Why does the emission wavelength of sodium appear to be an exception despite the trend for ionisation enthalpies among alkali metals? The wavelengths for emissions in the visible region for Group 1 elements is as follows:

$$
\begin{array}{lc}
\hline
\text{Element} & λ/\pu{nm} \\
\hline
\ce{Li} & 670.8 \\
\ce{Na} & 589.2 \\
\ce{K} & 766.5 \\
\ce{Rb} & 780.0 \\
\ce{Cs} & 455.5 \\
\hline
\end{array}
$$

Now, considering that the ionisation enthalpies (I.E.s) decrease going down the group, it should be relatively easier to remove an electron, which should reflect in the energy band gaps and ultimately in the wavelengths. Thus, there should be increase in this trend. However, $\ce{Na}$ is an exception. Why so?

I think the $\mathrm{d}$-electrons play a role in the increased band gap for $\ce{Cs}$, which shows in the trend. But I am missing something for $\ce{Na}$. Is its I.E. decreases because size outweighs nuclear charge along with fair amount of screening? But $\ce{Na}$ has a unique balance wherein size and screening are not enough to tip over.

On the NIST website you can look up the transitions corresponding to the wavelengths you list:

$$
\begin{array}{l c c c c}
\hline
\text{Element} & λ/\pu{nm}& |i\rangle & |f\rangle & \text{I.E./cm$^{-1}$}\\
\hline
\ce{Li} & 670.8 & 1s^22s & 1s^22p & 43\,487\\
\ce{Na} & 589.2 & 2p^63s & 2p^63p & 41\,449\\
\ce{K} & 766.5 & 3p^64s & 3p^64p & 35\,010\\
\ce{Rb} & 780.0 & 4p^65s & 4p^65p & 33\,691 \\
\ce{Cs} & 455.5 & 5p^66s & 5p^67p & 31\,406\\
\hline
\end{array}
$$

You see that the transition in Cs that you give in the table corresponds to a $7p\leftarrow 6s$ transition and that for consistency you should compare it with a $6p\leftarrow 6s$ transition which has a wavelength around 860 nm (no $d$ electrons involved).

The term values of the electronic states of multi-electron atoms are relatively well described by Rydberg's formula

$$
\tilde{\nu}_{n,\ell}=\text{I.E.} - \frac{\mathcal{R}_M}{(n-\delta_\ell)^2},
$$
with $\mathcal{R}_M$ the mass-corrected Rydberg constant and where the quantum defect $\delta_\ell$ depends on the orbital angular momentum of the electron $\ell$ and on the properties of the ion core. The transition $n'p\leftarrow n''s$ can therefore be approximated by
$$
\tilde{\nu}_{n',\ell=1}-\tilde{\nu}_{n'',\ell=0}=\frac{\mathcal{R}_M}{(n''-\delta_{\ell=0})}-\frac{\mathcal{R}_M}{(n'-\delta_{\ell=1})},
$$
and thus depend on the quantum defects for the $s$ and $p$ states. The quantum defects depend on the scattering properties of the Rydberg electron with the ion core and it is difficult to predict their values for different elements. It is therefore difficult to make quantitative predictions. Also for the heavier elements (higher nuclear charge $Z$), relativistic effects become more important, which influences the electronic structure.

Note that these lowest states are not really Rydberg states and the Rydberg formula is a bad approximation. For high values of $n$ the Rydberg formula gets much more accurate. In fact, it allows for a very accurate determination of the ionization energies of these elements.


What compound(s) contribute to bleach's smell and are they dangerous? Just a few hours ago, I had to deal with a mold problem on some untreated wood and was told by a hardware store worker that bleach was an effective way to deal with the problem, following up with some wood filler paste. I did as I was told and left the room airing out with a window open and two fans running for multiple hours. In fact, they are still running right now; I was surprised to find that the smell of bleach was still lingering in the room. I also noticed that some bleach had gotten onto my hands and they smelled of bleach even after multiple washings with soap and water.
What chemical(s) are causing the smell to linger, in the air as well as on surfaces? Are they particularly dangerous, or just relatively mild product(s) from the degradation? Is there a way to eliminate the smell without producing harmful products?
I used a spray bottle with undiluted bleach and did not wipe anything up with damp paper towels. Even so, I assume that it isn't really too dangerous, given that it is unlikely I used more than 10 mL of the stuff, and there is no way most of that 10 mL is still aerosolized and/or inside the room. Regardless, I have looked on other questions in this StackExchange, and a lot of them talk about the breakdown of bleach, but I just want to be certain the room is safe enough to sleep in tonight. Thanks!
EDIT: After removing the cloth used to clean the wood (I don't know why I left it in there), the smell is less pungent but still noticeable. Figured this might be useful information to include.

Bleach is alkaline sodium hypochlorite solution, that eventually soaked inside the wood and is gradually releasing gaseous chlorine.
It is merely just disturbing and eventually irritating mucous tissues in longer term, not really dangerous. Some chlorinated public pools have strong chlorine smell as well, being smelled from swimming suit for long time until washed in the wash machine.
A countermeasure could by applying suitable diluted reducing solution, like ascorbic acid, pyrosulfite, thiosulphate ( big crystals from photo fixers ) or similar, that would react with hypochlorite and chlorine.
Questionable can be applicability on wood surface for estetic reasons. But if bleach was applied, this should not make things worse.


Spatial Wavefunction Symmetries and Identical Particles I was reading this and it mentions in the 3-electron section, that for a spacial wave function to be symmetric under fermion swapping, it must be a function of even parity. Similarly for anti-symmetry under fermion swapping, it must be a function of odd parity.

It is not immediately obvious to me why parity symmetry $(-1)^l$ has anything to do with the bosonic or fermionic properties of the spatial wave function. So I suppose my questions are:

Is it true that spatial inversion is the same symmetry as swapping?

And if so, why?

When we talk about exchanging electron $i$ with electron $j$, we are actually changing the wavefunction according to

$$\Psi(..., x_i, ..., x_j, ...) \to \Psi(..., x_j, ..., x_i, ...).$$

The operation is taken by the parity operator $P$. Applying it twice would
return the wavefunction to its original form. So the following eigenvalue equation is satisfied

$$ P^2 \Psi = 1 \Psi. $$

Then the $\lambda$ eigenvalue of the $P$ operator ($P\Psi = \lambda \Psi$)
also satisfies an analogous equation

$$ \lambda^2 = 1. $$

$\lambda$ can only be either $\pm 1$. If it is positive the wavefunction is symmetric under swapping and if the eigenvalue is negative the wavefunction is antisymmetric under swapping. It can also be said that the symmetric (antisymmetric) wavefunction has even (odd) parity but do not confuse it with spatial symmetry.

It is assumed that fermions must be described by antisymmetric wavefunctions
and bosons by symmetric wavefunctions. Many have argumented this fact alluding
to the spin-statistics theorem but others have, as well, stated that such justification is not satisfactory. Besides, it is not completely clear why.


What is lowest energy state, how does an atom gain it and what does it have to do with hybridization? I understand that hybridisation occurs due to the atom wanting to exist in the most stable state possible, and the most stable state has the lowest energy. What I don't understand, however, is say, in CH4, the bond angle is 109°18', so how does this particular value mean that the atoms and bonds in methane are farthest from each other? And if they are indeed farthest from each other, how does that mean the CH4 molecule's in the lowest energy state?

VSEPR stands for Valence Shell Electron Pair Repulsion.  It's a complicated acronym, but it means something that's not difficult to understand. Basically, the idea is that covalent bonds and lone pair electrons like to stay as far apart from each other as possible under all conditions (to minimize electrostatic repulsions of different pairs of electrons) .  This is because covalent bonds consist of electrons, and electrons don't like to hang around next to each other much because they have the same charge. This, of course, allows the molecule to be in its lowest energy state.

This VSEPR thing explains why molecules have their shapes.  If carbon has four atoms stuck to it (as in methane), these four atoms want to get as far away from each other as they can.  This isn't because the atoms necessarily hate each other, it's because the electrons in the bonds hate each other.  That's the idea behind VSEPR.

Now, one problem with the whole VSEPR thing is that if you have four things stuck to carbon, for example, there are no orbitals that want to get 109.5 degrees apart from each other (109.5 degrees corresponds to the geometric maximum distance the atoms can get apart).  After all, s-orbitals go in a complete sphere (360 degrees) and p-orbitals are 90 degrees apart.

What happens instead of using s- or p- orbitals is that when covalent bonds are formed, the s- and p- orbitals mix to form something called hybrid orbitals.  "Hybrid" just means "mixture of two different things", and that's exactly what a hybrid orbital is.  When three p-orbitals with 90 degree angles combine with one s-orbital with 360 degrees, they average to form four sp3 orbitals with 109.5 degree bond angles.  Depending on the numbers of s- and p-orbitals that mix, you can get a bunch of different bond angles.


Is there any collection of charge density maps for small molecules? Any good data at all? Despite wide variety of existing experimental methods producing electron/charge density maps, I failed to find a good dataset of experimental data for small (1-3 atoms) molecules.

There are datasets for integral properties such as https://cccbdb.nist.gov/ . It is fairly trivial to find atomic spectra data on NIST site and with some effort it is possible to obtain electronic spectra data for some molecules (though apparently there is no good unified database).

Regrettably, I failed to find any collection of experimental charge density data for small molecules. Furthermore, on second though I cannot remember where charge density maps in QC textbooks come from. Calculated charge densities are fairly easy to obtains, but experimental charge densities are a different matter. In fact, "experimental charge density hydrogen molecule" search gave me zero relevant references.

Any suggestions?

My background is macromolecular crystallography; while experimental data was kept locked in the research laboratories in the 1980s, they are now widely available through the international protein database. For example, you can view the electron density of crambin at http://www.rcsb.org/3d-view/1EJG.

The database for smaller structures is the Cambridge Structural Database. They have implemented depositing experimental data (structure factors) in the 2000s, but I did not find an efficient way to search for those structures that come with structure factors.

Charge density maps are often visualized as 2D-sections of the experimental density minus the density expected from isolated atoms (deformation density). One software program to calculate these is MoPro. Another program is CrystalExplorer, which shows deformation densities in 3D. In all cases, you would have to have access to the structure factors to make your own figures.

  Is there any collection of charge density maps for small molecules? Any good data at all?

I don't think so.


Only-C chain molecules Are (cyclic) $\ce{C}$-chain molecules as in the picture likely to occur or do they exist at all?

 and so on...

EDIT:

I'm considering only double bonds between $\ce{C}$ atoms. This rules out most common allotropes.

The term you are looking for is "allotropes", and carbon has a whole bunch of them, but the three most readily spotted in nature are graphite, diamond and fullerene, however, none of them are simple chains such as those postulated in the answer.

Graphite exists in large honeycomb flakes held together on two axes by covalent bonds, and on the third by weaker Van der Waals forces, which is why they come off readily under mechanical shear, leading to their popularity in pencil lead.

Diamond are huge, multiply interlocked complexes of crystalline carbon, although often with impurities locked into their structure, perverting the crystal structure and yielding interesting colors as a byproduct.

