Chemical software for solving reactions I'm looking for a software which can solve chemical equations. For example something like this (given with result):

\begin{align}
    \ce{CH4 + 2O2 & → CO2 + 2H2O} \\
    \ce{C3H6 + HCl & → C3H7Cl}
\end{align}

Is there are such software? I'm studying chemistry as a hobby and want to check if I'm doing the reactions correct.

No, there is no such thing as chemical reaction are not that easily predictable. You can get result right in some cases, but it is impossible to count all factors.


Why would red fuming nitric acid become white fuming nitric acid if left out at low temperature? I'm reading this answer which refers to both red fuming nitric acid (RFNA) and white fuming nitric acid (WFNA).
Those two Wikipedia articles say:

Red fuming nitric acid (RFNA) is a storable oxidizer used as a rocket propellant. It consists of 84% nitric acid (HNO3), 13% dinitrogen tetroxide and 1–2% water. The color of red fuming nitric acid is due to the dinitrogen tetroxide, which breaks down partially to form nitrogen dioxide. The nitrogen dioxide dissolves until the liquid is saturated, and evaporates off into fumes with a suffocating odor. RFNA increases the flammability of combustible materials and is highly exothermic when reacting with water.
White fuming nitric acid (WFNA) is a storable liquid oxidizer used with kerosene and hydrazine rocket fuel. It consists of nearly pure nitric acid (HNO3). WFNA is commonly specified as containing no more than 2% water and less than 0.5% dissolved nitrogen dioxide or dinitrogen tetroxide.
WFNA as an oxidizer has somewhat less performance than red fuming nitric acid (RFNA) but is considerably safer (though extremely corrosive), as it has little to no dissolved nitrogen tetroxide, which is an extremely toxic and volatile chemical. If not inhibited, it will form nitrogen tetroxide on contact with most metals and some organic materials. WFNA can be converted from RFNA by simply leaving the RFNA out in low temperature for a couple of hours. (emphasis added)

I'm asking about that last sentence; why would red fuming nitric acid convert to white fuming nitric acid by leaving it out at low temperature?

You can convert RFNA to WFNA by removing dissolved nitrogen oxides ($\ce{N2O4}$ and $\ce{NO2}$) from the RFNA.

Given the volatility of $\ce{N2O4}$ and $\ce{NO2}$ it might be possible to just let the gases evaporate from the solution at low temperature. However, since the boiling point of $\ce{N2O4}$ if about $21.69~\pu{°C}$, I do not think that this is a very practical method.

Another path that is theoretically possible is the reaction of residual water with excess $\ce{N2O4}$ and $\ce{O2}$ from air to $\ce{HNO3}$.

$$\ce{2 H2O + 2 N2O4 + O2 -> 4 HNO3}$$

Again, to me this seems not to be a very plausible way. Since the Wikipedia article does not give a source for its claim, we cannot decide which of the two paths, if at all, is the correct one.


What's the solid product of the reaction of liquid ammonia with excess sulfur? What would happen when you mix liquid $\ce{NH3}$ and excess of sulphur?

The answer key states that a solid with $\ce{N}$ and $\ce{S}$, and a gas which gives a black precipitate with lead acetate is evolved.

Surely, $\ce{H2S}$ gas will evolve. What about the solid? Is it ammonium sulphide or ammonium sulphite?

I don't think this question has a simple answer.

This paper's first page (all I can see without a subscription) indicates that sulfur can dissolve in liquid ammonia, producing hydrogen sulfide (as the question correctly states) and... complicated stuff: S7N-, which disproportionates into S4N- and "an" ammonium polysulfide (maybe (NH4)2S6, maybe another composition).

The paper's first page also refers to the system's light sensitivity, so if you don't keep it in the dark, you may get different products.


In a nucleophilic substitution reaction, does the other atom to which the nucleophile is bonded matter? For example,

1-bromopropane + $\ce{NaSH ->}$ Propanethiol + $\ce{NaBr}$
vs.
1-bromopropane + $\ce{H2S ->}$ Propanethiol + $\ce{HBr}$

Which is more valid, and why? Or are they equally valid?

When undergoing an SN rxn, does the other atom to which the
  nucleophile is ionically bonded matter?

It can matter.  If the counterion shifts the covalent-ionic nature of the bond connecting it to the nucleophile, then the nucleophilicity of the nucleophile can change (increase or decrease).  However in your example, the counterions don't make a significant difference.

  Which is more valid, and why?

On paper, they are both valid but

$\ce{H2S}$ is very toxic and corrosive, so the first method using sodium hydrosulfide would likely be preferred.  However, in industry, where raw material cost is very important, both of these methods are successfully used.

However both methods suffer the same drawback - once some of the desired thiol is prepared, it can react further with another molecule of alkyl bromide to produce a thioether; usually a mixture of thiol and thioether is obtained.

In the lab, neither method is used

In the chem lab, the preferred method to prepare simple thiols is through the use of thiourea.  Note the resonance structures for thiourea (just disregard the lower half of the diagram) that place negative charge on the sulfur, making it a good nucleophile.

