How does salt and water promote corrosion? I was always wondering why iron only rusts when it gets wet. I know that for example aluminium oxidises on its own on regular basis.

I also heard many complains that salt used to remove snow from roads increases corrosion rate of car undercarriage.

Recently, I was researching how to produce iron oxide to improve rocket candy fuel - and among with oxidising substances, especially hydrogen peroxide, salt water was suggested.

What do these things have to do with aerial oxygen reacting with iron? I have also noticed that copper roofs turn green. Copper stuff kept away from rain doesn't do that so fast.

I think a tantalising clue can be found in the principle of cathodic protection used in industry.

Cathodic protection is used in the shipping industry to prevent iron/steel hulls from rusting. Essentially you make your hull into a cathode by connecting a (typically) zinc anode to the hull. How do we know the ship's hull is the cathode and zinc the anode?

And zinc in fact?

Lets go back to the standard reduction potential tables. The following reaction has a reduction potential of $\ce{-0.7618V}$:

$$ \ce {Zn^2+(aq)} + 2{e}  \ce{->} \ce{Zn(s)}  $$

Whilst the following has a reduction potential of $\ce{-0.44V}$:

$$ \ce {Fe^2+(aq)} + 2{e}  \ce{->} \ce{Fe(s)}  $$

Clearly the rusting process works in the opposite direction (and thus so do the above equations) which just means we reverse the voltage to get the oxidation value.

A higher reversed voltage means that that process has a greater tendency to occur, which means that zinc here is preferentially oxidised over iron. Hence if we connect an iron hull to a zinc rod, the zinc acts as the anode whilst the hull is the cathode i.e any $\ce{Fe^2+}$ directly formed by reaction with sea-water is pushed back to iron metal by the production of $\ce{Zn^2+}$, a process which liberates two electrons.

This should suggest that some electrolytic process contributes to salt water affecting the oxidation of iron. The most basic explanation is that the sea water acts as an electrolyte, which facilitates this process much more easily than, say, distilled water.

What does this mean however when we don't have an electrochemical cell as above, and only have a single electrode i.e. the ship's hull?

We might like to look up the standard reduction potentials for both chloride and iron, however we quickly see that this isn't the case; both chloride and iron metal have to lose electrons in our case, which isn't going to form an electrochemical reaction.

It's an interesting case that bleach (alkali salts of chlorite, $\ce{ClO-}$)undergoes the following reaction with iron:

$$ \ce {Fe(s)} + \ce{ClO^-(aq)} +\ce{H2O}  \ce{->} \ce{Fe^2+(aq)} + \ce{HClO2(g)} +\ce{2H+(aq)}  $$ which has a favourable overall voltage of around $\ce{2.11V}$.

So we've seen that chloride doesn't directly react with iron, but chlorite does. This is a possible route for corrosion, although it's not obvious by which natural process chlorite comes about.

Another point to note is that seawater just doesn't contain sodium chloride, but myriad other ions which can contribute to this corrosion process,especially sulphate.

$$ \ce {SO4^2-(aq)} + \ce{4H^+(aq)} + \ce{Fe(s)}  \ce{->} \ce{Fe^2+(aq)} + \ce{SO_2(g)} + \ce{2H2O(g)} $$ which occurs with a total voltage of $\ce{0.61V}$, which tells us that the chlorite process is much more favourable thermodynamically.

So in conclusion, there are many different reactions that can utilise seawater to oxidise iron—it's not incredibly clear which ones do contribute, and which ones don't, but the above provides the means for some educated guesswork, so to speak.

As a final point, iron can oxidise in air without water, it just takes a long time. The presence of an electrolyte will speed up the process however. Corrosion also needn't be a reaction with oxygen either, but any process which makes iron ions from iron metal.


Number of Stereoisomers of C(CHClCH3)4 I am having trouble understanding why the structure

$$\ce{C(CHClMe)4}$$

has only five stereo isomers. According to my textbook, the stereo isomers correspond to the following series of absolute configurations:

(R,R,R,R)
(R,R,R,S)
(R,R,S,S)
(R,S,S,S)
(S,S,S,S)

But my question is, does not the order in which these appear change the total? Isn't (R,S,R,S) different from (R,R,S,S)?

To answer your question, if you haven't figured it out by now, RRSS is the same as RSRS and it is the only one of the five that is optically inactive (Compare with nonactin). Structure A numbers the positions of chiral chlorine-bearing carbons. Structure B is RSRS and C is RRSS. Clearly, they are superimposable and, therefore, identical. A similar exercise with RRRS and RSSS will give a similar result.


Two ways to calculate dissolved oxygen To calculate the dissolved oxygen in water at saturation pressure at 25 °C, we do:

$$
D_\ce{O2} = K_\mathrm{H} \times M \times p_\mathrm{g}
$$

$$
D_\ce{O2}~[\pu{g/L}]= 1.26\times10^{-3}~[\pu{mol/L/atm}] \times 32~[\pu{g/mol}] \times 0.21~[\pu{atm}]
$$

I used $\pu{1.26E-3 mol/L/atm}$ for the Henry's constant for $\ce{O2}$ in water at 25C which is from Warneck and Williams (2012), which was in turn taken from Battino (1981). This returns a dissolved oxygen of $\pu{8.47 mg/L}$.

Benson (1980) reports dissolved oxygen for "a body of pure water is in thermodynamic equilibrium with an atmosphere of standard composition saturated with water vapor at a total pressure $P$" as 8.26 mg/L which is a difference of 2.5 %. The Benson paper suggests an error of around 0.02 %. Why is there a difference in these two methods? Is it due to different assumptions being made? Or perhaps rounding errors in how data is presented?

The 21.0% of oxygen by volume in air is from dry air. Data for dry air from webpage of Columbia University's Department of Earth and Environmental Sciences.

Note that the "% by volume"/100 for dry air also corresponds to the partial pressure in atmospheres. For air saturated with water at 25 C, the vapor pressure of water, according to Wikipedia, is 0.0313 atm. The second column is calculated by multiplying all dry air values by (1-0.0313)=0.9687.

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Gas} &\hspace{-5em}\text{dry air}&\ce{H2O}\text{ sat. air} \\
 & \text{% by volume} & \text{% by volume} \\
\hline
\ce{N2} &  78.084    & 75.640 \\
\ce{O2}  & 20.947    & 20.291 \\
\ce{H2O} & 0.0      &  3.13 \\
\ce{Ar}  &  0.934    &  0.905 \\
\ce{CO2} & 0.035    &  0.034 \\
\ce{Ne}  &  0.001818 &  0.001761 \\
\ce{He}  &  0.000524 & 0.000508 \\
\ce{CH4} &  0.00017  &  0.00016 \\
\ce{Kr}  &  0.000114 &  0.000110 \\
\ce{H2}  &  0.000053 & 0.000053 \\
\ce{N2O} & 0.000031 &  0.000030 \\
\ce{Xe}  &  0.0000087&  0.0000084 \\
\hline
\text{TOTAL}& 100.0027187 & 100.0026304 \\
\hline
\end{array}
$$

Now correcting the data from Warneck and Williams (evidently Warneck,  P.  and  Williams,  J.:  The  Atmospheric  Chemist’s  Companion:  Numerical  Data  for  Use  in  the  Atmospheric  Sciences, Springer Verlag, 2012) for the value of water saturated air

$\ce{O2} = \dfrac{20.291}{21}\times 8.47 = 8.18$ mg/L

This is much better agreement with the Benson (1980) data.

The OP's value of 1.26e-3 Mol/L/atm for the Henry's constant for O2 in water at 25C was evidently take from Table 8.23 of Warneck and Williams. The value also has a list error bracket of +/- 0.02.

$$\dfrac{0.02}{1.26}\times 100\text{%} = 1.6 \text{ %}$$

For the two dissolved oxygen values:

$$\dfrac{8.26-8.18}{8.26}\times100\text{ %} = -0.97 \text{ %}$$

so the values are within the error limits for Henry's constant.


Cleaning with Vinegar - what surfaces does it react badly with? Metal window sills ok? I'm cleaning up some mould around the home and I have some substance on my window sills which I'm not quite sure if it's dirt or mould.

I have read about using Vinegar to clean up mould, but an article I was reading awhile back said you shouldn't use Vinegar on certain things as it can eat away at it or something because it's acidic; I thought this may have been metal and/or painted substances but I may be wrong.

I know you shouldn't use it on granite, marble and natural stone, but was wondering if using it on the painted metal window sills is ok or should I use something like diluted Tea Tree Oil instead, or would that be too oily?

FYI I would be using White Vinegar.

Here is a pic of my window sills:

Avoid aluminum alloys when working with vinegar solutions!

With time, the pH effect of acetic acid can penetrate the protective oxide coating. The exposed Al will even react with water:

2 Al (s) + 6 H2O (l) --> 2 Al(OH)3 + 3 H2 (g)

and with the presence of alloy metals, galvanic corrosion (especially if the 'dirt' is rich in salts that can serve as a good electrolyte) which apparently accelerates the attack on the Aluminum.

To quote a source:

  corrosivity environment for aluminium in food industry are foodstuffs with pH 3 – 5, such as fruit juices, jams and acidic canned fruits or hot gravies, sauces as well as dressings, vegetables and fish pickled in brines with 1 –3 % salt [1]. In particular, acetic acid needs to be taken into corrosion consideration, due to its wide usage in food industry (vegetable and fish pickling), and its representative properties among acids and juices in fruits, vegetables and other organic materials that can corrode metals. Although aluminium has a good resistance to acetic acid solution at room temperature, aluminium can corrode in almost any concentration of acetic acid at any temperature if the acid is contaminated with the proper species [2].

Further comments related to the presence of select metals:

  Although aluminium has a good resistance to almost all the concentrations of acetic acid at room temperature, care must be taken that the metal is free of other impurities such as iron, copper, tin and lead even in traces [2]. With increasing purity of aluminium, its resistance to acetic acid solution increases, and 99.5% aluminium can be used for the majority of engineering purposes, but components added to its alloy can increase the corrosion of aluminium in acetic acid solution [14]. Since the metal corrosion occurs via electrochemical reactions at the interface between the metal and an electrolyte solution, electrochemical techniques are ideal for the study of corrosion processes.

And importantly, the effect of pH:

  It is well known that aluminium resistance is related to the thin and compact layer of naturally formed oxide on aluminium surface, but this oxide layer is stable only in pH range 4-8. Lower or higher pH values caused prominent destroying of protective layer and thus the significant metal dissolution [15].


Is plane of symmetry a necessary and sufficient condition for optical inactivity in coordination compounds? In organic chemistry, we have to check for all three types of symmetry in a compound to check its chirality, ie, plane, alternate axis and centre of symmetry. But in coordination compounds, do we need to check all the symmetries to check its chirality? The ligands are supposed to be achiral. The difference between the two being that in coordination compounds only one "chiral" center is present but in organic compounds it can be more than one.
All the achiral coordination compounds of any coordination number which I have come across have a plane of symmetry. But for those which don't have, I have to check whether it's image coincides with itself or not. So, is plane of symmetry a necessary and sufficient condition for achirality of coordination compounds?
This thing is necessary for my purpose, as I have to solve problems in limited time, so I cannot use the orignal method always.

