (top); however, when reacted with tin and hydrochloric acid only the nitro group is reduced (bottom). Why is this so?

I think that sodium borohydride is too weak a reducing agent to reduce the nitro group, but I am unsure as to whether this is correct. If Sn/HCl can reduce the nitro group then why doesn't it reduce the carbonyl too?

The key question is how the reduction by the two reagents proceeds.

For tetraborohydride the reduction proceeds as a standard nucleophilic addition of hydride ion to carbonyl group.

For tin/HCl the reduction proceeds as a series of one-electron transfers from metal surface mixed with protonation. It seems that the lowest free orbital in the compound belongs mostly to nitro group (which makes sense as it has a strong π-withdrawing effect, stronger than carbonyl group), so it is reduced to a nitroso group and then further via a hydroxylamine to an amine. It is possible to stop the reduction at intermediate stages under some conditions.

So the question is, why borohydride does not reduce the nitro group. It should be able to do so, as a strong reducing agent which is able to reduce iron from solutions. A possible answer is that extremely small size of nitro group with a net negative charge on the nitro group prevents formation of a stable adduct with hydride ion.


Molecular orbital diagram of a complex including an oxido ligand With highly oxidized metal centers, we expect that the hydroxido ligand's proton becomes acidic and will be lost to the surrounding media, creating an oxido ligand.

The oxo ligand is a $\pi$ basic ligand and we expect both $\sigma $ and $\pi$ donation into the metal center.

The metal's $\mathrm{t_{2g}}$ orbitals are lowered in energy by the presence of the oxido ligand's filled p orbital donating into the $\mathrm{t_{2g}}$. My question is regarding the symmetry of the ligand group orbitals. A p orbital does not have $\mathrm{t_{2g}}$ symmetry and as we know non-symmetry interactions are forbidden. Therefore I am struggling to understand the details of the oxido ligand.

My only explanation is that the complex's symmetry is reduced with the presence of an oxo ligand maybe from $O_\mathrm{h}$ to $C_4$ or similar. This would then change the symmetry of the $\mathrm{t_{2g}}$ orbitals on the metal center, but I really need some help explaining this!

Although chemists and chemistry students alike often indiscriminatorily use the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ descriptors for the d-orbitals of a d-block metal in coordination complexes, those designations are only truly correct in an octahedral environment. The vast majority of coordination complexes, however, are not octahedral but distorted octahedra, octrahedroids with non-symmetric ligands and others. This is also the case for oxido-ligand coordinated complexes that you are asking about.

To restrict the discussion to a single species (although it can easily be expanded to others) I will craft my answer explaining the situation in the vanadyl cation $\ce{VO^2+}$. I will discuss the pentaaquaoxidovanadium(IV) cation $\ce{[VO(H2O)5]^2+}$ as Ballhausen and Gray analysed.[1] I will label the direction of the oxido ligand $+z$ (in line with their scheme).

Constructing the molecular orbital scheme

The first important thing to note is that the complex indeed does not display octahedral $O_\mathrm{h}$ symmetry but $C_\mathrm{4v}$. That is great for us because it facilitates practically every step in the construction of molecular orbitals. A character table for the $C_\mathrm{4v}$ point group can be found in orthocresol’s collection (Note: MathJax-heavy link. Give it time to load). In the tables below I have given the different metal and ligand orbitals and their irreducible representations.

$$\textbf{Table 1:}\text{ irreducible representations of the metal orbitals in }\ce{[VO(H2O)5]^2+}\\
\begin{array}{cccc}\hline \text{irrep} & \text{s orbital} & \text{p orbital} & \text {d orbital} \\ \hline
\mathrm{a_1} & \mathrm{4s} & \mathrm{4p}_z & \mathrm{3d}_{z^2}\\
\mathrm{a_2} & & & \\
\mathrm{b_1} & & & \mathrm{3d}_{x^2 - y^2}\\
\mathrm{b_2} & & & \mathrm{3d}_{xy}\\
\mathrm{e} & & \mathrm{4p}_x, \mathrm{4p}_y & \mathrm{3d}_{xz}, \mathrm{3d}_{yz} \\ \hline \end{array}$$

$$\textbf{Table 2:}\text{ irreducible representations of the ligand (group) orbitals in }\ce{[VO(H2O)5]^2+}\\
\begin{array}{cccc}\hline \text{irrep} & \ce{O^2-} & \ce{$eq${-}H2O} & \ce{$ax${-}H2O} \\ \hline
\mathrm{a_1} & \mathrm{p}_z & {\sigma_1} \atop {\pi_1} & \mathrm{p}_z\\
\mathrm{a_2} & & \pi_2 & & \\
\mathrm{b_1} & & {\sigma_2} \atop {\pi_3} & \\
\mathrm{b_2} & & \pi_4 & \\
\mathrm{e} & \mathrm{p}_x, \mathrm{p}_y & {\sigma_3, \sigma_4}\atop{\pi_5, \pi_6, \pi_7, \pi_8} & \mathrm{p}_x, \mathrm{p}_y\\ \hline\end{array}$$

If required, I can supply images of the π and σ symmetric orbitals of the equatorial ligands once I am back at a PC that has ChemDraw (not before the 8th of January). It is not always fully obvious which orbitals interact with which ones; however, it is very obvious that the π2 ligand group orbital is nonbonding with respect to the $\ce{L\bond{->}M}$ bond.

In a second round, we need to consider the strength of interaction. The oxido ligand is much closer to the vanadium centre ($d(\ce{V=O}) \approx170~\mathrm{pm}; d(\ce{V\bond{<-}OH2}) \approx 230~\mathrm{pm}$) meaning that any interactions with that will be considerably stronger. (There are hints pointing towards an even more elongated axial aqua ligand, but that would be secondary.)

The resulting scheme you construct should show a strong interaction of the oxido ligands $\mathrm{e}$ orbitals ($\mathrm{p}_x$ and $\mathrm{p}_y$) and the metal’s $\mathrm{e}$ orbitals ($\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$). This can be interpreted as the oxido ligand donating both p-type lone pairs to vanadium; while the resulting bond is typically depicted as $\ce{V=O}$ a better depiction may be $\ce{{V^-}#O+}$. The other interactions of $\mathrm{e}$ orbitals are considerably weaker.

The orbital scheme as published

Ballhausen and Gray, who I mentioned previously, also included a quantitative scheme in their paper, albeit having omitted the ligand-metal π interactions with the exception of those of the oxido ligand.[1] Thus, their scheme is a little bit simpler than the one I would have drawn. The low-lying σ orbital is the oxido ligand’s; the orbitals labelled π are obviously the oxido ligand’s $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals. Due to the lack of aqua-π orbitals, the $\mathrm{d}_{xy}$ orbital appears unmodified. The $\mathrm{e}$ orbitals are raised notably, as is the $\mathrm{b_1}$ orbital ($\mathrm{d}_{x^2-y^2}$) due to the ligand’s σ interactions. Finally, the most destabilised metal d orbital is $\mathrm{a_1}$ corresponding to the $\mathrm{d}_{z^2}$ orbital.

The resulting scheme can be seen as an atypical Jahn-Teller distortion of an octahedral complex. In typical Jahn-Teller distortions, the $z$-ligands would be moved away resulting in a stabilisation of orbitals with $z$ contribution. Here, one $z$-ligand is moved closer resulting in a destabilisation of anything including a $z$ contribution.

Reference

[1]: C. J. Ballhausen, H. B. Gray, Inorg. Chem. 1962, 1, 111. DOI: 10.1021/ic50001a022.


What's happening at the beginning of a weak acid titration? At the beginning of a Weak Acid/Strong Base titration curve there is a sharp decrease in $[\ce{H+}]$ and a sharp increase in pH before the ratio between the weak acid and it's conjugate base becomes low enough that we have a buffer solution.

What's happening at that point?

My interpretation is that, given the balanced equation:

$$\ce{
CH3COOH~(aq) + H2O(~l) <=> CH3COO- ~(aq) + H3O+~(aq)
}$$

The addition of $\ce{OH-}$ would shift equilibrium to the left by Le Chatelier's principle making the newly generated $\ce{CH3COO-}$ react with $\ce{H3O+}$ to make more $\ce{CH3COOH}$, thereby increasing pH. After that, the $[\ce{H3O+}]$ is so low that the addition of the anion is not enough to keep shifting the equilibrium position to the left, so the remaining weak acid resists the changes in pH with its dissociation.

Does this make any sense?