(image source)

In the first step the thiourea reacts with the alkyl halide via an $\mathrm{S_{N}2}$ reaction to yield an isothiouronium salt, which is subsequently hydrolyzed by sodium hydroxide to generate the thiol. This synthesis can be carried as a one pot, two step reaction.

(image source)


What does this property of transition metals mean? what does it mean:

"Transition metals either have incompletely filled d subshells or readily give rise to cations that have incompletely filled d subshells.

I don't understand the portion starting from " or readily give rise to..."

Take the electronic configuration of copper (atomic number $29$) $\ce{[Ar] 3d^10 4s^1}$. According to the first portion of the definition, copper doesn't have incompletely filled subshells. But, copper readily gives rise to cation $\ce{Cu^{2+}}$ that has the electronic configuration $\ce{[Ar] 3d^9 4s^0}$ with incompletely filled d subshells. So, copper is considered as a transition metal.

Now, you can guess why zinc (atomic number $30$) is not considered as a transition metal.


Why is calcium fluoride insoluble in water? Why is calcium fluoride insoluble in water? $\ce{NaCl}$ is soluble, but $\ce{CaF2}$ is not. Why is this?

This answer to a question about the solubility of silver nitrate and silver chloride will give you a good starting point to answer your question.

In order to dissolve a salt, you have to break apart the ions and hydrate them in solution. You can use the enthalpies of hydration of the ions, and the crystal lattice energy of the solid, to predict which compounds will dissolve. There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, in this reference. The values for enthalpies of hydration and crystal lattice energies of sodium chloride and calcium fluoride can be found in standard chemistry texts or with an on-line search.

If you have more questions after looking at this material, come back and ask for clarification.


Will a Grignard reagent react with an epoxide or episulfide first? In this question, I am confused whether the Grignard reagent will attack the epoxide or the episulfide (thiirane) first. The solution says ring opening of the episulfide will occur due to greater strain.

There are quite a few reasons why the episulfide might open before the epoxide:

Sulphur is less electronegative than oxygen. Therefore, by Bent's rule , more atomic s-character will be directed by the ring towards sulphur in the episulfide as compared to the oxygen in the epoxide. An increased s-character will lead to a bigger increase in the $\ce{C-S-C}$ bond angle as compared to the $\ce{C-O-C}$ angle, and thereby increasing the angle strain in the episulfide relative to the epoxide
Sulphur will also tend to be a better leaving group, as basicity decreases down the group(in this case, group 16)
The carbon-sulphur linkage is weaker than the carbon-oxygen one

So, the mechanism should look like this:

Some points to note:

Sulphur compounds also tend to be pretty good nucleophiles due to their softness as per HSAB as elaborated here. Therefore, the $\ce{RS-}$ formed after ring opening will again attack on the epoxide to open the latter.
This will also lead to the formation of an  ethereal linkage in the final product, which will be quite favorable due to the high stability of $\ce{C-O}$ ether bonds


How to calculate the concentration based on dilution of a stock solution? One of my exam questions is about lipstick and calculating the amount of lithium in the lipstick by AAS. The determination took place by making standard solutions with known concentrations of lithium. For the stock solution $653.3~\mathrm{mg}$ lithium chloride, $\ce{LiCl}$, was dissolved in $100~\mathrm{ml}$.

$M(\ce{Li}) = 6.49~\mathrm{g/mol}$
$M(\ce{LiCl}) = 42.4~\mathrm{g/mol}$

From the stock solution the following volumes were pipetted into $100~\mathrm{ml}$ flasks: $3$, $5$, $8$ and $10~\mathrm{ml}$. Also $1~\mathrm{ml}$ $\ce{KCl}$ solution, $1.0~\mathrm{M}$, was added to prevent ionisation.

I calculated the concentration Lithium in the stock as follows:

The mass of Lithium in the weighed $\ce{LiCl}$ is equal to the molar mass of lithium divided by the molar mass of lithium chloride multiplied by the mass that was weighed. So in the form of an equation:

$$m(Li)=\frac{M(\ce{Li})}{M(\ce{LiCl})}\cdot m(\text{total})=99.9~\mathrm{mg}$$

Therefore the concentration of the stock solution is equal to the mass of lithium divided by the volume of the flask:

$$C(\text{Li, stock}) = \frac{m(\ce{Li})}{V(\ce{Li})}=\frac{99.9}{0.1}=999~\mathrm{mg/l}$$

The concentration of the standard solutions therefore would be:

$$C(\text{standard}) = C(\text{stock}) \times \frac{V(\text{stock})}{V(\text{flask volume})}$$

With the concentration of the standard in $\mathrm{mg/L}$ and the volumes of both stock and flash volume in $\mathrm{mL}$.

This would give concentrations of $29.9$; $49.9$; $79.9$ and $99.9~\mathrm{mg/L}$ to the standard solutions where $3$, $5$, $8$ and $10~\mathrm{ml}$ were pipetted into. The answer sheet however provides us with $300$, $500$, $800$ and $1000~\mathrm{\mu g/L}$. I suspect I've made a mistake in unit conversion somewhere, due to the factor 100 mistake in the answer, although the units don't correspond either.