Fullerene is a molecular allotrope, consisting (usually) of 80 carbon atoms arranged in a buckyball. Initially thought to be a lab-grown species, it has since been identified as a natural result of certain combustion processes.

Now, none of these are chains like what you describe; while some of the compounds you sketch out are likely possible and present, they are also likely to be highly reactive, perhaps even to the point of decomposing spontaneously.

While amorphous carbon chains may very well exist, without a crystalline structure to shield their electrons, their reactivity would likely be such that they would be unable to exist for very long at all.

Simplistically speaking, the long stable chains of carbon found in hydrocarbons aren't entirely characteristic of carbon, as such, but rather, characteristic of hydrogenated carbon.


Two chemistry inquiries Here is my second question, I know 1000 grams of water is 1 liter. But does the density affect the result? What if the substance is mercury?

When you add seed crystal in to a supersaturated solution, the solute comes out. Does this make it saturated? Is what came out the extra solute that make it supersaturated? Or is it all the solute? This is my first question.

$1000 \, \text{g}$ of water occupy a volume of (approximately) $1 \, \text{L}$ because the density of water is (approximately) $1000 \, \text{g/L}$.
If you had $1000 \, \text{g}$ of mercury, whose density is $13534 \, \text{g/L}$, it would occupy a volume of $0.0739 \, \text{L}$.

Usually you can only dissolve so much of a solute into a solvent until a certain threshold is reached.
When this threshold is reached you have a saturated solution.
After that point, it is thermodynamically more favourable (under the given conditions, e.g. temperature, pressure, etc.) for any extra solute not to get dissolved in the solvent but clump together and precipitate.
But if you use some special techniques you can prepare a supersaturated solution that contains more solute than it should, thermodynamically speaking.
A supersaturated solution is a meta-stable state, meaning that it is only a local energy minimum for the system, not the global minimum.
But it would like to get to the global minimum if it gets the chance.
The only thing preventing that, is a small energy barrier that has to be surmounted.
So, if you disrupt the supersaturated solution in any way, like shaking it or adding a crystal seed, the solution can surmount the energy barrier and attain its global minimum state, i.e. the saturated state.


Why is Aniline a poor substrate for Nitration reaction? I have been asked to find out which one of the following is a poor substrate for Nitration reaction:

Upon encountering this question, my first reaction was the option c, as I know that the nitration mechanism involves the substitution of a positively charged Nitronium ion ($\ce{NO2+}$) and hence a $\ce{-OCOCH_3}$ group that shows $\mathrm{-M}$ effect will not be helpful for that process as it reduces the electron density in the benzene ring.
But the solution states that the answer is aniline (given in option a). I do not understand how is that possible. Even in the case when we use mixed acid as the reagent, it will form para- and meta-substituted nitro compounds. But, I do not know why and how does it behave as a 'poor substrate' in any case of nitration.
Please explain.

In the case of option c, the $\ce{-OCOCH_3}$ group will not be showing $\mathrm{-M}$ effect as the ring is directly attached to the oxygen atom. The group rather shows $\mathrm{+M}$ effect because of the two lone pairs present on the oxygen atom, increasing the electron density in the benzene ring and thus making it a good substrate for a nitration reaction.
On the other hand, as nitration is generally done with mixed acid as the reagent, the $\ce{-NH_2}$ group in aniline takes up a hydrogen atom to become $\ce{-NH_3^+}$,  which is a ring deactivating group.
Therefore, among the options given, aniline is the poorest substrate for nitration reactions.


Which functional groups are not reduced by DIBAL.H? According to the table here, diisobutyl aluminium hydride can reduce amides to aldehydes but according to the data given to me by my teacher DIBAL.H cannot reduce:  amides, acids, isocyanides and nitro groups.

It's I also given in my notes that DIBAL.H is  parallel to $\ce{LiAlH4}$ as a reducing agent but it is more selective.

I would like to know if DIBAL.H is able to reduce these functional groups or not.

DIBAL-H does the following:

Acid chlorides to aldehydes (Fast)
$3^\circ$ amides to aldehydes
Nitriles to aldehydes
Ketones to $2^\circ$ alcohols
Aldehydes to $1^\circ$ alcohols. (Slower)

Also, it works best in these solvents: THF, toluene.
We need to avoid solvents like alcohols, amines, strong acids and halogenated compounds.


Is d3-acetic acid more acidic than acetic acid? The following table is from Solomons, Fryhle and Snyder Organic Chemistry, chapter 3, An Introduction to Organic Reactions and Their Mechanisms: Acids and Bases:

  Table 3.8 Isotopic substituents and inductive effect
  $$ \small
\begin{array}{ccc}
\hline
\text{Acid} & \ce{pK_a} & \text{Inductive effect of the substituent} \\
\hline
\ce{CH3-COOH} & 4.75 & \text{+I} \\
\ce{CD3-COOH} & 4.60 & \text{+I} \\
\hline
\end{array}
$$
  +I effect of deuterium is stronger than that of hydrogen. +I ($\ce{-CD3>-CH3}$)

Why is $\ce{CD3-COOH}$ more acidic than $\ce{CH3-COOH}$ despite being less electronegative? The book also said that $\ce{CD3} > \ce{CH3}$ in terms of + Inductive effect.

The following articles suggest that the data reported from Solomons, Fryhle and Snyder Organic Chemistry is likely incorrect.

Maya Paabo, Roger G. Bates and R. A. Robinson ("Dissociation of Acetic Acid-d3 in Aqueous Solution and Related Isotope Effects from 0 to 50°", J. Phys. Chem. 1966, 70, 2, 540-543) report that the $\mathrm{p}K_\text{a}$ of the deuterioacid is greater (less acidic) than that of the protio analogue.
See Table 1 on p. 134 of C. L. Perrin's chapter ("Advances in Physical Organic Chemistry", Volume 44, edited by John P. Richard) where it is also reported that the $\mathrm{p}K_\text{a}$ of the deuterioacid is greater (less acidic) than that of the protio analogue.
Rose McConnell, Walter Godwin, Brant Stanley, and M.Shane Green ("Acidity Studies of Deuterated Acids and Bases Commonly Used as Buffers in NMR Studies", Journal of the Arkansas Academy of Science, Volume 51 Article 21) also report (see Table 1 on p. 140) similar findings.

These findings indicate that acetic acid-d3 is less acidic than acetic acid-h3.  This would be consistent with the operation of a simple inductive effect where protium is more electronegative than deuterium (as expected).  Hence, the more electron withdrawing $\ce{CH3}$ group is better able to stabilize the electron rich carboxylate anion in the the ionized acid as compared to the less electron withdrawing $\ce{CD3}$ group.


Why does nitric acid have a different molecular structure from phosphoric and arsenic acids? I remember that this really bugged me as an undergrad, and upon revisiting some inorganic chemistry notes I realized I still don't have a satisfactory answer for it. The structure of nitric acid, $\ce{HNO3}$, is quite different from the structure of phosphoric acid, $\ce{H3PO4}$ and arsenic acid, $\ce{H3AsO4}$:

Something tells me there should be an easy, intuitive argument explaining this, perhaps something about p-orbitals being more diffuse in phosphorus and arsenic? I find it kind of difficult to analyze this as a sort of "trend", since other groups don't really seem to have well defined oxyacids for the first two or three elements.

There is no oxyacid for water, but sulphuric and selenic acids are similar in structure; there is no oxyacid for fluorine either, but the oxyacids of chlorine and bromine are also analogous. In the carbon group something kind of strange also happens, with silicic acid (I believe) not really being a well-defined compound but a plethora of different things with different relative stabilities.

Phosphorus and arsenic are much bigger atoms with lower electronegativity so they can fit more atoms around them and oxygen can bond more easily.

Under extreme pressure it is possible to make the orthonitrate ion from nitrate and oxide. It rapidly decomposes in contact with water or carbon dioxide.


What type of polyester is most commonly used for clothing? Many polymers have the name of the monomer in their names, e.g. polyethylene, polyvinyl chloride, polypropylene, polystyrene etc. But the ester in polyester refers to a group of chemicals (having an hydroxyl group replaced by an alkoxy group1) rather than a single chemical compound.

This article from Wikipedia states that polyethylene terephthalate (PET or PETE) is the most common type of polyester. PET is the plastic from which almost all plastic soda and beverage bottles are made. This leads me to wonder if the polyester from which clothes are commonly made are PET, and if not, what types of other polyesters (e.g. polyglycolide, polycaprolactone, polyethylene adipate, polybutylene succinate, etc.) clothes may be made from.I'm aware that most plastics can be stretched into long thin filaments/strands and wound into threads in a process similar to how natural-fiber threads are made. My question is specifically whether or not PET is commonly used as the polyester in clothes, and if not, what is? I've done some Google searches but can't find any references; maybe I just haven't gotten the search string right. Thanks.

https://en.wikipedia.org/wiki/Ester

Based on several sources (shown below), it appears that polyethylene terephthalate (PET) is in fact used in the production of fabrics, including textiles. PET is marketed under names such as Dacron, Terelene, or Lavsan by various companies depending on your location. As to if it is the only textile polyester, I was unable to be completely certain. However, it appears to be a major textile polyester at the very least, as 39 million megagrams were produced in 2008 for fabrics alone.

Though it was not part of your question, I thought that I should include the following, as I found it interesting:

As it turns out, not only is PET used as a fabric, but it also makes a very nice fabric that is highly resistant to wrinkling. This is due to the presence of an aromatic ring within the monomers, which makes it quite strong/stiff; especially in this fibre form. These wrinkle-resistant fabrics are appropriately known as "permanent press fabrics."

To find out more information, I have included some sources below:

https://en.wikipedia.org/wiki/Polyester
https://en.wikipedia.org/wiki/Polyethylene_terephthalate
https://www.britannica.com/science/polyethylene-terephthalate


Entropy increase of both system and surroundings In "why chemical reactions happen" the author (James Keeler) says that "a process which leads to an increase in the entropy of the system can be ... exothermic." Could someone please give me an example of a process which is exothermic and increases the entropy of the system? It seems strange to me how a system can release energy in the form of heat while also increasing its own entropy.

We don't need to go to nuclear fusion to find an example. Lots of chemical reactions meet your criteria of being both exothermic (releasing heat to the surroundings) and increasing the entropy of the system.
A reaction is spontaneous if the free energy change of the reaction is negative ($\Delta_rG <0)$.The free energy change of a reaction at constant temperature is equal to the sum of the enthalpy change ($\Delta_rH$) and the temperature multiplied by the negative entropy change ($-T\Delta_rS$):
$$\Delta_rG = \Delta_rH - T\Delta_rS$$
One way for $\Delta_rG$ to be negative is for $\Delta_rH$ to be negative and $\Delta_rS$ to be positive. In a reaction conducted at constant pressure, the enthalpy change $\Delta_rH$ is equal to the heat exchanged between the system (reactants) and the surroundings. When $\Delta_rH <0$, the reaction is exothermic.
To find examples that meen your criteria, you only need a source of standard entropy and ethalpy data so that you can calculate $\Delta_rH$ and $\Delta_rS$ for a reaction. There are many such examples of reactions that are exothermic and have a positive entropy change of the system. Below is a simple example:
$$\ce{C(s) + O2(g) -> CO2(g)}$$
Use the NIST Chemistry Webbook, here are the standard enthalpies of formation and standard entropies of formation for carbon, oxygen, and carbon dioxide:
$$\begin{array}{|l|c|c|} \hline
 & \Delta_fH^\circ\ \mathrm{(kJ/mol)} & \Delta S^\circ\ \mathrm{J\cdot K/mol} \\ \hline
\ce{C(g)} & 0 & 5.8 \\
\ce{O2(g)} & 0 & 205.15 \\
\ce{CO2(g)} & -383.52 & 213.79 \\ \hline
\Delta_r & -383.52 & 2.84 \\ \hline
\end{array}$$
At all temperatures, this reaction is exothermic and has a positive entropy change for the  system.