The number $\pi$ for bowhead whales† is still 3.1415... Same thing here.
Chirality of coordination compounds is the same thing as chirality of organic compounds. If all your symmetry elements are just rotations, you have it; if some are improper rotations, then you don't. It is as simple as that, and is not going to be any simpler.
Can you just check for the planes and call it a day? Well, you kinda can, and that will do, until you come across a compound which doesn't have a plane, but still is achiral because of some other symmetry element.
Are there such compounds? Well, that's like asking whether there is a place on Earth where three bricks stand on top of each other. You may google long and hard, or you may apply for a travel grant to visit countries with rich traditions of brickwork masonry... or you may just take three bricks and put them on top of each other. That's it; problem solved.
Here are my bricks. I claim that the tetrahedral $\ce{[Zn(NH3)4}]^{2+}$ has an inversion axis $\bar4$ and hence is achiral. You may point out that it has a plane as well. Fine. Let's give it some substituents: change each ligand to $\ce{N(CH3)3}$. What? Does it still have a plane, given that everything which rotates is rotated to the appropriate position? Fine. Let's give it different substituents. Change each ligand to $\ce{CH3NHC2H5}$. Where is the plane now?
So it goes.

† There was an old anecdote about an even more old paper on biophysics which was said to include some volume calculations and some crude approximations of a whale as a cylinder, then of course the number $\pi$ showed up, followed by the hilarious phrase along the lines of "The number $\pi$ for bowhead whales is about 3.14".


Which group does germanium belong to? $\mathrm{Z= 32}$

$\mathrm{1s^2\ 2s^2p^6\ 3s^2p^6d^{10}\ 4s^2p^2}$

According to me it belongs the $\mathrm{IV\ B}$ group since it has the $\mathrm{d}$ completed, but it belongs to $\mathrm{IV\ A}$. But we know that in the $\mathrm{A}$ group only the $\mathrm{s}$ and $\mathrm{p}$ elements belong. Why does this happen?

According to the IUPAC, the is not such thing as groups IV A and B anymore.
All the elements you are analyzing belong to the 14th group of the periodic table, each one in one period, from the 2nd to the 6th.

In regards to your other question, you need to look for a trend on the elements of the group.
What happens as you go down the group?
I think you've already figured out that the number of electrons around the nucleus significantly increases, and this has an important effect on the radius of the atom.


Is Pauling electronegativity well-defined? Pauling's definition of electronegativity defines the difference of electronegativity between two elements.

Suppose three elements A, B, and C. Then we can calculate $ϰ_a - ϰ_b$, $ϰ_b - ϰ_c$, and $ϰ_c - ϰ_a$ independently. Is the sum of the three values always zero? Why?

If you have a single, unique value of $\chi_i$ for each element $i$ then yes, the sum of the three values is always zero.

If you are assigning electronegativities that differ with oxidation state or chemical environment, the answer can be no (for example, if element B has a different electronegativity in an AB bond than it has in a BC bond).

The Pauling electronegativity difference is

$${\chi}_{a}-{\chi}_{b}=(1\ eV)^{-\frac{1}{2}}\,\sqrt{{E}_{d}\left( a,b\right) -\frac{{E}_{d}\left( a,a\right) +{E}_{d}\left( b,b\right) }{2}}$$

where $E_d(i,j)$ is the dissociation energy for a single bond between element $i$ and element $j$.
The definition assumes that you can consistently and accurately say that the square of the electronegativity difference between A and B is equal to the difference between the dissociation energy for AB and the average of the dissociation energies for AA and BB. If this isn't consistently true for each combination of three elements A, B, and C, your sum will be nonzero. Pauling fit his electronegativity values to the data to average out these inconsistencies.


How to calculate the average size of cubic crystals of KCl powder? The total area of $\pu{0.5 g}$ sample of $\ce{KCl}$ powder is measured to be $\pu{1620 cm2}$. If the powder consists of tiny cubic crystals, what is the average size of these crystals?

I'm studying for a physical chemistry final and am not sure how to do this practice question.
I know that the density is
$$\delta = \frac{m}{A},\label{one}\tag1$$
where $m$ is the mass and $A$ is the surface area, so do I just find
$$A=\frac{m}{\delta}$$
to find the average "size"?
Or do I use \eqref{one} and then find out what the unit cell structure of $\ce{KCl}$ powder is and use that to estimate what the surface area of the crystal is?

For a cube, the surface area is given by the formula $6a^2$ where $a$ is the length of one side. The volume is given by $a^3$. Surface area per volume is
$$\frac{6a^2}{a^3}= \frac{6}{a}.$$

We know the total surface area to be $\pu{1620 cm^2}$. We can calculate the total volume from the mass of the $\ce{KCl}$ and its density.

\begin{align}
  V &= \frac{m}{\rho}\\
  \rho(\ce{KCl}) &= \pu{1.98 g//cm^3}\\
  V &= \frac{\pu{0.5 g}}{\pu{1.95 g//cm^3}}\\
    &= \pu{0.3 cm^3}
\end{align}

An average crystal should have the same ratio as the total. So we set the two equal and solve for the length.

\begin{align}
  \frac{6}{a} &= \frac{\pu{1650 cm^2}}{\pu{0.3 cm^3}}\\
  a &= \frac{6 \times \pu{0.3 cm^3}}{\pu{1650 cm^2}}\\
  a &= \pu{9E-4 cm}
\end{align}


What aspects of a macromolecule/protein do the various contributions to its entropy relate to? I've come across a few different contributions to entropy in macromolecules such as proteins: configurational, conformational and vibrational. The problem is that I can't seem to find a consistent definition and there may be some overlap.

In addition, I've only really understood entropy to be made up of rotational, vibrational and translational components and I'm struggling to reconcile this with the contributions to entropy for proteins described in the literature.

I've only really understood entropy to be made up of rotational, vibrational and translational components

There is also electronic entropy, but it's a fun exercise to show that (at least for the hydrogen atom) the corresponding partition function $q_{\text{elec}} = 1$ at room temperature, and for all intents and purposes there is only population of a single state. So, the other 3 kinds of entropy (translational, rotational, vibrational) are those directly from a statistical mechanical derivation that are relevant for most (bio)molecules.

  I've come across a few different contributions to entropy in macromolecules such as proteins: configurational, conformational and vibrational.

Hopefully you agree that vibrational entropy in the context of proteins is the same as in the context of any other system in that vibrations can't be decomposed into other kinds of excitations (translational or rotational), regardless of whether or not the vibration is localized to a bond, highly delocalized, or quaternary structure subunits moving with a regular period.

Configurational and conformational entropy are being used slightly differently. From Wikipedia,

  Conformational entropy – is the entropy associated with the physical arrangement of a polymer chain that assumes a compact or globular state in solution.
  Configuration entropy - is the portion of a system's entropy that is related to the position of its constituent particles rather than to their velocity or momentum.

For proteins, I think the distinction between conformation and configuration is less clear than it is for more molecular systems. In my limited experience, a configuration of a protein is similar to a macrostate in statistical mechanics, which is distinguished from other possible configurations by some criteria related to the position of a protein's primary or secondary structure. More specifically, different techniques can be used for clustering the different protein conformations together into distinct configurations.

In these cases, the greater the number of conformations a protein can take, or the number of configurations it samples based on some metric, the greater the entropy. For something like an intrinsically disordered protein, its conformational and configurational entropy are likely to be higher than for a globular protein.


Frontier orbital analysis of zinc-catalysed carbonyl-ene reaction Q1. Is the $\sigma_\ce{C-H}$ bond a donor in this reaction? If not, then can we take the alternate combination (HOMO of alkene, LUMO of the $\ce{C-H}$ bond and, HOMO of the carbonyl)?
Q2. We can also draw the LUMO of the carbonyl group in the other possible manner(simply flip the shaded lobe on each atom, hence keeping the node). The reaction would then have to go through an antarafacial manner.
Why do we draw the LUMO in this manner?

First of all, a molecule only has one HOMO and LUMO. Don't confuse MOs with localized orbitals from VB theory.
You can basically approximate this mechanism with four MOs:

LUMO+1: Mainly $\pi^*_{C=C}$ with contributions from $\sigma^*_{C-H}$
LUMO: Mainly $\pi^*_{C=O}$
HOMO: Mainly $\pi_{C=C}$ with contributions from $\sigma_{C-H}$
HOMO-1: Mainly $p_O$ with contributions from $\sigma$-orbitals on the carbonyl carbon. We can regard this MO as one of the lone pairs of the oxygen.

The corresponding interactions will roughly look like this:

Hence, the HOMO - LUMO interaction explains the C-C bond formation, while the HOMO-1 - LUMO+1 interaction explains C-H bond breaking and O-H bond formation. Both HOMO and LUMO+1 also explain the shift of the double bond.
Now to your question:

Q1: Yes, $\sigma_{C-H}$ is part of HOMO, but has a lower contribution compared to $\pi_{C=C}$. So I would mainly call the C=C bond as donor. As you can see $\pi_{C=O}$ is not part of the FMOs, so your proposal makes little sense. It mainly contributes to more stable occupied MOs.
Q2: If I understand correctly, you propose an antarafacial [2+2], which is in principle symmetry allowed, but would lead to a very strained transition state with a high barrier. You are free to swap the sings on MOs, but as soon as interact with another MO, you take the one that leads to stabilization of course.

Edit: As reference, I got the FMOs from a HF/6-31G calculation on a slightly simplified system:


How does the Freezing Point fall in a solution? It is known that addition of a non-volatile solute to a volatile solvent(liquid) to give a solution reduces the Vapour Pressure of the solution (well solvent actually as only solvent is volatile).

This leads to the Elevation of Boiling Point and "Deppresion" of Freezing Point.

I am clear with elevation of BP. But it is said that a liquid freezes when the vapour pressure of the liquid phase attains the vapour pressure of its solid state.

So, how possibly does the freezing point falls? For instance let the Vapour Pressure of the solid state be at an arbitrary point "x"(at a temperature A Kelvin) and that of the liquid phase be at "y".

So when we add a solute to the liquid:

Vapour pressure of liquid(solution) = y - $\alpha $ (where $\alpha $ is the depression in the vapour pressure)

So now the vapour pressure of the liquid(solution) reaches "x" rapidly that is at a higher temperature or a temperature greater than A which is an elevation in the freezing point.

I am always confused with this part.

Focus more on free energies rather than on vapor pressures (which derive, ultimately, from free energies after all).  For a mixture of B (solute) in A (solvent), the entropy of mixing is $RT(x_A ln(x_A) + x_B ln(x_B))$, and the enthalpy of mixing will go approximately as $x_A x_B \Omega$, with $\Omega$ as a measure of the interaction of A and B. The entropy term will always result in a reduction of free energy at small $x_B$ regardless of the sign of $\Omega$, but in the case of salt in water $\Omega$ is negative, driving further solubility.