Well, I'll start by saying I redid your math, and you have rounding errors.  Using correct sig figs, your stock solution has 100. mg of Li in it, giving a concentration of 1.00E3 mg/L.  Those dilutions therefore should be 30.0 mg/L, 50.0 mg/L, 80.0 mg/L, and 100. mg/L.

You're correct aside from the rounding errors, and the answer key is wrong.  The question as you've written can't have those answers, as 300 μg/L is 0.3 mg/L, which is not the answer.


Why is the dissolution of AgCl considered endothermic? The solubility constant of $\ce{AgCl}$ at $25\mathrm{\,^{\circ}C}$ degrees is $1.6\cdot 10^{-10}$.

At $100\mathrm{\,^{\circ}C}$, the $K_{\mathrm{sp}}$ increases to $2.15\cdot 10^{-8}$.

Why is the dissolution of $\ce{AgCl}$ at $100\mathrm{\,^{\circ}C}$ considered endothermic? I would have thought it was exothermic.

The typical way of explaining this is using Le Chatelier's Principle. That is, we could consider the reaction,$$\ce{AgCl(s) + heat<=>Ag+(aq) + Cl^-(aq)}$$This then demonstrates that energy in the form of heat is required to have the reaction proceed. Thus, at a higher temerature, the reaction will shift to the right as predicted by Le Chatelier's principle.

There are obviously more thorough explanations possible, but this is a good place to start.


How do I print property matrices in Avogadro? I'd like to print the content of property matrices (i.e., the matrices shown after selecting View→Properties→X where X is one of the six options shown, including Angle properties..., Atom properties... and Bond properties...) in Avogadro, but I can't seem to find this option at Avogadro Wiki or other forms of Avogadro documentation.

e.g. this is the bond properties matrix.

I am not aware of a menu button to print these tables.

There is however a different way to get your hands on the numbers.

Avogadro features objects: molecules, bonds and atoms are instances of classes, whose properties can be read and processed by programming.
The easiest way is the build-in Python-Terminal.

(Note that I'm using Avogadro 1.1.1 on Linux in German, your menus will have different labels.)

In Avogadro, open a Python Terminal (In German, this is done via
Einstellungen -> Werkzeugleisten -> Python-Terminal, which should be something like Settings -> Tool bars -> Python terminal

When run in the Python terminal of Avogradro, the following snippet writes

atom index
atomic number
x, y, z coordinates

for all atoms of the currently opened molecule to a csv file into an existing tmp directory in the user's home.

import os.path
import csv

home = os.path.expanduser('~')
data_dir = 'tmp'
atom_table_name = 'atoms.csv'
atom_table_path = os.path.join(home, data_dir, atom_table_name)
# generate full path for file with atom data

with open(atom_table_path, 'wt') as fout:
    writer = csv.writer(fout)
    writer.writerow(('Index', 'Atomic number', 'X', 'Y', 'Z'))

    mol = Avogadro.molecule

    for atom in mol.atoms:
        index = atom.index + 1
        x, y, z = atom.pos
        writer.writerow((index, atom.atomicNumber, x, y, z))

EDIT

I mentioned the Python Terminal first because I didn't want to introduce too much different things at once.

Once you have played a bit with the build-in Python Terminal, you will want to go one step beyond and try a more comfortable IDE with logging, tab-completion, build-in help, etc. My suggestion is to try IPython!

With IPython (or any other IDE of your choice), you can make use of Avogadro through its API without actually running it.

Given that your data file propen.cml looks like this:

<molecule>
 <atomArray>
  <atom id="a1" elementType="C" x3="-6.736896" y3="1.198243" z3="-0.001854"/>
  <atom id="a2" elementType="C" x3="-5.625957" y3="2.233390" z3="-0.108371"/>
  <atom id="a3" elementType="H" x3="-6.748313" y3="0.730420" z3="0.988429"/>
  <atom id="a4" elementType="H" x3="-6.615420" y3="0.408802" z3="-0.751312"/>
  <atom id="a5" elementType="H" x3="-7.710837" y3="1.671184" z3="-0.163704"/>
  <atom id="a6" elementType="C" x3="-4.268024" y3="1.629800" z3="0.106393"/>
  <atom id="a7" elementType="H" x3="-5.801148" y3="3.017025" z3="0.637857"/>
  <atom id="a8" elementType="H" x3="-5.668681" y3="2.695849" z3="-1.101309"/>
  <atom id="a9" elementType="C" x3="-3.137190" y3="2.345874" z3="0.062502"/>
  <atom id="a10" elementType="H" x3="-4.203709" y3="0.562202" z3="0.306350"/>
  <atom id="a11" elementType="H" x3="-3.147760" y3="3.413685" z3="-0.133506"/>
  <atom id="a12" elementType="H" x3="-2.172936" y3="1.873521" z3="0.222912"/>
 </atomArray>
 <bondArray>
  <bond atomRefs2="a1 a2" order="1"/>
  <bond atomRefs2="a1 a3" order="1"/>
  <bond atomRefs2="a1 a4" order="1"/>
  <bond atomRefs2="a1 a5" order="1"/>
  <bond atomRefs2="a2 a6" order="1"/>
  <bond atomRefs2="a2 a7" order="1"/>
  <bond atomRefs2="a2 a8" order="1"/>
  <bond atomRefs2="a6 a9" order="2"/>
  <bond atomRefs2="a6 a10" order="1"/>
  <bond atomRefs2="a9 a11" order="1"/>
  <bond atomRefs2="a9 a12" order="1"/>
 </bondArray>
</molecule>