How come we can't use the equivalence point equation of NV=NV in this problem? So, the problem is you prepare to standardize a $\ce{Na2S2O3}$ solution. $\pu{32 mL}$ of $\ce{Na2S2O3}$ solution is titrated into $\pu{50 mL}$ of a $\pu{0.01 M} \ \ce{KIO3}$ solution to reach the equivalence point. They first titrate the $\ce{KIO3}$ solution until it loses color, then add a starch indicator until the reaction is complete. The reaction proceeds in these two steps:

$$\ce{IO3- + I- + H+ <=> I3- + H2O} \tag{1}$$
$$\ce{I3- + S2O3^2- <=> I- + S4O6^2-} \tag{2}$$

Determine the concentration of $\ce{Na2S2O3}$ solution at the beginning of the experiment.

The answer is $\pu{0.094 M}$, and basically you need to first balance out the two formulas to create a net ionic equation for the two reactions, and then you use the moles of $\ce{KIO3}$ (it's molarity times it's volume) and then the balanced stoichiometric ratio ($\pu{6 mol}$ of $\ce{S2O3^2-}$/$\pu{1 mol}$ of $\ce{IO3-}$) to get the moles of $\ce{S2O3^2-}$, which is the dissociated version of $\ce{Na2S2O3}$ in a $1:1$ ratio (dissolved $\ce{Na2S2O3- -> 2Na+ + S2O3^2-}$). Then you just divide the moles of $\ce{S2O3^2-}$ by it's volume converted to liters to get $\pu{0.094 M}$.

My question is, since we are looking at the equivalence point in a titration, why can't we use $N_\mathrm{acid} \times \pu{Volume_{acid}} = N_\mathrm{base} \times \pu{Volume_{base}}$, since equivalents of acids and bases are the same at the equivalence point? If we can, how would we do it?

My question is, since are looking at the equivalence point in a titration, why can't we use $N_\text{acid} \times \text{Volume}_\text{acid} = N_\text{base} \times \text{Volume}_\text{base}$, since equivalents of acids and bases are the same at the equivalence point? If we can, how would we do it?

First of all, I would suggest that you stick to the original method using moles and proper mole ratios.

Please forget about this normality equation because it is only historically important and rarely anyone uses this concept in 2020 for research or academic work. Note, industrial analytical calculations still use this equation for very routine work. Still if you want to know why this normality equation fails in your case is as follows.

0) The $N_aV_a = N_bV_b$ relation holds if and only if all the concentration units are in normality, where $a$ and $b$ can be acid-base or oxidizing-reducing agent pair. It will also work for precipitation titrations etc. Corollary: This equation fails in titration calculations when the units are in molarity.

(i) This is a redox titration not an acid-base titration.

(ii) Let us change the subscripts for the sake for indexing, $$N_\text{ox}V_\text{ox}=N_\text{red}V_\text{red}$$

You have $\pu{0.01 M}$ $\ce{KIO3}$, what will be its concentration in normality units?

For that you need to see how many electrons are gained in the half-reaction...

$\ce{KIO3}\longrightarrow\ce{I2}$; Balance the equation and see how many electrons are involved in the redox process. See the example here
Balancing this reaction

Multiply the number of electrons involved in the half reaction by the molar concentration to get normal concentration of potassium iodate.

Apply the above equation and you will get the correct result but in normality units i.e. thiosulfate concentration in normality units.


How to finish silanization process correctly? I have some 2mm glass beads and I'm doing silanization on them. For doing that, I have a small bottle that I fill 1/3 with glass beads and then pour silan solution on them and shake well. However, after 60 min that I drain the solution from bottle and as soon as I wash the beads with water, the beads are covered with a white greasy product which is hardly removed by water and finally I wash it by acetone to become transparent again.

Is this a correct method of silanization (and washing the products) to achieve the maximum hydrophobicity effect of silan or I should revise the method?

Sol-gel preparation techniques have many variations. The contact time -beads to the solution - determines thickness of the sol-gel layer. Here you can try to shorten the dipping (coating) time. Glass is usually dried at 60°C for 30 mins and then "cured" at 500 - 550°C form 5 s up to 2 hours. I would suggest to find out an article and follow its preparation method.


Why polymers have a range over which they melt instead of having a specific melting temperature? Does it have something to do with the fact that all chains are not of the same length?

No. Although we need to be careful in defining what is and isn't a "polymer". The nature of (nearly all) polymers is that they are (covalently) bonded (in at least one dimension) over large (relative to atomic diameters) distances. The nature of a liquid is that is exhibits no long range order, no long range "connectivity". There are, as far as I know, no polymer crystals, in their most ordered configurations they are quasi- or semi-crystalline, having disordered zones as well as often but not always having zones of some regularity/order. Often the degree of crystallinity of a polymer is determined by its processing history (temperature, pressure, and stresses applied to it), but even the most crystalline of polymers is not fully crystalline. What this means is that the transition between solid and liquid phases is not sharp; the disengagement of the polymer chains from one another takes place gradually with no sharp/distinct difference between the solid phase and the liquid phase. Perhaps a better way of saying that is that the polymer becomes more and more amorphous but continues to have some long-range order as each chain's motion relative to its neighbors increases. see https://en.wikipedia.org/wiki/Reptation and https://en.wikipedia.org/wiki/Amorphous_solid. Note that the presence of polymer  chains (assuming linear or nearly linear polymers) of varying MW increases the gradual transition from amorphous solid to polymer melt. If the amount of oligiomer is large enough, it may act as a solvent for the larger chains. Note also that the theory of glass to liquid and glass to solid phase transitions is an active area (not yet "settled science").


Water: What to use H3O+ or H+? Why can we use $\ce{H+}$ and $\ce{H3O+}$ interchangeably?

I have seen in many places that first the reaction is written including Hydronium ion but then in bracket its written that we can write it as $\ce{H+}$ for simplicity. I am not clear with this.
$$\ce{H2O +H2O <=> H3O+ + OH-}$$
OR
$$\ce{H2O(l) <=> H+(aq) + OH- (aq)}$$

Since your solvent is itself water, it makes no difference whether you use $\ce{H+}$ or $\ce{H3O+}$.
$\ce{H3O+}$ is basically the hydrated form of $\ce{H+}$. If you know, the oxygen atom in water contains two lone pairs. When it donates one of the lone pairs to the hydrogen atom which doesn't have any electrons, you get $\ce{H3O+}$.
So,
$\ce{H3O+}$ is not $\ce{H+}$
$\ce{H3O+}$ is $\ce{H+(aq)}$
This means that the aqueous form of $\ce{H+}$ is represented as $\ce{H3O+}$


How would I name O-methylations on sugars while retaining the sugar’s common name? Suppose I had a monosaccharide, e.g. D-arabinose. Suppose I do a few synthetic steps with it, and end up with the compound methyl (2S,3R,4R)-3,5-dihydroxy-2,4-dimethoxypentanoate. (as labelled by ChemDraw. If that is not the PIN, please correct.)

However, when writing my paper/thesis/lab report, I would love to derive the base name from D-arabinose for simplicity and consistency. The monosaccharide derivatives with a carboxylic acid in place of an aldehyde are -onic acids so my base name would be methyl D-arabinonate. But how would I correctly and unambiguously label the methoxy groups?

If I used 2,4-dimethyl, that would sound like this second compound:

And if I used 2,4-dimethoxy, wouldn’t that make it this one:

Is there a trick I am missing for clearly naming the compound as above with methyl D-arabinonate as a base name?

It is possible to name such compounds based on retained carbohydrate names in accordance with the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

P-102.5.6 Monosaccharide derivatives
P-102.5.6.1 O-Substitution
In order to maintain the integrity of structures and take advantage of retained names to imply the absolute configuration, O-substitution is allowed in carbohydrate nomenclature. Substituents replacing the hydrogen atom of an alcoholic hydroxy group of a monosaccharide or monosaccharide derivative are denoted as O-substituents. The substitution of an anomeric hydroxy group is discussed in P-102.5.6.3.2. The O-locant is not repeated for multiple substitution by the same atom or group. Number locants are used as necessary to specify the positions of substituents; they are not required for compounds fully substituted by identical atoms or groups.
P-102.5.6.1.1 O-Acetyl and O-alkyl functionalization.
For O-acyl derivatives, names with the acid component cited as a separate word ending in ‘ate’ after the monosaccharide name are preferred to names using O-acyl group prefixes. However when the ose ending is changed (e.g. to denote a glycosyl or an acid function having seniority over an ester) O-acyl prefixes are required. O-Alkyl derivatives are always expressed by prefixes.

Therefore, the name for the corrsponding D-arabinose derivative with methyl substituents replacing the hydrogen atoms of two alcoholic hydroxy group is 2,4-di-O-methyl-D-arabinose.


Le Chatelier's principle: what are the effects of decreasing volume? Which of the following is increased by decreasing the volume of the
  reaction system in the following reaction: $\ce{2H_2S(g) +3O_2(g)<=>
 2SO_2(g) +2H_2O(g)} + {\text{heat}}$

  I. Rate of Reaction

  II. Equilibrium concentration of reactants

  III. Value of $\ce{K_{eq}}$

My attempt: By the gas laws, decreasing the volume of the container will increase the pressure, so equilibrium should shift to the right, as there are less gases on the right. Thus III is true. Additionally, heat is also generated by the shifting of the equilibrium, and the increased pressure on the system due to the decreasing of volume would also speed up the reaction, so I is also true. II is not true because the Equilibrium concentration of the products increases, not the reactants. So my final answer is I and III. Yet the answer key says I only. Why is that?

I think a good place to start is with the idea that that $K_{eq}$ will not change in value unless the temperature changes (see the second answer here for a good explanation). With this in mind we can say that statement III is false because the temperature is not being changed.

Statement II is shown to be false via Le Chatelier's principle, as the equilibrium concentration of reactants should decrease rather than increase due to the decrease in volume because the products have fewer gaseous moles and so would exert less pressure on the container.

Statement I can be shown to be true via Le Chatelier's principle by the same logic as Statement II: for the reaction to shift towards the product, the rate of formation of product must increase.

Therefore, only Statement I is true.

You might wonder how $K_{eq}$ can stay constant, but the rate of forward reaction can increase. Its important to remember that $K_{eq}$ is the ratio of the forward and reverse rate constants, not the forward and reverse rates. The rate constants have no dependence on volume (or pressure) so they won't change.