Now, about those temperatures of phase transitions (which as you should recall occur when the free energies of the phases are equal, more fundamental than vapor pressures).  On the boiling end, water with salt in it has a lower free energy than water without salt - the boiling point of the salt water has to be higher than for pure water.  The presence of the solute makes it happy to stay liquid.

So, why isn't the freezing point raised in an analogous way? For pure A, of course, the freezing point remains that of pure water.  The entropy of mixing is equivalent for solid and liquid in this case, and it is not clear how different the excess enthalpy is going to be - so why does liquid continue to be stable?  Well, it only does while it has a higher concentration of salt in it than is in the solid.  It forms a classic eutectic point on the binary phase diagram (see, for example http://antoine.frostburg.edu/chem/senese/101/solutions/images/saltwater-phase-diagram.gif).  The only requirement of melting point depression is that water at a high concentration of salt has a lower free energy than the solid at a lower concentration of salt.  Given the outline of free energy above, this is likely to hold for some range of temperatures.

Again, remember that what is being lowered is the freezing point of water with salt in it - what starts freezing out is ice with a lower salt content in it.


Chromic acid oxidation of aldehydes and primary alcohols The answer has to be both A and D. Chromic acid is a strong oxidizer and will take both aldehydes and primary alcohols to carboxylic acids.

However the computer thinks otherwise. Am I just missing something really simple here?

There are many $\ce{CrO3}$ oxidation reagents (Jones, Sarett, Collins, PCC, etc.), however, most are selective and stop the oxidation of a primary alcohol at the aldehyde stage.  On the other hand, the Jones reagent is relatively unselective and will oxidize a primary alcohol to the corresponding carboxylic acid - if enough reagent is present. It takes two-thirds of an equivalent of Jones reagent to oxidize one equivalent of the alcohol to the aldehyde and an additional two-thirds equivalent to oxidize the aldehyde to the acid.  Also the Jones oxidation is run under acidic conditions (typically dilute sulfuric acid).

I would have though that A, B (the gem-diol will equilibrate with the aldehyde under acidic conditions) and D were correct answers.  It seems like the computer answer is either in error or it is a "trick" question (seems unlikely for computer-based learning).  The "trick" would be that since an excess of reagent was not specified only 1 equivalent is assumed.  Therefore there is not enough reagent to carry out two sequential oxidations; as a consequence only the aldehyde or gem-diol could be completely oxidized with one equivalent of reagent. But then A or B is not an acceptable answer either.


Is specific rotation defined for mixtures of enantiomers or meso compounds? I recently had to do an experiment where we had to make a racemic mixture of D and L complexes and then resolve the enantiomers, and measure the optical rotation of the racemic mixture, and the individual salts. After that we had to calculate the specific rotation in case.

However, when reading about optical activity, I came across the definition:

  For an optically active substance, defined by $\ce{[α]^θ_λ = α/γl}$, where α is the angle through which plane polarized light is rotated by a solution of mass concentration γ and path length l. Here θ is the Celsius temperature and λ the wavelength of the light at which the measurement is carried out.

Source: CRC handbook of Chemistry and Physics

Also, specific rotation is treated as an intensive property of a compound (i.e. one enantiomer). Given that, does it make any sense to calculate the specific optical rotation for a mixture of enantiomers (whether racemic or not)?

And applying the same logic, can we say that specific rotation has no meaning for a meso-compound ?

Nothing stops you from taking the equation and calculating a specific rotation of a mixture. Indeed, as far as I am aware when establishing a new asymmetric reaction, the e.e. is measured by something like chiral HPLC (which will unambiguously give two peaks that can be integrated and compared) while the specific rotation is measured for the resulting mixture of enantiomers as is. So the paper will say something like $99~\%$ yield ($98~\%\ \text{e.e.}$) but below there will be a standard calculation of $[\alpha]_\mathrm D^\pu{20^\circ C}$.

So obviously you can measure a specific rotation of the racemate or meso compound but whatever value you measure should fall into the error of the polarimeter so that the final value of $[\alpha]_\mathrm D^\pu{20^\circ C}=0$.


Why does the water in a Powerade bottle tastes a little like Powerade after many refills? I have a bottle of Powerade and after probably around 20+ refills of water, the new water I put in still tastes/smells a little like Powerade.

How does that happen?

Typically water bottles used in sports or biking are manufactured from high-density polyethylene (HDPE), which is essentially like a longer-chain version of paraffin wax. In short, it's basically a solid nonpolar hydrocarbon (with the exception of the Zeigler-Natta catalysts, which are only present in trace amounts).

Meanwhile, a quick stroll through Fenaroli's Handbook of Flavor Ingredients shows that many of the flavor compounds typically found in foods are either somewhat nonpolar, have nonpolar sections, or have (mostly unsubstituted) aromatic rings present. For example, ethyl valerate, one of the primary constituents of apple aroma, has a butyl chain, is fairly nonpolar, and is nearly insoluble in water.

However, it may be soluble in HDPE, and so it may infuse the bottle with the flavor agents so that the taste can remain for a long time.

Of course, doing a double-blind test may be your best bet for checking to see if this is a real phenomenon, or your imagination. But I've personally noticed this as well, so it could very well be real.


Building an incomplete redox half reaction/reduction I can't seem to figure out this redox reaction.

$$
\ce{CH3COO- + CO2 -> CH3CH2COO-}
$$

Is the charge on the carbon in the $\ce{CO2}$ supposed to match the charge on the carbon in the $\ce{CH3COO-}$? I am supposed to build a half reaction using $\ce{H2O}$ and $\ce{H+}$ as needed, but I can't seem to figure out how many electrons are actually transferred. What are the charges on the carbon?

It helps to use "fully compressed" formulae so you can easily count atoms and figure out how to balance carbons, oxygens, hydrogens, and charges.  That is, instead of

$\ce{CH3COO- +CO2 -> CH3CH2COO-}$

write it this way:

$\ce{C2H3O2- +CO2 -> C3H5O2-}$

Then it's easy to see that this is not a balanced equation.  Carbon atoms are the same on each side, but not oxygen or hydrogen atoms.  You need two more $\ce{O}$ atoms on the products side.  Those have to come from water since that is the only oxygen-containing species you are allowed to use.

$\ce{C2H3O2- +CO2 -> C3H5O2- + 2H2O}$

Now oxygen is balanced, but hydrogen is still imbalanced.  That can be rectified by adding $\ce{H+}$ into the equation as appropriate.

$\ce{C2H3O2- +CO2 +6H+ -> C3H5O2- + 2H2O}$

Now only charge is imbalanced, which can be fixed by adding electrons to complete the balanced half-reaction:

$\ce{C2H3O2- +CO2 +6H+ +6e- -> C3H5O2- + 2H2O}$


Effect of temperature on benzenediazonium during diazotization What will benezendiazonium ion become if it is not maintained at a temperature below $5\ °\mathrm{C}$ during diazotization?

Diazonium salts are prone to ejection of molecular nitrogen. The process is fairly slow for aromatic diazonium salts, so the salts are usually relatively stable around 0 celsius (alkyldiazonium salsts, though, are a completely different matter). In case of benzenediazonium resulting phenyl-cation is extremely active and will react with first nucleophile present, in many cases it is water, so the product would be phenol.


Why do pharmaceutical drugs have a expiration date? Every now and then you reach into the bathroom cabinet to find that some pill or ointment you where planning to use has gone past its expiration date.

I am interested whether this expiration date has any value. In a short search I found this article from Harvard medical school stating that some test found that 90% of all drugs where still perfectly active 15(!) years after the expiration date. The article then continues to give some vague statements on possible reasons for the expiration date (including financial reasons).

What I am wondering is whether there would be any chemical reasons for drugs to become less effective or even become harmful over time, which could warrant the existence of an expiration date?

As a specific example, in case it is difficult to answer this question in general, let's take an ointment meant to treat muscle pains which use Benzydaminehydrochloride as an active component.

It's a good question, and I cannot answer your specific example but can give some generic reasons why pharmaceuticals have an expiration date:

Degradation of the compound/formulation over time. This obviously comes to mind, but there is more to it than meets the eye. For example, you gave the example of an ointment and cited the active ingredient: however, the product you buy (and use) is more than just the active molecule. The whole formulation has to be stable, not only the benzydamine. The ointment is likely to be a complex emulsion, designed to target many different physical and chemical properties, including chemical stability, thermal stability, but also viscosity (so it's easy to apply), color, etc. I suppose in that particular case, the emulsion might be less stable than the active ingredient taken alone.
Also to note: many pharmaceuticals are sensitive to air, in many cases because they are unstable in the long term in presence of oxygen, or moisture. Such products are sold enclosed in a plastic, metallic or composite packaging that protects them. There, the limited stability in time might simply be due to the limit of stability of the (cheap) packaging itself: e.g., after some time of possible light exposure, plastic will become porous and let moisture in.

Being conservative, because we're talking about health.

Being conservative, because the company selling the product doesn't want to be liable.

Economics: it takes money to make absolutely sure the drugs are safe for a longer period of time. Are customers willing to pay more? Most people don't store drugs for use years later (nor is it recommended), so they won't see the point in paying more for longer expiration.


Mechanism of the Gatterman Reaction Aryl diazonium salt on treatment with copper and the corresponding hydrohalic acid, gives the aryl halide. This reaction is called the Gatterman reaction.

What is the mechanism of this reaction? Is it similar to Sandmeyer reaction?

The Gattermann reaction you mentioned in the question uses Copper powder and Hydrohalous acid to substitute diazonium group with a halide in an aromatic diazonium salt.
It uses $ArSN_1$  pathway, similar to Sandmeyer Reaction which uses Copper Halide and corresponding Hydrohalous acid instead of copper powder.
As diazonium group is a very good leaving group, it departs leading to a phenyl carbocation which is very unstable, this is in fact the slow step of the reaction.
The Nucleophile (Halide in this case) attacks the phenyl carbocation leading to the formation of Halobenzene(fast step).
Nitrogen gas and Hydrogen chloride are also formed.


Overlap integrals in Hückel theory In Hückel theory we are only interested in π systems, where $\mathrm p_z$ orbitals overlap. One of the approximations in Hückel theory is that the overlapping $\mathrm p_z$ orbitals are orthonormal:

$$
S_{ij} =
\begin{cases}
1\quad i = j \\
0\quad i \neq j \\
\end{cases}
$$

If orbital overlap leads to bond formation, how can the overlap integrals, between $\mathrm p_z$ orbitals, be equal to 0? Maybe I am missing something obvious, but doesn't a π-bond demand non-zero overlap between p-orbitals?

Hückel theory would still work if there would be overlap.

In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix equation can be directly interpreted in terms of the $\pi$ orbitals in the system.