you can load the CML file and iterate over the bonds to display the index numbers of the atoms in the bond and the bond length using:

import Avogadro as av

mol = av.MoleculeFile.readMolecule('propen.cml')
# don't forget to adjust the path for the data file!

for bond in mol.bonds:
    first = bond.beginAtom
    last = bond.endAtom
    print first.index, last.index, bond.length


How can absolute binding free energies be calculated? In the calculation of binding free energies, such as between a protein and a ligand, I learned that absolute values cannot be obtained from simulation (such as taught in this lecture, slide 2). Rather, with the appropriate reference state, only relative free energies can be calculated. However, there are a number of papers that discuss calculations of absolute free energies (ex. recent JACS paper). What does this mean? Are the absolute values--even among different papers--using some kind of general reference state?

Free energies are computed by transforming one state into another, e.g. by transforming ligand A into a ligand B (inside a protein in our case). This transformation yields the difference in binding free energies between ligands A and B: $\Delta \Delta G^{bind}_{A \rightarrow B}$. This is a relative binding free energy.

On the other hand, if ligand A is transformed into nothing, it's absolute binding free energy is calculated: $ \Delta G^{bind}_A$.

In both cases ($\Delta G$ and $\Delta \Delta G$), we are calculating differences between two states, and never calculating the absolute free energy $G$ of a single state. This is probably what you meant by: "only relative free energies can be calculated."

PS: since binding occurs in a solvent, it would be necessary to take ligand solvation and protein solvation into account, in order to properly describe the unbound state. For example, in the calculation of $\Delta G^{bind}_A$ described above, ligand A should appear in bulk water instead of disappearing completely.


Car Parinello advantages and disadvantages A look at wikipedia and some online sources state that Car Parinello is less computationally demanding than self consistent methods. The original journal "Unified approach for Molecular Dynamics and Density Functional Theory" by Car and Parinello state that their method is appropriate for

  (i)
  compute
  ground-state
  electronic
  properties
  of
  large
  and/or
  disordered
  systems
  at
  the
  level
  of
  state-of-the-art
  electronic
  structure
  calculations;
  (ii)
  perform
  ab
  initio
  MD
  simulations
  where
  the
  only
  assumptions
  are
  the
  validity
  of
  classical
  mechanics
  to
  describe
  ionic
  motion
  and
  the
  Born-
  Oppenheimer
  (BO)
  approximation
  to
  separate
  nuclear
  and
  electronic
  coordinates.

Both of these advantages arise from the fact that the Car Parinello method is computationally less demanding However, I would like to know about its disadvantages. There must be a reason why self consistent methods are still widely used. I would also like to know if there are more advantages than it being computationally less demanding.

Car Parinello is an Molecular Dynamics (MD) model. The MD models of chemicals I've worked with use a projection of the expected wave-function using classical mechanics to approximate what the atoms and molecules in the model are doing. Essentially assigning a force constant based on distance between the atoms in the system to approximate the energies involved in the simulation. This kind of simulation can give semi-qualitative numbers to energies of association or dynamics of a reaction in some cases where the motions of all the atoms are essentially classical in nature. However, if there is proton motion involved in the reaction coordinate or if you would like chemically meaningful energies of association MD models will not be workable for your goals. MD models can, typically, only get a "feel" for whether or not an association is possible or if a reaction could take place. And if there is a quantum mechanically allowed and classically forbidden step, say proton tunneling, the model will fail.

So MD models are great for very large or distributed systems. For example modelling a solute in solvent or large solid systems, both of which are beyond most ab initio Quantum Mechanics (QM) calculations due to the number of individual calculation steps needed for convergence. There are some "high level" (nit-picky) calculations that require precision beyond what a classically built model can give which is why DFT and ab initio calculations are still run today.

Now I have not ever used Car Parinello in my work, (I'm part of one of those nit-picky groups) so I cannot give any real numbers for the errors in this method as compared to other MD simulations. I would, however, say that the writer of the Wiki page is not un-biased (neither am I, see above regarding nit-picky group) and is trying to show how great this method is compared to other MD simulations.  This particular MD simulation seems to be using some SCF calculations and long range basis set functions for outer shell electrons and so I think it is mainly for use in solid ionic systems rather than the small isolated molecules my group typically models.

