Determining number of electrons transferred When determining the number of electrons transferred in a redox reaction is it the total in both half equations?

For example:

$$\ce{2I- + Zn^2+ -> I2 + Zn}$$

First we split it up into the two half-reactions and get the following

$$\ce{2I -> I2 + 2e-}$$

$$\ce{Zn^2+ + 2e- -> Zn}$$

So would the number of electrons transferred in the redox reaction be two or four?

When determining the number of electrons transferred in a redox reaction is it the total in both half equations?

In the total redox reaction that is properly balanced, the number of electrons transferred is equal to the stoichiometric coefficient of the $\ce{e-}$ species in either balanced half-reaction.

  So would the number of electrons transferred in the redox reaction be two or four?

Your half-reactions and total redox reaction happen to be balanced, to the number of electrons transferred in this case is two.


Why tertiary amines can't show chirality? Why is a tertiary amine with three different substituents not chiral? Doesn't the lone pair represent a fourth, different substituent?

The short answer to your question

Under conditions that inhibit inversion, an amine that has three different groups attached is chiral.

The above applies not just to tertiary but also to secondary amines.

Now, why inhibit inversion? Why does it make a difference?
As mentioned by Zhe (and mentioned in Newton's answer) in the comments:

Tertiary amines can definitely be chiral. It's just that they epimerize quickly, but they are definitely chiral when there are 3 different substituents. In addition, you can easily create a tertiary amine that does not invert by tying back the substituents.

If you notice, they are chiral for very short periods of time due to the inversion that takes place in amines. In regards to the time scale – Ammonia ($\ce{NH3}$) flips $4 \times 10^{10}$ times every second. Converting that into the time for one inversion that translates into an inversion every $\pu{2.5 \times10^{-11} s}$ which is $\pu{25 picoseconds}$. Such a scale of time cannot be discerned by any of our current methods to identify stereochemistry.
From http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch07/ch7-3.html

In this case, even though a chirality center is present in each molecule, the sample is optically inactive since the optical activity of the two extremes of inversion
averages out because they are enantiomeric.

For why umbrella inversion occurs, see this answer by SendersReagent

Now, what if we restrict this inversion by attaching free-rotation restricting groups? What happens then?
This restriction brings (or rather removes) a fact into the mix. There is no more inversion possible.  This is because rearrangement into $\mathrm {sp}^2$ cannot take place because of the groups added since it becomes sterically inhibited in becoming planar. An example of this would be quinuclidine which has no way of forming a planar structure.


Why do triple bonds have a low NMR value? Double bonds has a positive shift compared to single bonds. It seems like that there would be a trend but then the triple bond negatively shifts, despite having great electron density similar to the double bond. Any insight.

Chemical shifts arise due to differences in the local magnetic field in the different environments of a molecule. There are a number of contributing factors including the local diamagnetic contribution which is driven by the shielding effect of local electrons. Where there is functionality which generates electron circulation, a localised anisotropic magnetic field is created, and these can cause both downfield and upfield shifts, based on the position of the proton (nucleus) with respect to the functional group. These interactions, unsurprisingly, are called magnetic anisotropic effects.

The best example of this is the ring current effect arising from aromatic rings, with protons on the outside of an aromatic ring experiencing large downfield shifts, while protons either above the ring, or on the inside of the ring experiencing upfield shifts.

With double and triple bonds, the shielding cones are different in shape, and traditionally are drawn as shown below (you might find some better diagrams with a web search). Areas marked (+) increase the shielding (leading to decreased chemical shift) and areas marked (-) decrease the shielding (increasing chemical shift).

However, it is important to recognise that chemical shifts are not only influenced predominantly by electronics; sterics have a very significant contribution to chemical shifts also, often more so than electronics. A very good (recent) paper (Baranac-Stojanovic, M RSC Adv., 2014,4, 308-321) discusses traditional magnetic anisotropy models, and presents some very good examples that highlight the effects of steric influence on chemical shifts.


How much energy does it cost to have electron configurations that are not in accordance with Hund's rules? What is a ballpark figure for the difference in energy for an atom that follows Hund's rule vs one that has two electrons with opposite spins?  I'd be interested to know carbon and nitrogen.  Is there a bigger difference when there are two electrons with one spin and one electron with the opposite spin vs two electrons with opposite spins?

Each energy level of a given electron configuration is described by a  Russell-Saunders term symbol, assuming LS coupling. The ground state term symbol is predicted by Hund's rules. Tables of atomic energy levels identified by their term symbols have been compiled by NIST. So, the only thing you have to do, is to compare the energies of the terms describing the electron configurations you are looking for with their respective ground states.

In the following pictures the notation $m_{S} = + \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \uparrow$ and $m_{S} = - \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \downarrow$ will be used for the spin quantum number $m_{S}$.

The data for carbon can be found here. The ground state for carbon is ${}^{3}\mathrm{P}_{0}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{3}\mathrm{P}_{2}$ or ${}^{1}\mathrm{S}_{0}$ therefore I will give the values for both of them (for comparison: the thermal energy at room temperature is about $200 \, \mathrm{cm}^{-1}$).

The data for nitrogen can be found here. The ground state for nitrogen is ${}^{4}\mathrm{S}_{3/2}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{2}\mathrm{D}_{5/2}$ or ${}^{2}\mathrm{P}_{3/2}$ therefore I will give the values for both of them.


Influence of pH on precipitate crystallite sizes? It's normally assumed that a higher pH of solution would slow nucleation rates and hence lead to larger crystal formation. I've found literature that it is due to solid-liquid inter-facial tension. Which is a concept that I do not fully understand.

In my experiments, higher pH solutions led to the formation of smaller crystallites. I can't make sense of why its contradicting the literature. Here are the images from my experiments:

My samples are precipitates produced from mixing aqueous $\ce{CaCl2}$ +$\ce{MgCl2}$ and $\ce{Na2CO3}$ solutions. The elevated pH images were produced through the addition $\ce{NaOH}$ to the previously mentioned mixture.

If you have any comments or ideas, it would be greatly appreciated. I am a engineering student who is trying his hand at inorganic chemistry.

Thanks in advance.

At low pHs there is little "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Most of the carbonate species are dissolved $\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$. Thus the precipitate forms slowly and you get relatively large crystals.

At high pHs there is a lot "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Thus the precipitate forms relatively rapidly and you get relatively small crystals.

I have no idea what literature would lead you to believe otherwise.


What is the chemical equation, along with the change in heat of the crystallization process of Sodium Acetate This is the reaction found in chemical hand warmers, and I can't figure out the equation and the change in heat. Any help would be greatly appreciated

There is no chemical reaction.

The heat is released as the latent heat of crystallisation of a supercooled solution of sodium acetate trihydrate; you can obtain similar effects with sodium sulfate decahydrate or sodium thiosulfate pentahydrate.

When heated up, the hydrate dissolves in its own crystal water, which absorbs thermal energy similar to melting; when cooled down, the solution becomes supercooled.

Their state is called the metastable state: this means that it is not thermodynamically stable, but conditionally stable for kinetic reasons, as the crystallization requires a trigger.

The triggering can be done by, for example: insertion of a small crystal of the same salt, bending a steel chess (used in pocket heaters) or vigorous shaking.


How to calculate the mass of an element in a sample of a compound? What calculations are needed to express the grams of an element from a compound dissolved in aqueous solution? For example, I have a $0.5\,\mathrm{M}$ glucose aqueous solution and I want to know how many grams of $\ce{O}$ are in there.

I have tried to calculate in this way:

Molecular weight of glucose = $180.159 \mathrm{\frac{g}{mol}}$
Atomic weight $\ce{C} = 72$; $\ce{H} = 12.09$; $\ce{O} = 96$;
I am considering $0.5\,\mathrm{M}$, so in $1\,\mathrm{L}$ I have $90.08\,\mathrm{g}$ of glucose
At this point I divide the molecular weight of glucose by the atomic weight of oxygen $\frac{90.08}{96}$, but I don't think it is correct.

As suggested in the comments, keep track of the units!  If your answer doesn’t make physical sense, you made a mistake.

A thorough approach to this problem is to:

  calculate how much glucose you have.
  calculate the fraction of oxygen in glucose.
  multiply the (unitless) fraction found in (3) by the total glucose calculated in (1).

Using the values (and associated units) given in your problem:

  $0.5 \pu{\frac{mol}{L}} \text{ glucose}\times \pu{1 L} \text{ glucose}\times \pu{180.16 \frac{g}{mol}}$ glucose = $\pu{90.08 g}$ glucose.
  $\frac{\pu{96 g mol^-1} \ce{O}}{\pu{180.16 g mol^-1} \text{glucose}} = 0.533$
  mole fraction of $\ce{O}$ / glucose.
  $0.533 \times \pu{90.08 g}$ glucose = $\pu{48 g} ~\ce{O}$

Of course there are many shortcuts you could take or even different approaches altogether, but I like to parse a problem like this down stepwise in a way that shows what you are doing and facilitates keeping track of units.  This makes troubleshooting much easier and systematic if you find you've made a mistake or want to walk through it a second time to verify what you've done (as you definitely should!).


Chemical reaction in homemade deodorant Which ingredients in my homemade deodorant recipe are causing it to froth and overflow its container? Is there something I can add to stop this reaction?

Ingredients:

Magnesium oil (made from magnesium chloride flakes and distilled water)
Baking soda
Witch hazel
Aloe Vera gel
Arrowroot powder
Essential oils (lemon, orange blend and lavender)

The most likely culprit is something acidic reacting with the baking soda (sodium bicarbonate).

Try mixing separately:

Baking soda and witch hazel
Baking soda and aloe vera gel

My guess is that one of these two has an acidic buffer in it that's causing the baking soda to turn into a science fair volcano on you.

You could try adding some calcium carbonate to the mix (crush up an antacid tablet, or buy powder from somewhere). $\ce{CaCO3}$ is more basic than $\ce{NaHCO3}$, so it might neutralize enough of the acid that it avoids the foaming.

Sodium carbonate, often sold as "washing soda", is essentially a more-basic version of baking soda and might also work.  You have to be careful not to add too much $\ce{Na2CO3}$, though, because it could make the product too alkaline and thus irritating to the skin. In particular, do not substitute washing soda for all of the baking soda—only add the amount of washing soda needed to prevent the frothing.

There are harsher chemicals that you could use to neutralize the acidity, but I wouldn't recommend trying them in a household project like this.


Colorimetric test to measure dissolved oxygen in salt water Are there any simple colorimetric tests for measuring the dissolved oxygen in salt water, which do not involve titration like in Winkler method ?  i.e. Are there any chemical reagents, which when added to a salt water sample, would cause a change in the colour of the solution , giving us the measure of DO ?

Regards.

Unfortunatly I don't think there is an easy colorimetric technique available.

There are reagents that are strongly colored and oxygen-sensitive.  One exampleis reduced methylviologen radical.  That species can be prepared by mixing methylviologen with sodium dithionite in aqueous solution.  The one-electron reduced species is intensely blue.  It is rapidly oxidized by oxygen, leading to a loss in color.  In theory, you could use the degree of this color loss to assay DO.