There is the opposite view in Extended Huckel Methods whereby the assumption is that the overlap entries depend upon the off-diagonal elements. In ordinary Huckel theory, the diagonal elements are represented as
\begin{align}
\alpha_{ii} &= \langle \psi_i | \hat{H} | \psi_i \rangle \\
\hat{H} &= -\frac{\hbar^2}{2}\nabla^2 + V
\end{align}

But in the extended thoery, the offdiagonal elements are not denoted by $\beta_{ij} = \delta_{ij}$ but more as

\begin{align}
\beta_{ij} &= - \frac{C}{2}\Big \langle \psi_i \Big |-\frac{\hbar^2}{2}\nabla^2 + V \Big | \psi_j \Big \rangle\\
&= - \frac{C}{2} \left(\alpha_{ii} + \alpha_{jj} \right)S_{ij}
\end{align}
Where $S_{ij}$ is the usual notation for the overlap between the corresponding atomic orbitals.

Here $C$ is a constant, usually taken equal to $7/4$ and has units of energy.


What reaction takes place here? Upon reading the given question, I realized that:

The first reaction is an aldol condensation reaction between the given compound and formaldehyde.

So the product after 1st step is:

Now, the second part has thrown me off. The current ketone has no $\alpha$ hydrogen left. Condensation reactions are out of the question.
Cannizzaro reaction can not take place as this is a ketone (and not an aldehyde) also the medium is now acidic.
However, I saw one possibility of formation of hemi-acetals.
(I did not think about full acetals, as such a structure was not present in the given options to this question)

Which gave me the following product:

However, the answer provided seems to be not this.
The given answer is this:

Can someone give a suggestion that where I want wrong and/or how could I reach the answer and also what was wrong in my thinking as well?

This question is taken from the JEE(A)-2016 [Paper - 2]

I think a cross Cannizzaro with the Ketone and Formaldehyde takes place. The ketone give the alcohol and Formaldehyde formiate. In a second step a full acetal will be created with formaldehyd and now the Diole.
Sorry I cannot add comments here. So use answer function.


Are we wrong about Bakelite formation mechanism? In the reaction of phenol with acetyl chloride the oxygen of the phenol attacks the acetyl chloride. In Bakelite formation we know that pi electrons attack the carbonyl carbon of formaldehyde.My question is shouldn't the oxygen atom of phenol attack the formalaldehyde just as in the reaction of phenol with acetyl chloride?

C-acylation : acylating agent (acyl chloride or anhydride) and $\ce{AlCl3}$
O-acylation : acylating agent (acyl chloride or anhydride).This is the one which you have talked about in the question.

Since the formation of bakelite is by Lederer Manasse reaction which proceeds through electrophilic aromatic substitution on phenol, its path is similar to that in C- acylation and not O-acylation.
Hence, the polymerisation of phenol and formaldehyde takes place either from two places:

For para substitution end:

For ortho substitution end:


Potential energy curve for dissociation of cation of carbon dioxide Can anyone suggest me some literature where I can find the potential energy curve or the electronic states for singly ionized $\ce{CO2}$ ($\ce{CO2+}$) dissociating into singly charged oxygen ($\ce{O+}$) and carbon monoxide ($\ce{CO}$).

Few of the literatures are here:
http://iopscience.iop.org/article/10.1088/0953-4075/47/12/124025/meta

https://pubs.acs.org/doi/pdf/10.1021/jp907351s

https://aip.scitation.org/doi/10.1063/1.444413


Joules to kJ/mol conversion In my AP Chem class we are working on testing Hess's Law and conducted three reactions. Then enthalpy changes of the 1) $\ce{NaOH + HCl}$ and 2) $\ce{NaOH + NH4Cl}$ to predict the enthalpy change of 3) $\ce{HCl + NH3}$.

I used the equation $q = C_pm\Delta T$. We determined the $q$-values using a method given to us by our teacher. She told us to assume a denisty of $1.03\ \mathrm{g/mL}$ for all solutions, we then multiplied the density by $50\ \mathrm{mL}$ to determine the grams of each reactant present. That value is the $m$ term. We used $4.18\ \mathrm{J/(g\ ^\circ C)}$ for the specific heat of water. The temperature change for the first reaction was $14.7\ \mathrm{^\circ C}$, the second was $1.2\ \mathrm{^\circ C}$, and the third was $9.8\ \mathrm{^\circ C}$.

The first reaction yielded a q value of $6328.938\ \mathrm J$, the second yielded $516.648\ \mathrm J$, and the third yielded $4219.292\ \mathrm J$.

The next step would be to convert this $q$ values into $\mathrm{kJ/mol}$. I consulted my lab partner who did the following operation:

$$\mathrm{kJ/mol} = 6328.938\ \mathrm J \times 1\ \mathrm{kJ}/1000\ \mathrm J \times x/0.2\ \mathrm{mol}$$

In his math, there was no numerator on the $0.2\ \mathrm{mol}$ term. The mol value he used came from there being 0.2 moles of the reactants present. I believe his math to be incorrect, but as my teacher is out sick and there are no other chemistry teachers present, I cannot determine the correct conversion. I believe it to be:

$$\mathrm{kJ/mol} = 6328.938\ \mathrm J \times 1\ \mathrm{kJ}/1000\ \mathrm J \times 6.022^{23}/0.2\ \mathrm{mol}$$

Both of the conversions yield values with significant % error. Any assistance is appreciated.

Accepted book enthalpies are in the units $\mathrm{kJ/mol}$.

Simply convert $\mathrm J$ to $\mathrm{kJ}$ by dividing by $1000$. Since $1\ \mathrm{kJ} = 1000~\mathrm{J}$.

The moles of dilute aquaous solutions is best determined by volume (L) times concentration (mol/L).$$n = V \times c$$ You would use the moles of the product. If there is, in fact, "$0.2\ \mathrm{mol}$ of the reactants present" then there would be $0.2\ \mathrm{mol}$ of the products present. since the chemical reactions are $\ce{1 mol + 1 mol -> 1 mol + 1 mol}$.  (You are not interested in the number of atoms per mole, so you would not use Avogadro's number.)

$$q= C_\mathrm{sol} m\Delta T$$ where $C_\mathrm{sol}= 4.06~\mathrm{J/(g \ ^\circ C)}$ (Specific heat capacity for dilute aqueous solutions)

$$q= C_\mathrm{sol} m\Delta T / n_\mathrm{product}$$  ($n = \text{moles product}$)

Remember that Hess's Law states that

  if a reaction can be expressed as the sum of two or more reactions, then the heat flow for the overall reaction is equal to the sum of the heat flows.

Hence, the sum of the first 2 reactions should be equal to the 3rd reaction.

$\displaystyle \ce{NaOH + HCl -> NaCl + H2O}$
$\displaystyle \ce{NH4OH + NaCl -> NaOH + NH4Cl}$  (Reversed and Enthalpy sign changed)
$\displaystyle\ce{HCl + NH4OH -> NH4Cl + H2O}$

Note how $\ce{NaOH}$ and $\ce{NaCl}$ cross out leaving the third reaction.


Is there an opposite to shielding effect? I recently read about shielding effect and lowering of effective nuclear charge  due to penetration of other electrons. I wonder while doing calculations involving Slater's rules the electrons from outer shells are neglected.
I think they can reverse the shielding produced by inner shell electrons.

Slater's rules are an attempt to lump the effect of all other electrons on the wavefunction, and thereby other properties such as energy, of an electron (described by a hydrogen-like wavefunction). The effect of electron-electron repulsion is modeled indirectly by saying that inner shell electrons effectively screen the attractive nuclear charge sensed by electrons further away from the nucleus. For simplicity screening is modeled as a one-sided effect: electrons further from the nucleus are not expected to significantly shield those closer to it, since an outer electron is close to the nucleus less frequently than an inner one (or, using more accurate language, the density near the nucleus of the outer electron is lower).
Slater's rules amount to a method of estimating appropriate exponents in a hydrogenic approximation of the electron wavefunction. They are semi-empirical, useful as a guide to explain why certain trends are observed, and fit data because they contain "fudge-factors" (parameters obtained from fits to data, not from fundamental theory). Still, that they work at all indicates what amazing insight Slater had in selecting the shape of the function and using the method to predict a number of atomic properties across the periodic table. If you haven't you should look at his original paper (Ref. 1), it's impressive and accessible. This is also fairly well explained in a Wikipedia article.
Reference

Slater, J.C. Phys. Rev. 36 (1): 57–64. doi:10.1103/PhysRev.36.57.


Denaturing of Proteins and Nucleic Acids - Effect of Temperature (Heating vs Cooling) Heating a protein/nucleic acid will disrupt inter-molecular and intra-molecular forces in the tertiary structure. It will also interfere with the shape of the active site if the protein.

Does this apply with cooling also?

Complimentary nucleic acids are held together mainly by hydrogen bonding, so I'm not sure if they are able to be cold denatured, but cold denaturation is a well known phenomenon for proteins. Here is a decent review of the processes involved. In short, most proteins' tertiary structure is dominated by hydrophobic interactions and the energy barrier to unfolding a protein is generally quite small. The stabilizing effect of the hydrophobic interactions decreases with temperature and at a certain point, is insufficient to keep the protein folded. For many proteins, the cold denaturation temperature is below 0 °C, so it's difficult to study. Generally, studies on these proteins are done by adding chemical denaturants or changing the pressure of the system to shift this temperature higher, having the risk of conflating this deliberate denaturation with the temperature effect. Of course, not all proteins are stabilized this way and things like metallothioneins probably don't denature in this way as their tertiary structure is maintained by coordinating metal ions.


Neuter Reactants: What do they do? So imagine a hypothetical chemical equation

$$
A + B \rightarrow C + B
$$

It is obvious from the equation that $B$ is not consumed during the reaction. If $B$ actively take part in the reaction, if it participates in the transition state and isn't just sitting there in the environment, I classify $B$ as a Neuter Reactant.

Under these defitions I would like to know:

Is a Neuter Reactant always a Reaction Inhibitor or Catalyst?

or are there instances of neuter reactants $B$ such that their reactions $A+B \rightarrow C+B$ have a similar rates to the straight reaction $A \rightarrow C$?

There are reactions similar to the one you show, which is straightforward catalysis, these are autocatalytic reactions such as $\ce{A + B\to 2B}$ or $\ce{A + 3B\to 3B}$ which lead to interesting non-linear behaviour. Of more importance is the Michaelis-Menten scheme  $\ce{E + S\leftrightharpoons ES \to P + E}$ where E is an enzyme , S a substrate, ES a complex of both, and P the product.