For DFT (or essentially all modern QM calculations) the electronic wave function is approximated using basis functions. These are pre-programmed functions that can be mixed together with little computational effort to approximate the true wave function. The basis sets used by Car Parinello are plane wave basis function which means the wave function repeats at long range and allows (forces) assignment of electrons far removed from one another. This means that a self consistent state is quickly reached for solid crystalline systems.

My personal bias (use a grain of salt for this section) is against such basis sets due to the fact that it forces a pseudo-crystalline pattern on the system (at least at long range). For most chemically interesting systems it is the oddities on a surface or the internal structure which give rise to chemical change. And so, this model may fail to represent these systems well due to the choice of basis set. Once again, not having used this model I cannot say it would fail to represent these interesting features, I just doubt it would give reasonable energies or wave functions after completing the calculations.


pKa of phenols using UV-Vis spectroscopy I have done an experiment where I dissolved four different substituted phenols in acidic, basic, and buffer solutions, and recorded the UV-Vis spectra for each. I'm supposed to pick a wavelength where the absorbances for the protonated, normal and dissociated forms are very different. I then use these equations given to me to calculate the $\mathrm{p}K_\mathrm{a}$ of that phenol:

$$x = \frac{A_\text{buffer} - A_\text{acid}}{A_\text{base} - A_\text{acid}}\tag{1}$$

where $x$ is the mole fraction; $A$ is the absorbance in corresponding medium, and

$$\mathrm{p}K_\mathrm{a} = \mathrm{pH} - \log\left(\frac{1-x}{x}\right)\tag{2}$$

I've calculated the $\mathrm{p}K_\mathrm{a}$ values of each phenol, but none of them are close to literature values. I'm not sure what I've done wrong. Can anyone help explain these equations and any ways I could have messed up?

To quote the key points of an easy to read publication by Salgado and Vargas-Hernández (doi 10.4236/ajac.2014.517135, open access):

All starts with the dissociation equilibrium between the acid $\ce{HA}$ and its anion $\ce{A-}$ for which you write

$$\ce{HA + H2O <=> H3O+ + A-}$$

By consequence of this, the recorded total absorbance $A_\mathrm{t}$ for system, assuming only the acid and its anion absorb light in the region inspected, equates to

$$A_\mathrm{t} = d(c_{\ce{HA}} \cdot \varepsilon_\ce{HA} + c_\ce{A-} \cdot \varepsilon_\ce{A-}) = d \cdot c_\mathrm{t} \cdot \varepsilon_\mathrm{t}$$

with $c_\mathrm{t}$ as total concentration; $\varepsilon$, the absorption coefficient; and $d$, the optical path length.

For one, you record a spectrum at low pH to convert all of your sample into the acid form ($c_\mathrm{t} = c_\mathrm{HA}$, and $c_\ce{A^-} = 0$) where all absorption is due to the acid.  Equally, you record a spectrum at a pH this high that all or your sample is deprotonated (then  $c_\mathrm{t} = c_\mathrm{A^-}$).  You need then spectra between the two extrema, with some of $\ce{HA}$ in presence of some of $\ce{A-}$, because the proportion of the concentrations of acid to its anion may be expressed by

$$\frac{c_\ce{A-}}{c_\ce{HA}} = \frac{[\ce{A-}]}{[\ce{HA}]} = \frac{A_\mathrm{t} - A_\ce{HA}}{A_\ce{A-} - A_\mathrm{t}}$$

This is why you search for the isobestic point, where the absorption spectra recorded do not vary while varying the pH value of your analyte, and where $\varepsilon_\ce{HA} = \varepsilon_\ce{A^-}$.

(credit: doi 10.4236/ajac.2014.517135)

Because the equilibrium between the acid and its anion may be described by

$$\log{K_\mathrm{a}} = \log[\ce{H3O+}] + \log{\frac{[\ce{A^-}]} {[\ce{HA}]}} $$

and the definition of $pK_\mathrm{a} = - \log{K_\mathrm{a}}$ you may substitute absorptions by concentrations to yield

$$\log{\left(\frac{A_\mathrm{t} - A_\ce{HA}}  {A_\ce{A-} - A_\mathrm{t}}\right)} = \mathrm{pH} - pK_\mathrm{a} $$

Hence, plotting $\log{\left(\frac{A_\mathrm{t} - A_\ce{HA}}  {A_\ce{A-} - A_\mathrm{t}}\right)}$ in function of $\mathrm{pH}$ yields a slope with an intercept of $y = - pK_\mathrm{a}$.

(credit: loc. cit.)

Addition:
With WebPlotDigitizer it was possible to  convert above mentioned figure 8 into numbers, separate for the basic (b), the acidic (a) and the intermediate / about neuter run at pH = 7.7 (n).  (The result, a .csv file, is deposit here.)  For a wavelength of about $550\,\mathrm{nm}$ the readout of Abs equals to 1.9935 (b), 0.6502 (n) and 0.2241 (a), and the computation

$$ pK_\mathrm{a} = 7.7 - \log_{10}{\frac {0.4534 - 0.2984} {1.2121 - 0.4534}} $$
$$ pK_\mathrm{a} = 7.7 - \log_{10}{\frac {0.1550} {0.7587}} $$
$$ pK_\mathrm{a} (550\,\mathrm{nm}) = 8.39 $$

By similar token, $pK_\mathrm{a} (604\,\mathrm{nm}) = 8.20$ (with Abs of 1.9935 (b), 0.6502 (n), and 0.2241 (a)); or $pK_\mathrm{a} (432\,\mathrm{nm}) = 8.51$ (with Abs of 0.3411 (b), 0.8198 (n), and 0.8940 (a)).