Unfortunately this is not correct.

  $$\ce{CH3COOH~(aq) + H2O(~l) <=> CH3COO- ~(aq) + H3O+~(aq)}$$

If you have a pure acetic acid solution the equation you state governs the pH of the solution.

If you are now adding a strong base like $\ce{{}^{-}OH}$, you are removing the hydronium ion from the right hand side of the equation. According to Le Châtelier, the equilibrium adapts and shifts more to the right, i.e. more acetic acid dissociates. As you remove $\ce{H3O+}$ the pH increases.
Another way of looking at this is, that you remove acetic acid from the left hand side (similar to what you stated).
$$\ce{H3CCOOH + {}^{-}OH <=> H3CCOO- + H2O}$$
According to Le Châtelier, the equilibrium (in the first equation) has to shift to the left, to regenerate more acetic acid, removing once again $\ce{H3O+}$ and the pH increases.

In any case you choose to look at, you increase the concentration of acetate in the solution and another equilibrium comes into play.
$$\ce{H3CCOO- + H2O <=> H3CCOOH + {}^{-}OH}$$
This equilibrium is essentially responsible for raising the pH.


What does this curve over the C-B bond mean? I have come across this notation in a question I am trying to answer and am unsure of what exactly it means, can anyone clarify what this represents?

That ‘curve’ is a bracket, and the whole thing is a shorthand notation for ‘two of the same group’.

Thus, the molecule could be termed bis-(3-methylbut-2-yl)borane.


Assigning Peaks and Determining J Values for Mata-Disubstituted Benzene I need to determine as many J values as I can for 3-methoxybenzaldehyde. The peaks overlap a lot, and I don't see the expected ddd or dt pattern that I would expect for the hydrogen on carbon 6. The HSQC I was given has the clearest peaks for the H NMR, so I attached that. Maybe someone with more NMR experience can help me tease apart the signals?

There's almost nothing you can do in this situation, unfortunately. Because of strong coupling and other issues, aromatic peaks often do not conform to the so-called $n+1$ rule, which makes it highly non-trivial to extract coupling constants. It is probably possible to "fit" coupling constants to the multiplet shape by simulating the spectrum for a given set of $J$'s and seeing which set best matches the observed spectrum.

The HSQC trace is unfortunately not useful in this regard. At first glance, it may look simpler than the 1D proton spectrum you undoubtedly have. But that is because HSQC spectra are invariably recorded with low resolution in the proton dimension. The main reason for that is because of the high-power 13C decoupling that is performed during the acquisition. Acquisition times therefore need to be kept to approximately 100 ms, or there is a serious risk of damaging the spectrometer. On the other hand, a 1D proton spectrum doesn't utilise 13C decoupling, and can have acquisition times of several seconds.

Assuming the spectral width is constant, a shorter acquisition time means poorer resolution (this is a property of the discrete Fourier transform).


How does pH affect the spontaneity of biochemical processes? A decrease in $\mathrm{pH}$ increases the hydrogen ion concentration, thereby decreasing $Q,$ and decreasing Gibbs free energy as mathematically expressed:
$$\mathrm dG =\mathrm dG^\circ + RT\ln Q$$
This means that the more acidic the environment is, the more spontaneous the reaction is. But don't biochemical processes operate at approximately neutral $\mathrm{pH} = 7,$ so enzymes function optimally (does not denature)? But why does the equation favor spontaneity at (for example) $\mathrm{pH} = 3$ than $\mathrm{pH} = 7?$
For specific case, such as the biological oxidation of ethanol? Does it mean that it will thrive at acidic environment based on the equation? But how can we explain for the specificity of biological reaction operating at their optimal pH ranges?
For example, in the stomach it requires an acidic environment so the equation holds true. But what about for other biochemical processes that operate at neutral pH (the equation cannot explain it)? Or is there actually another equation to utilize to explain how spontaneous biochemical reactions can be?

Note that pH effect is reaction specific.
If a reaction is e.g. $\ce{A + B + H+ <=> C + D+}$ , or $\ce{A + B- <=> C + D + OH−}$,
then lowering pH supports the spontaneity.
If OTOH, the reaction is $\ce{A + B + OH-  <=> C + D−}$ , or $\ce{A + B+ <=> C + D + H+}$,
then lowering pH goes against the spontaneity.
It can be generalized by applying the Le Chatelier's principle, saying systems adapts to external changes by such equilibrium shifts that partially decreases the system chage caused by the external factor.
Lowering pH supports spontaneity of reactions consuming H+ and vice versa.
Increasing temperature prefers endothermic reactions consuming heat and vice versa.
Increasing pressure prefers reactions decreasing the total molar amount of gaseous components and vice versa.

Note that biological systems are much more complex than chemical systems.
Lowering pH may support thermodynamic spontaneity of some particular reaction. But an enzyme catalyzing its kinetic rate may start loosing its activity, get denatured, cells may stop producing the enzym, cells may stop overall life functions, or may die.


Is the largest number of molecules in 36 g of water or 54 g of dinitrogen pentaoxide? Recently my teacher asked us which of the following has the largest number of molecules? He gave us two options:

$\pu{36 g}$ of $\ce{H2O}$
$\pu{54 g}$ of $\ce{N2O5}$

I'm stuck, because both of them are having two moles of each of the respective molecules. And technically, the number of molecules equals the amount of substance multiplied by Avogadro's number.

So shouldn't $\pu{2 mol} \times N_\mathrm{A} = N(\text{molecules})$ be the same number of molecules for both of them?

But yet, the teacher said that water is the correct answer.

And I'm still not able to figure out why?

I'm stuck, because both of them are having two moles of each of the respective molecules.

Are you sure?

\begin{align}
M(\ce{H2O})  &= \pu{18 g mol^-1} &
\implies \pu{36 g} &\mathop{\hat{=}} \pu{2 mol} \\
M(\ce{N2O5}) &= \pu{108 g mol^-1} &
\implies \pu{54 g} &\mathop{\hat{=}} \pu{0.5 mol} \\
\end{align}


Strong acid equilibrium So a teacher at school told me that strong acids do not create an equilibrium in general, while weak acids do (thats the difference, strong acids namely ionize completely). However, I lately stumbled upon a passage reporting that when experimenting with extreme pH, so extreme concentrations of acids, strange things start to occur and strong acids are not able to ionize completely, resulting in an equilibrium. I would explain this by the extreme concentration -> less water molecules in the vicinity of acid to react -> less chance to ionize.
But then I thought, are strong acids in fact not also creating an equilibrium, but because they react so strongly, they immediately release their proton. The conjugate bases are extremely weak, but they should be able to pick up a proton right? They simply become and acid, and release it immediately again. Please clarify this entire situation for me, thanks already.

Whether a substance is a strong or weak acid/base depends on the solvent it's in. For example, $\ce{HNO_3}$ is a strong acid in water. This is partly because $\ce{HNO_3}$ has a relatively high tendency to donate a proton (the conjugate base $\ce{NO^{-}_3}$ is quite stable and therefore has a low tendency to accept a proton back), and partly because water has a certain basic character, accepting protons to create $\ce{H3O+}$. However, the same $\ce{HNO_3}$ in pure liquid $\ce{H3CCOOH}$ (also known as glacial acetic acid) is actually a weak acid. This change in character is largely due to the fact that $\ce{H3CCOOH}$ is a much, much weaker base than water, forming the $\ce{H3CCOOH^+_2}$ ion with significantly more reluctance. Even though $\ce{HNO_3}$ can easily lose its proton, $\ce{H3CCOOH}$ is even moreso unwilling to accept it, and thus $\ce{HNO_3}$ becomes a weak acid. An amusing situation arises in pure liquid $\ce{H_2SO_4}$. Adding $\ce{HNO_3}$ there, we find that $\ce{HNO_3}$ actually behaves as a weak base. $\ce{H_2SO_4}$ itself also has a high tendency to donate protons, and of the two possible reaction products from protonation, $\ce{H_2NO^+_3}$ is somewhat more stable than $\ce{H_3SO^+_4}$.

Hence, acidity is dependent on the solvent, not only on the solute. A strong acid such as $\ce{HNO_3}$ in aqueous solution at high dilution is completely dissociated, as we expect of a strong acid. However, if more and more water is removed, then the solution starts to behave differently. Even though $\ce{HNO_3}$ has a strong tendency to protonate water, you can have a solution that contains more moles of $\ce{HNO_3}$ than water. Then it becomes obvious that not all $\ce{HNO_3}$ molecules are deprotonated. Practically all of the water left in the solution will be in the form of $\ce{H_3O^+}$, but there will be left over undissociated $\ce{HNO_3}$, so most of the excess stays undissociated, except for a small amount that undergoes self-protonation, generating a bit of $\ce{H_2NO^+_3}$.