The problem in real life with applying this reagent to DO determination is that oxygen in the air will also react, very quickly.  So you would have to put your seawater sample under an inert atmosphere (and also prepare the reduced MV radical solution under an inert atmosphere).  That is very inconvenient, and could create problems where your seawater sample outgases oxygen into the anaerobic environment before or during measurement.


Choose electrolyte for electroplating I am planning to try electroplating a steel ring with copper, than nickel, and than finally gold.  It seems to me that this would be as simple as using a bath of HCl for electrolytes and attaching the object to be plated to the cathode and the metal to be plated to the anode.  I though I would then simply run a current through the setup.  However, I have seen several "recipes" for electroplating using other electrolytes such as copper and nickel sulfate or dissolving the gold in aqua regia or cyanide before the plating.  Is this necessary, as the metal to be plated will dissolve into solution when electricity is run through the circuit anyway.  I would prefer to buy fewer chemicals if possible and reduce the complexity of the process, while obtaining the best results.

There are two issues here:

First, you will be oxidizing water. You would end up making oxygen in most of those cases, before oxidizing your metal and making the cation you need.

Second, even in cases where your metal will oxidize before water, the concentration of the metal cation in the solution will be low and variable. This will create problems with controlling rates of deposition and uniformity of deposition.

That is why the electroplating baths rely on the already dissolved cation and not the metal as the source. This way you have lots of the cation you need in the solution to work with and you can put some inert counter electrode.


HCP unit cell c-axis Today I was asked to show that the perpendicular height, $H,$ of a regular tetrahedron is related to the length of a side, $L,$ of the tetrahedron, by $H = \sqrt{(2/3)} \cdot L$. I did this no problem.

I was then asked:
Explain how this is relevant in determining a relationship between the unit cell parameters, $a$ and $c$ of a hexagonal-close packed crystal structure: $c = 1.633 · a$.

Could anyone possibly help me out here?

A picture is worth a thousand words (see the two tetrahedra heights making up the vertical c-axis?):

The HCP packing has layers of closest-packed spheres in an ABAB... pattern. The distance between layers is one height of a tetrahedron, and you need two layers to get a repeating structure along c, i.e. the unit cell parameter.
(picture found at https://chemistry.stackexchange.com/a/111921)


Does temperature affect the %oxygen in air by volume? (even though oxygen density varies inversely with temperature according to P/RT=n/V) I want to investigate how temperature affect the concentration of oxygen in air as it will affect how much oxygen we inhale per breathe. Oxygen by percent composition in air is 20.95%. However, the oxygen concentration/density in air varies due to the expansion of gas particles and varies inversely wth temperature according to P/RT=n/V (where n/V is the density of gas).

I did an experiment that involves the rusting of steel wool to determine the percent oxygen in air by volume (determined by volume of water rose in a test tube divided by the total volume of air in the test tube). However, I'm not sure if the % oxygen I determined is a measure of the oxygen density (if so I can convert the percentage to ppm).

My initial hypothesis is that as there are less oxygen molecules in hot air, water level will rise less since there is less oxygen reacting with the steel wool (compared to cold air). Hence lower % oxygen concentration. Vice versa for cold air.

Then later, I'm thinking that as oxygen molecules (or any gas particles) are heated, they move quicker so each oxygen molecule will occupy more volume. Vice versa if the gas particles are colder. However, since there are more oxygen molecules in cold air than warm air, the increase in number of oxygen molecules will occupy more volume and make up for the smaller volume each oxygen molecule occupy by slower movement.

Therefore, the % oxygen in air by volume I determined by the volume of rising water level over the volume of air, will theoretically be the same and hence it can't be used to determine the oxygen density in air.

But then I also saw a picture of a vernier O2 probe with percentage O2 reading (which im not sure if it's percent by volume or percent composition) that can be converted to concentration in ppm by a click of a button. I then did some research and found out percentage concentration (by mass or by volume) can be converted to ppm (which will be a measure of oxygen density).

No, the concentration does not depend on density in a well-mixed ideal gas.

The key principle here is that the atmosphere is close to being an ideal gas. In an ideal gas the components are well mixed and the composition will not change unless there is a chemical change that acts selectively on one of the components.

The ideal gas equation applies equally to each component of the mixture of gases that is air. If you warm a vessel of fixed size, the pressure goes up. If you warm the mixture in a variable volume, the volume increases. But nothing in either process causes the composition of the gas to change, so the concentrations of the components will remain the same.

You seem to be applying the gas equation to just the oxygen and ignoring the other components (Nitrogen, argon etc.). Yes the warmer air will have a lower density, but the composition will be unaltered. So the relative amount of oxygen in the air remains the same. Warm air has an overall lower density (which is why hot air balloons work) and the volume density of all the components will decrease. So the percentage of oxygen per litre will be lower but so will the percentage of Nitrogen. Mostly when measuring gas composition chemists talk about the concentration relative to all the components in the mixture not to the volume (and in an ideal gas, the volume/pressure/temperature relationship is the same for all components). So, trivially, the volume density of all components is lower when the gas is warmer at constant pressure. There is no selective change for just one component.


Isomerisation of alkanes under Lewis acidic conditions My textbook says:

  n-Alkanes on heating in the presence of anhydrous aluminium chloride and hydrogen chloride gas isomerise to branched chain alkanes.

But no mechanism is given. After a little research, I found this: https://goenkanscienceblog.wordpress.com/2015/01/29/suggested-mechanism-of-isomerisation-of-alkanes/

However, it doesn't explain the role of the HCl gas. Is the mechanism correct and if so, what is the role of HCl?

The mechanism for butane is shown here, instead of a general n-alkane.

Initiation

As the comments have mentioned,  it probably begins by the protonation of butene impurities.

The Lewis acid, $\ce{AlCl3}$, stabilizes the chloride ion from $\ce{HCl}$ when the $\ce{H}$ is used to protonate the butene.

A secondary carbocation is formed, which undergoes methyl shift to form a primary carbocation. I have no idea why a primary carbocation can be formed.

Propagation

An alkane attacks the carbocation, and supplies a "hydride" to the carbocation to complete the carbocation.

Meanwhile, another carbocation is formed which also undergoes methyl shift to regenerate the carbocation at the beginning of this step.

Termination

I am going out of my way here to suggest a isobutyl chloride (1-chloro-2-methylpropane) termination product. I can't find other Lewis bases to complete the carbocation for the termination step.


Is viscosity proportional to the number of hydroxyl groups? Is the following statement always true?

  The more $\ce{-OH}$ functional groups in the molecule, the more is its viscosity?

I think it is true, because it is known that weak intermolecular forces lead to lower viscosities and strong intermolecular forces lead to higher viscosities
and because the more $\ce{OH}$ functional groups in the molecule will give rise to a molecule having stronger molecular forces. As the case in glycerol ($\ce{C3H8O3}$) and water ($\ce{H2O}$)

Therefore the statement must be true. What do you think?

It is a reasonable rule of thumb, but certainly not always true.

Compare for example the viscosities of Dodecane ($\ce{C12H26}$, $\mu=1.374 \ce{\; mPa\cdot s}$, no $\ce{OH}$ groups) and Ethanol ($\ce{C2H5OH}$, $\mu=1.095 \ce{\; mPa\cdot s}$, 1 $\ce{OH}$ group) (source). There you can see that ethanol has a higher number of $\ce{OH}$-groups, but a lower viscosity.

That said, I do think you are on to something regarding the strenght of intermolecular interactions, it is just not purely the number of $\ce{OH}$-groups that matters for that. The length of the carbon chain for example (like in the example I mention above) also plays a role.


A lot of confusion in 1st law of thermodynamics My sir told me that Total energy of system = K.E + P.E in starting.
Then Change in energy = F(External force on body ) * displacement of walls.
Then from here , change in energy = W+q.
(Don’t understand how we reached till here )
This change in energy is also called change in  internal energy.
Then it should be equal to Kinetic + potential energy of the molecules inside the system that we are considering .
Q1 is that if it is equal to kinetic + potential energy, then in case of ideal gases.P.E = 0.So K.E will be equal to change in energy $\delta U$.
Q2 is that if it change in energy , then it should be change in kinetic energy as well right.
Ok then my sir also tells me that when a reaction happens in non ideal gas scenario , example of 1mole N2 + 3 mole H2 = 2mole NH3.Now my sir says , there will be a lot of potential energy here.So we can’t use the formula kinetic energy = q+w here.
When Na(s) in converted to Na(g) we consider , delta H of sublimation of Na and then potential energy change I.e internal energy change = Na+(g) + e. Is this internal energy change $\delta U$.
All these doubts arrived from this question.
Because here they ask for internal energy change , so does it mean that it is $\delta H_1$ + $\delta H_2$ + Internal energy of 1 + internal energy of 2.
I am sorry if this in not written in good manner but am so much in tension right now that I just gave out everything I knew

I'm glad that you have got good concepts but the problem is you're unable to connect the discontinuous dots. See, in order to answer your first two questions I would like to bring attention towards the difference of microscopic energy (energy possesed by the molecules present inside the system) and macroscopic energy (energy you get to see in the system). As a example, a cup of coffee is placed over ground all will say it has zero K.E and 0 P.E. But the molecules present inside the system posses both K.E and P.E. You are right with the fact that change in internal energy for ideal gas means change in K.E but the change in K.E means, difference between the total K.E(summation of K.E of every molecules present in the system) of the sytem before and after the action. In case of non ideal gas, here both P.E and K.E both will contribute.
In case of sublimation of Na, here solid is changing to gas you can't neglect volume change also you have to keep it in mind that change in enthalpy is equal to change in internal energy if the PV-work is zero and the process is isobaric.


What alcohol concentration makes bathroom cleaner "disinfecting"? There are many do-it-yourself cleaner solutions out there, using a combination of vinegar, alcohol (in a "cheap vodka" concentration), dish soap, baking soda and essential oils. I like these recipes for two reasons, first for cost and second for health (sometimes my toddler helps me clean his bathtub, and I wouldn't want him to use too harsh stuff).

However, many of these recipes claim to be "disinfecting", and as far as my understanding goes, alcohol has that effect, as has stuff that offsets the pH-level (such as the acidity of vinegar). But given that this stuff is just sprayed onto the surface, can a mixture of 2 parts water 1 part vodka really be concentrated enough to be "disinfecting"? And if you mix baking soda and vinegar, won't that bring you closer to a neutral pH, i.e. offset any "disinfecting powers"?

Is there some sort of guideline as to how much alcohol/other agent should be in the final solution for it to be of good use in cleaning a bathroom?

According to this Center for Disease Control document:

  Ethyl alcohol (70%) is a powerful broad-spectrum germicide and is considered generally superior to isopropyl alcohol. Alcohol is often used to disinfect small surfaces (e.g. rubber stoppers of multiple-dosemedication vials, and thermometers) and occasionally external surfaces of equipment (e.g. stethoscopes and ventilators).