What is the relationship between emission and fluorescence? A molecule exposed to a photon with some energy is put into an excited state, and emits a photon of some energy when it returns to ground state.
The photon that provided the energy for excitation corresponds to a particular wavelength, at which we see a peak in an absorption spectrum.
The photon that was released when the molecule returned to its ground state also corresponds to a particular wavelength, at which we see a peak in an emission spectrum.
Fluorescence is when a molecule emits a photon of lower energy when it returns to ground state than what it absorbed initially to be put in an excited state. That photon corresponds to a particular wavelength, at which we see a peak in the fluorescence spectrum.
But isn’t the emission of a lower energy photon the case for all molecules, since all molecules lose some energy in form of vibration and rotation before they reach ground state? How is fluorescence distinguished from emission?

You can google all of this for more detail...

Basically an "emission spectrum" is the most general term for detecting photons as a function of energy or wavelength. All the term implies is that your measuring photons.

A "fluorescence spectrum" is a little more specific. It means that a photon was used to excite the atom/molecule and that the atom/molecule decayed very rapidly with the emission of a photon of longer wavelength (lower energy).


Why is the more electronegative atom lower in energy in an MO diagram? Let's take $\ce{NO}$ for example. Why is $\ce{O}$ lower in energy? I asked my teacher and he said that because it is more electronegative and this lowers the bonding energy of the MO. But why?

I am summarizing from the excellent comments while adding some glue:

The convention is that lower electron energy means electrons that are more tightly bound. An energy of zero would mean a free electron (not bound in the electrostatic field of a nucleus).

As you go across from left to right in a period of the periodic table, the number of protons in the nucleus increases and the number of electrons bound (in an atom) increases likewise. In the simplest explanation of electron energies, you conceptually separate the electrons into inner electrons and valence (or outer shell) electrons, and treat the inner electrons and the nucleus together as an entity with some charge (the effective nuclear charge) that the outer electrons "see".

Comparing oxygen to nitrogen, the effective nuclear charge increases by one (same number of inner electrons, one more proton in the nucleus), so valence electrons of oxygen are more tightly bound than valence electrons of nitrogen. Thus, they appear lower in an MO diagram.

Not the OP's question, but for completeness: If you go from one period to the next (compare F with Na, say), the effective nuclear charge decreases (because we now also count the n = 2 electrons as inner electrons), and valence electrons of Na are less tightly bound than those of F. Most atomic properties change in a periodic pattern when you plot them against atomic number for the same reason (e.g. ionization energies, atomic radii)


Grignard Reagent in THF vs in Diethyl ether While learning about the Grignard reagent I was taught this by my teacher, but I couldn’t justify why it was so..
I cannot see how Grignard reagent would react differently in basically an ethereal solvent..
So i needed a justification as in why this was happening.
Thanks

THF has a number of differences with diethyl ether:

Oxygen non-bonding doublets are more available for bonding with magnesium because the 2 ethyl groups are not freely rotating around (you may think of THF as diethyl ether with its arms tied in its back).
THF boils at 66°C (ether 34.6°C). Because the reaction is usually carried out so to maintain a gentle reflux, the higher boiling temperature of THF will speed up the reaction, and solubilize reactants/products better.

As @Alchimista pointed out, THF is just one of the possible solvents you can use if ether fails or gives poor yields. You can also use higher boiling ethers like dimethoxyethane.

Note: this is a collective answer asked by other users. I have been harassed. Call the police :-)


How are odor and electronic structure related to aromaticity? I know that cyclic planer molecules with $2n + 1$ $\pi$-electrons are aromatic. Also, the color of a compound is due to electronic transitions. But how is it related with aromaticity (smell)? Is there any such relation or is it the chemical property of any compound (matter)?

If it is chemical property of matter then what is its cause?

P.S. My question isn't about nomenclature, its about what is in benzene or any other aromatic compound that make it aromatic i.e. their smell? And what specifically is the roll of electrons in their smell?

I can address the part of your query pertaining to odor and nomenclature (more accurately, I can link the UC Davis ChemWiki page that does so and quote it here):

Benzene derived products are well known to be pleasantly fragrant. For this reason, organic compounds containing benzene rings were classified as being "aromatic" (sweet smelling) amongst scientists in the early 19th century when a relation was established between benzene derived compounds and sweet/spicy fragrances. There is a misconception amongst the scientific community, however, that all aromatics are sweet smelling and that all sweet smelling compounds would have a benzene ring in its structure. This is false, since non-aromatic compounds, such as camphor, extracted from the camphor laurel tree, release a strong, minty aroma, yet it lacks the benzene ring in its structure (See figure 1). On the other hand, benzene itself gives off a rather strong and unpleasant smell that would otherwise invalidate the definition of an aromatic (sweet-smelling) compound. Despite this inconsistency, however, the term aromatic continues to be used today in order to designate molecules with benzene-like rings in their structures. For a modern, chemical definition of aromaticity, refer to sections Aromaticity and Hückel's Rule.

You will find (in the linked section on Aromaticity) that presently:

The three general requirements for a compound to be aromatic are:

The compound must be cyclic

Each element within the ring must have a p-orbital that is perpendicular to the ring, hence the molecule is planar.

The compound must follow Hückel's Rule (the ring has to contain 4n+2 p-orbital electrons).

Putting it all together: The historical naming convention for molecules that smelled good is responsible for the term aromatic. It turns out that many of the molecules that fall into the aromatic category happen to be rather stable/unreactive. These molecules happen to have certain physical characteristics which, taken together, are indicative of a molecule being classified as aromatic or not (based on present-day definitions not relating to odor/smell).


Why can copper have multiple coordination numbers? Is there a way to explain why the coordination number of for example $\ce{Cu^2+}$ can be either 4 or 6?

the coordination number of for example $\ce{Cu^2+}$ can be either 4 or 6.

Allow me to preface this answer by debunking that myth.

Balamurugan and Palaniandavar published a set of interesting sterically strained trigonal planar copper(I) complexes, among them 1. Under cyclic voltammetric conditions, they observe two distinctly different oxidation potentials, one of which they attribute to the oxidation to the corresponding trigonal copper(II) complex. Unfortunately, they do not report isolation.[1]

Tetracoordinated copper(II) species are not as common as hexacoordinated ones, but they are also rather well-known. For example, Kitajima et al. synthesised the structurally interesting complex 2, a formal copper(II) and boron(III) binuclear complex. Naturally and as expected for tetracoordinated copper(II) complexes, its structure is tetragonal around both the copper and the boron centres.[2]

Slightly more interesting is the second complex Kitajima et al. were able to crystallise: it is the formal addition of a DMSO solvent molecule to complex 2 to give complex 3. Here, copper(II) is pentacoordinated and equally interestingly the coordination sphere represents a square pyramidal structure, not the usual trigonal bipyramid.[2]

I’ll skip hexacoordinated complexes, because they are damn common and just note that everybody who witnessed complex chemistry at any educational institution has come in contact with $\ce{[Cu(H2O)6]^2+}$ and $\ce{[Cu(NH3)4(H2O)2]^2+}$.

Finally, Martínez-Bulit et al. were able to synthesise an interesting heptacoordinated copper(II) complex 4. Unfortunately, the complex did not crystallise like the corresponding cobalt(II) and zinc(II) counterparts so we are left with analogy to the other syntheses, MS and IR data for establishing it’s structure. In figure 2, you can see a depiction of the coordination sphere.[3]

A variant of heptacoordinated, albeit only characterised for copper(I) to the best of my knowledge, is $\ce{[CuCp(TMS-C#C-TMS)]}$, complex 5.[4] Cyclopentadienyl — even though it is a $\unicode[Times]{x3b7}^5$ complex — should be characterised as pentacoordinating because there are five coordinating atoms. Likewise, acetylene is dicoordinating for a total of 7. It should be possible to extend the concept of sandwhich complexes to $\ce{[CuCp2]}$ 6; however, that complex is currently unknown outside of calculations.[5]

Figure 1: Structures of the complexes mentioned above.

Figure 2: Three-dimensional representation of 4’s ligand sphere.

Now why does copper(II) assume these structures? Well, you can argue things like complex stability, Jahn-Teller distortion, steric strain, electronic effects and much more. In the end, it all boils down to: if the MO scheme of a complex provides sufficient stabilisation with respect to the individual components, the complex is viable to be formed. If removing a ligand will substantially destabilise the complex, it is likely to remain inert. Unfortunately, I was unable to find a linear copper(II) complex (or any linear copper complex for that matter); however, I would assume it to exist if a ligand provided a large cavity with one coordination site and only a single anion was present to coordinate the other. Or maybe not even anions but neutral donors; the anion being $\ce{F- . SbF5}$, a very non-coordinating ion. I’m also sure that the cuprocene sandwhich complex will be synthesised given sufficient time. That would extend copper(II)’s coordination count range from 2 to 10 — always given beneficial circumstances for a certian coordination number.

So tl;dr: There is no simple explanation for the choice of exactly that coordination number.

Note that in aquaeous solution the ligand sphere of the seemingly so stable complex $\ce{[Cu(H2O)6]^2+}$ is in constant fluctuation with dissociating and reassociating ligands. Otherwise, there would be no rapid exchange mechanism that would allow the formation of $\ce{[Cu(NH3)4(H2O)2]^2+}$ upon addition of concentrated ammonia. Compare iron(II): while the hexaammin complex $\ce{[Fe(NH3)6]^2+}$ is undoubtedly stable and characterised,[6] you won’t be able to get it by reaction $\ce{[Fe(H2O)6]^2+}$ with ammonia. Instead, $\ce{Fe(OH)3}$ will precipitate, demonstrating that ligand exchange in iron(III) complexes is slow with respect to deprotonation, while in copper(II) complexes it is reasonably fast.

References:

[1]: R. Balamurugan, M. Palaniandavar, Inorg. Chem. 2001, 40, 2246. DOI: 10.1021/ic0003372.

[2]: N. Kitajima, K. Fujisawa, Y. Moro-oka, J. Am. Chem. Soc. 1990, 112, 3210. DOI: 10.1021/ja00164a052.

[3]: P. Martínez-Bulit, A. Garza-Ortíz, E. Mijangos, L. Barrón-Sosa, F. Sánchez-Bartéz, I. Gracia-Mora, A. Flores-Parra, R. Contreras, J. Reedijk, N. Barba-Behrens, J. Inorg. Biochem. 2015, 142, 1. DOI: 10.1016/j.jinorgbio.2014.09.007.

[4]: D. W. Macomber, M. D. Rausch, J. Am. Chem. Soc. 1983, 105, 5325. DOI: 10.1021/ja00354a023.

[5]: D. W. Clack, K. D. Warren, Inorg. Chim. Acta 1978, 30, 251. DOI: 10.1016/S0020-1693(00)89045-3.

[6]: G. L. Schimek, J. W. Kolis, Chem. Mater. 1997, 9, 2776. DOI: 10.1021/cm970095c.


Capillary action only works with glass tubes? I tried to reproduce the capillary action and measure how high the water rises. So far, I've tried unsuccessfully with a regular drinking straw and a very narrow plastic tube with an internal diameter of about 0.5mm. Does the tube have to be glass for it to work?

Few things in chemistry are truly unique. The tube does not have to be glass; it only has to be wettable, which glass typically is and plastic typically isn't.