All in all, in extremely concentrated solutions, even strong acids/bases will not show 100% dissociation, and saying that "strong acids completely ionize" is just a rule of thumb which shouldn't be taken too far.


What is J in the rigid rotor model? In my class notes regarding the rigid rotor model, J is described as simply the angular momentum (replacing roe or p) and is also referred to as the angular momentum quantum number.

Are these two thing synonymous? If so in what way? If not, what do each mean and when is it appropriate to use which one?

TL;DR It depends on context and is unfortunately something that isn't going to change in the near future. However, the more you understand angular momentum, the easier it becomes to figure out the meaning from the context.

Quantum mechanical angular momentum is a tricky topic and it is not helped by some books which use confusing nomenclature. In QM, angular momentum is a vector quantity (much like in classical mechanics).* This is denoted with various letters;$^\dagger$ we will stick with J for the time being:

$$\text{angular momentum: }\mathbf{J} = (J_x, J_y, J_z)$$

Alternatively, you may also see vectors denoted with an arrow sign: $\vec{J}$.

All three components of the (QM) angular momentum are observable quantities. They are more commonly referred to as projections of the angular momentum onto an axis, which means the same thing. In any case, each one is associated with a Hermitian operator :

$$\text{ang. mom. projection operators: }\hat{J}_x, \hat{J}_y, \hat{J}_z$$

Apart from the $x$-, $y$-, and $z$-components of the angular momentum, the magnitude of the angular momentum is also an observable. This is similar to the length of the vector $\mathbf{J}$, and classically we would write $|\mathbf{J}|^2 = J_x^2 + J_y^2 + J_z^2$. The corresponding quantum mechanical operator is usually written without the boldface as

$$\text{ang. mom. magnitude operator: }\hat{J}{}^2 = \hat{J}{}_x^2 + \hat{J}{}_y^2 + \hat{J}{}_z^2$$

So, now, we have four unique operators: $\hat{J}{}^2, \hat{J}_x, \hat{J}_y, \hat{J}_z$. Because they do not all commute with one another, we usually focus on two of them, which are (conventionally) $\hat{J}{}^2$ and $\hat{J}_z$, and ignore the other two.

The quantum mechanical states of interest are those which are simultaneous eigenstates of these two operators, namely $\hat{J}{}^2$ and $\hat{J}_z$. It turns out that these eigenstates have the properties

\begin{align}
\hat{J}{}^2|\psi\rangle &= s(s+1)\hbar^2 |\psi\rangle \\
\hat{J}_z|\psi\rangle &= t\hbar |\psi\rangle \\
\end{align}

where $s$ is constrained to be a non-negative integer $0, 1, 2, \ldots$ and $t$ is constrained to be $-s, \ldots, -1, 0, 1, \ldots, s$. Different eigenstates will have different values of $s$ and $t$, which immediately suggests that we should label the eigenstates $|\psi\rangle$ by the values of $s$ and $t$ that apply to them. Constants such as $s$ and $t$ which can only take on certain values are called quantum numbers.

We can therefore associate the eigenstate $|\psi_{s,t}\rangle$ with the angular momentum $\sqrt{s(s+1)}\hbar$ (because the angular momentum squared was $s(s+1)\hbar^2$). We could write this as:

$$J_s = \sqrt{s(s+1)}\hbar$$

which says that the (magnitude of the) angular momentum $J$ of a state with quantum number $s$ is equal to $\sqrt{s(s+1)}\hbar$. In the rigid rotor system, we ignore $\hat{J}_z$ and instead look at another observable, which is the energy. The energy of a state with quantum number $s$ is:

$$E_s = \frac{\hbar^2}{2I}s(s+1)$$

where $I$ is the moment of inertia. So far, so good. But confusingly, instead of using innocuous labels like $s$, the convention is to use $J$ instead. And $t$ is replaced with $M$ (sometimes $M_J$), which at least is slightly less confusing.

\begin{align}
J_J &= \sqrt{J(J+1)}\hbar \\
E_J &= \frac{\hbar^2}{2I}J(J+1)
\end{align}

And therefore we have a situation where $J$ can mean several things, and it completely depends on context. To be extra clear, I'll colour the $J$ which means "the value of angular momentum" in red, and the $J$'s that refer to the quantum number in blue:

\begin{align}
\color{red}{J}_\color{blue}{J} &= \sqrt{\color{blue}{J}(\color{blue}{J}+1)}\hbar \\
E_\color{blue}{J} &= \frac{\hbar^2}{2I}\color{blue}{J}(\color{blue}{J}+1)
\end{align}

and the corresponding eigenvalue equations (recall that the operator for energy is always denoted $\hat{H}$, for the Hamiltonian):

\begin{align}
\hat{\color{green}{J}}{}^2|\color{blue}{J}\rangle &= \color{blue}{J}(\color{blue}{J}+1)\hbar^2|\color{blue}{J}\rangle \\
\hat{H}|\color{blue}{J}\rangle &= \frac{\hbar^2}{2I}\color{blue}{J}(\color{blue}{J}+1)|\color{blue}{J}\rangle
\end{align}

Here the green colour indicates an operator; the red in the previous equation was not an operator, but rather the quantity of the observable. Some texts will help you out by putting a hat on operators, but often it is omitted (either because authors are lazy or because it gets really cluttered after a while), so be careful!

* Technically a pseudovector.

$^\dagger$ $J$ is usually used for the total angular momentum of a system. In classical mechanics, there is only really one type of angular momentum, but in quantum mechanics there is orbital angular momentum and spin angular momentum.

Orbital angular momentum is the one which is much more closely related to the classical idea of angular momentum (for example, the orbital angular momentum operators are directly derived from $\mathbf{L} = \mathbf{r} \times \mathbf{p}$), and is typically denoted with $L$, not $J$. The quantum number is usually (helpfully) denoted $l$, although some books will also use $L$.

Spin angular momentum, on the other hand, is denoted with $S$. It has no classical analogue and has even weirder behaviour in that the associated quantum number $s$ can take on half-integer values, i.e. $0, \frac 1 2, 1, \frac 3 2, \ldots$

In quantum mechanics, the total angular momentum $J$ is the sum of both these two contributions. In the rigid rotor model we are not concerned with the spin of the particle(s), so $J$ is simply equal to the orbital angular momentum $L$, and there is no ambiguity about which is which. But if you start thinking about particles like nuclei or electrons, then it becomes important to distinguish between the two.


Can fluorine act as the central atom in interhalogen compounds? Why can't fluorine be the central atom in inter-halogen compounds?

A $\ce{F-F}$ bond is weaker than a $\ce{F-X}$ bond and thus, fluorine should be happy to form inter-halogen compounds. But, why doesn't it act as the central atom?

In my textbook, the reason given is because of the high electronegativity of fluorine atoms. But, how does that affect fluorine being the central atom or not?

If fluorine is the central atom, it can draw electrons from other less electronegative halogens and be happy. Does this lead to instability of the compound?

Why can't fluorine be the central atom in inter-halogen compounds?

First off, fluorine can be at the "center of things."  Examples would include the strongly hydrogen bonded hydrofluoric acid

and the very relevant example of the trifluoride anion

$$\ce{[F-F-F]^-}$$

The trifluoride anion example is critical as it demonstrates that fluorine can be hypercoordinated (hypervalent).  Of course, in this last example we've "tricked" the fluorine into occupying the middle position, so your question remains (slightly altered), why doesn't fluorine occupy a central position more often?

In order to answer this question, two general cases need to be considered,

The anionic situation $\ce{[X-F-X]^-}$ vs.  $\ce{[X-X-F]^-}$
The cationic situation $\ce{[X-F-X]^+}$ vs. $\ce{[X-X-F]^+}$

where X is Cl, Br or I.

In the case of the anion, Lewis dot structures suggest a negative charge on the central atom.

However, computational analysis of the electron distribution in the triiodide anion (see p. 156) suggests that the central atom is positively charged (+0.27 electron), while the terminal atoms are negatively charged (-0.63 electrons on each terminal atom).  Assuming that the same general trend would be applicable to mixed trihalide anions, then in the case of $\ce{[X-F-X]^-}$ we would be placing positive charge on highly electronegative fluorine atom instead of the larger, more polarizable (e.g. more capable of bearing a positive charge) bromine, chlorine or iodine atom.