Although 70% alcohol is ideal, another CDC article states that it can be effective at 50%.  According to this article:

  In the healthcare setting, "alcohol" refers to two water-soluble chemical compounds—ethyl alcohol and isopropyl alcohol—that have generally underrated germicidal characteristics.  FDA has not cleared any liquid chemical sterilant or high-level disinfectant with alcohol as the main active ingredient. These alcohols are rapidly bactericidal rather than bacteriostatic against vegetative forms of bacteria; they also are tuberculocidal, fungicidal, and virucidal but do not destroy bacterial spores. Their cidal activity drops sharply when diluted below 50% concentration, and the optimum bactericidal concentration is 60%–90% solutions in water (volume/volume).

So, alcohol has certainly been shown to be a powerful, broad-spectrum antimicrobial agent, though not at the dillution you proposed. I would just use the straight vodka if it were greater than 50% alcohol, or probably beter (cheaper) yet would be denatured alcohol if that's available where you live.


Do weak acid/weak base neutralisation reactions go to completion? I'm wondering if reactions that involve a weak acid and a weak base go to completion. For example, say we have equal amounts and equal volumes of acetic acid and ammonia being mixed together. Would this neutralisation reaction go to completion or would it reach a state of equilibrium?

$\ce{NH_3 + CH_3COOH <=> NH_4^+ + CH_3COO^-}$

Instinctively, I'm thinking that the reaction would be an equilibrium, since proton transfer should be able to occur between the products ($\ce{NH_4^+ + CH_3COO^-}$) to reform the reactants. Can anyone clarify if this is true?

Weak acids and bases do indeed remain in an equilibrium state with a certain concentration of the free acid and the free base alongside a certain concentration of the deprotonated acid anion and the protonated base cation.

In general, you can consider any acid-base reaction to be essentially the following:

$$\ce{HA <=> H+ + A-}\tag{1}$$

Which in turn means that the $\mathrm pK_\mathrm a$ value is defined as in $(2)$:

$$K_\mathrm a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{2}$$

Reactions $(1)$ and $(2)$ are usually written out in water with hydronium in place of a naked proton but the same principle applies. Of course, a base reaction can be considered essentially the reverse, meaning that the acid constant typically given is actually $\mathrm pK_\mathrm a (\ce{HB+})$ rather than anything directly corresponding to the free base. If we now take the reaction of a (weak) acid and a
(weak) base, we get equation $(3)$ and the equilibrium constant as given in $(4)$:

$$\begin{align}\ce{HA + B &<=> A- + HB+}\tag{3}\\[0.7em]
K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\end{align}$$

We can now perform simple mathematics with $(4)$ to arrive at the modified equation $(5)$ as below:

$$\begin{align}K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\\[0.7em]
&= \frac{[\ce{A-}][\ce{HB+}][\ce{H+}]}{[\ce{HA}][\ce{B}][\ce{H+}]}\\[0.7em]
&= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{HB+}]}{[\ce{B}][\ce{H+}]}\\[0.7em]
K &= \frac{K_\mathrm a(\ce{A})}{K_\mathrm a (\ce{HB+})}\tag{5}\end{align}$$

We have thus arrived at a way to determine the equilibrium constant of the proton transfer reaction just from the acidity constants of both participants. If both acids weak and have similar $\mathrm pK_\mathrm a$ values — as ammonium and acetic acid do — the equilibrium constant will be close to $1$ and thus both sides of the equation will have similar concentrations.

Using $(5)$, you can also predict a ‘degree of completion’ for any acid-base reaction.


How do these cascade reactions work? A. $\ce{Cr2O7^2-}$  B. $\ce{H+}$  C. $\ce{OH-}$  D.
  $\ce{Cr^3+}$  E. $\ce{Ba^2+}$

  Which of the above is the appropriate ion for each blank in the
  following series of reactions?

  $$\begin{array}{ccc}\ce{OH-} + (5) \xrightarrow{} & \ce{CrO4^2- + H+}
 & \\  & \ce{v} (6) & \\  & \ce{BaCrO4 (s)} & \ce{<=>[(7)][(8)]} &
 \ce{Ba^2+ + Cr2O7^2- + H2O}\\  & \text{(yellow precipitate)} & & \ce{v
 SO2 (g), H+} \\  & & & \ce{BaSO4(s)} + (9) + \ce{H2O} \end{array}$$

This question can also be viewed at https://i.stack.imgur.com/i08sR.jpg

So while studying, I came across this reactions' question, which I believe is called a 'cascade reaction'. I have no idea how to do it. I can figure out a few parts, like ($6$) is obviously $\text{E}$, but how do you do the others?

how do you do the others?

You do it like a detective. Even the slightest details matter, and you've to link a lot of concepts together.

The main theme you're missing in your question is this equilibrium:

$$\ce{ CrO4^2- <=>[\ce{H+}][\ce{OH-}] Cr2O7^2- }$$

Dichromates are supposed to be stable in an acidic medium. So using this you can figure out that ($7$) is $\ce{H+}$ and ($8$) is $\ce{OH-}$. Again for ($5$) you're adding a base which is converting it to a chromate. So ($5$) should be a dichromate $\ce{Cr2O7^2-}$.

So we're only left with ($9$) and $\ce{Cr^3+}$ and they are bound to go with each other. Since this is not an exam here's a reasoning too,

$\ce{K2Cr2O7}$ is a strong oxidizer in acidic mediums. If it's going to oxidize others, it will itself get reduced. To be precise here's the reaction that's taking place:

$$\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$$


Why is borole not considered aromatic I came across a problem which stated that borole isn't aromatic. I thought that borole is similar to pyrrole so it should be aromatic, but that isn't the case. Can I know why isn't it aromatic?
I was also wondering if we replace the nitrogen in pyridine with boron then will the compound remain aromatic?

A very basic way to do this would be to draw a diagram that shows the p-orbitals going above and below the plane.
For Borole, the bonding orbitals would look like as shown below (taken from the wikipedia page on Borole):

Natural Bonding Orbitals of Borole. Structure optimised using ORCA BP86-D3BJ and def2- TZVPP basis set. The calculated occupencies of the obitals going from left to right are $0.13$, $1.9$ and $1.9$ respectively.

[emphasis mine]
The point to be taken here is the sentence in bold, the occupany in the empty p-orbital of boron [left most diagram and first value given] is small compared to the other two π-bond orbitals. This means that there is no complete delocalisation in borole. Only the double bonds are in conjugation.
A reason for this is that the orbital of Boron is an empty p-orbital and so the total number of electrons that can potentially delocalise is $4$ and so not of the form $4n+2$. If anything, the compound would become "anti-aromatic" if conjugation were to happen and so that delocalisation doesn't take place.
Now, on your second question, the lone pair of nitrogen in pyridine doesn't contribute to the aromaticity since the lone pair is out of plane of the ring. So adding placing boron shouldn't affect the aromaticity.


Are empty orbitals degenerate? This is apparently a true statement:

"In a hydrogen atom, the 2s and 2p subshells have the same energy."

Why is this the case? Why would they be of the same energy? Wouldn't the 2p be slightly higher in energy than the 2s?

My chemistry teacher Mr Ong Chiau Jin answers:

  Why are all the orbitals that have the same principal number in
  Hydrogen degenerate?

  In hydrogen, all orbitals with the same principal quantum number 'n'
  (1,2,3...) are degenerate, regardless of the orbital angular momentum
  quantum number 'l' (0,1...n-1 or s,p,d,). However, in atoms with more
  than one electron, orbitals with different values of l for a given
  value for n are not degenerate. Why is this? Surely the radial
  distribution functions are similar for hydrogen (in that there's still
  penetration of orbitals and so on). Or is it that orbitals with
  different values of l are degenerate for a value of n greater than
  would be occupied in that particular atom's ground state?

  The answer is right there in your question. It's only the interaction
  of multiple electrons in an atom like He, Li, Be, etc. that makes the
  different angular momenta wave functions differ in energy. Consider
  this. For the one electron system, why should a p or d orbital differ
  in energy from an s? What makes them differ? In the multi-electron
  case, the p orbitals have different spatial extent, different angular
  components, so the electron density caused by an electron in that
  orbital will interact differently with the other electrons. In other
  words, you need to have more than one electron for the "shape" of the
  p and d and f orbitals to matter to the other electrons. In the H
  atom, there's only one electron, so there's no electron-electron
  repulsion to differentiate the s, p, and d orbitals.

This may not be the perfect answer but it does capture the essence of the perfect answer, which is the idea of multi-electron interactions. Hope this helps.


How to make aluminium monostearate? Can I make aluminium monostearate $(\ce{C18H37AlO4};$ $\pu{344.472 g mol-1})$ by reaction between stearic acid $(\ce{C18H36O2};$ $\pu{284.484 g mol-1})$ and aluminium hydroxide $(\ce{Al(OH)3};$ $\pu{78.00 g mol-1})$ or by adding more aluminium hydroxide in aluminium tristearate?

Aluminium hydroxide isn't a partilarly reactive or soluble compound, so you probably don't want to directly mix it with stearic acid; also, you'd likely obtain a tristearate this way.
According to a procedure described in US3056819A patent, a synthetic path would be to obtain aluminum dihydroxychloride $\ce{Al(OH)2Cl}$ first, and then add stearic acid and sodium hydroxide (please disregard full stops after units, this isn't a standard notation these days):

  (1) 2.7 g. (0.1 mole) of metallic aluminium is dissolved in 50 cc. of 2N-HCl (0.1 mole) under heating to obtain a solution of a basic aluminium chloride whose formula is $\ce{Al(OH)2Cl}.$

  (2) To an aqueous solution of 47.58 g (0.1 mole) of potassium alum $(\ce{AlK(SO4)·12H2O})$ is added a solution of sodium bicarbonate to form aluminium hydroxide.
  The aluminium hydroxide is dissolved in 50 cc. of 2N-HCl with stirring and the solution is heated for 1 hour at 100 °C. to give a solution of a basic aluminium chloride whose formula is $\ce{Al(OH)2Cl}.$

  (3) To a solution of 24.2 g. (0.1 mole) of hydrous aluminium chloride in 100 cc. of water is added gradually 16.8 g. (0.2 mole) of sodium bicarbonate with stirring and the solution is heated for 1 hour at 100° C. on a water-bath.

  Any of the solutions of the basic aluminium chloride produced by the above three methods is diluted with water to 200 cc. and added to a solution of 28.45 g. (0.1 mole) of stearic acid and 4 g. of sodium hydroxide (0.1 mole) in 1000 cc. of water, and the mixture is heated on a water bath until the product precipitates.
  After washing with water to remove chlorine ion, the product is extracted with acetone in Soxhlet's extractor to remove free stearic acid, obtaining 33 g. of aluminium monostearate whose formula is $\ce{Al(OH)2(C17H35COO)}.$

References

Hideo, T. Aluminium Soaps and Their Production. US3056819A, October 2, 1962. (PDF)


Ramachandran Plot for Phi and Psi angles of peptides Why do we consider the normal and extreme limits for Ramachandran plots to be less than the Van Der Waal distances between two atoms?