In the tubes made of non-wettable materials, capillary action works the other way: the liquid will go down rather than up, akin to Hg on the picture provided.

 (source)

So it goes.


Compute minimum energy paths from arbitrary positions on the potential energy surface The minimum energy path is a useful concept in understanding chemical reaction paths. As I understand it, such MEP's are most often computed to verify that a transition state structure connects two minima of interest. Several algorithms exist for computing MEP's, but I suppose all have in common that they are able to follow the gradient.

I am in a situation where I think it would be useful to follow the gradient, but not from a TS structure. Can this be done? I think the IRC functionality of software expects to follow the gradient in two directions, and demand the reaction vector.

Is there a simple and common way of computing MEP's from arbitrary positions on the PES? (I will use Molcas). What would happen if I just do a geometry optimization from the arbitrary position? It is my understanding that geometry optimization algorithms displace the atoms in the direction of the force, and in that way a lower energy is guaranteed for each successive step. But will this actually follow the gradient?

There are some possibilities here:

The negative of the gradient of a function always points to the direction of optimal descent (for a discussion, see here).
A method that simply follows the gradient is called first order.
As permeakra pointed out, extrapolation techniques are commonly used for optimizations.
This is because we take finite steps in the PES and this introduces the risk of "jumping" over the minima.
Better approximations are either second order in nature (e.g., the Hessian matrix is calculated in each step) or quasi-second order (the Hessian matrix is approximated or approximately updated).

In order to simply follow the gradient you want the steepest descent-like, first order method with probably a very small maximum step and this can be accomplished by setting the Hessian or the Hessian approximation equal to the identity matrix.
In ORCA (I have never used Molcas, sorry) for example you may use (source):

! Opt
%geom
  inhess unit
  update noupdate
  step qn
end

permeakra also cited the Nudged Elastic Band, which is used for finding minimum energy paths connecting two structures.
This might be very useful and is available for instance in VASP and in NWChem (free).
You may also want to take a direction from one Hessian eigenvector.
This is commonly associated with a molecular vibrational degree of freedom.
This may be accomplished with NWChem.

If you just want to go down the PES from an arbitrary geometry without any jumping, solution 1. is probably the one you want.


Any suggestions how to solve this? $m(\ce{FeSO4. $x$~H2O})=20.0\:\mathrm{g}$

$N(\ce{H2O})=3.03 \cdot10^{23}$

$x=?$

What I've done:

(1.)

  \begin{aligned}
   m (\ce{H2O})&= \frac{N(\ce{H2O}) \cdot M(\ce{H2O})}{N_{A}}\\
   &= \frac{3.03 \cdot10^{23} \cdot 18.02$\:\mathrm{g/mol}}{6.02 \cdot 10^{23}/\:\mathrm{mol}}\\
   &= 9.06987\:\mathrm{g}
  \end{aligned}

(2.)

  \begin{aligned}
   m(\ce{FeSO4})&= m(\ce{FeSO4} \cdot x \ce{H2O}) - m(\ce{H2O})\\
   &= 20.00\:\mathrm{g} - 9.06987\:\mathrm{g}\\
   &= 10.9301\:\mathrm{g}
  \end{aligned}

Any suggestions what to do next?

Calculate $n(\ce{FeSO4})$ with $M(\ce{FeSO4})\approx 136\:\mathrm{g/mol}$

   $n(\ce{FeSO4})\approx0.08\:\mathrm{mol}$

Calculate $n(\ce{H2O})$

   $n(\ce{H2O})\approx0.5\:\mathrm{mol}$

Set them into relation $n(\ce{FeSO4}):n(\ce{H2O})= 1:x$

   $x\approx 6$


What exactly is a mole? Upon coming back to University I've started studying basic chemistry again. I've always had trouble with the concept of "moles". According to my study book,

  1 mole of any element or compound is equal to its molecular weight in
  grams.

I understand that part but later in the paragraph it also says:

  One mole of any substance always contains exactly the same number of solute particles, that is, $6.02 \times 10^{23}$ (Avogadro's number). So whether you weigh out 1 mole of glucose (180 g) or 1 mole of water (18 g) or 1 mole of methane (16 g), in each case you will have $6.02 \times 10^{23}$ molecules of that substance.

If 1 mole equals the molecular weight of a compound, then how come 1 mole of glucose doesn't equate to 1 molecule of glucose? Basically, why is 1 mole of glucose $6.02\times10^{23}$ molecules instead of 1 molecule?

The first thing to realize is that "mole" is not a mass unit. It is simply a quantity - a number - like dozen or gross or score. Just as a dozen eggs is 12 eggs, a mole of glucose is $6.02 \times 10^{23}$ glucose molecules, and a mole of carbon atoms is $6.02 \times 10^{23}$ carbon atoms. "Moles" are only associated with mass because individual objects have mass, and thus a mole of objects also has a certain mass.

So why $6.02 \times 10^{23}$? What's so special about Avogadro's number? Well, nothing, really, it just makes calculations work out nicely. Avogadro's number is defined as the number of $\ce{^{12}C}$ atoms which weigh 12 g. So it's effectively a ratio: how many times larger is a gram than an atomic mass unit? If you have one atom of $\ce{^{12}C}$, it weighs 12 amu. If you have $6.02 \times 10^{23}$ of them, they weigh 12 g. If you have one molecule of glucose, it weighs 180 amu (or thereabouts). If you have $6.02 \times 10^{23}$ of them, they weigh 180 g (or thereabouts). -- This is just like if one egg weighs 60 g (on average), then a dozen of them weigh 720 g (on average), and if one cup of flour weighs 120 g (on average), then a dozen cups of flour weigh 1440 g (on average). The only difference is that dozen is defined forwards ("a dozen is twelve"), whereas a mole is defined "backwards" ("a dozen is the number of 60 g eggs that are in a collection of eggs that weighs 720 g.")

This convenience definition - pegging the value of Avogadro's number directly to the difference in scale between the amu and the gram - is what is probably throwing you. Moles are not a mass unit, but the definition is intimately tied to mass units. The equivalence in numbers at the atomic scale (amu) and at the macroscopic scale (grams) can also result in chemists playing fast and loose with terminology, quickly working back and forth from atomic to macroscopic scale, without a necessarily clear distinction between the two.


Acidity of aldehydes and ketones and enolate formation In most academic courses you'd learn that aldehydes are more acidic (lower $\text{p}K_\text{a}$) than ketones due to the lower electron donating effect of the proton compared to the alkyl group of the ketone.

However, when talking about condensations, for example, it's not a weird choice to form a ketone enol over an aldehyde enol on the basis of "thermodynamic stability due to a more substituted alkene under thermodynamic control". Yet, $\text{p}K_\text{a}$ is the measurement of such stability, and thermodynamically the aldehyde enol is more stable. Adding to this, a following argument is that "aldehydes react much quicker towards nucleophilic attack than ketones". I agree with this, but weren't we just discussing thermodynamic control?

The questions are then:

Is it valid to use the "more substituted alkene" to argue a step on a mechanism under claims of thermodynamic control?
Can this type of argument be coupled with a kinetic one like the example provided as well? I.e. is it valid to mix thermodynamic and kinetic arguments one in the same? (e.g. this intermediate forms because it's more thermodynamically favored, and the next one forms because it is kinetically favored).

(I realize question 2 may be broader than this, but I'm trying to keep this in the sense of mixed aldol condensations)

I would like to first comment on the reasoning given as to why aldehydes are more acidic than ketones. The usually provided reasoning is, as stated here:

H atoms are regarded as having no electronic effect : they don't withdraw or donate electrons.
Alkyl groups are weakly electron donating, they tend to destabilise anions (you should recall that they stabilise carbocations). This is because they will be "pushing" electrons towards a negative system which is unfavourable electrostatically. Hence, the anion of a ketone, where there are extra alkyl groups is less stable than that of an aldehyde, and so, a ketone is less acidic.

We need to be clear of what we are measuring when we measure the acidity of an aqueous solution of an aldehyde. It is important that we recall what is mentioned in this post, on the acidity of benzaldehyde, as it relates to the acidity of aldehydes in aqueous solution. Most aldehydes exist in aqueous solution with a significant proportion of their hydrated forms, which are geminal diols. Hence, it is very likely that the $\mathrm pK_\mathrm{a}$ values we have obtained are for the $\ce {-OH}$ protons of the hydrate and not for the $\ce {C-H}$ protons at the $\alpha$ position of the aldehyde.
Let's take a look at acetaldehyde. Ethanol has $\mathrm pK_\mathrm{a}$ value of $15.9$ and the hydration equilibrium constant for acetaldehyde is $\ce {1}$, as mentioned here. Based on the guideline that geminal diols have $\mathrm pK_\mathrm{a}$ $2.5$ units less than that of the monohydric alcohol, the predicted $\mathrm pK_\mathrm{a}$ value for acetaldehyde is $13.4$. The experimental measurement gives $13.57$, as seen from here, which is very close to the value determined via calculation with the guidelines. This closeness of the values suggests that the hydrate is indeed formed and the acidity that we are measuring for the aqueous aldehyde is indeed that of the hydrate.
If this is what we are measuring, then clearly the reasoning highlighted in the box may not even be correct. In fact, we can't even be sure that the $\ce {\alpha}$ $\ce {C-H}$ proton of the aldehyde is more acidic than that of a ketone since we won't be able to measure their acidities, at least not in water.
With the clarification I have made, I hope that there is no longer any inconsistency in the issues you have mentioned. In particular, it is correct to say that under thermodynamic control, the intermediate with greater stability (e.g. the enolate with the more substituted alkene) is formed. Regarding your 2nd question, I am unsure of what you mean but I will just say that kinetic and thermodynamic control refer to different reaction conditions resulting in the reaction undergoing different pathways to thus form different products.
As an extension, we can also consider if the $\mathrm pK_\mathrm{a}$ values given for the ketones are representative of the acidity of the $\alpha$ protons or that of the $\ce {-OH}$ protons of the hydrate. Let us consider acetone, with $\mathrm pK_\mathrm{a}$ given as $\ce {19.16}$ here. The hydration constant for acetone is $1.4 \times 10^{-3}$ and the $\mathrm pK_\mathrm{a}$ of 2-propanol is $16.5$ (as seen here). Based on the same sequence of calculations and the same guideline, the predicted $\mathrm pK_\mathrm{a}$ value is $16.9$. The significant difference between the predicted value and the experimentally measured value suggests that the acidity of the acetone is not just due to that of the hydrate $\ce {-OH}$ protons and that there may be significant deprotonation of the $\alpha$ protons taking place.
An important question that we would like to know the answer to is what then is the $\mathrm pK_\mathrm{a}$ value of an $\alpha$ proton of an aldehyde. Well... It seems like due to hydration of the aldehyde, we would not be able to directly measure the acidity of this proton. However, one would reasonably estimate that it is close to that of the ketone $\alpha$ hydrogens. Again, that is something we may not be able to measure directly for the same reason. An educated guess would be in the range of $18-22$ as it needs to be more lower than the $\mathrm pK_\mathrm{a}$ of esters but higher than that of alcohols.