In the case of the cation, similar arguments come into play.  Again, calculations for a variety of trihalide cations (mixed and homogenous) suggest substantial positive charge on the central atom (this time in agreement with Lewis Dot structures). The same arguments made above would operate here to destabilize structures with a central fluorine atom. This analysis also found extremely long (less stabilizing) bonds resulted when fluorine (or the most electronegative of the 3 atoms) occupied the central position.

Arguments that don't explain the infrequent occurrence of fluorine at a central position in inter-halogen compounds include:

the lack of "d-orbitals"; as other's have commented, this is an old, invalid argument; the existence of $\ce{F3^-}$ demonstrates that fluorine is capable of hypercoordinate bonding without the need for d-orbital involvement
inefficient 2p-3p overlap; $\ce{FCl}$ has a higher bond strength and lower heat of formation than $\ce{Cl2}$


Is enthalpy additive in mixing? When mixing two fluids of different composition at constant pressure when no heat can escape, is the enthalpy additive?

If not, how to compute the enthalpy of the mixture resulting from mixing two fluids with given composition after equilibrium is re-established?

The enthalpy is not additive under isobar, adiabatic conditions for non-ideal mixtures.

The change in the enthalpy is called the excess enthalpy (enthalpy of mixing). It can be computed with models such as the Margules, van Laar, Wilson, NRTL, and UNIQUAC.

These model give equations for the excess Gibbs energy. The excess enthalpy can be derived from these as described in 6.2 Fundamental Relations of Excess Functions in Molecular Thermodynamics of Fluid-Phase Equilibria (3rd Edition). (F'x already referred to this excellent book.)

One has to be careful with the conflicting sign convetions in the literature.


Reaction of Cu in hydrochloric acid bath with exposure to air over several months? I assume that copper will be slowly oxidized by oxygen, dissolved in water, forming copper(I) oxide (as stated here):
$$\ce{4 \overset{\pm0}{Cu} + O2 -> 2 \overset{+I}{Cu}_2O}$$

Reaction rate might be affected by the hydrochloric acid (cf. here)

And some of the copper could actually be converted to $\ce{CuCl2}$ (cf. here):
$$\ce{2HCl (aq) + \overset{\pm0}{Cu}(s) -> \overset{+II}{Cu}Cl2(aq) + H2(g)}$$

But what will happen to $\ce{Cu2O}$?

Will it degrade to $\ce{CuO}$, which then forms $\ce{CuCl2}$? That would mean that one should find higher concentrations of $\ce{CuCl2}$ in the solution, as time goes on.
Will it not react any further but instead get solved in the hydrochloric acid solution, because it is soluable in acids?
Or will something else happen?

Note: The concentration of hydrochloric acid solution is around 10 %.

Related Questions:
Copper Doesn't React with Hydrochloric acid
Copper and hydrochloric acid

In the presence of excess hydrochloric acid, copper oxides are not the immediate reaction product of copper metal and oxygen (e.g. from the air).  The air does oxidize the copper...

$$\ce{Cu + 1/2 O2  + 2 H+ -> Cu^{2+} + H2O}$$

... but the final oxidation product is copper(II) ion, not copper(I).  (Copper(I) is formed initially but is more prone to air oxidation than copper metal is, so it is relatively quickly converted to copper(II).)  Additionally, in the presence of chloride anions, the copper(II) ions are preferentially complexed by chloride.  The low concentration of hydroxide means that hydroxide complexes are not favored, and chloride is a better ligand than water for copper.

The actual complexes formed vary in their stoichiometry and their color.  $\ce{CuCl2(aq)}$, $\ce{CuCl3^{-}(aq)}$ and $\ce{CuCl4^{2-}(aq)}$ are all possible:

$$\ce{Cu^{2+} + n Cl- -> CuCl_n^{(n-2)-}(aq)}$$

$$n \in \{2, 3, 4\}$$

Nurdrage made a nice video showing how the reaction of $\ce{HCl}$, air, and copper metal leads to copper chloride.  It's the first of the three methods he presents for making copper chloride.  As he notes, the reaction is very slow.
$%edit$


Titration curve equation/function How can I create a graph of a titration curve (acid-base, strong or weak, not something specific), with an equation or a function. In other words, what is the equation that describes such a curve that could be used in graphing software to recreate the curve without actually needing any experimental values? Also, if they exist, what are the parameters of the function (Molarity? Volume? Etc.)

If $1$ liter $\ce{HCl} \pu{1 M}$ is gradually neutralized by adding $\pu{x mol }  \ce{NaOH}$, without change of volume, the pH of the obtained solution is given by : $$\mathrm{pH} = - \log(\frac{1-x}{2} + \frac{1}{2} \sqrt{(1-x)^2 + 4·10^{-14}})$$


How do you create primary amines from alcohols? By a primary amine I mean a compound where the hydroxyl group of the corresponding alcohol is replaced with an amine group.

In the conversion of primary alcohols to primary amines

$$\ce{R-CH2OH -> R-CH2NH2}$$

direct alkylation of ammonia normally is the last thing you want to do in the lab.

Under conditions where $\ce{OH}$ is a good leaving group, i.e. in acidic medium, the nucleophilicity of ammonia is reduced due to protonation. Moreover, there's little chance to prevent further alkylation of the initial product.

More promising strategies are:

Gabriel synthesis: Hydroxy-halogen-exchange, followed by reaction of the primary halide with phthalimide and subsequent hydrazinolysis
Mitsunobu reaction ($\ce{Ph3P, DEAD}$) of the primary alcohol using hydrazoic acid $\ce{HN3}$ as a nucleophile, followed by reduction of the alkyl azide
An (almost-one-pot) combination of the Appel and the Staudinger reaction

Heat the primary alcohol ($\ce{R-CH2OH}$) with $\ce{Ph3P}$  and bromotrichloromethane $\ce{CBrCl3}$ in benzene, then cool down to RT (room temperature) to yield $\ce{R-CH2Hal}$
Add $\ce{NaN3}$,tetrabutylammonium bromide and N,N-dimethylformamide and reflux again to obtain the azide $\ce{R-CH2N3}$, then cool down to RT, pour on water, extract with benzene and dry the organic layer
Add triethylphosphite $\ce{(EtO)3P}$ to the organic layer, stir at RT to yield the iminophosphorane $\ce{R-CH2N=P(OEt)3}$
Hydrolyze the iminophosphorane by refluxing with hydrochloric acid, then neutralize to release the amine


Why are isomers of parathion less active acetylcholinesterase inhibitors than paraoxon? Parathion itself has been found to be a very weak inhibitor of acetylcholinesterase. It normally requires metabolic activation and the conversion into paraoxon in the body to actually start exhibiting anticholinesterase activity. However, the study cited below found that active isomers of parathion, O,S-Diethyl O-4-nitrophenyl thiophosphate and O,O-diethyl S-4-Nitrophenyl thiophosphate (Compounds II and III in the cited paper, respectively), are less active inhibitors of AChE than paraoxon. I would have thought that the isomers of parathion would be more active than paraoxon due to thioethyl and 4-nitrothiophenyl being better leaving groups than 4-nitrophenyl, and thus the two active isomers of parathion would more readily form an adduct with the enzyme. I would presume that they also wouldn't need metabolic activation due to already having an oxygen atom double-bonded to the phosphorus. However, my prediction does not seem to hold true. So, why exactly are the active isomers of parathion (Reference numbers 7 and 8 in the table in the cited study) less active AChE inhibitors than paraoxon (Reference number 9 in the table in the cited study)?
Reference
Diggle, W. M., & Gage, J. C. (1951). Cholinesterase inhibitionin vitrobyOO-diethylO-p-nitrophenyl thiophosphate (Parathion, E 605). Biochemical Journal, 49(4), 491–494. doi:10.1042/bj0490491

Background
Paraoxon is an irreversible inhibitor. It forms an adduct with the catalytic residue Ser203 of AChE, along with release of p-nitrophenol as leaving group. The resulting covalent phosphoester conjugate is quite stable, but also reversible when using an appropriate oxime before the complex ages. See Franklin et al. 2016 for more details.
To facilitate the discussion below, here are the key molecules:

Regarding the OP's questions
Parathion is known to be metabolized to Paraoxon in the liver, likely though the action of cytochrome P450. (Chambers et al., 1991)
A phosphorothioester bond is indeed expected to be more susceptible to hydrolysis than a phosphoroester bond. Such a P-S labile linkage is ~106-fold more labile than P-O at neutral pH, according to Kuimelis-RG and McLaughlin-LW, 1995. However, Xu-Y and Kool-ET, 1998 noted the importance of unfavorable interactions of a studied enzyme with the sulfur atom, “because of the increased bond lengths or relatively poor hydrogen bond accepting ability of the sulfur relative to oxygen.” Indeed the P-S bond length is ~0.4 Å longer than P-O bond. (Knight et al., 1991)
The enzyme’s active site environment, that is beyond how good the leaving group is may help better understand the inhibition results of the molecules listed above.
Structural environment in AChE
The co-crystal structure of AChE with paraoxon (see Franklin reference above) deposited with RCSB as entry 5HF5.pdb provides the context of how the residual adduct binds to the AChE active site. The Franklin paper provides nice descriptive panels of this structure, but I thought I’d prepare the following image to help give more context of this interaction. In the following panel of the co-crystal structure, the residual adduct covalently linked to Ser203 is colored with cyan carbons:

a) The oxygen of P=O accepts a Hydrogen bond from multiple amides. Given that a sulfur atom does not form as good Hydrogen bonds as an oxygen, then this enzymatic environment will not provide as good stabilization to a sulfur as it does to an oxygen. In addition, as a sulfur atom has a larger radius than an oxygen atom, there may not be enough room for a sulfur atom. These two factors may begin to explain why parathione is not as good a substrate as paraoxon.
b) One of the ethoxy oxygen accepts a hydrogen bond from His447. Having a sulfur atom at that position, such as with Compound II, would bring it too close to the imidazole, leading to steric hindrance. In addition, such a sulfur atom would not form as a productive hydrogen bond as the oxygen. Similarly, the other ethoxy forms close van der Waals interactions with Phe297 and Trp236, there may not be enough room to accommodate a larger sulfur atom. This may begin to explain why Compound II is not as good as paraoxon.
c) Similar to the above, the environment may be too restrictive to accommodate Compound III for the productive formation of an adduct.
In summary, the immediate environment of the active site may be more optimized for paraoxon than the thiol analogs; this may help explain the published SAR of the active isomers of parathion.
Edited for clarity.


A metal, M, forms two fluorides containing 40.43% F and 50.44% F by mass. Identify the metal A metal, $\ce{M}$, forms two fluorides containing 40.43 % $\ce{F}$ and 50.44 % $\ce{F}$ by mass. Identify the metal.

I get that you can find out how much of the metal reacts with 1 mole of fluorine atoms in the 40.43 % one. This can be calculated to be 27.99 g.

Then from that you can work out that 28.49 g of fluorine reacts with the metal in the 50.44 % one and that that is about 1.5 moles of $\ce{F}$.

I'm a bit lost after that. How might you continue from here?

So the law of multiple proportions conveys that the fluorides contain fluorine in a 1.5:1 or 3:2 ratio. This means the possibilities are

$\ce{MF2}$ and $\ce{MF3}$, or
$\ce{MF4}$ and $\ce{MF6}$

Since there is twice as much fluorine, the mass of $\ce{M}$ in $\ce{MF2}$ must be 56 g. The next one is 112 g. If there are no errors in what was said, the metal must be iron or cadmium. What makes one a better candidate over the other?

You need to use the general concepts of the law of definite proportions and the law of multiple proportions to solve this:
Sample 1 is 59.57% $\ce{M}$ and 40.43% $\ce{F}$.
Sample 2 is 49.56% $\ce{M}$ and 50.44% $\ce{F}$.
Assume $\ce{M_{x}F_{y}}$ for sample 1 and $\ce{M_{a}F_{b}}$ for compound 2.
Let $m$ by the molar mass of $\ce{M}$. We have
$$\frac{19.00y}{mx + 19.00y}=\frac{40.43}{100}$$
and
$$\frac{19.00b}{ma + 19.00b}=\frac{50.44}{100}$$
We can rearrange the equations to:
$$m\frac{x}{y}=28.0$$
and
$$m\frac{a}{b}=18.7$$
Note that $$\frac{28.0}{18.7}\approx\frac{3}{2}$$
so we can start by trying:
$$m\frac{3}{1} = 28.0$$
and
$$m\frac{2}{1} = 18.7$$
This is weird because the compounds would be $\ce{M3F}$ and $\ce{M2F}$.
Fortunately, as long as the proportions work, it's fine, so divide both fractions by 6:
$$m\frac{1}{2} = 28.0$$
and
$$m\frac{1}{3} = 18.7$$
This gives $m=56.0$, fairly close to the molar mass of iron which is $55.9$.
Cadmium is a bad candidate, since that would be $\ce{CdF4}$ and $\ce{CdF6}$.


Hydrolysis of benzalchloride What should be the mechanism for hydrolysis of benzalchloride to benzaldehyde?

Here

B denotes Brønsted base
and HA denotes acid


Why do the first and second ionizations of a diprotic acid produce the same hydronium ion concentrations in solution? I was going through Oxtoby Chemistry when I got confused by the following passage in the textbook. It said that "two simultaneous equilibria are involved in the ionization of a diprotic acid such as carbonic acid":

I understand the chemical species for both the first and second ionizations barely dissociate, which means the hydronium ion production is very little.

However, what makes hydronium ion production identical (i.e. important point #1)? Shouldn't the second ionization occur to a lesser extent than the first ionization? Thanks.

This is saying both the $\ce{H2CO3}$ and the $\ce{HCO3-}$ are both donating $\ce{H+}$ to the water to form $\ce{H3O+}$. What it means when it says "The $\ce{[H3O+]}$ in the two ionization equilibria are one in the same" is to say that both of these reactions are occurring simultaneously in the same system, and that there is only one measure of the concentration of $\ce{H3O+}$ for the system.

It is not saying that both species are donating $\ce{H+}$ to the same extent. As a matter of fact, the $\ce{H+}$ being donated by $\ce{H2CO3}$ will keep the the $\ce{HCO3- + H2O -> CO3^{2-} +H3O+}$ shifted almost entirely to the left, meaning $\ce{HCO3-}$ will be donating an extremely small fraction of the $\ce{H+}$.


Thin Layer Chromatography & Stationary Phase There are two types of stationary phases used in thin layer chromatography. One stationary phase is made of aluminium oxide. Which of the following molecules would be expected to have the smallest $R_f$ using aluminium oxide as the stationary phase on the glass slide?
(a) $\ce{CH3CH2CH2CH2CH2COOH}$
(b) $\ce{CH3CH2CH2CH2CH2OH}$
(c) $\ce{CH3CH2CH2CH2NH2}$
(d) $\ce{CH3CH2CH2CH2CH3}$

$\ce{AL2O3}$ is a non-polar stationary phase, therefore if the molecule is also non-polar it would display high adsorption  to the stationary phase, and therefore a smaller $R_f$ value.
The answer is (a).
Isn't (a) polar?
Wouldn't this increase the $R_f$ value, because it doesn't adsorb to the non-polar stationary phase? Can someone please explain, correct me?

Aluminum Oxide is more polar than Silica Gel and polar compounds do not migrate far from the starting point.

Thin Layer Chromatographic Analyses

7. Thin-Layer Chromatography


In an emission spectrum, the limit of convergence at higher frequency corresponds to the first ionization energy This is one of the syllabus points in the IB syllabus 12.1. We are advised to use the formula for the energy to remove one electron:

$$E = hν$$

which apparently is the same as

$$E_2 - E_1$$

where $E_2$ is the energy of the convergence limit and $E_1$ is the energy of the energy level where you remove the electron from.

However, isn't $ν$ the frequency of the emitted radiation when the electron falls from $E_2$ to $E_1$ (so with energy same as the energy difference between $E_2$ and $E_1),$ and not the frequency with the energy $E_2,$ i.e. the convergence limit?

I think I've misunderstood something and I'd appreciate if someone could explain this to me. Please use simple language because I'm only in high school.

This is an excellent question! Just to elaborate upon what @porphyrn and @M. Farooq have said, imagine that the universe is entirely empty except for one sodium atom and one photon (particle of light) that has a wavelength of $\pu{242 nm}$. If the photon bumps into the sodium atom, it might just bounce off: this is called scattering. But another big possibility is that the photon will be absorbed (i.e., ‘taken in’, as it were) by sodium’s 3s valence electron. The energy of a $\pu{242 nm}$ photon is $E_{\pu{242 nm}} = h\nu = \frac{hc}\lambda$, where $h = \pu{6.626 x 10^{-34} Js}$ and $c = \pu{2.9979E8 m/s} = \pu{2.9979E17 nm/s}$. This energy is $\pu{8.21E-19 J}$.