A remarkable aspect of the Ramachandran plots is that the implementation of a hard-sphere potential with flexibility limited to the two backbone dihedrals proved sufficient to largely explain the conformational range sampled by real polypeptides. Take for instance the following map from a seminal article by Ramachandran (from Ref. 1; dashed line encloses the partially allowed region):

The map was generated with the following set of restraints:

In the article Ramachandran et al. explain how they selected the range of vdW distances used to generate their maps:

The next step towards working out the allowed conformations is to choose a set of permitted minimum contact distances between the different types of atoms. Only those conformations (4,4') which do not have any of the contact distances less than these minimum values now become allowed. Two such sets are given in Table II, termed "normally allowed" and "outer limit" contact distances. These are essentially the same as those given by Ramachandran, Ramakrishnan, and Sasisekha-ran (1963) arrived at from a study of the contact distances observed in structures of various organic compounds. Contact distances inbetween the two limits have also been observed in actual crystals, but not as frequently as the normally allowed values.

van der Waals radii are meant to define distances of closest approach. They are useful to identify conformational regions forbidden due to steric clashes. However, tabulated vdW radii are representative values. In real molecules the internuclear potential can vary somewhat depending on atomic types and the local electronic environment.
In addition, molecules are fairly flexible and can to a degree circumvent potential clashes by changing internuclear distances and angles, but a rigid model does not capture that. That a simple rigid model suggests a steric clash, because one or more vdW distance restraints are violated, does not guarantee that a conformation (phi/psi pair) is forbidden. Therefore, if you use a rigid model to generate a Ramachandran plot, that is, if only phi and psi are allowed to change, then you should allow "wiggle room", for instance by setting a lower threshold for the allowed values of the vdW distances. The use of "tolerances" in vdW radii should therefore not be surprising.
Close contacts can be energetically costly even with compensating rearrangements, so the associated configurational space - marked as "partially allowed" in a traditional diagram - is more sparse. In a modern approach a more descriptive potential for the protein backbone is employed, and ideally degrees of freedom other than the dihedral angles are relaxed, and energy isolines are displayed. The following, from Ref. 2, shows roughly how such a diagram might look:

References

G.N.Ramachandran, C.M. Venkatachalam, and S.Krimm. Stereochemical Criteria for Polypeptide and Protein Conformations II. Helical and Hydrogen-Bonded Polypeptide Chains. Biophysical Journal  6 (1966) 849-872.
ibid, 909-933.


What does ethylene glycol decompose to when exposed to air for an extended period of time? I came across this article on wikipedia on ethylene glycol stating that it decomposes fairly quickly. I was under the impression that ethylene glycol was dangerous to let leach into soil, but I think I'm mistaken as wikipedia states

  it [ethylene glycol] breaks down in air in about 10 days and in water or soil in a few weeks. It enters the environment through the dispersal of ethylene glycol-containing products, especially at airports, where it is used in deicing agents for runways and aeroplanes.

What does it decompose into ($\ce{CO2 + H2O}$?) and what is the mechanism?
Does this mean that a tub of ethylene glycol exposed to air will eventually contain only water?
Is the reaction the same when ethylene glycol penetrates soil around runways?
What is the activation energy for the decomposition and at what temperature will it no longer decompose?

This image highlights the oxidation process of ethylene glycol fairly well

( Taken from Ethylene glycol: properties, synthesis, and applications )

Since it has little tendency to evaporate, unless it is kept in direct sunlight, it would probably oxidize to a mixture of the above liquid compounds (glyoxal, glyoxalic acid) and not directly to water. If it is kept in sunlight, the reaction would be photochemically catalysed and a it may degrade to carbon dioxide faster, but according to theoretical principles you would not get much water.
Ethylene glycol is relatively hydrophilic and also has a low tendency to separate from water, so the only major degradation process available to it in soil would be biodegradation by either aerobic or anaerobic process, which is NOT the same as atmospheric oxidation1
Not sure about reaction energetics regarding the same

Most of the danger regarding glycol spills (and other chemicals in the environment) comes from absorption by other organisms; since glycol is comparatively less volatile and more hydrophilic than, say ethanol, it has a greater tendency to be absorbed by organisms.

1 Taken from William R. Roy, in Handbook of Antistatics (Second Edition), 2016


Can I reactivate a chemoluminescent reaction? I have created a firefly jar with glow sticks to sell at a school fair. The firefly jar lasts only for an hour or so. I have heard that glow sticks can be reactivated. The firefly jars contain the chemicals found in good sticks. Is there any way to make the jar glow again?

Short answer: No, you can't!

Longer answer: Still no, and this is why: The reaction between the oxalate and hydrogen peroxide is irreversible. Both compounds are consumed in the reaction and there is no way back!

The only compound that survives is a fluorescent material, that absorbs in the UV, emits in the visible range and gives the colour that you finally observe.

As Ben Norris pointed out in his comment, your only options are:

add fresh diphenyloxalate and hydrogen peroxide (which again will be consumed) or
irradiate the fluorescent compound that has survived with a UV lamp


Why is silver nitrate used for cyanide titration? Or more to the point, since silver isn't exactly cheap: Why can't cuprous nitrate ($\ce{CuNO3}$) substitute for the pricier ($\ce{AgNO3}$) in cyanide titration?

I assume that free cyanides bind eagerly with all Group 9-11 metals, since they're used industrially to leach the heavier ones.

I'm guessing that there's something relatively quirky about the Period 4 metals Co, Ni, and Cu that prevents their use as a titrant of aqueous cyanide: E.g., it appears that nickel nitrates are very hygroscopic.

Due to the hardness of copper ions relative to silver ions, copper ions will bind hydroxide more readily in the presence of cyanide than silver ions do. Once ligated by $\ce{OH^-}$ ions, $\ce{Cu^{1+}}$ will oxidize to $\ce{Cu}^{2+}.$

The softer acid, $\ce{Ag^{1+}}$, doesn't do this nearly as quickly, especially not in the presence of a softer ligand like cyanide.


Approach to model solvent in a lattice: e.g. H2O in NaCl If a lattice contains a solvent, lets say $\ce{NaCl}$ and water of crystallisation, what is the condition for a solvent molecule to change it's location?

From my current point of view I assume that such a water molecule is bound to a lattice ion by ion-dipole forces. It vibrates at that location until, randomly, the particle energy exceeds the binding energy. Then it goes to a new location, in the direction that minimizes Gibbs free energy.

How is the water molecule energy distributed? Currently I use Maxwell-Boltzmann distribution. However, Maxwell-Boltzmann is for non-interacting particles and therefore not quite a good fit for molecules that interact with a ion. Further, my chemistry book suggest that ion-dipole bonds have a binding energy of >$50~\mathrm{kJ~mol^{-1}}$, which will in my calculations rarely allow a jump at all.

I ask this question because I'm working on a particle-drying simulation. I intend to use a Random Walk simulation to obtain diffusion coefficients under special circumstances: e.g. if the lattice salvation behaviour is somehow weird. The model must not be physically correct in all details. Even bigger "model" error will most likely be small in the simulation results.

Note: I asked this question already here but have not received an answer. As suggested there, I open it in Chem.SE.

This is a bit outside my realm of expertise - you'd want to look at molecular dynamics (MD) simulations for more details. Let me take your question in a few parts.

1) A water molecule in a lattice is likely bound by electrostatic (and hydrogen-bond) interactions. So your general assumption is a good one.

2) MD simulations usually have a "thermostat" depending on the type of ensemble used. Your thermal energy will be distributed into different degrees of freedom, such as rotations, translations, and vibrations. So I'd work out the amount of translational free energy from statistical mechanics, and then use the electrostatic interactions to determine the path for jumps. (Your main question here seems like the activation barrier, since each lattice site should have the same energy, right?)

3) If the moves are rare, you might want to look into "rare event sampling" methods, which preserve the correct statistics, but allow you to more efficiently run your model.

Many of these techniques are already implemented in simulation codes. You might want to look at something like LAMMPS which can handle different levels of particles (e.g., atoms, molecules, crystal domains, etc.)

Hope that helps - I can only talk in broad generalities here.


Why is phenolphthalein at pH 10 non-planar? Phenolphthalein is colorless when pH = 0-8.2 and pink when pH = 8.2-12.

Why does the ring bearing the carboxylate (COO-) group not lying on the same plane as the other 2 rings?

Image source: https://www.learner.org/courses/chemistry/images/lrg_img/Pink.jpg

Ok, let's look at this structure. I've drawn/calculated (B3LYP/6-31G) a 3D model of this structure and optimized the geometry.

So what do we see here? Actually no two rings are in one plane, but two are nearly planar (~40° of twist) while one is basically at 90° twist. Let's first look at the one with the carboxy group which is twisted a lot. The carboxy group is in ortho position. If we twist the ring to be in a plane with one of the other rings the oxygen would collide with a CH group. The carbons and oxygens are quite big, so is the repulsion between them and the angle has to be quite big for it to work out. Here's a picture of those groups colliding (I did put it at some degree angle because it's easier to see), keep in mind that this representation of atoms is not the truth. The electron "cloud" around them is actually much larger.

Now why do the other two rings have an angle of about 40° between them? If they are planar they can have conjugation which is favorable. Now What happens if we put them into one plane?

You can see that the two hydrogens in the top get very close. In fact, too close. So there are two ways to avoid this. Twist the two rings or make the angle around the central carbon atom between the rings bigger. In a sp2 carbon like we got here the optimal value would be 120°, we could make it bigger, let's say 145 degree:

Now the hydrogens would not clash any more, but it turns out this costs much more energy than we would lose if we twist the two rings to avoid the hydrogens clashing together. What happens in the end is that this angle gets a little bit bigger (~124° instead of 120) and we get that twist between the aromatic rings which is still low enough for good conjugation between them (see first structure).


What is the density of compressed natural gas (CNG) used in vehicles? I would like to know the density of the compressed natural gas used for natural gas vehicles in order to complete the formula in this Mathematics Stack Exchange post.

The density of CNG at the pressure of 200 bars (which is optimized for use in vehicles) is 435 $\frac{kg}{m^3}$. However, it's (not usually) 175 $\frac{kg}{m^3}$ in some vehicles in some countries. Note that it differs according to your home country. Source 1 and source 2.


How does the reaction Mg + O₂ → MgO occur? $\ce{Mg + O2 → MgO}$

Magnesium, oxygen and magnesium oxide form the above equation.

How?
What do I need to learn to do this? atomic numbers/ valencies?

First of all, do you know that chemical equations are means to express chemical reactions?

Second, this equation represents the reaction of magnesium with oxygen. Magnesium and Oxygen are the reactants and Magnesium Oxide is the product.

The arrow indicates the direction of the reaction.

To write basic chemical equations, you only need to know your reactants, your products and how to write the symbols for them. Usually, you can 'guess' what the products can be from generic patterns or mechanism, but I don't think you will be there yet (or required to remember them) soon if you just got started on them.

To get the molecular formula or ionic formula in this case, you need to know the valencies of the elements constituting that compound. For instance here, Magnesium has valency 2, Oxygen has valency 2, so $\ce{MgO}$ is the formula for Magnesium Oxide.

For example, Sodium Oxide, Sodium has valency 1, Oxygen has valency 2, so $\ce{Na_2O}$ is the formula for Sodium Oxide.

Iron (III) Oxide becomes: $\ce{Fe_2O_3}$ (Iron valency = 3; Oxygen valency = 2)

Magnesium Chloride becomes $\ce{MgCl_2}$ (Magnesium valency = 2; Chlorine valency = 1)

Does this answer your question or were you looking for something different?