Why aren't the Anton computers used more widely? The Anton computers are purpose-built by DE Shaw research for molecular dynamics simulations. They achieve performance up to an order of magnitude faster than can be achieved with commodity hardware, but don't seem to be used all that much by the wider computational chemistry community. This is despite the fact that limitations on simulation timescales is a major impediment to using simulations to guide experiments, leading to all kinds of intricate theoretical methods to deduce longer-timescale properties from relatively short simulation trajectories.

Having read a bit about the Anton machines, which have been around for over a decade now, I'm perplexed as to why they haven't been more widely used. Is it because they are too inflexible, not available to enough researchers, not helpful in solving scientific problems, or is there some other reason I'm missing?

Protein dynamics occur in multiple different time scales. I myself is a person using MD simulation for my research. One reason I am not so much interested in ANTON is not it's cost, limited resource etc. It's rather because one extremely long simulation wouldn't help much at least for my research.
Protein's thermodynamics as well as kinetics all depend on free energy landscape which is a priori unknown. To directly compute free energy from simulation, you generally need multiple order of magnitude faster machine even than ANTON. However, there has been many techniques such as Umbrella Sampling that helps resolve this issue on computers in our lab.
I would appreciate it if I can run my Umbrella Sampling in ANTON but it would be so much waste of time to wait on the line if I can run the same simulation for a week in my workstation.
Another reason I can guess why ANTON has not been widely used is because of its specificity. ANTON does not run typical MD simulation packages such as GROMACS, NAMD, AMBER etc. All packages use same algorithm to run the simulation but the commands, file structures etc, are all different. Typical MD user would prefer to use package that he/she is familiar with. I am not sure how much I would have to study to get used to ANTON MD package but why bother if I am not desperate?
So in summary, only those users who are interested in actual trajectories of protein and water in millisecond time scale would want to use ANTON but not many others. Other than such purpose, there are many techniques and algorithms that would work better than running a single extremely long simulation.


Where does computational chemistry overlap with organic chemistry? I was wondering what research exists that combines computational chemistry and organic chemistry.

For the past three years, I've been using Amber and Gaussian (in hopes) to identify and understand trends in intramolecular interactions. This really just serves as an example of my past research which incorporated both physical chemistry and computational chemistry.

I would like to know what kind of research could be done that combines both organic chemistry with computational chemistry.

I've Googled computational research in organic chemistry, but haven't found much information. What are some examples of projects that have been done in organic chemistry that rely heavily on the computational side of things?

The first thing that comes to mind: read the backlist of Henry Rzepa's and (especially) Steven Bachrach's blogs.  Depending exactly on what resources you have available, there are lots of things they do on the computational side of organic chemistry that you could at first imitate and then use as a springboard to innovate.

In particular, I've found Bachrach's posts on reaction dynamics to be especially fascinating.  This is the phenomenon where the topography of the potential energy surface near a transition state is such that the reactions don't always follow the minimum-energy path due, effectively, to the inertia of the nuclei as they cross the TS.

I also have a personal interest in Bader's Quantum Theory of Atoms in Molecules, which uses the topology of the electron density to divide molecules up into atoms, where each atomic volume is enclosed by a 'zero-flux surface' in the electron density.  (There's some ongoing controversy over the theory, but it makes good sense to me, for whatever that's worth.)  Bader's book is a good place to start; or, you can search the journal literature for Richard Bader and a whole bunch of examples will surely turn up. I engaged with Dr. Rzepa last year in the comment thread of a post of his about electrides that touched on QTAIM, the electron localization function (ELF), non-covalent interactions (NCI), etc.

Finally, I've been very curious lately about 'charge-shift bonding', a relatively newly described class of bond where the nuclear framework tries to cram so many electrons into a small space that in the end Pauli repulsion drives them apart, causing what might otherwise appear to be a covalent bond to instead take the form of a resonance between two ionic structures.  (The electronic structure of $\ce{F_2},$ for example, is better described as a balanced resonance between $\ce{F- \!- F+}$ and $\ce{F+ \!- F-}$ overlaid atop a very weak covalent interaction, than as a purely covalent $\ce{F\!-F}$ species.) Search the literature for Sason Shaik's work for more information and examples.

Other things that I've come across that have seemed interesting include carbenes, cyclic alkynes, unusual Diels-Alder reactions, and fullerene systems.

I would also subscribe to, e.g., Computational Chemistry Highlights, CCL, and Jan Jensen's Computational Chemistry Daily -- the posts and/or questions that come across there might give you some inspiration.


Why is zinc blende not directly reduced in smelting furnaces? My question can also be generalized as, why are some sulfide ores (like $\ce{ZnS}$) roasted first to generate oxides, and only then reduced in a blast furnace or similar furnace?

The standard free energy of formation of most of the sulphides are larger than those of $\ce{CS2}$ and $\ce{H2S}$. Therefore, these sulphides are most stable than those with carbon or hydrogen. So, neither carbon not hydrogen can reduce metal sulphides to metals.

But standard free energies of formation of oxides are much lower than those of $\ce{SO2}$.  Therefore, oxidation of metal sulphides to metal oxides is thermodynamically favorable. Also, the oxides are easily reducible.


what is the basic factor which determines the order of ionization energies of K,Li,Ca,S? There was a question which was asking about the order of ionization energies of
K,Li,Ca,S.
The electronic configuration of all the elements were as follows

Li:${1s^2 2s^1}$
K :${1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$
Ca :${1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$
S :${1s^2 2s^2 2p^6 3s^2 3p^4}$

From the electronic configurations:

Potassium should have lowest ionization energy as it is very easy to remove electron from 4th shell.
Then it should be followed by Calcium as its nuclear charge increases from going across the period
Then it should be followed by sulphur as it is more easier to remove the electron from 3rd shell than from 2nd shell.
Then it should be followed by Lithium as it has lowest principle quantum number(n) so it is more difficult to remove an electron from 2nd shell than from 3rd and 4th shell.

So according to my calculation the order should be  K < Ca < S < Li.
But the correct answer in my textbook is K < Li < Ca < S. If I am wrong anywhere then what factor would help in determining the Ionization energies of the following elements, and what would be the exact order of I.E for these elements.

Looked up ionization energies (kJ/mol) from Wikipedia

K  = 418.8
Li = 520.2
Ca = 589.8
S = 999.6

so the book answer is correct...

Ok, here is best hand waving explanation that I can give. The general trend...

Li and K are both alkali metals. As the atomic number goes up the shells get further away from the nucleus and the inner electrons shield the valence electron. So in order of increasing ionization energy

Li > Na > K > Rb > Cs > Fr

So to start we easily get Li > K.

Ca and K are in the same period in the periodic table. Generally as you go up in atomic number in a period the ionization energy will increase. So Ca > K.

Sulfur is a non-metal in the upper right hand of the periodic table so it should have the highest ionization energy.

So we're down to either S > Ca > Li > K or S > Li > Ca > K.

  To me giving either of these answers wouldn't be embarrassing.

If you go to the Bohr model of the hydrogen atom then energy of the $n$th shell is

$\mathrm{E}_n  \propto \dfrac{Z^2}{n^2} $

If we assume that the 2s and 3s electrons are perfectly shielded by all the electrons in the lower shells, then

$ \mathrm{E_{Li}} \propto \dfrac{1}{2^2} = 0.25$

The two Ca 3s electrons being equal the charge is 2

$ \mathrm{E_{Ca}} \propto \dfrac{2^2}{4^2} = 0.25$

So I would have guessed S > Li > Ca > K

Now I know that is wrong. I know that the "problem" is with the shielding of the electrons. Below is a section of the periodic table with ionization energies listed.

I just don't know how to explain the "right" answer without hand waving about shielding.

I don't know how you'd rationalize all the relationships between the group 1 and 2 elements.

If you assume the Bohr model as noted above then you end up with this which correctly predicts the trend down a group but overestimates shielding of the inner electrons.


Boiling point elevation for a mixture of ethanol and water Properties such as boiling point, freezing point and vapour pressure
of a pure solvent change when solute molecules are added to get
homogeneous solution. These are called colligative properties.
Applications of colligative properties are very useful in day-today
life. One of its examples is the use of ethylene glycol and water
mixture as anti-freezing liquid in the radiator of automobiles. A
solution M is prepared by mixing ethanol and water. The mole fraction
of ethanol in the mixture is 0.1.
Given:
Boiling point elevation constant of water =0.52 K kg/mol Boiling point
elevation constant of ethanol = 1.2 k kg/mol
Boiling point of water =373 K
Boiling point of ethanol =351.5 K
The boiling point of this solution is?

Now,
I am aware that for a solution of a non-volatile solute in a solvent
The elevation in boiling point is given as
$\Delta\mathrm{T}=im\mathrm{K}_{b}.$
In this case,which is the solute and which the solvent?
The material that I am studying from considers ethanol to be the solute and water,the solvent.And thus,calculates the elevation in boiling point of water.
Therefore,the new boiling point is higher than that of pure water
My problem with this is,since the boiling point of a mixture of ethanol and water,for all compositions,lies between that of water(maximum) and ethanol(minimum), how can a mixture of water and ethanol of 0.1 molarity (by ethanol) have a higher boiling point than that of water?

Are you sure that is the right equation to answer the question? It only holds true for non-volatile compounds. You should be able to use Raoult's Law. Perhaps this helps http://www.chemguide.co.uk/physical/phaseeqia/idealpd.html


Why can't bicyclic structures undergo E2 elimination? Bicyclic structures like norbornane don't undergo E2 eliminations, but why is this the case? I understand that it can't do E1 as it can't form a stable carbocation (the bridgehead carbons that is) but I don't see the issue for E2. Can you explain?

Elimination via a E2 mechanism cannot take place at the bridgehead carbon of a bicyclic compound due to the lack of the required trans geometry between the leaving group at the bridgehead carbon and any of the six β-hydrogens.

On the other hand, as you have correctly observed, a E1 mechanism has to be ruled out as well due to the strain at the bridgehead carbocation. In fact bicyclic structures prevent the tertiary bridgehead carbon becoming planar, which implies that is very high in energy because the nonplanar structure forces the cation to be an empty "sp3 orbital" instead of an empty "p orbital". This observation was made for the first time by Bredt, therefore is generally known as "Bredt's rule".


Addition of a noble gas - Le Chatlier So, regarding a generic reaction (for the sake of argument, let us assume that it is completely gas phase), we have Le Chatelier's Principle. Let us use this example:

$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \quad\quad \Delta H < 0$$

Now, we have the questions: oxygen gas is added, heat is added, a noble gas is added, etc.

For the addition of oxygen gas, we obviously shift right, and for heat, we shift left.