Is the solution isotonic, hypertonic, or hypotonic? The problem is shown below. I don't understand how can the tonicity of a solution be determined if the solution has multiple solutes dissolved in it.

Do I take the total sum of the mass of solutes and divide it by the volume of the solution in both outside and inside of the cell (total concentration) and determine the osmosis direction based on whichever is greater? Or should it be done in another way?

If it could be of use, here are the questions that were asked below the picture and their answers. This question was taken out of a test-bank for the 8th edition of the Campbell biology textbook.

Osmotic pressure for non-electrolytic solutes is given by

$$\pi = CRT$$

where $C$ is the effective concentration of all the solutes. In our case, with multiple solutes, we simply add all their concentrations to obtain the effective concentration. This gives us

$$
\begin{align}
\pi_\mathrm{cell} &= 0.05RT\\
\pi_\mathrm{environment} &= 0.03RT
\end{align}
$$

Since $\pi_\mathrm{environment} < \pi_\mathrm{cell},$ The environment solution is hypotonic.


Is the reaction of silver nitrate and copper endothermic or exothermic? Is the following reaction endothermic or exothermic and why?
$$\ce{2 AgNO3(aq) + Cu(s) -> Cu(NO3)2(aq) + 2 Ag(s)}$$

I have a feeling that it is exothermic due to the fact that it is spontaneous and will end in around 30 minutes (which I am guessing is because it will run out of energy to burn) and no source of energy is there for it to gain energy from, but is this right?
I found on the internet, that I was right about the reaction being exothermic as $\Delta_rH^\circ= \pu{-147.39 kJ}$, but my real question is why? Is its spontaneity the only means of support for this answer?

You may already know that the world is not about enthalpy but about Gibbs free energy. A reaction is sponanteous if the change Gibbs free energy is negative.

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

As you see, enthalpy and entropy play into this. Looking at your reaction, we note that we are taking two ions out of a solution to release one into solution. Compare the following, slightly different description:

$$\ce{2 Ag+ (aq) + Cu (s) <=>> 2 Ag (s) + Cu^2+ (aq)}$$

From an entropic point of view, this is bad. Rather than having two dissolved ions, we have one on the products’ side. Therefore, we can assume $\Delta S < 0$, therefore $T \cdot \Delta S < 0$. Negating it gives a positive value.

But the reaction is spontaneous, so $\Delta G < 0$. This can only be true if $\Delta H$ is sufficiently negative to counteract the loss of entropy. Thus we conclude that $\Delta H < 0$ and the reaction is exothermic.


How many liters of hydrogen gas is produced per gram of aluminium? How many liters of $\ce{H2(g)}$ at STP is produced per gram of $\ce{Al(s)}$ consumed in the following reaction?  $$\ce{2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)}$$
  Express your answer to four significant figures and include the appropriate units.

My attempted solution is included below:

$$ 1.00~\mathrm{g~Al} \left( \dfrac{1~\mathrm{mol~Al}}{26.9815385} \right)
\left(\dfrac{3~\mathrm{mol~H_2}}{2~\mathrm{mol~Al}} \right)
\left( \dfrac{22.4~\mathrm{L}}{1~\mathrm{mol~H_2}}\right) = 1.245~\mathrm{L}$$

This is being marked as the wrong answer.  Where is my error?

Your work is correct.  I have to assume that this is being graded by a computer, so trivial differences between your answer and the answer that it is expecting matter.

Possible source of error 1: Ideal Gas Law

Instead of using $22.4\ \mathrm{L/mol}$ they may have been expecting you to use the ideal gas law.
$$PV = nRT$$
$$101.3\ \mathrm{kPa}\times V = 0.05559\ \mathrm{mol} \times 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 273.15\ \mathrm K$$
$$V = 1.246\ \mathrm L$$

Possible source of error 2: Units

The computer may be looking for units in liters per gram of aluminum, something like:

$$\frac{\mathrm L}{\mathrm g} \ \text{or}\ \frac{\mathrm L}{\mathrm g\ \ce{Al}}$$

I doubt that this is the case, but it's possible.

Possible source of error 3: Typo

Either you or the person who set up the quiz made a mistake and typed in the answer incorrectly.


Single-molecule magnet with electrically-controlled permeability: How does the Titan Shield from Deus Ex work? In a recent installment of Deus Ex game series there is an augmentation called "Titan Shield" (it has nothing to do with $\ce{Ti}$ element):

A neodymium skin underlay matrix built of nano-meshed rare earth magnets and powered with hook-ins to the Biocell electrical system, the TITAN skin augmentation can be activated at will and then dissipated instantaneously.
When activated, dimorphic magnetorheological fluid is ejected from tiny nozzles installed throughout the skin. Simultaneously, the neodymium underlay electrifies, causing the fluid to solidify and seal the user inside an iron shell, effectively protecting them from all physical damage. The shield is resistant to gunfire and all forms of conventional explosives, but is vulnerable to EMP-based weaponry.

Titan Shield in action. From the official trailer of Deus Ex: Mankind Divided.
The second part regarding dimorphic magnetorheological fluid (MRF) looks quite plausible: such smart fluids, combining micro-sized spherical iron particles with nano-sized iron wires in order to prevent sedimentation and increase its responsiveness to external magnetic field are well known and can be prepared relatively easy [1].
Also, MRFs are used in vehicle suspension systems for more a decade now and they can withstand physical impact and dampen incoming momentum [2]. Typical design suggests putting a piston rod with MRF inside a coil which is upon electrifying controls solidification [3].
But the first part regarding "nano-meshed rare earth magnets" that can be triggered via sending electrical impulses caught me off guard. I thought that as long as conventional lanthanide-based magnets are permanent and have constant permeability in absence of external magnetic field, nanoscaled/single-molecule magnets are also permanent and cannot be used to dinamically change rheological properties of magnetorheological fluid.
Are there any existing examples of single-molecule magnets with permeability that can be controlled via altering current flow just like in case of electromagnets? If not, is there an explanation why it is not possible?
And yes, I do ask for this:)
References

Ngatu, G. T.; Wereley, N. M.; Karli, J. O.; Bell, R. C. Smart Mater. Struct. 2008, 17 (4), 045022. DOI: 10.1088/0964-1726/17/4/045022.
Gołdasz, J.; Sapiński, B. Insight into Magnetorheological Shock Absorbers; Springer, 2015. DOI: 10.1007/978-3-319-13233-4.
Liu, Q.; Jing, T.; Mo, A.; Xu, X.; Zhang, W. IEEE, 2015,; pp 2495–2500. DOI: 10.1109/ROBIO.2015.7419714.

Is there any existing examples of single-molecule magnets with permeability that can be controlled via altering current flow just like in case of electromagnets? If not, is there an explanation why it is not possible?

Short answers: no, and not yet.
Longer answer, and a warning that I'm using recent results from my group as references:

Yes, single molecule magnets exist, but... The field of single molecule magnets is hot, but the working temperatures of single molecule magnets are still below liquid nitrogen. Things seem to be improving, but they do not seem to be improving a lot. See the plot below for how the maximum temperature at which long-term magnetic memory has been reported has improved over the years (for single-ion magnets based on lanthanides, which are the best single molecule magnets right now). While there is no consensus that room-temperature single molecule magnetism is impossible, it has not happened yet, and there is no explanation why it could not happen in the future.

Maximum magnetic hysteresis temperature vs year when the Single Molecule Magnet was reported. Plot generated dynamically by the SIMDAVIS (AGPL-3.0) web app: https://rosaleny.shinyapps.io/simdavis_dashboard/ [1]

Yes, magnetism in molecules can be controlled via electricity, but... In this work, and in several related works, Wernsdorfer and coworkers show electric control (and electric read-out) of the magnetic state of a single molecule magnet. [2] However, this happens at temperatures below liquid helium, and the machinery needed is only able to do this for a single molecule at a time. Furthermore, the device is so delicate that hundreds of attempts are needed to trap the molecule in the right configuration. If the device heats up by accident, the molecule is lost forever and one needs to start from scratch. Again, there is no solid reason why this could not be achieved at some point in the far future. As promised, an alternate way of coupling the magnetic (spin) state of molecules to electric current, reported recently by my group.[3] In this approach, molecular helicity can be used to generate magnetic current (like a molecular electromagnet), and in principle this could eventually be employed to control the magnetic state of a single molecule magnet (again, at some hypothetical point in the far future).

To complete the answer, while neodymium-based alloys behave very well, neodymium-based single molecule magnets are, so far, greatly outperformed by dysprosium, terbium and even erbium (see figure below). Since there are fundamental reasons for this, it is more likely that any future technology resembling the one described in the question will be based on dysprosium (as of may 2021 holding the record for highest temperature single molecule magnet) or terbium (the ion employed in the cited examples about electrical control of magnetism in molecules).

Maximum magnetic hysteresis temperature vs lanthanide ion. Plot generated dynamically by the SIMDAVIS (AGPL-3.0) web app: https://rosaleny.shinyapps.io/simdavis_dashboard/ [1]
References

Duan et al. Data mining, dashboard and statistical analysis: a powerful framework for the chemical design of molecular nanomagnets. Preprint at https://arxiv.org/abs/2103.03199 (2021).
Godfrin et al., Operating Quantum States in Single Magnetic Molecules: Implementation of Grover’s Quantum Algorithm,  Phys. Rev. Lett. (2017) 119, 187702, https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.187702
Torres-Cavanillas et al., Reinforced Room-Temperature Spin Filtering in Chiral Paramagnetic Metallopeptides, J. Am. Chem. Soc. (2020), 142, 41, 17572–17580, https://pubs.acs.org/doi/abs/10.1021/jacs.0c07531]


Why is a temporary name given to an element with an atomic number above 100? All the elements with an atomic number more than 100 are given temporary names by IUPAC according to nomenclature rules. For example, element 101 was temporarily  named "Unnilunium" until they give it a permanent name (101 is now known as Mendelevium).

But why does the IUPAC give a temporary name to elements?

The systematic element names refer to a temporary name and symbol for hypothetical and newly synthesized elements. Because of naming controversies in the past, these names were introduced by the IUPAC to discuss the elements without ambiguity. After the discovery is confirmed, they receive a permanent name.

The IUPAC states in the red book this controversy as follows:

  In the past, some elements were given two names because two groups claimed to have discovered them. To avoid such confusion it was decided in 1947 that after the existence of a new element had been proved beyond reasonable doubt, discoverers had the right to suggest a name to IUPAC, but that only the Commission on Nomenclature of Inorganic Chemistry (CNIC) could make a recommendation to the IUPAC Council to make the final decision.