To put this into perspectivive, the arithmetical mean of the four methods compared in Salgado's paper equals to $pK_\mathrm{a} = 8.277$.  Possibly, the difference between their result and the «recuperation»  here were smaller if either a) WebPlotDigitizer were used with a lesser data increment, b) using an interpolation / smoothing function among the data extracted by the program, c) figure 7 were reconstructed from their figure 4.


Exactly how the subscripted numbers are given to the different types of boron sheets (borophenes)? I was going to start a project on Borophene, and hence I was going through a review journal, 2D Boron: Boraphene, Borophene, Boronene, where they have said many types of borophenes had been proposed. They have been given different greek letters like α, β, γ (β12), δ and χ(chi) to classify different types and they also classified those types into different sub-types by using subscripted numbers like α1, α2, α3 or β1, β2, β3, β12 etc.
At first, I thought it was coming from the hexagonal hole density (η) term which is defined by -
η = (a number of hexagonal holes/a number of atoms in the original triangular sheet). But later I found that it's not the case (considering all the types).
My question is - how exactly these subscripted numbers are given to sub-classify the different types of borophene?
Different types of Borophene(name) and their properties

P.S. - All the numbers next to the greek letters will be in subscript.

All the subscripted numbers are to denote different phases.
In the physical sciences, a phase is a region of space (a thermodynamic system), throughout which all physical properties of a material are essentially uniform. Distinct phases may be described as different states of matter such as gas, liquid, solid, plasma or Bose-Einstein condensate.
Distinct phases may also exist within a given state of matter. As shown in the following diagram for iron alloys, several phases exist for both the solid and liquid states.
For Borophene, α, β, χ, δ and ψ are different types of structures based on their coordination number(CN) like,

α (CN = 5,6)
β (CN = 4,5,6)
χ (CN = 4,5)
δ (CN = 3,4,5)
ψ (Single valud CN)

Now each different structures can have more than one phases. When any particular type of structure has more than one phases, those are denoted by numbers.
For example, α has 7 different structures. Now the first one is named α. After that total of 6 phases are found. One of which is a very similar structure with α. Hence it was named α'. The difference between α and α' is that there is no corrugation in α whereas α' has a corrugation resulting in better stability than α. However, the other 5 phases are named accordingly from α₁ to α₅ based on their earlier prediction. You can find the different phase structures of α Borophene from the following image -
Ref:

https://en.wikipedia.org/wiki/Phase_(matter)
Xiaojun Wu, Jun Dai, Yu Zhao, Zhiwen Zhuo, Jinlong Yang, and Xiao Cheng Zeng. Two-Dimensional Boron Monolayer Sheets. ACS Nano 2012, 6, 8, 7443–7453. https://doi.org/10.1021/nn302696v


Why can Grignard's Reagent be stored in ether? Grignard's Reagent will generate a strong base and nucleophile, $\ce{R-}$. Can this not show nucleophilic substitution at the C atom attached to the O atom of the ether, causing $\ce{RO-}$ to depart as a leaving group? Since $\ce{RO-}$ is a better leaving group than $\ce{R-}$ this reaction should be feasible. Additonally, $\ce{R-}$ is a very good nucleophile.

The free nucleophile $\ce{R-}$ never really exists in a Grignard reagent. That's why we write it $\ce{RMgX}$. Ethers are used as a solvent as the oxygen coordinates to the $\ce{Mg}$, improving solubility and stability.

Regarding your mentioned substitution reaction, it probably would happen if a Grignard reagent was left long enough. There's a reason you make Grignard reagents immediately before use. They don't last if left on the shelf!

But Grignard reagents are usually reacted with carbonyl compounds. These are much more electrophilic than ethers, because the oxygen atom is pulling electrons away from one carbón atom instead of spreading its pull between 2, and because a double $\ce{C=O}$ bond is inherently weaker than two $\ce{C-O}$ bonds. Hence the desired reaction happens much faster than the undesired one.


What gives rise to the color of sapphires? Here is what I know:

Sapphires are composed primarily of $\ce{Al2O3}$ (in pure form colorless) as well as Fe and Ti (trace impurities responsible for the color). I know that the proportion and identity of trace elements gives rise to various colors but I just want to focus on the combination that gives blue right now.