Then the sodium’s 3s electron would have just enough energy to escape, i.e., ‘break free’, from the sodium atom. The result would be a sodium ion, $\ce{Na+}$, and an electron, $\ce {e-}$, that were free from one another: the sodium atom has been ionized, i.e., has become an ion, so the energy of the $\pu{242 nm}$ photon is the ionization energy of the 3s electron in sodium. Using the $h$ and $c$ values, the required ionization energy, for one sodium atom, is $\pu{8.21 x 10^{-19} J}$. For a mole of sodium atoms, multiply by Avogadro’s number, resulting in $\pu{494 kJ/mol}$.

So where does the infinity enter the picture? Well, a positively charge sodium ion and the electron that escaped from it attract: opposite charges attract. The Coulombic force of attraction is an inverse square law, similar to gravity, so total absolute freedom would mean that the electron was infinitely far away from the sodium ion and had zero kinetic energy, i.e., zero energy of motion. But, as a practical matter, the separation can be relatively small, not infinity! As a practical matter, it is known experimentally that an isolated sodium atom, doing its thing in a vacuum, can be ionized by shining $\pu{242 nm}$ light on it.

In your example, suppose a sodium ion, in vacuum, happened to recombine with a free electron. Assume the sodium ion and electron are moving very slowly relatively, so there is negligible excess kinetic energy: we don't want the electron to 'harpoon' the poor sodium ion! The electron could drop down into any empty orbital of the sodium ion. Potentially, that is an infinite number of possibilities, hence M. Farooq's answer (and sweet Grotian diagram). The result would be release of a photon with wavelength longer (numerically 'larger') than $\pu{242 nm}$. This excited (it has more energy than a 'ground state') sodium atom must eventually drop to the 'ground state' (the only state where it is indefinitely stable) by emitting one or more additional photons or perhaps getting 'mugged' of its excess energy, via collision with something else. This is collisional deactivation.

But what if the electron dropped all the way down, in one fell swoop, to the 3s orbital? Then the resulting sodium atom would be in its 'ground state', i.e., lowest energy (and only indefinitely stable) state, and the emitted photon would be at $\pu{242 nm}$.


What is the meaning of overbased additive in lubricants? From the google search that I did, I do not understand what overbased means.

Any comment,link or answer would be useful

In detergents, by overbasing, it is meant that a stoichiometric excess of the metal be present over that required to neutralize the anion of the salt. The excess metal from overbasing has the effect of neutralizing acids which may build up in the lubricant. A second advantage is that the overbased salt increases the dynamic coefficient of friction. Preferably, the excess metal will be present over that which is required to neutralize the acids.


Why does Carnot Engine have the curves as isothermal and adiabatic? Why exactly does the Carnot engine have two isothermal and two adiabatic curves? (My book says that “a process is quasi-static if the temperature difference between the system and the reservoir is infinitesimal. Hence it is isothermal. For the connecting two isotherms ;only reversible adiabatic is suitable. If(say) it were isochoric then we shall need a series of reservoirs in the temperature range defined by the two reservoir and the sink” .I don’t get it. Please help.

By definition, a Carnot cycle is comprised of 2 reversible isotherms and 2 reversible adiabats.

Regarding your question about isochoric heating or cooling, if you are going to carry this out reversibly, then you can't do it with just one constant temperature reservoir.  For the heating or cooling to be reversible, the temperature difference between the system and the reservoir(s) must be tiny over the entire path.  To do this, you need a sequence of constant temperature reservoirs running from the starting temperature to the ending temperature.  After the system has equilibrated with each reservoir in the sequence, you move on to the next reservoir.  That way, the temperature difference between the system and the reservoir is tiny along each step of the process path.


Does the prefix of an inverted unit also get inverted? What is the energy of radiation that has a frequency of $\pu{2.51 \times 10^11 ms-1}$?
(a) $\pu{1.66 \times 10^-19 J}$ (supposedly correct)
(b) $\pu{1.66 \times 10^-22 J}$
(c) $\pu{7.92 \times 10^-37 J}$
(d) $\pu{1.66 \times 10^-25 J}$

My argument was that: since $\pu{Hz} = \pu{s-1}$, then $\pu{ms-1} = \pu{mHz}$. So, I divided $2.51 \times 10^{11}$ by $1000$ and solved the question normally (using $E = h\nu$). But my professor said that since $\pu{m}$ is a prefix, it should follow whatever is in front of it, so $\pu{ms-1}$ will become $1/\pu{ms}$, then you multiply it by 1000 to change it to Hz. Is that the correct way to do it?

You should read $\pu{ms-1}$ as $(\pu{ms})^{-1}$ and not as $\mathrm{m}(\pu{s-1})$. Your teacher is correct.

  [...] since $\pu{Hz} = \pu{s-1}$ then $\pu{ms-1} = \pu{mHz}$

This is incorrect.

If something happens once per hour, it happens less often than when it happens once per minute. This is because an hour is longer than a minute. In a similar manner, if something happens once per second, it happens less often than when it happens once per millisecond. This is because a second is longer than a millisecond.

So $\pu{ms-1} = \pu{kHz}$, i.e. the frequency of something happening every millisecond is larger than the frequency of something happening every second.


How do I calculate lattice points per unit volume? My professor was explaining simple cubic, body centered cubic, and face centered cubic lattices and I didn't quite understand the calculations below:

Simple Cubic: $8 \cdot 1/8 = 1$ lattice point per unit volume
Body Centered Cubic: $(8 \cdot 1/8) + 1 = 2$ lattice points per unit volume
Face Centered Cubic: $(8 \cdot 1/8) + (6 \cdot 1/2) = 4$ lattice points per unit volume

Can someone explain the reasoning behind these values in the equations?

Atoms on the corners, edges, and faces of a unit cell are shared by more than one unit cell.

An atom on a face is shared by two unit cells $\implies$ half an atom belongs to each unit cell.

An atom on the edge is shared by four unit cells

and an atom on the corner is shared by eight unit cells

What does this have to do with the equations?

Well lets say we have an sodium atom that is crystallized in a simple cubic unit cell, then there would be an sodium atom on each of the eight corners of the cell. However, only $\frac{1}{8}$ of these atoms can be assigned to a given cell. Hence the simple cubic structure is

$$8 \, \text{(corners)} \times \frac{1}{8}=1 \, \text{atom}$$

if sodium formed a body centered cubic structure, there would be two atoms per unit cell (as seen in the top diagram). It isn't shared with an other cells. Therefore the body-centered cubic structure is given by

$$8 \, \text{(corners)}\, \times \frac{1}{8} + 1 \, \text{(body)}= 2 \, \text{atoms}$$

now finally if sodium was a face-centered cubic structure the six atoms on the faces of the unit cell would contribute three net sodium atoms, for a total of four atoms per unit cell. Hence the face-centered cubic structure is

$$8 \, \text{(corners)} \times \frac{1}{8} + \Big{(}6 \, \text{(faces)} \times \frac{1}{2}\Big{)}= 4 \, \text{atoms} $$

Hope this helps


Half-life and shelf-life of second-order reaction For integrated rate laws, I have attempted to find the shelf-life and half-life of second-order reaction:

However, the answers I have obtained (which are labelled with a red cross) are very different from the correct answers (bottom-right corner). I am wondering which parts were done incorrectly.