As this text suggests, this is a highly political process, that can be dragged out quite easily. The systematic nomenclature and symbols for new elements are stated in IR-3.1.1 as follows:

  Newly discovered elements may be referred to in the scientific literature but until they have received permanent names and symbols from IUPAC, temporary designators are required. Such elements may be referred to by their atomic numbers, as in ‘element 120’ for example, but IUPAC has approved a systematic nomenclature and series of three-letter symbols (see Table II).2 The name is derived directly from the atomic number of the element using the following numerical roots:
  0 = nil
  1 = un
  2 = bi
  3 = tri
  4 = quad
  5 = pent
  6 = hex
  7 = sept
  8 = oct
  9 = enn
  The roots are put together in the order of the digits which make up the atomic number and terminated by ‘ium’ to spell out the name. The final ‘n’ of ‘enn’ is elided when it occurs before ‘nil’, and the final ‘i’ of ‘bi’ and of ‘tri’ when it occurs before ‘ium’.
  The symbol for the element is composed of the initial letters of the numerical roots which make up the name.
  Example:
  1. element 113 = ununtrium, symbol Uut

As for the transfermium elements, their names have been finalised only in 1994 (Pure Appl. Chem., 1994, Vol. 66, No. 12, pp. 2419-2421).

However, for Mendelevium the first publication already suggested the name and a symbol. It was in use before officially recommended by the IUPAC (A. Ghiorso, B. G. Harvey, G. R. Choppin, S. G. Thompson, and G. T. Seaborg, Phys. Rev. 1955, 98, 1518).


Transition state and intermediate very close to each other I've been studying a particular reaction with DFT, where a transition state and the following intermediate are very close to each other. The change in geometry is rather small and therefore the electronic energies differ only by about ~ 1-2 millihartrees. Still, the TS has the higher electronic energy (everything else would be surprising?), but adding thermal corrections to both structures leaves the intermediate about 2 kJ/mol higher in energy.

Does that make sense? Or is there something I did not consider in my calculations?

1 Kcal/mol is indeed quite low. This is around 1 hydrogen bond energy. In general - compute frequencies using freq command in Gaussian or whatever you are using. You should have a single negative frequency. It should be relatively low (abs value compared to lowest positive frequencies). Visualize it using Moleken or a similar tool. If movement is along the reaction coordinate then you are good.

Also, compare all bond lengths and angles to see what changed. Did bonds along the reaction coordinate change the most?

Please use Kcal/mol instead of Hartrees. I realize that Gaussian prints out
Hartrees, but it is easy enough to convert to Kcal/mol.


Low Energy Nuclear Reactions I was interested in knowing if Low Energy Nuclear Reactions are practical or just mythical. What is the current stance of the scientific society about their feasibility?

The stance is currently an even mix of hogwash conspiracy theories, pseudoscience and scams.

Even if the field had merit - which it most likely doesn't, the number of anti-scientific "papers" on the subject has ruined it for a long time.


Does carbon monoxide form or carbon dioxide forms when carbon reduces metal oxide? I would like to discuss this in context of ellingham diagrams.
When $\ce{C -> CO}$ line is below metal oxide line, CO is produced.
Similarly, in other case $\ce{CO2}$ is produced.
But when both lines are below metal oxide, is carbon monoxide produced or carbon dioxide?

If both carbon monoxide and carbon dioxide are thermodynamically favored, then carbon monoxide comes "first" in the sense that the metal oxide first oxidizes carbon to $\ce{CO}$.  This is because the reaction is carried out at relatively high temperature where any potential carbon dioxide would itself react with carbon to form the monoxide.  If more oxide is available after all the carbon is converted to $\ce{CO}$ and further oxidation to $\ce{CO2}$ is favorable, the latter reaction then occurs.


Does the number of H+ ions in solution go up on dilution of a weak acid? In my textbook, a footnote says:

  In case of weak acids, on dilution the total number of $\ce{H^{+}}$ ions in solution increases because dissociation of the weak acid increases

This didn't make sense to me however. Consider the general dissociation reaction of a weak acid:

$$\ce{HA (aq)  + H_2O \rightleftharpoons H_3O^{+} (aq) + A^{-}(aq)}$$

Now, the Chatelier's principle tells us that if we add more reactants ($\ce{H_2O}$), the reaction will proceed more to the right, and so the dissociation of the acid will increase, but the part about the total number of $\ce{H^{+}}$ ions in water being higher was still unclear to me.

Delving into the math a bit, we can see that

$$K_a = \frac{[\ce{H_3O^{+}}][\ce{A^{-}}]}{[\ce{HA}]}$$

Since the concentrations of the two products are equal,

$$[\ce{H_3O^{+}}] = \sqrt{[\ce{HA}]\cdot K_a}$$

Now, let us assume that we have a $1 \ M$ solution of the acid in question dissolved in 1 litre of water, plugging this into the equation, we have a $\sqrt{K_a}$ molar solution of $H^{+}$ ions. Since only 1 litre of solvent is there, we have $\sqrt{K_a}$ moles of ions.

Now, say we dump 1 litre of water into this solution. The concentration of our acid will be halved. Plugging this into the equation, we get the concentration of $\ce{H_3O^{+}}$ ions as approx. $0.7 \sqrt{K_a} \ M$. We have two litres of solvent, so we have $1.4 \sqrt{K_a}$ moles of ions, which is considerably higher. So, the numbers all work out, but what's the intuition behind it?

Let's say we have got a solution of an acid with a pK$_a=5$. The resulting pH dependent on the concentration of that acid would be like:

$\hskip2in$

Examples: $0.1~\ce{M}$: pH$=3.0$; $0.001~\ce{M}$: pH$=4.0$; $0.00001~\ce{M}$: pH$=5.2$; $0.0000001~\ce{M}$: pH$=6.8$

Depending on the pH resulting from these example concentrations we can have a look at percentage composition of the whole acid base system. (Degree of association $\alpha'$ in blue, degree of dissociation $\alpha$ in light orange.)

$\hskip2in$

Examples: pH$=3.0$: $\alpha\approx 1.0~\%$; pH$=4.0$: $\alpha\approx 9.5~\%$; pH$=5.2$: $\alpha\approx 61.8~\%$; pH$=6.8$: $\alpha\approx 98.4~\%$

One can see that low acid concentrations lead to higher pH values than higher acid concentrations which automatically result in an increasing degree of dissociation.

So for me this is no big intuition thing but rather a small sequence of actions.

Lets say it in math:

The proton concentration for your system can be described by $$c(\ce{H3O+}) = \sqrt{c_0(\ce{HA})~k_a+k_W}$$ (adding $k_W$ for very low concentrations) and the equation for the degree of dissociation is $$\alpha = \frac{k_a}{k_a+c(\ce{H3O+})}$$ which can be combined to give $$\alpha\left(c_0(\ce{HA})\right)=\frac{k_a}{k_a+\sqrt{c_0(\ce{HA})~k_a+k_W}}$$

One could expand this system a little more (using the exact pH equation including the self-ionization of water) and come up with a thorough scheme like the following.

$\hskip2in$


Vapour pressure and Raoult's Law The following question from my textbook assignment reads

A mixture contains $1$ mole volatile liquid $A$ ($P_\mathrm A^\circ = \pu{100 mmHg}$) and $3$ moles of volatile liquid $B$ ($P_\mathrm B^\circ = \pu{80 mmHg}$). If the solution behaves ideally, the total vapour pressure of the distillate is?

Using Raoult's Law:
\begin{align}
  P_\text{total} &= P_\mathrm A^\circ \times \chi_\mathrm A + P_\mathrm B^\circ \times \chi_\mathrm B\\
  P_\text{total} &= \pu{100 mmHg} \times 0.25 + \pu{80 mmHg} \times 0.75\\
  P_\text{total} &= \pu{85 mmHg}
\end{align}
But the answer is $\pu{85.88 mmHg}$.
I tried searching online for an explanatory solution but all of them seem to use the mole fraction of solute in vapour phase without any explanation. I figured it was due to the word distillate. Can someone explain why do we need to use mole fraction of the vapour phase and not of the solute in solution?

During the process of distillation, we condense the vapours (B) of the mixture (A) to give a mixture (C) that has the same mole fractions as that of the vapour (B) of mixture (A).
In order to solve this question there would be three steps.

Firstly, finding the vapour pressure of the initial mixture.
Second, we find the mole fraction of components in the vapour phase of this mixture.
The third and final part would be to find the vapour pressure of the distillate leading to our answer.

However, finding the actual composition of the vapours here, would be the biggest challenge in this question.
Step $1$
In order to find the vapour pressure, we use Raoult's Law which mathematically states:
$$P_T = \sum{P^\circ_i\chi_i}$$
Since this is a two-component system, the total pressure would be expressed as a sum of two terms for each component A and B.
$$P_T = P^\circ_\mathrm a \chi_\mathrm a +P^\circ_\mathrm b \chi_b$$
Substituting the values given in the question, we get $P_T= \pu{85 mmHg}$
Step $2$
The next step is to find the composition in the vapour phase. For this, we use the  property that the partial pressure of a component in the vapour phase is equal to the contribution of the same component in the liquid phase. Mathematically, it is equivalent to stating:
$$P_Ty_\mathrm a= P^\circ_\mathrm a \chi_a$$
Here, $y_a$ is the mole fraction of component A in vapour phase. Solving for $y_\mathrm a$, we get that $y_\mathrm a = \frac{5}{17}$, which implies $y_\mathrm b = \frac{12}{17}$.
Step $3$
The final is exactly the same as the first step, wherein you use Raoult's law to find the final vapour pressure of the distillate (mole fraction of components equal to that mole fraction of vapour in initial mixture).
Solving for the vapour pressure, you get $P_T^{'} = \pu{85.88 mmHg}$, which is the answer provided.


Naming a ring with an alcohol group? Is it necessary to include the "1" in the name to indicate that the OH group is at the 1st carbon? Isn't it kind of assumed?

Is it necessary to name a compound "2-(3-methylbutyl)-1-cyclopentanol"?

Or is "2-(3-methylbutyl)cyclopentanol" sufficient?

Thanks!

It depends on the priority of the substituents on the ring. If the highest-priority group is the alcohol group, then the numbering of the carbons in the ring starts there - so the 1 is implicit, and you do not need to state it. See 2-(3-Methylbutyl)cyclohexanol as an example.

Note that the alkyl group position is assigned a number relative to the position of the -OH group. However, if there is a higher-priority group than -OH then the position of the alcohol will be assigned a number relative to this higher-priority group; see 2-hydroxy-3-chlorobenzoic acid.

In your example, the 1 is not necessary.


Is a liquid in a container always in equilibrium with its vapour? This is essentially a question about the meaning and significance of the term vapour pressure (or vapor pressure if you're American). From what I understand a liquid in a container will have a certain amount of its vapour above it exerting a certain pressure: the vapour pressure - which is dependant on the identity of the substance and the temperature of the system only. The point at which the ambient pressure equals the vapour pressure the liquid boils - the liquid is in equilibrium with its vapour/gas phase (the line in the diagram separating liquid and gas).

However, this had me slightly confused because in the situation initially described (at a point somewhere within the liquid region - NOT on the phase boundary) some vapour existed above the liquid in the container and thus there must be some kind of equilibrium already existing. So, when the ambient pressure equals the vapour pressure nothing has changed in the sense that equilibrium remains present it's just that the pressure is no longer significant enough to facilitate the liquid.