The explanation that I could find online says that the reason we see blue is because of intervalence charge transfer between the iron and titanium ions. Specifically Ti(IV), Ti(III), Fe(III), and Fe(II) are involved. The ground state configuration is controversial but is presumed to be both ions in their 3+ state. It could also be Ti(IV)-Fe(II). The first part of my question is what is the exact equation for the process occurring? I could not find a redox potential for the reaction $\ce{Ti(IV) + e^- -> Ti(III)}$ so I'm not sure which direction the electrons flow. The half reaction $\ce{Fe(III) + e^- -> Fe(II)}$ has a positive reduction potential of +0.77 V. I assumed that a photon provides enough energy for the fourth ionization energy of Ti (4174.65 kJ/mol) and that electron is transferred to Fe(III). I don't think this can be correct though because that energy would be in the ultraviolet region on the J/atom scale. This is the same for the third ionization energy of Fe. I was expecting to see something closer to yellow light region ~575nm. So from this its clear that I'm probably missing the idea regarding the physics occurring here.

The second part of my question has to deal with color subtraction in general. Why does a sapphires absorption of yellow light lead to blue color in the first place? I've seen an explanation that you imagine white light as a mixture of only the primary colors red, blue, and green light. Also its stated that the secondary colors are formed as follows: red + green = yellow, blue + green = cyan, and red+ blue = magenta. Then if a sample absorbs for example yellow light (R+G) then only blue light remains from the original R+G+B=W combination. This makes sense at face value for me but I wanted to go a bit further. I assume the origin of this color addition and subtraction theory is that humans have cones specific for detection of red, blue, and green light (with a range of wavelengths of course to cover the whole spectrum). My difficulty with the color subtraction answer for the color origin is the fact that our cones detect a range of wavelengths. If the sapphire only removes a single wavelength of light close to 575nm (yellow), how is the effect so dramatic to make the stone blue? Won't the stimulation of the cones not have changed dramatically?

Feel free to answer as in depth as possible! Thanks-

The blue of sapphires is produced by the elements titanium and iron that are both present. There is a process called charge transfer causing the transfer of an electron between Ti$^{4+}$ and Fe$^{2+}$ to Ti$^{3+}$ and Fe$^{3+}$ to fit together with the Al$^{3+}$ of the corundum lattice. Those ions give the blue sapphire its colour.

Note that the ions in a crystal are different from isolated ions in the gas phase, they tend to have larger absorption bands instead of sharp lines. Also the distances in the lattice are important and may modify the colour, anisotropy can lead to colour dichroism, the sapphire may be blue viewed from one side and green from another.


Size of atomic shells Is the distance of some $n^{th}$ shell ($n$ $\neq$ outermost shell) from the nucleus of an atom different for different elements?

If so, then how much is the difference or how could we calculate it?

Yes, the size* of the $n$-th shell will differ between elements. The reason is that the nuclear charge will shrink each shell as it goes up. Additional electrons cannot fully compensate this effect because they do not perfectly shield the core.

The difference can be measured (see below) or calculated using quantum chemistry with flexible basis sets. Results from the latter can be validated against X-ray spectroscopy. However, even the Bohr model will yield different radii for the first shell when increasing the nuclear charge.

*The definition of the size is not trivial. One can use the van-der-Waals radius, ionic radius or determine the isosphere that contains e.g. 90% of the electron probability density as long as one is consistent about it.


Material that absorbs only in NIR and emits in visible This Thorlabs VRC4 card here uses a material that absorbs only NIR and emits in visible. The spec sheets don't list what chemical this is. Do you know materials that show similar emission properties even with weak laser excitation?

Such materials not only exist but are common

Most people will have seen an actual demonstration of materials that exploit the phenomenon of a substance that absorbs NIR light and emits visible light. Most Green laser-pointers are based on their use.

These usually rely on a semiconductor laser outputting near-infra-red (NIR) light at 1063nm wavelength which then passes through a crystal of KDP (mono potassium phosphate) or KTP (potassium titanyl phosphate) which emits green light at 532nm (which is the second harmonic of the original laser frequency). The mechanism involves non-linear optical processes (which usually require a fairly high intensity of light so won't happen as easily as phosphorescence, for example).


Why can the acidic strength between H2S and HCl be explained on the basis of electronegativity but H2S and H2O can't? Consider simply aqueous solutions of $\ce{H2S}$  and $\ce{HCl}$ . It is pretty much known that $\ce{HCl}$ is stronger than $\ce{H2S}$ in water. The reason behind that first is the electronegativity of $\ce{Cl}$ is higher than $\ce{S}$, so, in a polar medium, it becomes easier for the $\ce{H-Cl}$ bond to break than the other one.  More ever , $\ce{Cl^-}$ ion being smaller in size is extensively hydrated in water than $\ce{HS^-}$ ion, which also favours in the increased acidity of $\ce{HCl}$.

But now consider acidity of $\ce{H2S}$ and $\ce{H2O}$ in water. If we apply the similar reasons as before, we will ultimately conclude water is  a stronger acid than $\ce{H2S}$. But actually it’s the reverse. The only fact supporting the actual phenomenon is the weaker bond energy of $\ce{H-S}$, but if we apply this to the earlier case of $\ce{HCl}$ and $\ce{H2S}$, it will again contradict the actual phenomenon.

Why is this the case ? If my reasoning is incorrect, what is the actual reasoning that will support the correct acidity in both the cases ?