For a noble gas, I would expect no shift to happen. Indeed, this is what the SAT's seem to say, and I completely agree. However, my teacher seems to have the conception that when one adds a noble gas to a reaction such as above, you decreases molecular space, and the reaction does shift/speed up.

I have a test tomorrow, and if she gives me a question like this, I am planning to write that there is no shift or rate change. Now, if she marks this wrong, I need a formal argument to support my claim.

Am I correct in saying that with the addition of a noble gas, there is no change in either equilibrium shift or rate change? Why?

See, in an equilibrium reaction in gas phase at constant temperature, noble gases can be added in two ways―

At constant volume in the system
At constant pressure in the system.

Effect of adding noble gases at constant volume

At constant temperature and volume, if noble gas is added, total number of molecules,i.e., amount of the element increases (or you can say, 'number of moles' increases, though it's not correct, have you ever heard of number of kilograms?) in the system. As a result, also total pressure of the system increases, but there is no change in partial pressure of the reactants and products (because total pressure increases and mole fraction decreases). So, there is no effect on equilibrium by adding noble gases in a system at constant temperature and volume.

Effect of adding noble gases at constant pressure

At constant temperature and pressure, addition of noble gas to the system leads to increase in volume (cause, amount of the element increases). As a result, partial pressure of the reactants and products decreases (because total pressure constant but mole fraction decreases) as well as the sum of partial pressure of the reactants and products also decreases. Here, according to Le Chatelier's Principle, the equilibrium would be shifted in that direction where number of molecules or volume increases.


How to determine toxicity of a substance? I often read description of various compounds or elements as being poisonous. Now I know that through experience we have discovered the toxicity of various substances like lead, chlorine or phosgene etc. We know nitrogen isn't poisonous as its present in the atmosphere etc.

What I wanted to ask is whether there is a perfectly analytical or experimental way through which chemists determine whether a newly discovered substance is or isn't poisonous? Or is it always found out only by exposure to humans (which I don't think is always possible)? Or perhaps whether such an information can't always be deduced?

I think animal testing is a possible (but immoral) method but again some substances could be harmful to them but harmless to us.

No, there is no way to determine toxicity to humans with a "magical method".

Most of our knowledge about toxicity is extrapolated from animal testings and from reported accidental exposures in human.

Also, Nazis carried out a large number of human testing on Jews, and this is not limited to toxicity.


Is iron in the brown ring compound in a +1 oxidation state? In the standard brown ring test for the nitrate ion, the brown ring complex is:
$$\ce{[Fe(H2O)5(NO)]^{2+}}$$

In this compound, the nitrosyl ligand is positively charged, and iron is in a $+1$ oxidation state.

Now, iron has stable oxidation states +2 and +3. Nitrosyl, as a ligand, comes in many flavours, of which a negatively charged nitrosyl is one.

I see no reason why the iron doesn't spontaneously oxidise to +3 and reduce the $\ce{NO}$ to −1 to gain stability. But I don't know how to analyse this situation anyway. I think that there may be some nifty backbonding increasing the stability, but I'm not sure.

So, why is iron in +1 here when we can have a seemingly stable situation with iron in +3?

According Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited

The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$

For many years it was thought that iron was reduced to $\ce{Fe^{I}}$ and $\ce{NO}$ oxidized to $\ce{NO+}$, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin $\ce{Fe^{III}}$ ($S=5/2$) antiferromagnetically couples with $\ce{NO-}$ ($S=1$) for an observed spin of $S=3/2$.


Are there any major exceptions when comparing electron affinity? I was tasked with figuring out whether carbon or nitrogen has a more negative electron affinity value. I initially picked nitrogen, just because nitrogen has a higher $Z_\mathrm{eff}$, creating a larger attraction between electrons and protons, decreasing the radius, causing a higher ionization energy, and therefore decreasing the electron affinity value, but I was actually wrong, and the solutions manual explains it as this:

  "As you go from C to N across the Periodic Table you would normally expect N to have the more negative electron affinity. However, N has a half-filled p subshell, which lends it extra stability; therefore, it is harder to add an electron."

Are there any major exceptions to the rules when comparing electron affinity? I'm hesitant to use nitrogen as an exception, because I don't know how far it extends. If nitrogen has a more positive EA than carbon, does that also extend to boron, aluminium, or phosphorus?

I later found that this also applies when comparing silicon and phosphorus. The explanation given was the same.

What exceptions should be noted when comparing electron affinities? Are there any at all? And how far does the exception with atoms with half-filled p subshells extend?

This exception rule is actually orbital filling rule. For two electrons to be in same orbital they need to have different spins (Pauli exclusion principal). This electron pairing requires additional energy and thus it is easier to add electrons if there are free orbitals. When element has a half-filled p sublevel all 3 orbitals have one electron and pairing takes place (difference between energy levels of 2p and 3s is greater than electron pairing energy).

Electron pairing effects have significant impact to physical properties of coordination complexes (like color and magnetic properties).


Why isn't Indigo soluble in water? Why isn't Indigo soluble in water? I mean, it has 2 H-N groups, shouldn't it then be able to hydrogen bond with water?

True, Indigo can form hydrogen bonds, but with what? Is it water? Nope.

It makes hydrogen bonds with itself.

This makes the molecule incapable of bonding with water. Not only that, but the molecule itself is quite symmetric, with oppositely oriented polar bonds, that cancel out each others' dipole moments. Notice the molecule's similarity to an ace of diamonds:

This type of symmetry is called center of symmetry as it is about it's center we can observe the symmetry.

So we got it, it can't associate itself with water, and it's nonpolar. What do we expect? It doesn't dissolve in water.


Calculation of dispersion energy in Gaussian 09 How can I calculate the dispersion energy using Gaussian 09? Where is it printed in the output file when using wb97xd/cc-pvdz sp? If it isn't printed by default, what flags can be used to turn printing on?

The value of the empirical dispersion correction is printed when the print level is turned up using #p in the route line. For any method that contains an empirical dispersion correction, search for R6Disp or Dispersion energy, where other parts of the line will vary depending on the type of correction (-D2, -D3, or -D3(BJ)):

R6Disp:  Grimme-D2 Dispersion energy=       -0.0174233223 Hartrees.

This line is not printed with # or #n at the start of the route section.

For wavefunction methods that include dispersion via electron correlation, the dispersion contribution is non-separable. For a method like B2PLYP-D3, the printed "Dispersion energy" is the empirical correction, not the MP2 contribution!


why lithium is less reactive than sodium? Lithium lies above sodium in a group and is also smaller in size. According to periodic trend reactivity decreases from left to right in period and down the group.

Reactivity decreases down the group for halogens and increases down the group for alkali metals.

As you have said, Lithium is smaller than Sodium, the valence electron on sodium is further away from the positive nucleus and therefore it has a weaker electrostatic attraction and is more easily lost in a chemical reaction, making it more reactive. This is due to the shielding effect of the second energy level (electron shell) which makes the valence electron on Sodium further away from the nucleus and hence the larger atomic radii.


What is the difference between these two formulas for pH of weak acids? In the German Wikipedia article on $\mathrm{pH}$, I found the following formula for calculating the $\mathrm{pH}$ of weak acids (which are there defined of having $4.5 < \mathrm{p}K_\mathrm{a} < 9.5:$
$$c(\ce{H3O+}) = c^\circ\cdot\sqrt{K_\mathrm{a}\cdot c_0/c^\circ}\label{eqn:1}\tag{1}$$
I am a bit confused about this as in school, I learned that the $\mathrm{pH}$ (or more precisely, the concentration of $\ce{H3O+};$ one would have to take $-\log c(\ce{H3O+})$ to get $\mathrm{pH})$ for weak acids is
$$c(\ce{H3O+}) = \sqrt{K_\mathrm{a}\cdot c_0}\label{eqn:2}\tag{2}$$
I see that these formula are somewhat similar, but they aren't identical. The Wikipedia formula \eqref{eqn:1} includes $c^\circ,$ of which I could not find out what it is (it also appears in the equations for strong and very weak acids).
So I have two questions:

What is $c^\circ$?
Are these formulae different and if yes, how?

The purpose of this $c^\circ$ is to ensure dimensional consistency. Keep in mind that Wikipedia can be edited by anyone, although the content is often of very good quality, sometimes the volunteer writer misses some points and assumes that the reader might be aware of his/her symbols. In your link, the writer does not explicitly define $c^\circ$.
The German version is ensuring that the equilibrium constant is dimensionless and all the quantities which are mathematically operated upon are dimensionless numbers. Therefore $c^\circ$ must be 1 mol/L, if $c_0$ has molar units.
$$c(\ce{H3O+}) = c^\circ\cdot\sqrt{K_a\cdot c_0/c^\circ} \tag{1}$$
Note the quantity under the square has been made dimensionless. In order to attach a unit to the square root term, $c^\circ$ has been multiplied.
If you want to dig deeper, there is something called quantity calculus, which I quote again from my previous answer

Are the units of mole of oxygen molecules the same with the units of mole of nitrogen molecules?
Calculus here is not the integration / differentiation, but rather the
Latin calculus implying a method of calculation.
There is a very nice article "Quantity Calculus: Unambiguous
Designation of Units in Graphs and Tables" by Mary Anne White in the
Journal of Chemical Education. Please read this if you are seriously
interested. Search on Google Scholar and it is free to download from
there.
In quantity calculus Each physical quantity as the product of a
numerical value and a unit:
physical quantity = numerical value × unit
This approach was introduced by British scientists and many leading
physicists used it. Now there is there is nothing less or nothing
more. Therefore your ambiguity arises from introducing another factor
such as "oxygen" or "nitrogen". The unit mol does not know whether it
belongs to oxygen or nitrogen.
As explained in the comments, suppose we write L symbolizing the
height of a tree, then I can only write, L = 10 m. For mathematical
purposes, I will not introduce "tree" anywhere in this equation. The
tree is already incorporated in L (in your mind) but not in the
mathematical equation. One can also write L/m =10. Now you have a pure
number on both sides.


Working out pH of solution when excess NaOH is added In a pH titration, $30\text{cm}^3$ of $\ce{NaOH}$ is added to $\pu{20 cm3}$ $\ce{CH3COOH}$. The concentrations of both solutions was $\pu{0.5 mol dm-3}$. $K_\mathrm{a} = \pu{1.7 x 10^{-5}}$ for $\ce{CH3COOH}$. Calculate the pH of the solution after all the $\ce{NaOH}$ is added.

Clearly, $\pu{10 cm3}~\ce{NaOH}$ is in excess. So moles $\ce{NaOH}$ in excess $= \pu{0.005 mol}$. Then in the mark scheme, it says $$\left[\frac{\ce{NaOH}}{\ce{OH^-}}\right]=0.005 \times \frac{1000}{50} = \pu{0.1 mol dm-3}$$ and I don't understand where this part has come from.

The excess NaOH is, indeed 0.005 moles. The total final volume is 30 ml + 20 ml = 50 ml = 0.05 L. Thus, the final concentration is 0.005 moles/0.05 L = 0.1 M