Is what I've said above correct? Mainly in the definition of vapour pressure being the pressure of the vapour above the liquid AT ANY POINT in the liquid region on the phase diagram and also whether the vapour pressure is just a function of temperature.

The definition of vapor pressure means that for any temperature and pressure within the limits of a liquid system, there is a certain amount of vapor present above the liquid. The liquid and vapor at this point are in equilibrium because the vapor pressure does not change over time for that given temperature and pressure. When the liquid starts boiling, the system is undergoing a phase transition and in not at equilibrium because the system volume is changing and energy is continually being added to raise the system's temperature. Of course, we know the temperature does not start to change until vaporization has completed.

Vapor pressure of any given pure liquid is normally a strong function of temperature and a weak function of pressure. If you took a beaker of a pure liquid and put it inside a container that could support a vacuum, then as the pressure external to the beaker starts to decrease, more molecules from the liquid phase have enough energy to escape the liquid phase and exist in the vapor phase. If you continued developing more vacuum, the liquid would boil. This is seen on the phase diagram as moving straight down from somewhere in the liquid region until you reach the gas-liquid equilibrium line.


Why is carbon dioxide considered a Lewis acid? $\ce{CO2}$ is considered a Lewis acid. How it is an acid? According to Lewis: “species that accept an electron pair are acids”. But $\ce{CO2}$ can't accept electron pairs because oxygen and carbon both are electron sufficient (have complete octets). If the reason is within a lone pair of electrons on oxygen then ammonia ($\ce{NH3}$), which also has a lone pair of electrons, should be acidic, too. But ammonia basic in nature.

So why can carbon dioxide be considered a Lewis acid?

If you've learned how to draw resonance structures, they will help you greatly here (and in general). In the attached figure, 1 is the relevant resonance, implying the carbon in $\ce{CO2}$ is partially positive and therefore may be prone to accept electrons (see the resonance hybrid).

In theory, any Lewis base (electron-donor) can react with $\ce{CO2}$ (Lewis base abbreviated B in 2 & 3). We can draw the reaction mechanism using the charged resonance structure (2) or the neutral, most "stable" structure (3). In reality, these are not two discretely different forms but just "snapshots in time" of a true $\ce{CO2}$ molecule, so it doesn't really matter which structure we as humans choose to represent on paper.

I hope you can see how $\ce{CO2}$ is a Lewis acid because it is able to accept electrons. Although in the principal resonance structure you see all atoms have full octets of electrons, the double bonds can easily break, rearranging the electrons. If full octets prevented atoms from receiving electrons, a great percentage of all known reactions would not occur. The key is that the electrons in the octet must have an alternative place to go if new electrons are donated to the molecule (in this case, the electrons migrate to the $\ce{O}$ atom). In more advanced chemistry, we would say $\ce{O}$ acted as an electron sink in this process.

It is interesting to note that any $\ce{C=O}$ structure can react as a Lewis acid in exactly the way your $\ce{CO2}$ molecule did, and this reaction of the $\ce{C=O}$ group (known as a carbonyl functional group) is among the most important patterns in the biochemical reactions occuring in our bodies.


When does hydrogen bonding happen? What kinds of molecules will experience hydrogen bonding (inter and intra)? Why does it occur for some H atoms but not others?

Most people seems to think that hydrogen bonds are limited to $\ce{X-H\bond{~}Y}$ interactions, where $\ce{X}$ and $\ce{Y}$ are heteroatoms.

In the crystal, a $\ce{C-H}$ of an aromatic system or a $\ce{C=C}$ double bond can serve as a donor as well and interact with heteroatoms, such as oxygen atoms in $\ce{C=O}$ or $\ce{C_{arene}-O-C_{alkyl}}$.

These interactions can form complex networks of rings and strands.

We've examined and published that a while ago for a couple of different phthalimides, such as N-(4-methoxybenzyl)phthalimide or N-allylphthalimide, and some more.

While the articles are behind a paywall (they were published before Acta Cryst. E went open source), the CIF files can be downloaded and played with/examined.

A crucial parameter for hydrogen bonds is distance, if the acceptor is too far away, there is (obviously) no hydrogen bond.

As a rule of thumb, $2.3\,\mathrm{\mathring{A}}$ is a distance where hydrogen bonds can be expected.


Product of treating 1,3‐dibromocyclobutane with alcoholic KOH at elevated temperatures Passage-II

  $$
\ce{
A ->[\text{1.}~\ce{NH2-}(\pu{2 mol})][\text{2.}~\ce{CH3-I}(\pu{2 mol})]
B ->[\ce{HgSO4}, \ce{H2SO4}][\ce{H2O}]
C ->[\text{dil.}~\ce{OH-}][Δ]
D [Mixture]
}
$$

  31. "A" gives

  A) Lucas test
  B) Silver mirror test
  C) White precipitate with ammoniacal $\ce{AgNO3}$
  D) Yellow precipitate with ammoniacal cuprous chloride

  32. How many isomeric products are existed in the mixture "D"?

  A) 2
  B) 4
  C) 6
  D) 8

What would be 'A' in the reaction? Any help would be deeply appreciated.

I know that alc. KOH does elimination, but I was confused whether it would eliminate one bromine each from two rings and create a bond between both the rings or it would simply eliminate both bromines from same ring and create two conjugated double bonds (cyclobut-1,3-diene).

EDIT: There is one more possibility: both the bromines get eliminated and a diagonal bond is formed in the ring (cyclobutane with any two diagonal carbons joined).

The two equivalents $\ce{HBr}$ are eliminated from two different molecules, one from the starting material and one from an intermediate. In the first step, $\ce{HBr}$ is eliminated under cleavage of the cyclobutane ring, which is favored by the release of ring strain. Elimination of $\ce{HBr}$ from 1-bromo-1,3-butadiene finally yields but-1-en-3-yne.


Why does reduced mass help when talking about two body problems? Firstly I would like to say that despite the fact that you may be thinking that this is a physics question and should probably be asked on that part of the website, I came across reduced mass for the first time in Chemistry - physical and inorganic - and it is used a lot in Chemistry so I thought it suitable for this part of the site.

With that said, I will now elaborate on my question. I hear people say that reduced mass ,$\mu$, is used for two body problems, but what sort of problems? Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?

I can follow the derivation (which I shall go through shortly) in a mathematical sense but I can seem to be able to grasp the ideas behind why the steps were taken.

Derivation:

Step [1] - From Newton's laws of motion: $$F_{12}=-F_{21}$$ It follows that: $$m_1a_1=-m_2a_2$$
Step [2] - Simple rearrangement of the result from step 1 to make $a_1$ the subject (it seems a reasonable thing to do but I don't know why it's done). $$a_1=-\frac{m_2}{m_1}a_2$$
Step [3] - The relative acceleration, $a_{rel}$, can be found by subtracting the acceleration of one body from the other (I get how this is done and get that it would give the relative acceleration BUT I don't know why you would need the relative acceleration). $$a_{rel}=a_1-a_2=(1+\frac{m_1}{m_2})a_1=\frac{m_2+m_1}{m_1m_2}m_1a_1=\frac{m_2+m_1}{m_1m_2}F_{12}$$
Step [4] - According to Wikipedia the result from the last step can be used thus (I have no idea what's going on in this step):
$$\frac{m_2+m_1}{m_1m_2}F_{12}=\frac{F_{12}}{\mu}$$ Thus: $$\mu=\frac{m_1m_2}{m_2+m_1}$$

Again, I can do the maths but my problem here is conceptual I believe. I have included the derivation in my question because I believe that it's possible that if you were able to shed some light on why particular steps were taken that this may help.

You're transforming a problem in terms of two particles with individual masses and accelerations into a problem with a single fictitious particle. That simplifies the problem and lets you separate the relative motions and interactions of the particles from overall translation of their center of mass.

In step 3, you need the relative acceleration $a_{rel}$ because that's the acceleration the single fictitious particle has.

In step 4, you're calculating the mass $\mu$ of the fictitious particle.

Before, you had $F_{12} = m_1 a_1$ and $F_{21} = m_2 a_2$ (two forces, two masses, two accelerations). Now you've got $F_{12} = \mu \ a_{rel}$ (one force, one mass, one acceleration). The price is that we've thrown away information about the overall translation of the system through space; we're only looking at the relative motions of the two particles.

The coordinates of the fictitious particle are relative or internal coordinates $$x = x_2-x_1,\quad y = y_2-y_1, \quad z = z_2-z_1$$
This is useful because the potential energy $V$ of the interacting particles depends only on these relative coordinates. Making use of reduced masses and relative velocities and coordinates cleanly separates internal motion from translation of the system as a whole. The classical energy of the two body system will be
$$E = \left[\frac{1}{2\mu}(p_x^2+p_y^2+p_z^2)+V(x,y,z)\right] + \left[\frac{1}{2M}(p_X^2+p_Y^2+p_Z^2)\right]$$
where $p_i$ means momentum for motion along the $i$-th coordinate. The first term in brackets is the energy of our fictitious particle with mass $\mu$. The second term is the energy of a second fictitious particle with total mass $M=m_1+m_2$ and coordinates  $X$, $Y$, and $Z$ located at the center of mass of the system (e. g. $X = (m_1x_1 + m_2x_2)/M$).

In chemistry, we don't care about the second term. For example, in an atom, we'll assume that the nucleus is clamped in place so we can just look at interactions within the atom.


How to make cyclohex-4-ene-1,2-diol from cyclohexa-1,4-diene? As the question says, I'm supposed to make cyclohex-4-ene-1,2-diol using cyclohexa-1,4-diene:

I can use any inorganic reactant, any solvent, and any organic molecule with 3 or fewer carbons. I am not sure how to go about this question, as I do not know how to use regional-specific reactions. Any help, even just mentioning a reaction mechanism I can use for this would be greatly appreciated.
I could put the cyclohexa-1,4-diene on $\ce{KMnO4}$ for a catalyst and $\ce{H2O}$ with $\ce{HO-}$ as solvents in order to break the double bond and replace it with the $\ce{HO-}$ molecules. Problem is that this would affect the molecule as a whole – not just a specific side which is what I need.
I also thought of using bromine as something temporary, since they are easy to remove however, again, this affects the whole molecule.
I also thought about Diels-Alder reaction, but we haven't learned anything about that so I do not know how to use it- much less on a cyclic molecule.
I have also seen the osmium tetroxide oxidation of alkenes. However, I still have the problem that I can't control this reaction to only affect one side.

I think the easier way of doing this is to epoxidise the diene with 1 eq of a peracid epoxidation with peracids . The standard reagent m-Chloroperoxybenzoic acid (MCPBA) breaks your rule of organic molecules with 3 carbon atoms or fewer, but you can also do it with peracetic or pertrifluoroacetic acid preparation of peracetic acid. You will inevitably get some di-epoxide, but this will be separable by chromatography. The mono-epoxide can then be reacted in mild aqueous acid to give the diol. You could also carry the mixture through to the aqueous acid treatment as the tetraol will be far more water soluble than the diol and can be removed by an aqueous wash.