If fluorine has a lower electron affinity than chlorine, why does it have a higher ionization energy? I have read that fluorine has a lower electron affinity than chlorine despite its lower atomic radius because its electron cloud is extremely dense. If this is the case, shouldn't the ionization energy of fluorine be lower than that of chlorine, since fluorine's dense electron cloud should eagerly repel its electrons?

Even more confusing is that electron affinity can also be defined as the ionization energy of the singly charged anion form of an atom, i.e. the ionization energy of $\ce{Cl-}$ is equal in magnitude to the electron affinity of $\ce{Cl}$. Therefore, it seems that the trends for ionization energy and electron affinity should always match, which is not the case with fluorine and chlorine.

To quote chemguide:

  The first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.

That means, the ionization energy of fluorine is the energy of the following reaction:

$$\ce{F -> F+ + e-}$$

The ionization energy of chlorine is the energy of the following reaction:

$$\ce{Cl -> Cl+ + e-}$$

Since the most loosely held electron in fluorine is closer to the nucleus than chlorine, it takes more energy to remove the electron from fluorine than in chlorine.

The ionization energy of chloride ion ($\ce{Cl-}$) is the energy of the following reaction:

$$\ce{Cl- -> Cl + e-}$$

The electron affinity of $\ce{Cl}$ is the energy of the following reaction:

$$\ce{Cl + e- -> Cl-}$$

As a result, their energies are equal in magnitude but opposite in sign.

Why does chlorine have a higher electron affinity than fluorine?

To quote this answer:

  Fluorine, though higher than chlorine in the periodic table, has a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.

As can be seen in the second photo, this effect seems to be true for Period 2 elements.

(source: lardbucket.org)

(source: lardbucket.org)


Disregarding negative externalities and other barriers, what are some fuels that have a higher power density than gasoline? Gasoline’s power throughput per kg is $12\,200\:\mathrm{Wh/kg}$ (Wikipedia.org). Are there any other fuels that have a higher power output that are not considered viable solutions as they are difficult to handle or have negative externalities?

Also what characteristics of their molecular structure give them the high energy throughput?

I would suggest you take a look at what they use as rocket fuel.  There are a lot of fuels with higher power densities but they wouldn't work well on the highway.

The Navy has had great success with powering vessels on nuclear energy.  That is the ultimate top of the viable fuel density list.  Maybe one day fusion will become viable, then it will be at the top of the list.

As for the second part of your question, the characteristics of chemical fuels have to with free radicals.

Any viable fuel needs to be readily split into component parts (triggered by a spark or sudden compression or the like) that then combine in other configurations in an exothermic reaction so you get a net energy output.


How to calculate the pH of the neutralisation of HCN with excess of KOH? Write the reaction between $\pu{0.05 mol}$ $\ce{HCN (aq)}$ and  $\pu{0.08 mol}$ $\ce{KOH (aq)}$ with $V = \pu{500 mL}$ and then calculate the pH.

So I wrote the reaction:
$$\ce{HCN (aq) + KOH (aq) <=> KCN (aq) + H2O}$$

Begin:
\begin{align}
  n_\mathrm{begin}(\ce{HCN}) &= \pu{0.05 mol}\\
  n_\mathrm{begin}(\ce{KOH}) &= \pu{0.08 mol}\\
  n_\mathrm{begin}(\ce{KCN}) &= -\\
  n_\mathrm{begin}(\ce{H2O}) &= -\\
\end{align}

End:
\begin{align}
  n_\mathrm{end}(\ce{HCN}) &= -\\
  n_\mathrm{end}(\ce{KOH}) &= \pu{0.08 mol} - \pu{0.05 mol} = \pu{0.03 mol}\\
  n_\mathrm{end}(\ce{KCN}) &= \pu{0.05 mol}\\
  n_\mathrm{end}(\ce{H2O}) &= \pu{0.05 mol}\\
\end{align}

The product of the reaction is a salt of a strong base ($\ce{CN-}$) completely dissociated in water:
$$\ce{KCN (aq) -> K+ (aq) + CN- (aq)}$$

And
$$\ce{CN- (aq) + H2O -> HCN (aq) + OH- (aq)}$$

Knowing this I calculated the
$$[\ce{KCN}] = \frac{\pu{0.05 mol}}{\pu{0.5 L}} = \pu{0.1 M};$$

Then I used the equation:
$$[\ce{OH-}]_1 = \sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}} \cdot [\ce{KCN}]} = \pu{1.27x10^-3 M}$$

Since I still have some $\ce{KOH}$, it dissociates completely creating $\ce{OH-}$ ions, so:
\begin{align}
  [\ce{OH-}]_2 &= [\ce{KOH}]\\
  [\ce{OH-}]_2 &= \frac{\pu{0.03 mol}}{\pu{0.5 L}} = \pu{0.06 M}
\end{align}

Finally
\begin{align}
[\ce{OH-}] &= [\ce{OH-}]_1 + [\ce{OH-}]_2 \\
&= \pu{1.27x10^-3 M} + \pu{0.06 M} = \pu{0.06127 M}\\
\ce{pOH} &= -\log[\ce{OH-}] = -\log(0.0617) = 1.213\\
\ce{pH} &= 14 - \ce{pOH} = 12.787\\
\end{align}

Is it right to calculate the $[\ce{OH-}]$ in this way?

I based my calcualtions on this equation:
Knowing that the amounts of substance in moles are defined by $Z = Z_1 + Z_2$ in $K$ liters of solution, we can consider this concentration to be
$$[X] = \frac{Z}{K} = \frac{Z_1 + Z_2}{K} = \frac{Z_1}{K} + \frac{Z_2}{K} = [X_1] + [X_2]$$

While you've generally done a good job, you've made an important error. The hydrolysis of the cyanide ion should be written as an equilibrium reaction:

$$\ce{CN-(aq) + H2O(l) <=>  HCN(aq) + OH-(aq)}$$

That's because $\ce{HCN}$ is a very weak acid and its conjugated base a weak base (homework: look up the $K_b$ of $\ce{CN-}$).

Now, as you duly noted, there's excess $\ce{KOH}$ in the solution, which dissociates completely. The $\ce{OH-}$ resulting from that dissociation now push the equilibrium:

$$K_b=\frac{[\ce{HCN}]\times[\ce{OH-}]}{[\ce{CN-}]}$$

to the left. As a result, the contribution of the $[\ce{OH-}]$ coming from the hydrolysis of the cyanide is negligible and can be ignored. Only the $[\ce{OH-}]$ from the excess $\ce{KOH}$ is to be counted. So $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$.

We can confirm that numerically by assuming $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$ and with:

$$\frac{[\ce{HCN}]}{[\ce{CN-}]}=\frac{K_b}{[\ce{OH-}]}\approx 0.0027$$

So neglibibly little cyanide is present as the acid $\ce{HCN}$.

As regards the matter of adding molarities, in some instances that is allowed.

Say we add $n_1$ moles of $X$ to $V\ \mathrm{L}$, that would give a molarity $M_1=\frac{n_1}{V}$. At a later stage we add $n_2$ moles of $X$, that molarity would be $M_2=\frac{n_2}{V}$.

The total molarity would be:

$$M=M_1+M_2=\frac{n_1}{V}+\frac{n_2}{V}=\frac{n_1+n_2}{V}$$

But if we were to mix volumes of solutions it would be:

$$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$$

Now  they are no longer simply additive.


Determination of molecular formula of hydrocarbons by combustion analysis So, I've this question in my textbook:

  $\pu{20 mL}$ of unknown gaseous hydrocarbon reacts with excess oxygen gas. It decreases by $\pu{30 mL}$ upon cooling down. It decreases farther $\pu{40 mL}$ after it passes through an alkali solution. What is the formula of the hydrocarbon?

There is an answer provided, and I get the part where they calculate the "$x$" or how many carbon atoms one molecule of the hydrocarbon has. But, I cannot understand how they calculate the number of hydrogen atoms. There is a strange line,

  $$\pu{20 mL} + 20(x + y/4)~\pu{mL} - \pu{20x mL} = \pu{30mL}$$

The statement is:

  combustion of hydrocarbon and the following cooling down of the resultants mean the decreased volume of water vapor is $\pu{30 mL}$.

I do not get this line. Do they mean that the combined volumes of the reactants and the resultants are the same? That's an incorrect inference, isn't it?

P.S. Sorry for poor translation. The question is in my native language.

The combustion can be described by the following reaction:

$$\ce{C_xH_y (g) + ($x + y/4$) O2 (g) ->[\Delta] x CO2(g) + $y/2$ H2O (g)}$$

All products are gaseous at first, but as the system cools down to the room temperature, water condenses attributing to the decrease in $\pu{30 mL}$. Further decrease in $\pu{40 mL}$ is due to reaction of alkali with the remaining gaseous product, $\ce{CO2}$:

$$\ce{CO2 (g) + 2 M^IOH (aq) -> M^I2CO3 (aq) + H2O}$$

From the ideal gas law it's obvious that amounts of gaseous substances are proportional to their volumes:

$$n = \frac{p}{RT}V \quad\to\quad n\propto V$$

Let's rewrite the first equation with assigned volumes:

$$\ce{\underset{\pu{20 mL}}{C_xH_y (g)} + ($x + y/4$) O2 (g) ->[\Delta] \underset{\pu{40 mL}}{x CO2(g)} + \underset{\pu{30 mL}}{$y/2$ H2O (g)}}$$

Since $n\propto V$, and for the balanced chemical equation amounts of reactants and products are equal, so are their volumes. Hence:

  Q: Do they mean that the combined volumes of the reactants and the resultants are the same? That's an incorrect inference, isn't it?

A: Yes; no, this is correct for the given reaction.

That "strange line"

  $$\pu{20 mL} + 20(x + y/4)~\pu{mL} - \pu{20 mL} = \pu{30mL}$$

is a direct consequence of that as it can also be rewritten as

$$\underset{\ce{C_xH_y}}{\pu{20 mL}} + \underset{\ce{O2}}{20(x + y/4)~\pu{mL}} = \underset{\ce{CO2}}{\pu{20 mL}} + \underset{\ce{H2O}}{\pu{30mL}}.$$

However, as this is a single equation with two unknowns, it only illustrates the preservation of the total amounts of products and reactants. In order to find the formula of unknown hydrocarbon, you just have to equate corresponding coefficients with the ratio between $\ce{C_xH_y}$, $\ce{CO2}$ and $\ce{H2O}$ ($20:40:30 = 1:2:1.5$):

$$x = 2$$
$$y/2 = 1.5 \to y = 3$$

which results in empirical formula $\ce{C2H3}$. In order to find out the molecular formula you would also need auxiliary data such as molecular weight of the hydrocarbon.


How to use dispersion correction with TPSSh in Gaussian 16? A colleague of mine suggested to use TPSSh as a functional. Since I am applying Grimme's D3 dispersion correction by default, I was surprised that it wasn't implemented in the standard density functionals, cf. Gaussian Manual (Copy @ The Internet Archive).