Your mathematics is correct until you calculated for specific cases. Let's go back to your $n$th order version:

$$\frac{[\ce{A}]^{1-n}}{1-n} = -k_nt + \frac{[\ce{A}]_\circ^{1-n}}{1-n} \ \text{where } n \ne 1 \tag1$$

At this point, since you are working on second order kinetic, it is easy if you substitute $n = 2$ in equation $(1)$:

$$\frac{[\ce{A}]^{-1}}{-1} = -k_2t + \frac{[\ce{A}]_\circ^{-1}}{-1} $$

Once simplify, it becomes:

$$\frac{1}{[\ce{A}]} = k_2t + \frac{1}{[\ce{A}]_\circ}   \tag2$$

For self life, substitute $t = t_{10\%}$ and ${[\ce{A}]} = 0.9{[\ce{A}]_\circ}$ in equation $(2)$:

$$\frac{1}{0.9[\ce{A}]_\circ} = k_2t_{10\%} + \frac{1}{[\ce{A}]_\circ}  $$

$$\therefore \; t_{10\%} = \frac{1}{k_2}\left(\frac{1}{0.9[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.9} - 1 \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{9} \right) = \bbox[yellow]{\frac{0.11}{k_2[\ce{A}]_\circ}}$$

Similarly, for half life, substitute $t = t_{50\%}$ and ${[\ce{A}]} = 0.5{[\ce{A}]_\circ}$ in equation $(2)$:

$$\frac{1}{0.5[\ce{A}]_\circ} = k_2t_{50\%} + \frac{1}{[\ce{A}]_\circ}  $$

$$\therefore \; t_{50\%} = \frac{1}{k_2}\left(\frac{1}{0.5[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.5} - 1 \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{0.5}{0.5} \right) = \bbox[yellow]{\frac{1}{k_2[\ce{A}]_\circ}}$$


Products of the o-Toluidine oxidation with glucose oxidase and peroxidase Could anybody describe which product is formed in the following reaction (o-Toluidine oxidation)?

$$\ce{H2O2 + o-Toluidine_{uncolored} - Peroxidase -> H2O + 1/2 O2 + o-Toluidine\; Oxidate_{colored}} $$

Note: Hydrogen peroxide is one of the products of glucose oxidation to glucolactone by glucose oxydase.

The product most likely is 2-methylnitrobenzene (3), which means that the amino group ($\ce{-NH2}$) in o-toluidine (1) is oxidized to a nitro group ($\ce{-NO2}$). The oxidation proceeds via 2-methylnitrosobenzene (2).

The formation of 2-methylnitrobenzene can be detected photometrically at $\lambda$ = 400 nm, where the aniline does not absorb.


Resonance - Legal Moves My teacher insists that we can push electrons either from a bond to an atom's valence, or vice versa. Bond to lone pair; lone pair to bond, he insists.

However, what about bond to bond?

Is there a pedagogical reason he might be avoiding bond-to-bond movement? This is a legal move.

Or does he simply not see bond-to-bond movement as a separate entity; could bond-to-bond electron pushing be construed as simply a bond to lone pair and lone pair to new bond move?

Resonance is a concept that originates from the incapability of the original Lewis structure to describe delocalised systems. The formulation of these resonance structures are a representation of extreme states that are possible. The true bonding pattern lies between all possible resonance structures.

Consider benzene, where we, among others, consider two major resonance structures. In fact the electron structure is well known to be a delocalised $\pi$ system, therefore better described by the right figure.

In this particular case, electrons appear to be "pushed" from one bond to another. This is not the case, as electrons are occupying one (or more degenerate) orbitals the whole time.

So when it comes to resonance structures, there are no clear rules, what is allowed and what is not allowed, however it has to be a possible representation of the actual bonding pattern. Resonance is a mere concept of describing a very complicated bonding pattern in simpler terms.


What does a change in temperature do to the concentration of gases at equilibrium? I am confused about a specific part of Le Chatelier's Principle. My question revolves around a homework assignment question:

$$\ce{2SO2(g)  + O2(g) <=> 2SO3(g)} \qquad\Delta H^\circ = -198.2\:\mathrm{\frac{kJ}{mol}}$$
Comment on the changes in the concentrations of $\ce{SO2}$, $\ce{O2}$, and $\ce{SO3}$ at equilibrium if we were to increase the temperature of the system.

Now I understand that increasing the temperature of an exothermic process will cause the equilibrium to shift towards the left (in the direction of the endothermic reaction); however, what does the temperature do to the concentrations of the gases?  I think it does nothing, but I am not sure.

You are correct in saying that the increase in temperature causes the equilibrium to shift to the left, in the direction of the endothermic reaction. As the reaction continues in the endothermic direction, the concentrations of $\ce{SO3}$ will decrease in favour of the evolution of more $\ce{SO2}$ and $\ce{O2}$, due to the additional heat available.


Effect on mass when decreasing pressure in equilibrium reaction For example, if we have the reaction:
$$\ce{A + B <=> C}$$
And the volume is decreased, the pressure would increase. Hence, according to Le chatelier's principle, the system will partially oppose the change by favoring the forward reaction to produce more $\ce{C}$. I have a few questions:

Would the mass of $\ce{C}$ increase?

I think yes, since the forward reaction is favored

Would the concentration of $\ce{[A]}$ be greater when compared to the original concentration of $\ce{[A]}$?

I think yes, since the volume is decreased.

Would the amount of $\ce{A}$ be reduced?

I think yes, since it's being consumed for the forward reaction.

This is an approach to Question 2.
The reaction equilibrium condition for the system is:  $$\frac{V}{RT}\frac{n_C}{n_A n_B}=K_P\tag{1}$$Let $n_{A0}$, $n_{B0}$, and $n_{C0}$ be the number of moles at equilibrium for each species when the volume is equal to $V_0$.  Then we have
$$\frac{V_0}{RT}\frac{n_{C0}}{n_{A0} n_{B0}}=K_P\tag{2}$$.  For a small change in volume dV, we can write the new number of moles of A, B, and C at equilibrium as
$$n_A=n_{A0}-dn\tag{3a}$$
$$n_B=n_{B0}-dn\tag{3b}$$
$$n_C=n_{C0}+dn\tag{3c}$$Combining Eqns. 1-3 yields:$$\left(1+\frac{dV}{V_0}\right)=\frac{(1-dn/n_{A0})(1-dn/n_{B0})}{(1+dn/n_{C0})}$$And, linearizing this equation with respect to dn yields:$$\frac{dV}{V_0}=-\left(\frac{1}{n_{A0}}+\frac{1}{n_{B0}}+\frac{1}{n_{C0}}\right)dn$$This equation indicates that if dV is negative, then dn is positive, and the reactants decrease while the product increases.
If we apply this same methodology to the concentration of A, we obtain
$$dC_A=-\frac{n_{A)}}{V_0}\left(\frac{dn}{n_{A0}}+\frac{dV}{V_0}\right)$$Combining this with the previous equation then yields:
$$\frac{dC_A}{dV}=-\frac{n_{A0}}{V_0^2}\left[\frac{1/n_{B0}+1/n_{C0}}{1/n_{A0}+1/n_{B0}+1/n_{C0}}\right]$$The term in brackets is always positive, so, when the volume decreases, the concentration of A increases at all points along the equilibrium contour.  This conclusion is independent of the specific value of the equilibrium constant.


What is the final temperature of each substance upon absorbing 2.25 kJ of heat? Suppose that $\pu{28 g}$ of silver is initially at $\pu{29.0^\circ C}$. What is its temperature upon absorbing $\pu{2.25 kJ}$ of heat?

My work: $2250 = 0.240 \times 28 \times t - 29$

Solve for $t$ and I get $\pu{340 ^\circ C}$. But the correct answer is $\pu{370 ^\circ C}$ and I have no clue how to get there.

You are almost correct, however, to calculate these values you need to put a bracket around he $t-29$ term (which is the $\Delta T$ term from $Q = mc \Delta T$). Hence, from your working out step:

$$2250 = 0.240 \times 28 \times (t - 29)$$

You'll need to expand the brackets.


How would you make/produce sulfuric acid with tools only the romans had? I have been reading about sulfuric acid and its other compounds the romans had, yet I don't understand how they where getting sulfur rich compounds and how they convert it to sulfuric acid. Was it from the volcanoes around them, or was it from sulfur rich rocks/mineral deposits?

There are several naturally occurring sulfate minerals such as copper(II) sulfate (pentahydrate mineral name: chalcanthite) that will decompose into $\ce{SO3}$ when heated to a few hundred degrees, which produces sulfuric acid when dissolved in water.
These compounds were known as vitriols due to their glassy appearance and sulfuric acid was known as oil or spirit of vitriol since it was derived from these minerals.
Chalcanthite seems to be a natural oxidation product of copper deposits.
Copper(II) sulfate in particular was known as Roman vitriol, but I don't know if this means that that was the primary source for the Romans.

In any case, ancient civilizations did not have any means of mass producing sulfuric acid and industrialized processes involving sulfur dioxide from the combustion of elemental sulfur arose much later.


What is the minimum mass of iron(III) chloride required to be added until the dark blue precipitate first appear? How do I solve this question?

  The presence of iron(III) ion is confirmed in a qualitative analysis by adding potassium hexacyanoferrate(II) ($\ce{K4[Fe(CN)6]}$) solution.