Leaving group ability and solvent's effect Regarding leaving groups and solvent effects (SN1 and SN2 reactions), I would like to ask some questions.
As the nucleophilic character is also influenced by the solvents, there is some indication (to me at least) that the leaving group can be influenced as well. Unfortunately, there is not much information found. Let us consider the halides, the quality of the leaving group is $\ce{I-} > \ce{Br-} > \ce{Cl-} > \ce{F-}$ as the larger ions are more polarizable, more stable and less basic.
So imagining a polar protic solvent, wouldn't the solvation stabilize $\ce{F-}$ via hydrogen bonds? I totally understand, that fluorine forms tighter bonds with carbons, so it naturally is a bad leaving group.
Considering the Finkelstein reaction, the driving force of the reaction is the formation of an insoluble salt, so basically the solvent influences the leaving group ability.
So, can I generally say that the solvent can theoretically influence the leaving group ability? Is there some general thumb rule I can use?
Until now, I haven't seen any reaction, where $\ce{Cl-}$ is preferred over $\ce{I-}$ as a leaving group except for the Finkelstein reaction.

The solvent generally doesn't affect the leaving group ability of an anion. But coming to your Frinkelstein case, it a special one cause here some precipitate is being formed (yes, that's correct that it is due to lower solubility product of say $\ce{NaCl}$ in acetone). As soon as some ppt. is formed, those atoms lose their ionic identity, settle at the bottom and don't interfere in the solvent as solute and so their concentration decreases in the solvent. Due to Le Chatelier's principle, the reaction goes forward to a great extent. This is also the case for other super leaving groups like gases (for eg. $\ce{N2}$, $\ce{H2}$).


Which half cell equation do you switch when calculating E° cell? Given the following standard potentials calculate the overall $E^\circ_\mathrm{cell}$:
$$
\begin{align}
\ce{Pb(SO4)(s) + 2 e- &→ Pb(s) +SO4^2-(aq)} &\quad E^\circ &= \pu{-0.356 V} \\
\ce{Pb^2+(aq) + 2 e- &→ Pb(s)} &\quad E^\circ &= \pu{–0.125V}
\end{align}
$$

I know the answer should be $\pu{0.231 V}$. I was told that $E^\circ$ cell always has to be positive. I tried doing this:
$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{0.231 V}$$
The only way I get the correct answer is if I do not switch the negative sign of the standard reduction potential of the first equation to a positive sign. Is that correct? I thought you always had to switch signs if you reversed the equation?

Now, lets first clear this up. $E_\mathrm{cell}^\circ$ can be negative, it just means that the reaction won't spontaneously occur. This means that an external current is required to be applied to push the electrons in that direction for the reaction to occur. When $E_\mathrm{cell}^\circ > 0$, that just means that reaction occurs spontaneously. So in voltaic or galvanic cells, $E_\mathrm{cell}^\circ > 0$.

Now to calculate $E_\mathrm{cell}^\circ$, the formula is:
$$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} + E_\mathrm{ox}^\circ \mathrm{(anode)}$$
As you correctly stated, if you switch the direction of the reaction (so it goes from an oxidising to reducing half equation), the sign of $E^\circ$ changes. Hence the equation for $E_\mathrm{cell}^\circ$ can also be written as:
$$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} - E_\mathrm{ox}^\circ \mathrm{(anode)}$$
This is the formula that you have used and hence this is why you get the correct answer when you do this.


What happens when we heat an atom? Question is simple.. If we take an atom of any element and then supply heat energy to it then what will happen?

What I thought is that in the beginning, energy (quanta; due to excitement of electron and then moving back to ground state) should be  evolved and then electron should be emitted... Am I right?

If we take an atom of any element and then supply heat energy to it
  then what will happen?

A single atom can move in three dimensions (x, y, z) through space, it cannot rotate about an external axis like a diatomic molecule can, nor can it vibrate.  As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space.  If you supply enough energy, then the electrons surrounding the atom will absorb that energy and undergo electronic transitions from the ground state to various excited states.  This process can be diagrammed using a Morse potential diagram , I've drawn a simple one below.  The diagram is meant to show promotion of an electron from the ground electronic state to higher and higher electronic levels as more energy is applied to the system.  Note that as the electron is promoted to higher energy levels, the average separation distance between it and the nucleus increases.  If enough energy is supplied, an electron will dissociate from the atom and we will be left with an ion.  Continue to supply even more energy and more electrons will be removed from the atom. Pumping even more energy into the system will ultimately lead to transformations in the nucleus of the atom.


Grammar of elimination reactions I would like to double check the English grammar of a sentence of an elimination reaction, but I cannot find an answer.

I'd normally say "the elimination of A with X as the leaving group", where A is the initial compound.

So my question is saying "the elimination of X from A", correct? I've never been told not to, but I feel it is wrong chemically, so I would change it to "the leaving/loss of X from A", which unfortunately sounds casual.

EDIT.
So based on the answer below I was totally wrong —I am glad I asked!— and to recap with A as reactant and X as leaving group and optional parts in parenthesis:

✓ → ✗ — the elimination of A (with loss of X)
✗ — the elimination of X (from A)
✓ — the loss of X (from A)
✓ → iffy — the leaving of X (from A)
✓ → ✗  — A gets/is eliminated
✗ — X gets/is eliminated from A
✓ — A undergoes elimination (with loss of X)

"A gets eliminated" or "the elimination of A" is just plain wrong, typical sloppy lab slang by many (too often native?) speakers. Very common, but still terrible.

As you say, the small molecule X gets eliminated from substrate A. Chemical English is still just English, and that is the correct way to put the case in question into words in this language.

"A undergoes elimination" (short for "A undergoes an elimination reaction) is a widely used, and imo also correct, variant.


Can hydrazine be protonated twice? First, $\ce{N2H4}$ is a base when it reacts with water, it accepts a proton.

$$\ce{N2H4 + H2O -> N2H5+ + OH-}$$

Can I follow up the reaction with this?

$$\ce{N2H5+ + H2O -> N2H6^{2+} + OH-}$$

Yes, it is possible, but only to a very small amount.

The $\mathrm pK_\mathrm a$ of the hydrazinium (2+) ion in water is $-1$ (CRC Handbook of Chemistry and Physics, Ed. D.R. Lide, CRC Press, Boca Raton, FL, 2005), so the equilibrium

$$\ce{N2H5+ + H2O <<=> N2H6^2+ + OH-}$$

is almost completely on the left side.


How are hormone levels measured? I have been searching google and didn't found any links that describes how  this process is performed - I mean the chemical procedure, not the process of taking the blood from the vein.

The steps for hormone level measurement generally follow a similar sequence.  For the most accurate analysis liquid chromatography-mass spectroscopy (lc-ms) is typically used.  The liquid chromatography separates the sample into fractions, conditions are selected such that a clean, separate peak is seen for the hormone being evaluated.  The purified hormone derived from the liquid chromatography step is then introduced into the mass spectrometer for quantification.

The general process involves:

sample collection: a sample of saliva, blood or urine is collected
addition of a known quantity of hormone to act as an internal standard.  Often a $\ce{^{13}C}$ labeled hormone is used (commercially available).  The parent ion for the labeled standard will occur one mass unit higher than the parent ion for naturally occurring $\ce{^{12}C}$-containing collected sample.  If the peak height of the sample in the mass spec is half as large as that of the standard, then the concentration of hormone in the collected sample is half that in the standard.
chemically separate the hormone from any large binding proteins
isolate the fraction containing the hormone using liquid-liquid extraction
run a second liquid-liquid extraction to remove highly polar materials
run lc-ms analysis on sample that has now been purified and enriched in hormone

Here is a link to a detailed procedure for testosterone analysis.

To keep things understandable I have simplified the description of some of the
steps in the process. For example lc-tandem mass spectrometry is often used to improve accuracy.

Research today in hormone quantification is focused on reducing cost and turn-around time.  A significant amount of work is particularly being done on analyzing the raw collected sample without further purification.


Why the ionization energy of Au is more than that of Cu? As we go down the group from 3d to 5

d, size increases.

Vanderwaal radius of Cu and Au is 140pm and 166pm respectively. Since Cu is smaller in size, ionization energy of Cu must be greater.

Then why is the ionization energy of Au greater than that of copper?

Smaller size does not necessarily correlate with higher ionisation energy. The trends in 1st IE going across Period 2 offer a simple counterexample.

A better (but not 100% foolproof) guideline is to use the orbital energies (see: Koopmans' theorem). The higher the energy of the orbital from which the electron is removed, the lower the IE.

In the case of gold versus copper, the first ionisation corresponds to the removal of a 6s or 4s electron respectively. In gold, the 6s orbital is relativistically stabilised, leading to a larger than expected ionisation energy. See also Geoff Hutchison's answer here.


Smallest whole number coefficient? What is the smallest whole number coefficent of $\ce {Cr^{2+}}$ when the equation is balanced?
$$\ce {Cr^{2+}_{(aq)} + Al_{(s)} -> Cr_{(s)} + Al^{3+}_{(aq)}}$$

I don't understand, how can this be balanced? And why would 3 be the smallest coefficient?

You need to factor in the electrons that are being moved around

Chromium is doing this:
$\ce {Cr^{2+}_{(aq)} + 2 e- ->  Cr_{(s)}}$

Aluminum is doing the reverse of this:

$\ce {Al^{3+}_{(aq)} + 3 e- -> Al_{(s)}}$

Now, like a math equation flip around the aluminum reaction.
And balance out the electrons

$\ce {[Cr^{2+} + 2 e- ->   Cr_{(s)}] \times 3}$

$\ce {[Al_{(s)} -> Al^{3+} + 3 e- ] \times 2}$

$\ce {3Cr^{2+} + 6 e- + 2 Al_{(s)} -> 3Cr_{(s)} + 2 Al^{3+} + 6 e-}$

Hope this helps


Force Fields and units I am a math student trying to implement a simple force field in python for determining the energy of a small molecule and trying to minimize it in order to find a stable configuration. I was wondering if there is a prefered system of units to be used am currently using the SI system of units however when i calculate Van der waals energies the numbers are quite difficult to deal with . Any reference of basic force fields would be much appreciated

Atomic units are the standard in almost any ab-initio (LAMMPS, NWChem, Orca, QChem) and/or classical force field (LAMMPS, MPMC, CHARMM, AMBER, GROMACS etc.) modeling code because you end up computing numbers much closer to 1.