Since I might not be the only one to come across that: How can I use dispersion correction with TPSSh (or other non-implemented functionals) in Gaussian 16?

In principle the following route section is in use:

#P TPSSh/def2TZVP           ! verbose, method/ basis set
EmpiricalDispersion(GD3BJ)  ! Switch on D3 with Becke-Johnson damping
opt(MaxCycle=100,VeryTight) ! Run optimisation
int(ultrafinegrid)          ! use finer grid for dfa

Related Error is if attempted:

 R6DS8: Unable to choose the S8 parameter, IExCor= 2523 IXCFnc= 35 ScaHFX=  0.100000 IDFTD=4
 Error termination via Lnk1e in [...] g16.b01_bin/l301.exe at Thu Oct  4 18:59:24 2018.

But it could also be the S6, a1, or a2 parameter. If zero-damping is used, then it could also be S6R.

The answer is quite simple and already given in the manual:

  You can use this empirical dispersion method with other functionals via the IOps(3/174-178) (S6 should be 1.0).

Parameters and the standalone program can be found can be found on the DFTD3 homepage at Mulliken Center for Theoretical Chemistry. Here are links via The Internet archive to the tables for zero-damping and BJ-damping.

The Overlay3 manual can be found on gaussian.com, or The Internet Archive:

  IOp(3/174)

  S6 scale factor in Grimme’s D2/D3/D3BJ dispersion.

0          Default (see subroutine R6DS6).
-1         Set S6 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

  IOp(3/175)

  S8 scale factor in Grimme’s D2/D3/D3BJ dispersion.

0          Default (see subroutine R6DS8).
-1         Set S8 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

  IOp(3/176)

  SR6 scale factor in Grimme’s D2/D3/D3BJ dispersion.

0          Default (see subroutine R6DSR6).
-1         Set SR6 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

  IOp(3/177)

  A1 parameter in Becke-Johnson damping for D3BJ and XDM.

0          Default (see subroutine R6DABJ/XDMABJ).
-1         Set A1 to 0.
NNNNNNNN   A value of NNNNNN/1000000.

  IOp(3/178)

  A2 parameter in Becke-Johnson damping for D3BJ and XDM.

0          Default (see subroutine R6DABJ/XDMABJ).
-1         Set A2 to 0.
NNNNNNNN   A value of NNNNNN/1000000 Ang.

For TPSSh the route section should therefore look similar to the following:

#P TPSSh/def2TZVP           ! verbose, method/ basis set
EmpiricalDispersion(GD3BJ)  ! Switch on D3 with Becke-Johnson damping
IOp(3/174=1000000)          ! Set s6 to 1.000
IOp(3/175=2238200)          ! Set s8 to 2.2382
IOp(3/177=452900)           ! Set a1 to 0.4529
IOp(3/178=4655000)          ! Set a2 to 4.6550
opt(MaxCycle=100,VeryTight) ! Run optimisation
int(ultrafinegrid)          ! use finer grid for dfa


Evaluate mass of salt needed to add to a buffer solution knowing only pH This is an exercise taken from an old exam, I'm struggling with the resolution.

  To a solution of a generic weak acid $\ce{HA}$ were added $\pu{2.40 g}$ of a potassium salt of the $\ce{KA}$. The solution has $\mathrm{pH} = 4.8$. Evaluate how many grams of salt are needed if we want to shift the solution to a $\mathrm{pH} = 5$.

Neither $K_\mathrm{a}$ nor $K_\mathrm{b}$ are given, nor anything. The only thing I got is just confusion.

I think there is all the data you need. $\mathrm{pH}$ of a buffer formed by a weak acid $\ce{HA}$ and its potassium salt $\ce{KA}$ can be found as

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{KA})}{C(\ce{HA})}}$$

On the other hand

$$C(\ce{KA}) = \frac{m(\ce{KA})}{M(\ce{KA})V}$$

where $m$ and $M$ are mass and molecular mass of $\ce{KA}$; $V$ is volume. In general, first equation can be rewritten as

$$
\begin{align}
\mathrm{pH}_i &= \mathrm{p}K_\mathrm{a} + \log{C_i(\ce{KA})} - \log{C(\ce{HA})} \\
&= \mathrm{p}K_\mathrm{a} + \log{m_i(\ce{KA})} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})}
\end{align}
$$

Rearranging:

$$\mathrm{pH}_i - \log{m_i(\ce{KA})} = \mathrm{p}K_\mathrm{a} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})} = \mathrm{const}$$

so that now we can equate conditions for both solutions and find the mass:

$$\mathrm{pH}_1 - \log{m_1(\ce{KA})} = \mathrm{pH}_2 - \log{m_2(\ce{KA})}$$
$$\log{\frac{m_2(\ce{KA})}{m_1(\ce{KA})}} = \mathrm{pH}_2 - \mathrm{pH}_1$$
$$m_2(\ce{KA}) = m_1(\ce{KA})\cdot10^{\mathrm{pH}_2 - \mathrm{pH}_1} = \pu{2.40 g}\cdot10^{5.0-4.8}\approx\pu{3.80 g}$$

So, in order to achieve $\mathrm{pH} = 5.0$, one has to add $\pu{3.80 g} - \pu{2.40 g} = \pu{1.40 g}$ of potassium salt.


What's the bonding in the diborane(6)- hydrogen chloride dimer? I was researching and Googling and researching again, about this very interesting molecule: Diborane.

Then I faced a very interesting structure, something that I have never seen before. Something that turns up lesser than 50 results in Google, most of which are behind paywalls, and thus leave me in pain of not understanding this species. I'm talking about diborane - hydrogen chloride.

Source

So, I'd like to understand this species better.

The bonding certainly doesn't seem ionic. What is the attraction force between $\ce{HCl}$ and $\ce{B2H6}$? There must be one or they wouldn't have written these two compounds together.

Edit: I'm specifically looking for the hydrogen chloride - diborane(6) dimer you can see in the image. And the Cl is pointing towards the H in diborane, so this isn't about dipole/induced dipole interactions. You can find more info about it in here. Interestingly, in calculating molar mass, both the $M$s of $\ce{HCl}$ and $\ce{B2H6}$ are taken into account.

I find this question to be quite interesting as it clearly pertains to non-covalent interactions.

I have analyzed two conformations of the diborane-HCl dimer system, one with the chlorine on HCl orientated toward diborane (HCl-Dib) and one with the chlorine oriented away from the diborane (ClH-Dib).  For comparison, I have similarly analyzed the Water-HCl dimer with the chlorine oriented toward the two hydrogens on water.

Methodology

Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with Gaussian 09 using Grimme's B97D density functional along with the 6-31+G(d',p') basis set.  All geometries were characterized to C$_{2V}$ symmetry.  Symmetry adapted perturbation theory (SAPT) was implemented at the SAPT2+3(CCD) level using the Psi4 software package.  These computations implemented Dunning's heavy-aug-cc-pVTZ (haTZ) basis set (i.e. aug-cc-pVTZ for non-hydrogen atoms and cc-pVTZ for all hydrogen atoms).  Finally, natural charges were quantified using the Natural Bond Orbital (NBO) analysis routine available in Gaussian 09 (see http://nbo6.chem.wisc.edu/ for more details).

Results

Both diborane-containing complexes were minima on the B97D potential energy surface.  The H$_2$O-HCl dimer is a higher order saddle point exhibiting two imaginary modes of vibration.  The interaction energy ($E_{\mathrm{int}}$) can be quantified by attractive and repulsive contributions, the former consisting of electrostatics ($E_{\mathrm{elst}}$), induction ($E_{\mathrm{ind}}$), and dispersion ($E_{\mathrm{disp}}$) and the latter consisting of exchange ($E_{\mathrm{exch}}$). The table below summarizes the SAPT results.

Clearly, the dominant attractive force in each case is dispersion.  The ClH-Dib complex exhibits the strongest interaction energy and contains a dipole that points from the diborane monomer to the chlorine.  The HCl-Dib complex has an interaction energy that is about 1 kcal mol$^{-1}$ smaller in magnitude and exhibits a dipole pointing toward the diborane monomer.  The analogous H$_2$O-HCl system is qualitatively similar to the HCl-Dib complex in terms of the magnitudes of the components to the interaction energy.  Surprisingly, this dipole-dipole interaction is slightly weaker (by about 0.1 kcal mol$^{-1}$) than that seen in the dipole/induced-dipole system of HCl-Dib.

I have included below figures for the HCl-Dib, ClH-Dib, and H$_2$O-HCl complexes, respectively, with corresponding natural atomic charges shown on each atom.

Figure 1 - HCl-Dib

Figure 2 - ClH-Dib

Figure 3 - H$_2$O-HCl

Interestingly, when chlorine is pointed toward diborane (Figure 1), the natural atomic charges for diborane remain rather symmetrical and diborane barely exhibits an individual dipole moment.  The H$\cdots$Cl separation between the two monomers is about 3.4 Angstroms, a long-range interaction.  However, when chlorine is pointed away from diborane (Figure 2), the H$\cdots$H interaction is much smaller, exhibiting a separation of about 2.2 Angstroms.  When considering water, the separation increases once again (Figure 3) to about 3.2 Angstroms.

Conclusions

The dominate attractive force to the interaction energy for these dimer systems is dispersion, though we see a competing contribution from electrostatics in ClH-Dib (Figure 2).  Surprisingly, the orientation of the halogen significantly affects the geometry and energetics of these systems.  When chlorine is oriented toward the two hydrogens, a weaker interaction and a larger inter-monomer separation is seen.  However, when hydrogen is oriented toward the two hydrogens, the interaction energy increases in magnitude and the inter-monomer separation decreases.

Q/A

I find it more convenient to address some of Martin's comments within the answer itself (sometimes responses can become quite long).

Which compound is lower in energy?

Based on absolute energies, H$_2$O-Cl is the lowest in energy of the three dimers (by about 23 Hartrees).

How stable are they wrt decomposition?

The interaction energy will give you this information (defined as the energy difference between the interacting monomers in an optimized dimer geometry and infinitely separated monomers adopting the geometry they have in the dimer complex).  Given that the magnitude of $E_{\mathrm{int}}$ is less than 2 kcal mol$^{-1}$, I would say that these are not very 'stable'.

Have you tried asymmetric states?

No.  I do not believe breaking symmetry in the diborane-containing complexes will change the energy/geometry appreciably.

I find it quite surprising, that the BH--ClH distance is much longer (0.6A) than the sum of the respective vdW radii, while the BH--HCl distance is about the sum of the vdW radii. Still, in both cases there should be (about) no contribution from the orbitals. How do the bond lengths change in the two conformations?

In the diborane-containing complexes, the bond lengths deviate by about 0.01 Angstroms (so... no change).

Can you make a rough comparison between the B--Cl distance in the two cases?
They appear to be quite similar.

In HCl-Dib, the B$\cdots$Cl separation is about 3.79 Angstroms.  In ClH$\cdots$Dib, the B$\cdots$Cl separation is about 3.86 Angstroms.  Similar they are indeed.