  A dark blue precipitate, iron(III) hexacyanoferrate(II) ($\ce{Fe4[Fe(CN)6]3}$), is formed.

  What is the minimum mass of iron(III) chloride (in grams) required to add to a $\pu{200 cm3}$ of $\pu{0.0010 mol dm-3}$ potassium hexacyanoferrate(II) solution until the dark blue precipitate first appears?

  solubility product of $\ce{Fe4[Fe(CN)6]3} = \pu{3.30 x 10^-41 mol^7dm^-21}$
  relative molecular mass of $\ce{FeCl3} = \pu{162.3 g/mol}$

  Choices:

  $\pu{1.35 x 10^-8 g}$
  $\pu{4.38 x 10^-7 g}$
  $\pu{5.48 x 10^-7 g}$
  $\pu{2.19 x 10^-6 g}$

This is what I came up with just now :

  $K_\mathrm{sp} = \ce{[Fe^3+]^4 [Fe[CN]6^4-]^3}$

  $\pu{3.3x10^-41}$ = $\ce{[Fe^3+]^4[0.001]^3}$

  $\ce{[Fe^3+]^4}$ = $\pu{3.3 x 10^-32 mol dm^-3}$

  $\ce{[Fe^3+]}$ = $\pu{1.35 x 10^-8 mol dm^-3}$

  Mass of $\ce{FeCl3} = \pu{1.35 x 10^-8 mol dm^-3 \times 0.2 dm^3 \times 162.3 g/mol}$

  = $\pu{4.38 x 10^-7 g}$

  I am still relatively new to this topic so I am not sure if I am doing this correctly. Any help will be appreciated. Thank you !

Yes that is right.

However I would not round intermediate results to avoid round off errors. In fact I'd carry two extra significant figures for intermediate calculations and only round off final answer. (Easy now with a calculator, not so easy to do when I was doing these problems with a slide rule...)  So I'd get Fe as $1.3478\times10^{-8}$ molar. Using that number I get 4.3749893530170255598967635197789e-7 as the final answer which rounds to 4.37e-7.

The problem's answer is really wrong since the concentration of the hexacyanoferrate(II) is only given to two significant figures. So the answer shouldn't have more than two either. If the problem wanted three significant figures then the hexacyanoferrate(II) should have been specified as 0.00100 molar.

You have to realize that significant figures are a crude way to do error propagation. So two significant figures for 0.0010 molar means +/- 0.00005 molar, so a +/-5% specification. However if the answer to two significant figures is 9.9 then that means +/- 0.05 which only has an error of +/-0.5%. So although crude using significant figures keeps you from stupidly assuming that all the figures in the calculator's answer, like 4.3749893530170255598967635197789e-7, are significant.


Can the first ionization energy for a specific element be calculated? I'm taking a first-year Inorganic Chemistry course in college, and stumbled upon this problem:

  For $\ce{Mn}$ and $\ce{P}$, calculate the first ionization energy and explain the difference between the obtained values.

Now, my course materials (written by our professor) apparently say that the first ionization energy can be computed using this formula. I say apparently because the "course materials" are poorly written PowerPoint slides with loads of unexplained terminology and abbreviations, so I may be misunderstanding.

$$
\text{IE} = \frac{Z_\mathrm{eff}^2}{n_\mathrm{eff}^2} \cdot 13.6~\mathrm{eV}
$$

where $Z_\mathrm{eff}$ is the atomic number minus the differentiating electron's shielding constant (calculated using Slater's rules) and $n_\mathrm{eff}$ is pulled out of a table of values (see here).

Unfortunately, it only seems to produce (somewhat) numerically correct results for $\ce{H}$ and $\ce{Li}$, and stops being useful for comparing them at $Z=5$. I wrote an implementation of the algorithm in Mathematica and compared the results it produces to the first ionization energies returned by ElementData (suitably converted from molar to regular), and it's a very rough approximation.

Much too rough, in fact, to be usable for the problem above: it returns $47.38$ instead of $7.43$ for $\ce{Mn}$ and $45.71$ instead of $10.48$ for $\ce{P}$. It sorts them the wrong way around, in other words.

My question is, what am I missing? As far as I can tell from Wikipedia, there's no way to simply calculate ionization energies - they have to be measured. Is there some obvious detail I haven't spotted? Another way to interpret the problem text? Some elementary knowledge I don't have?

I tried to find out empirically whether your formula makes sense. The data I used was effective nuclear charges from this Wikipedia page (always the minimal value, Pt was discarded), and ionization energies already gathered for the answer to this question on this site. The principal quantum number was also manually added to the data set.

I explored the data in many ways and tried to find a simple expression which could give an estimate for where the ionization energy should lie.

First, I looked at what your formula would give:

It is quite obvious that your formula describes two atoms perfectly and is absolutely terrible for the overwhelming majority of elements in the periodic table.

I present to you the best guess so far, based on simple manual adjustments and some exploration of the data:

$$ E_\text{ion} / \mathrm{eV} = 6 \left(\frac{Z_\text{eff}}{n^{1.5}} \right)^2 + 4$$

Below is the graphical representation:

The error will still be large for many atoms, I estimate the median error to be about 2 eV. But if you want to use a simple formula to calculate the first ionization barrier, this would be the one I would use.

The data exploration was done in Python and is recorded in this iPython notebook. Feel free to have a look at it and even come up with a better model.


Melting points of NaCl and RbCl Why is the melting point of $\ce{NaCl}$ more than that of $\ce{RbCl}$ ? Using Fajan's Rule of polarizing power, $\ce{RbCl}$ has more ionic character and thus should have a higher melting point.

No doubt about it: it is due to the ionic radii and thus the distance between ions. The Rb-Cl bond is not as strong because of the distance between them. Na has an E.N. value of 0.9 Paulings while Rb is at 0.8; thus no significant difference in the magnetic forces of attraction.

LiCl has an even lower melting point due to excessive polarization (this is when Farjan's rules come into play), and the result, along with electronegativity again, is still an ionic compound, but with more covalent character.


Does the reaction of sulfur and potassium nitrate involve production of sulfur trioxide? I've heard that oxidising sulfur with potassium nitrate yields $\ce{SO3}$ gas, but can't find a definitive answer anywhere.

Is sulfur trioxide produced? More generally, what's the full reaction?

Generally a mixture of sulfur, potassium nitrate with charcoal or an organic carbon compound is used in gunpowder mixture.

$$\ce{3C + S + 2KNO3 → 3CO2 + N2 + K2S}$$

(Source)

There are many questions in chem.SE related to gunpowder and explosives which are all based on this reaction. You can find them of help.

What is the oxidation mechanism of gunpowder?
What causes KNO3 to decompose into KNO2 and Oxygen in gunpowder?
Why can sulfur act like a catalyst?
Write the chemical equation of this reaction:


Why does ionization energy increase as we go from left to right in a period? Why does ionization energy increase as we go from left to right in a period?

In my textbook, the explanation is as follows:

"This is consistent with the idea that electrons added in the same principal quantum level do not completely shield the increasing nuclear charge caused by the added protons. Thus electrons in the same principal quantum level are generally more strongly bound as we move to the right on the periodic table, and there is a generally more increase in ionization energy values as electrons are added to a given principal quantum level."

This, however, doesn't make any sense to me. From what I understand, in a stepwise ionization process, it is always the highest-energy electron (the one bound least tightly) that is removed first. So when we go more to the right in the periodic table, there are more electrons, and thus much more possibility for these to shield the outer electron from attraction to the nucleus. This would make it easier to remove that outer electron, and so I would say the opposite: the ionization energy should lower when we go to the right.

So can someone explain to me why this is not so? Thanks for any clarifications.

Crash Course on Ionization Energy:

As we all know, atoms prefer a full valence shell. So as we go right in a period, we are increasing $e-$. And also ADDING PROTONS. Because we are adding protons, the size of the atom gets smaller because the nuclear charge will be more powerful. Adding protons in a period trumps the addition of electrons. At the end of the day, we have a small atom with many electrons in it's valence shell that does not want to let go of them. Especially the Noble Gasses and Halogens.
Ionization energy decreases as we move down a group because:

As we move down, a new full energy level is being added. More electrons means more repulsion.
This creates the shielding effect where the addition of the shells, shields the outer electron from receiving the nucleic charge. NOTE: Here, however the addition of another energy level trumps the addition of protons.

This is just a piece of the whole picture.