Things can get bad very quickly for small values in python. To prevent this, I would use atomic units, and wherever arithmetic error occurs, try using the decimal type, instead of default float, via float -> string -> decimal:

from decimal import *

energy = Decimal(str(3.4280284e-19))

Here is a sample file I've used for atomic units in python:

pi = 3.14159265359
kb = 1.38064852e-23 # boltzmann / J/K
ab = 5.2917721092e-11 # bohr radius / meters
hbar = 1.054571726e-34 # planck / Js
h = hbar*2*pi
me = 9.10938291e-31 # electron rest mass / kg
e = 1.602176565e-19 # electron charge / C
ke = 8.9875517873681e9 # coulomb constant / kgm^3 s^-2 C^-2
Eh = 4.35974417e-18 # hartree energy / J
c = 299792458 # speed of light, m/s
cs = 340.29 # speed of sound, m/s
t = hbar / Eh # time / s
v = ab*Eh/hbar # velocity m/s
F = Eh/ab #force
T = Eh/kb # temp
P = Eh/(ab**3) # pressure
Ef = Eh/(e*ab) # electric field
Ep = Eh/e # electric potential
u = e*ab # electric dipole moment


"Encyclopedic" / semi-exhaustive reference work on Chemistry I'm buying my brother a present—It's the holiday season after all. He's about to start his academic career in chemistry. I would like to give him a book on chemistry, but not something that is considered "chem 101", "popular", or too specialized and probably off-topic. I am a mathematician myself, and in my field we have the beautiful "Princeton companion to mathematics". It is edited by a Field's medallist, has contributions of many high-profile mathematicians, and is both incredibly varied and in-depth. It contains many, many beautiful phenomena, but is by no means a "popular book" or an easy read. I was wondering if anything like this exists in Chemistry. So to clarify: the book would ideally introduce an advanced reader to several concepts in other branches of chemistry than the person's speciality, in a way that is appealing to a serious academic. Many thanks!

One of the most comprehensive descriptive chemistry books is Holleman and Wiberg's Inorganic Chemistry. It fits your encyclopedic criterion with about 2000 pages with all major atomic structure concepts plus descriptive chemistry of the elements. It is being published for more than 100 years and originally it was in German.
Another classic, again of about >1000 pages was Glasstone's "A Textbook of Physical Chemistry". It is slightly dated (1940s) but very exhaustive in terms of covering all the classical concepts to be ever encountered in physical chemistry.


Orbital correlation diagram for the cleavage of cyclopropylmethylene carbene In the last exam we were asked to rationalize the mechanism of the cyclopropylmethylene carbene cleavage into acetylene and ethylene.

In the lecture we discussed pericyclic reactions (cycloadditions, electrocyclic reactions, sigmatropic reactions, ene reactions). I don't think it's a con-/disrotatory ring opening, because in my view only a $\sigma$-plane is conserved over the entire reaction coordinate. What type of ring opening is this; is it a concerted or radical based C-C bond cleavage? Because there is highly probable an elongation of the bond length (as drawn in the stretched cyclopropane moiety), I don't assume it to be concerted but rather radical based.

I tried to classify the reaction in terms of symmetry along its entire reaction coordinate. Because from the task, a singlet carbene, I know it's a $sp^2$ hybrid, so I strongly assume the empty p-orbital to be in conjugation with the Walsh orbitals, more precisely with the Möbius type orbitals.

The problem is very similar to cyclopropylcarbinyl cation:

In the exam I was asked to draw the orbital correlation diagram. For me it looks like two radicals forms the double bond of ethylene and the "biradical" like centers recombines with the "biradicaloid" (in $sp^2$ orbital). I call a carbene a diradicaloid or biradicaloid because the two radical electrons share the same orbital.

To correlate reactants with products I need to combine orbitals with the same symmetry. We only draw the orbitals which are involved either in bond braking or in bond making. How should I account for the biradicaloid singlet $sp^2$ orbital in the orbital correlation diagram? (It is forming the triple bond of acetylene, so there must be a change of hybridization.) For me it looks like to have no symmetry (neither A' nor A'' because it lies in the mirror plane, because it is eclipsed with the $\alpha$-carbon-H.

If it's a radical mechanism I assume the Möbius orbital of A'' symmetry (depicted in green) to correlate with the $\pi^*$ of ethylene.

I assume, the cleavage happens preferently with light, because in this way I can put a photon from A'' (green) into the orbital of A'' symmetry on the reactant site and therefore lowering the activation energy for the reaction.

I assume the orbital correlation diagram to look something like this:

There are some points unclear to me:

Has the empty p-orbital A'' symmetry ?
For me it looks somehow ugly to correlate orbitals which don't have symmetry labels (I think this procedure is not allowed and I need to build a proper set.) How can I build a linear combination in this particulate case, so that the orbital correlation from red to red makes sense?
There is a "$\pi^*$ orbital acetylene,2" unused (not correlated). Did I miss an orbital from the reactant site? (because there is one orbital too many on the product site)

What is the type of the ring opening; is it a concerted or radical
  based C-C bond cleavage?

This is a chelotropic reaction, a subset of cycloaddition\fragmentation reactions.  As to whether it proceeds by a concerted or radical mechanism, well, you haven't presented any experimental data that can help us decide.  For example, if the reaction had been run with the cis-2,3-dimethylcyclopropane analogue and both cis- and trans-but-2-ene were found among the products, we could argue for a biradical, two-step (non-concerted) process.  When we create an orbital symmetry diagram, that just allows us to make a prediction as to whether or not we expect the reaction to be concerted.

  I strongly assume the empty p-orbital to be in hyperconjugation with
  the walsh orbitals

There are a number of different conformations available to the molecule, but I agree with you that having the carbenic p-orbital aligned with the plane of the cyclopropane ring (the "bisected" conformation) should be favored.  Just as the bent carbon-carbon bonds with high p-character in a cyclopropane ring stabilize the empty p-orbital in a cyclopropylcarbinyl cation (your drawing above), the same thing should happen here with the carbene's empty p-orbital.

  I was asked to draw the orbital correlation diagram.

I have difficulty reading your diagram, so I created a new orbital correlation diagram.

Let me say a few things about my figure in order to make it easier to understand.

The rear cyclopropane bond is bisected by the plane of the screen, this plane also contains the cyclopropyl carbon connected to the carbene center.
The carbene p-orbital is perpendicular to the plane of the screen, the carbenic $\ce{sp^2}$ orbital lies in the plane.
The acetylenic sigma bond lies in the plane of the screen just as the sigma bond connecting the cyclopropyl carbon to the carbenic carbon does.
There are two pi bonds in the acetylene, Px is perpendicular to the plane and Pz is contained in the plane of the screen.
This plane of the screen is the plane of symmetry that we will use to determine the symmetry of the various orbitals and bonds involved in this transformation.
I haven't redrawn the Walsh cyclopropane orbitals, we can refer to your drawing up above.
$\mathrm{\sigma_1}$, $\mathrm{\sigma_2}$ and $\mathrm{\sigma_3}$ refer to the 3 bonding Walsh orbitals (cyclopropane) that you've drawn above.
I haven't drawn any of the antibonding orbitals, as we'll see below, we don't need them in this case.
The empty carbenic p-orbital and $\ce{sp^2}$ orbital should be around E=0, but I've drawn them slightly below for clarity.
The "A" and "S" labels tell us whether the orbital is antisymmetric or symmetric with respect to our preserved symmetry element, the plane of the screen.

Now, if you look at the reactant occupied orbitals and their symmetries, we have 3 orbitals with S symmetry and 1 with A symmetry.  In the products we also have 3 occupied orbitals with S symmetry and 1 with A symmetry. So there is a smooth correlation between reactant orbitals and product orbitals.  This analysis suggests that the thermal reaction can proceed in a concerted fashion.

Miscellaneous

Just because a reaction is allowed to proceed in a concerted fashion does not mean that the reaction will occur.  The most common reaction of cyclopropyl carbenes is ring expansion to the corresponding cyclobutene.  Other concerted reactions may be preferred because they happen to be of lower energy.
Of course, if a reaction is permitted in the forward direction, then it is also permitted in the back direction.  But when we heat ethylene and acetylene we don't generate cyclopropyl carbene - why not?  Well, these are all equilibria and the forward reaction is extremely exothermic.  We remove a highly strained cyclopropane and a high energy carbene and replace them with pi bonds.  Therefore we would expect the back reaction (combination of ehtylene and acetylene) to be extremely endothermic.  The products are much more stable than the reactants and we don't expect to see a significant concentration of reactants at equilibria.  It would be interesting to heat ethylene and acetylene to high temperature and see if any cyclobutene (or its ring opened product, buta-1,3-diene) is formed.  If these products are observed they would be suggestive of the presence of cyclopropyl carbene which then underwent the ring expansion reaction to cyclobutene.

Response to OP's comments:

  How can we see in general from an orbital symmetry correlation diagram
  if a reaction is concerted or not?

If all of the reactant bonding MO's correlate with product bonding MO's, then the reaction is allowed to proceed in a concerted fashion.  If some reactant bonding MO's correlate with antibonding product MO's, then the reaction is not allowed to proceed in a concerted fashion.

In the current example, the reactant bonding (occupied) MO's $\sigma_1,~ \sigma_2,~ \sigma_3$ and the carbene $\ce{sp^2}$ correlate with product MO's $\sigma$, ethylene $\pi$, acetylene $\pi_x$ and acetylene $\pi_z$ respectively.  All ground state reactant orbitals correlate with ground state product orbitals.  Therefore the reaction is allowed to proceed in a concerted manner.

  Is it okay to say biradicaloid instead of carbene?

No, not really. "Biradical" is generally used in cases where the electrons are on different atoms.  Like when we break a bond in cyclopropane we have a 1,3-biradical.

  In the orbital correlation diagram I would predict the acetylene a
  little bit lower because of more s character (sp) than ethylene (sp2).
  May be the energy difference is roughly as high as the difference
  between carbene p orbital A and carbene sp2 orbital.

That may be, but in both ethylene and acetylene the pi electrons (which is what we are talking about) are in pure p orbitals.  My guess was that they would have roughly the same energy.

  For the retro cheletropic reaction I assume a photolytical pathway

The retro reaction, like the forward reaction, should be thermally allowed and photochemically forbidden.

  According to the paper, I've found here...

I can only see the abstract, but from what I read, I don't think this is a reaction involving a photochemically produced carbene in an excited state.  I think that they are photochemically decomposing the diazo compound and produce the same carbene they generate thermally at higher temperatures.  The photochemical route just allows them to generate the same ground state carbene at very low temperatures.


What is an entropic trap? Can someone explain the concept of entropic traps, or provide good sources to look up? The term is not present in any chemistry books I own, and I cannot find online sources explaining it.

I think this should address your query:

  The separation of macromolecules such as polymers and DNA by means of electrophoresis, gel permeation chromatography or filtration exploits size-dependent differences in the time it takes for the molecules to migrate through a random porous network. Transport through the gel matrices, which usually consist of full swollen crosslinked polymers depends on the relative size of the macromolecule compared with the pore radius. Sufficiently small molecules are thought to adopt an approximately spherical conformation when diffusing through the gel matrix, whereas larger ones are forced to migrate in a snake-like fashion. Molecules of intermediate size, however, can get temporarily trapped in the largest pores of the matrix, where the molecule can extend and thus maximize its conformational entropy. This 'entropic trapping' is thought to increase the dependence of diffusion rate on molecular size. Here we report the direct experimental verification of this phenomenon. Bragg diffraction from a hydrogel containing a periodic array of monodisperse water voids confirms that polymers of different weights partition between the hydrogel matrix and the water voids according to the predictions of the entropic trapping theory. Our approach might also lead to the design of improved separation media based on entropic trapping. – Source

I don't know if you needed a textbook definition, but thought you'll get a more elaborate view from this.

This is a schematics of an entropic trap device(ref: Anal. Chem. 2002, 74, 394-401)