Can weathering change the composition of something so that it appears different? There's a murder trial ongoing in Ireland at the moment and one of the pieces of evidence is a spade found at the scene along with the skeletal remains. It was contended by the prosecution that this spade was belonging to the defendant but while they were similar

  "forensic evidence found the paint splatters on the spade were not the
  same as on the [defendant's] garden fence"

My question is: can something left out in the open be changed either chemically or structurally by the environment such that it appears to be similar but not the same? If so why would this not have been called out as a caveat by the expert forensics witness?

The specifics of the chemical tests weren't gone into in the media (sample article here) so I don't know how accurate it was.

Yes, weather can and probably will change the composition of materials and make them different.

However, forensic analysis usually takes that into account. There are some properties that might change because of weathering (you shouldn't compare water content, for example), but others won't. According to the sample article, the difference was found in the chemical composition of the paints.  Since details were not disclosed, one can only guess what these differences are, but some possibilites are:

Presence or absence of metals in the paints
Substances that wouldn't appear or disappear because of wheatering
Different isotopic ratios of elements
Traces of the solvent indicating they were not the same
Probably others

Assuming the analysis were performed correctly, it is usually not hard to say two samples are different. On the other side, when the results are similar, you can hardly say they are equal, but only that they match and that no differences were found.


Correlating the decrease in orbital radius and penetrating power with the increase in l value I have read that s-orbitals have a stronger penetration effect as compared to p-orbitals and p-orbitals have a stronger penetration effect as compared to d-orbitals, etc.

Therefore, the electrons in an s-orbital are closer to the nucleus of the atom and their ionisation energies are greater than for an electron in a p-orbital. (Example: This is what we use to explain why the energy of ionisation decreases from $\ce{Be}$ to $\ce{B}$).

Everything made perfect sense to me until I read:

  The orbital radius slightly decreases with an increase of $l$ value.

Since s-orbitals penetrate more, shouldn't that mean that being closer to the nucleus, they have smaller radius as compared to an orbital with higher $l$ value, say, the p-orbital?

How can we justify this? Or are these two pieces of information not related in this way?

tl;dr

The statement that "the orbital radius slightly decreases with the increase of $l$ value" is true only for orbitals with the same value of $n$ (shell). The penetration of the nucleus by an electron is measured by the relative electron density, which depends on both shell ($n$) and subshell ($l$) of an electron in an atom.

What is Orbital Penetration:

  Penetration describes the proximity of electrons in an orbital
  to the nucleus. Electrons which experience greater penetration
  experience less shielding and therefore experience a larger Effective
  Nuclear Charge ($Z_\text{eff}$), but shield other electrons more effectively. Electrons in different orbitals have different
  wavefunctions and therefore different distributions around the
  nucleus. However, contrary to what many think, penetration is not the
  outer electrons penetrating through the shield of the core electrons.
  It is actually just how well the electrons feel the nucleus. This is
  similar to the idea of outer electrons penetrating, but not the same.
  They are not the same because the core electrons have more penetration
  than the outer electrons since they (the core electrons) feel the
  strongest pull.

Clarification

As you can see from the definition above, the penetration of a nucleus by an electron is measured by the relative electron density near the nucleus of an atom. It is essentially, how effectively electrons can get close to the nucleus.  For instance, below are the electron probability densities for s orbitals and a p orbital.

Nodes are regions of zero electron probability (white areas),
The orange color corresponds to regions of space where the phase of the wave function is positive, and the blue color corresponds to regions of space where the phase of the wave function is negative.

In a multi-electron system, the penetration of the nucleus by an electron is measured by the relative electron density near the nucleus of an atom for each shell and subshell of an electron. For example, the 2s electron is penetrating the nucleus of an atom more than the 2p electron, because the 2s has more electron density near the nucleus than the 2p electron. The penetration power of an electron, in a multi-electron atom, is dependent on the values of both the shell (n) and subshell (l) of an electron in an atom.

For the same shell value (n) the penetrating power of an electron follows this trend in subshells:

$$\ce{s > p > d > f}$$

And for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend:

$$\ce{1s > 2s > 2p > 3s > 3p > 4s > 3d > 4p > 5s > 4d > 5p > 6s > 4f ...}$$


Why is boric acid soapy to touch? Bases are soapy to touch because of soap formation on contact with lipids present on our skin. If this is the case then why and how is boric acid soapy to touch?

From what I've been able to find, it is similar to how graphite acts as a lubricant. Boric acid is flat and it hydrogen bonds to form layers of flat networks. These slide on each other like layers of graphite do, reducing friction.


How long do the bonds in a molecule hold under nominally stable conditions? Disclaimer: I am not a chemist

I was just drinking my Ice Age mineral water which is 'untouched since the Ice Age' and wondered if the bonds formed in each molecule have remained intact for all those years.

In a liquid that is just lying there underground I'm guessing that there is some, if not very energetic, movement of the molecules. What happens when they bump into each other? Might an atom become dislodged and bind with another molecule?

I have no idea but the thought interested me.

Water with its fluctuating hydrogen bond networks is a special case. If you feel for it, have a look at Water clusters: Untangling the mysteries of the liquid, one molecule at a time, but it doesn't answer your question ;-)

It is more interesting to have a look at paleochemical/geobiological findings.

A complex organic molecule, such as 2$\alpha$-methylhopane was isolated from

  2700-million-year-old shales from the Pilbara Craton, Australia
  (Source)

Have look at the website of Roger Summons at MIT and his freely available publications, such as The Exceptional Preservation of Interesting and Informative Biomolecules.

In short, bonds can last a pretty long time!


Why hydrogen bonding in some acids make them a stronger acid when it is present even before deprotonation? If we take the example of salicylic acid, hydrogen bonding is present in the acid as:

Even after deprotonation, it has intramolecular hydrogen bonding as:

My doubt:
p-hydroxybenzoic acid($\mathrm pK_\mathrm a = 4.54$) is less acidic than o-hydroxybenzoic acid($\mathrm pK_\mathrm a = 2.97$), the reason being hydrogen bonding in the latter after deprotonation.
If the same reason is accounted for the acidity of Salicylic acid (o-hydroxybenzoic acid) then why is the H-bonding before the deprotonation not considered (which should decrease the acidity of the compound)?

Some of the conformers of salicylic acid are given below, all of which are conformers planar in configuration.$\mathrm{^{[1]}}$

Although the compound can adopt several conformations the first conformer(in the first image, the structure on the left) is of the lowest energy and has a strong intramolecular hydrogen bond. The second conformer(first image, right structure) is the conformer with the second lowest energy as described in the published paper as given below:

The molecule can adopt several conformations, although only one of them is highly populated and was assigned in IR experiments. The spectrum was recorded by Fiedler et al., $\mathrm{^{[2]}}$ in tetrachloride solution and indicated a strong intramolecular H-bond. The deviation of the OH
−1
stretching frequency from that in phenol was 395 cm . The theoretically calculated deviation is
359 cm−1 at the B3LYP/6-311+G(d,p) level. The lowest-energy conformer is planar, the phenolic OH is a H-bond donor to the carbonyl oxygen of the syn carboxylic group. The =O...H distance and the =O...H–O bond angle were calculated at 176 pm and 145°, respectively. The second-most-stable conformer is higher in energy by 14.3 kJ/mol, where the phenolic OH is the H-bond donor to the syn carboxylic OH. Similar conclusions were drawn by Yahagi et al.,$\mathrm{^{[3]}}$ by interpreting the gas-phase IR frequencies of the phenolic OH.

Much more convincing is the fact that the lowest energy conformer has a fractional population distribution$\mathrm{^*}$ of nearly 99.7% which further supports the argument regarding the higher acidity of salicylic acid in comparison to its para isomer. The higher acidity of the compound thus is contributed by the first conformer. The lesser interaction of the carboxyl group’s acidic proton and the stability of the conjugate base due to hydrogen bonding thus increases the acidity of the compound. Regarding detailed mechanism of acidity of salicylic acid you may refer this and this post in chemistry SE. You may further note that the other structures might have decreased the acidity of the compound if their population distributions would have been higher. The same has been mentioned in the published paper as given below:

In a former calculation by Nagy et al.$\mathrm{^{[4]}}$, the two conformers above were found also to be the most stable with MP2/6-31G*//HF/6-31G* energy separation of 13.7 kJ/mol and free energy difference of 12.1 kJ/mol at T = 298 K. All other conformers are much higher in free energy, supporting the estimate of Fiedler that the population of the lowest-energy form is 99.7%.

Notes and References

Nagy, Peter. Competing Intramolecular vs. Intermolecular Hydrogen Bonds in Solution. Int. J. Mol. Sci. 2014, DOI: 10.3390/ijms151119562
Fiedler, P.; Böhm, S.; Kulhánek, J.; Exner, O. Acidity of ortho-substituted benzoic acids: An infrared and theoretical study of the intramolecular hydrogen bonds. Org. Biomol. Chem. 2006, 4, 2003–2011. DOI: https://doi.org/10.1039/B601875K
Yahagi, T.; Fujii, A.; Ebata, T.; Mikam, N. Infrared spectroscopy of the OH stretching vibrations of jet-cooled salicylic acid and its dimer in S0 and S1. J. Phys. Chem. A 2001, 105, 10673–10680. DOI: https://doi.org/10.1021/jp0126199
Nagy, P.I.; Dunn, W.J, III.; Alagona, G.; Ghio, C. Theoretical studies of the 2- and 4-hydroxybenzoic acids with competing hydrogen bonds in the gas phase and aqueous solution. J. Phys. Chem. 1993, 97, 4628–4642. DOI: https://doi.org/10.1039/B601875K
$\mathrm{*}$ Note that here fractional population distribution refers to the population of each conformer in the compound. You may find more information about it on the Wikipedia article.


Why are noble gases stable? Generally, noble gases are very unreactive. Why?

I understand that it has to do with electron shielding prospective electrons that could join the atom or leave and the energies associated with both actions, but why are 5 electrons in the p orbital (to take Fluorine as an example) so much worse in shielding the nuclear charge than 6 (Neon)?

What IS electron shielding, anyway? I understand it in the classical lie-to-children way, but not satisfactorily.

It's all about the energy levels of electrons. Electrons exist in different energy levels and each energy level can only be occupied by a specific number of electrons (google quantum theory).

Example: Neon has an electron configuration of $1s^2 2s^2 1p^6$, which simply means that it has a full $1s$ orbital (the maximum is 2 electrons as denoted by the superscript), and has a full valence shell which is the $2s$ and $1p$ orbitals (which is the valence shell for molecules in row 2).

Now for neon to lose an electron (to bond) it would have to take an electron out of a full orbital ($1p$), which is what makes molecules reactive (a non-full valence shell), and would just take the electron right back.

Think of Ne$^+$ as more electronegative (wanting electrons) than elemental Fluorine, so it will pull an electron from almost anything and go back to its elemental form.

If this answer is not satisfactory then I can explain further, without diving head first into quantum theory.


How to determine which ions will pair to form a compound in an aqueous solution? Let us assume we have an aqueous solution with the ions $\ce{A+}$, $\ce{B+}$, $\ce{C-}$ and $\ce{D-}$. Which ions will form a bond and why? Will the compound formed be $\ce{AC}$ and $\ce{BD}$ or $\ce{AD}$ and $\ce{BC}$?

Is there any general rule to identify the compounds we have at the end?

Your question arises because you believe that ions in an aqueous solution remain bonded with each other. That isn't true though! In fact, ions in an aqueous solution freely roam around throughout the solution.

Note that most ionic compounds have ions bonded to each other in solid state only. Once you dissolve that ionic compound in water, the ions quickly dissociate (the ionic solid dissolves) and become solvated (surrounded by water molecules).

So, to answer your current question: no, there won't be any compound formed, and all the ions would be separate from each other.

Though, another possibility is of precipitation. In that case, the ions involved ($\ce{AD}$ for example) would separate out of the solution and settle down as a solid. Indeed, there may be more than one possible precipitates. So, to decide between any given number of compounds, as to which precipitates first, we'll need to check if their ionic product exceeds their $K_\mathrm{sp}$. The compound for which this happens earlier, due to lower $K_\mathrm{sp}$, would precipitate earlier.

Bonus content: Your question would make more sense if the ions were given in gas phase instead. Now, when the oppositely charged ions would come together, they would stabilize each other via the large amount of electrostatic energy released, and thus forming a stable compound.

Whether $\ce{A+}$ would combine with $\ce{C-}$ or $\ce{D-}$, would depend on their energy of interaction $\left(=\displaystyle -\frac{Kq_1q_2}{r}\right)$. More the energy released for a particular bond, more would be the probability of formation of that bond.


Is associative ligand exchange possible with octahedral complexes? Can associative ligand exchange produce an octahedral intermediate as long as the initial complex has 16 electrons (unsaturated)?

Not sure. all I ever see is a square planar turning into a trigonal bipyramidal intermediate?

It could — and there isn’t even a general restriction on the electron count.

Any pentacoordinated species can easily be expanded to a hexacoordinated one. It is just a question of the ligand approaching from the correct angle. In fact, I would even want to prefer associative ligand exchange for pentacoordinated species over dissociative since there is less reorganisation involved (only two ligands need to shift positions compared to at least three in dissociative).

Also, it should be mentioned that the number of electrons in each structure does not have to be 18 or less. Many (pseudo)octahedral complexes such as $\ce{[Co(H2O)6]^2+}$, $\ce{[Cu(NH3)4(H2O)2]^2+}$ or $\ce{[Ni(H2O)6]^2+}$ have more than 18 electrons and are fully comfortable with it.


What is known about the phase of matter at very high temperature and enough pressure? Are there theoretical reasons for or against the existence of materials that will not become solid at some fixed, sufficiently large temperature when enough pressure is applied? Or is this even the normal state of affairs?

The usual phase diagrams typically only give a finite portion of the thermodynamical state space, probably as more cannot be experimentally tested.

But theory might tell us what happens, one way or another.

There is a point in which it is too hot for a molecule to remain bonded and will break apart, no matter how much pressure is applied. Now with more pressure the electrons and nuclei will start free flowing in what is known as a plasma. And add a little more heat and fission happens.

So basically the reason they don't show you any more of the curve because your molecule under enough pressure and heat is no longer a molecule.


Why are tertiary carbocations the most reactive, if they're already stable? This may seem silly, but doesn't it seem weird for a compound that's stable (in this context, the tertiary carbocation) to be the most reactive?

I mean, wouldn't it be the least, given that it's already stable and wouldn't want to leave that stable configuration. On the same lines, the methyl carbocation sounds like the most reactive due to its high instability, right? wouldn't it want to react in order to form a stable compound?

This is an excellent question. Please correct me if I'm wrong, but I think this is what you are grasping at:

First, it is true that tertiary carbocations are generally more stable than primary carbocations (and secondary carbocations) due to having more inductively donating alkyl groups. The hyperconjugative effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations.

Second, transition states involving tertiary carbocations as opposed to primary carbocations are more favorable precisely because tertiary carbocations are more stable than primary carbocations.

Note that this does not mean that tertiary carbocations are more reactive. We don't generally say that something which exists in a transition state is reactive/unreactive. One reason is that the transition state complex cannot be directly captured or observed. Another reason is that it is the transition state ... of course it's going to be reactive - of course it'll change very rapidly!

We may however say that reactants and products are either reactive or unreactive. So if something goes through a transition state involving a tertiary carbocation, it might well be more reactive than something which goes through a transition state involving a primary carbocation.

Of course, before labeling anything as reactive or unreactive, you'd do well to note exactly what you mean by reactivity. To put it lightly, there are myriad reactions. An alkyne such as acetylene is reactive with regard to combustion; acetylene torches are commonplace. However, acetylene isn't going to be reactive with regard to, say, backside attack (for a variety of reasons).

Ultimately I think your issue lies with semantics rather than the chemistry ... still, a good question!


Grignard reaction with cycloheptatriene I'm having trouble with the following set of reactions and their products.

Product A is achieved by exploiting the unique aromaticity of cycloheptatriene cations. When A  is treated with the Grignard reagent $\ce{MeMgI}$, a mixture of B and C is obtained. Treating A with catalytic amount of $\ce{HCl}$ in $\ce{CCl_4}$ furnishes isomer A’, which is again converted into a mixture of B and C upon treatment with Lewis-acidic Grignard reagents.

For A I got the following structure, since I could not figure out how the aromaticity of the first structure is affecting the reaction.

This cannot be correct, since the grignard reaction would produce only one product and not a mixture of two.

Having done some reasearch online on the resonance structures of cycloheptatriene, I thought of the following structure for A. But this would not produce two different products either if I am not mistaken.

The product A, shown below, is formed by first reducing the ketone to a secondary alcohol with $\ce{LiAlH_4}$ and then by adding a methyl group to form an ether using the reagent $\ce{CH_3I}$.
$\hskip2in$
Next the grignard reaction with product A will form two products. The grignard reagent $\ce{MeMgI}$ can attack A in a $1$,$2$ or in a $1$,$4$ mode, due to the unique aromaticity of cycloheptatriene, giving a mixture of products B and C, the structures of which are shown below.
$\hskip1in$


How often is the IUPAC Goldbook Updated? I was wondering how often does the IUPAC update its Gold Book, ie, IUPAC Compendium of Chemical Terminology - the Gold Book.

It poped in my mind when I saw this question about jellium aromaticity, that the term should be there. However, it was not. Whereas, the term has been used several times in the linked references (1 2), which I like to consider credible.

So, I was wondering how up-to-date the book is and how much can one rely on it for latest information.

PS While I understand that this question is primarily and heavily opinion based, I think it is equally relevant to all of us as we delve further in the chemical world.

As various users have contributed to the arrival of this answer, it goes to the  community.

Firstly, the IUPAC is the authority of chemistry, which decides on and accepts terminology for international compatibility so that information can be unambiguously expressed by researchers, scholars, educators and students. So, it can be finally concluded that unless it is IUPAC approved, it is not official terminology.

Secondly, since the IUPAC is an enormous organisation, dealing at a global level, it can be understood that changes will be a little delayed as the information has to be checked for authenticity and a common end has to be reached. For example, it took quite some time to decide on the newcomers of the Periodic Table(Elements of Atomic Number 113-118).

Thirdly, a lot of terminology has a very limited scope and can be at the very least left out. For example, the iupac tag was dropped due to its limited scope.(See comments above).

Lastly, I thank the community for the comments which have fabricated the "bonds" of this answer.


What is it about the relationship between the Fe²⁺ and Fe³⁺ in magnetite that makes it diamagnetic? Hematite is composed of $\ce{Fe2O3}$, and is paramagnetic, whereas magnetite is $\ce{Fe3O4}$ and is diamagnetic.  Magnetite's nature is due to the presence of both $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ (Wikipedia even goes so far as to call it $\ce{FeO}\cdot \ce{Fe2O3}$, but explains that this is not a "solid solution").

I can surmise that in the crystalline solid, there are divalent and trivalent cations dispersed within.  I can also assume that there might be some sharing of electrons between the oxygens and each of these cations.  One would assume this also happens within the hematite crystal, so what is it about "tossing" the divalent cation into the mix of the crystalline structure that makes the magnetite diamagnetic?

Does the charge differential between the two cations cause a permanent dipole, if so, why don't all of the small dipoles simply cancel each other out?

Magnetite has a spinel structure with two types of $\ce{Fe}$ sites: octahedral and tetrahedral, respectively in green and orange below.

(in this structure, oxygen atoms are the small red balls).

This structure is called a spinel, with $\ce{Fe^{2+}}$ ion in tetrahedral coordination and $\ce{Fe^{3+}}$> ions in octahedral sites. The coupling between atoms from these two lattices are superexchange interactions, which result in an antiparallel alignment of spins. However, because the magnetic moments of the two lattices are not equal, their addition results in a net magnetic moment, even though they are of opposition alignment. This is called ferrimagnetism, and in practice most of its macroscopic properties are similar to ferromagnetism, though they stem from a completely different microscopic magnetic ordering.


What is the difference between SBA-15 and MCM-41? I'm trying to understand the difference between these two mesoporous materials.  They both have the same polys-siloxane backbone and surface silanol groups, the same hexagonal pores, and yet I don't understand why they are different materials.

The difference between your two materials is primarily in pore geometry: for MCM-41, the pores are hexagonal; for SBA-15, they are round. Further, there are apparently differences in wall thicknesses.

  Using high-resolution transmission electronic micrograph (HR-TEM) observation, one can clearly see the pore geometry of the MCM-41 and SBA-15 mesoporous silicas to determine that their pore shapes are hexagonal and round, respectively. With the perpendicular orientations of the nanochannels to the electron beam, parallel line images of the (100) and (110) repeating spacings were observed. In the SBA-15 mesoporous silicas, there are byproducts of the granular silica and disordered mesostructures, attributed to the weak hydrogen interactions between Pluronic 123 blockcopolymer and the silica species. There are also many different and significant +π disclination defects in SBA-15 and MCM-41 surfactant-silica composites. The SBA-15 with a thicker silica wall is more stable under irradiation by high-energy electron beams compared to MCM-41, which has thinner wall thickness. Some carbon nanostructure impurities were found in some carbon films on the metal grids.

Reference: Lin, Tang, and Lin. Detailed Structural Characterizations of SBA-15 and MCM-41 Mesoporous Silicas on a High-Resolution Transmission Electron Microscope. Journal of the Chinese Chemical Society 49(6) 981-988 (2002). Abstract here.


Mass number of isotopes of copernicium denoted with 'm' On this website:

http://en.wikipedia.org/wiki/Isotopes_of_copernicium

there is a sentence "This isomer is unconfirmed" as a footnote explaining two isotopes.
What do they mean by 'isomer' in this context?

Is it possible that an element may have different spatial distribution of neutrons and protons that could have an impact on its stability?

The 'm' does indeed stand for metastable. The word 'isomer' in this sense refers to nuclear isomers, in which at least one of the nucleons is in an excited state. Think of it as somewhat analogous to an electron in an excited state, but occurring within the nucleus. Just like electrons release photons when an excited state decays to the ground state, when nuclear isomers decay to the ground state (which is different than 'normal' nuclear decay like $\alpha$ or $\beta$ decay), a gamma ray is release.