You are right that electronegativity is an important consideration but another factor that is also very important is size of the atom. Although there is a decrease in electronegativity from $\ce {O}$ to $\ce {S}$, the size of the atom which the charge is mainly centred on is now larger. This allows the charge to be distributed over a larger volume of space, decreasing inter-electronic repulsions, providing greater stabilisation of the negative charge.

However, the greater size decreases charge density, decreasing the strength of ion-dipole interactions between the anion and the surrounding water molecules. Thus, the extent of solvation decreases for the larger anion.

Electronegativity is defined as the ability of the atom to attract a bonding pair of electrons. It merely serves as a proxy for the ability of the atom to stabilise negative charge. It should not be viewed as the most important factor that influences the atom's ability to stabilise negative charge.

Also, mentioned in your answer, the strength of the bond formed with the larger atom with more diffuse orbitals decreases. A bond that is more easily cleaved would facilitate bond cleavage and thus, favour the acid with the larger atom bonded to $\ce {H}$ as the stronger acid.

Consolidating the above discussion, there are really two factors on each side of the table below. Perhaps, the distribution of charge increases very significantly and the bond energy also decreases significantly while the decrease in extent of solvation and the electronegativity exhibit a relatively less significant effect. It would be great if someone is able to quantitatively show this...

\begin{array}{|c|c|c|c|} \hline
\text{Favouring acids with larger atom} & \text{Favouring acids with smaller atom} \\ \hline
\ce{Distribution of charge} & \ce{Extent of solvation}\\ \hline
\ce{Bond strength} & \ce{Electronegativity}\\ \hline
\end{array}


I am confused about the stability of carbanion III and IV in the image I am confused about the stability of carbanion III and IV in the image, I think III is more unstable because of electron donating effect (+R) of  $\ce{-OCH3}$ group but my book says III is more stable than IV. In compound III, $\ce{-OCH3}$ would donate its lone pair and would make the carbanion unstable but no such effect happens in compound IV. $\ce{-OCH3}$ in compound III would also have -I effect to stabilize the carbanion but I have read that for $\ce{-OR}$ group, +R effect is more than its -I effect i.e., its electron donating nature is more than its electron withdrawing nature through inductive effect, +R means electron donating by resonance (In $\ce{-OR}$, $\ce{R}$ is any alkyl group)

I think in 3, carbanion is resonance stabilised. While in 4, C=C-C=O are in conjugation, thus will not stabilise carbanion.


Maximum Pressure of Covalently Bonded Material At high enough pressure, atomic material no longer exists, for example under the conditions of a neutron star.  Is there a different maximal pressure above which covalent bonds no longer exist?

With the twist of thumbscrew, a hand-held diamond anvil cell will exert 300 GPa (43,500,000 psi).  No chemical bonding problem there. If you stack stuff - planetary and stellar cores - it gets more interesting, depending on composition and temperature.  Temperature matters.

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

Magnetic fields exert pressure.  How much magnetic field diddles with chemical bonding to give you a pressure estimate?

http://en.wikipedia.org/wiki/Magnetar
http://en.wikipedia.org/wiki/Magnetic_pressure
  10^10 teslas is 4.0×10^25 J/m^3 E/c^2 mass density more than 10^4 times that of lead.

That'll put some hurt on chemical bonding.  Scale down to 347 kJ/mole-bond multiplied in a molar volume of diamond.  Atomizing diamond requires about 800 kJ/mol per atom against cohesive energy.  The strongest chemical bonding interaction is roughly 3 eV.  Then you can pontificate with a Lennard-Jones potential.

http://en.wikipedia.org/wiki/Lennard-Jones_potential


What transition metal ions form without oxidant in water pH 7? I am not sure how many transition metals will ionize without oxidants, where could I learn this?

If you are counting lanthanides as "transition metals", then look at europium whose reactivity with water is similar to calcium.  Although the product listed in the reference is the trivalent hydroxide $\ce{Eu(OH)3}$, europium has a relativy stable +2 oxidation state and so the reaction msy proceed via a relatively soluble divalent hydroxide or cation ($\ce{Eu(OH)2, Eu^{2+}}$).  The combination of a highly electropositive element, possibly with formation of a relatively soluble, low oxidation state hydroxide/cation makes europium more reactive to water than most transition metals.


Remove only salt from salt water I have a bottle of salt water I collected on a local beach. I want to see what is in it, apart from water and salt. If I boil the whole bottle and wait for it to evaporate, mostly salt will remain. I want to see everything except salt that is still in the pot after I boil the water. Is this possible with common household items and commonly available (and safe) chemicals?

Disclaimer: I know nothing about chemistry.

You could remove the bulk of the sodium chloride (and most other salts) by evaporation and crystallization. Most of the impurities you are looking for are going to stay in solution or suspension when you filter off the salts.

You could also use dialysis to remove the remaining salts: this requires a bag or length of thin-walled tubing made of a common polymer such as cellulose acetate. Tie off one end (the bottom), put the concentrate from the above step into the tubing, and tie off the top end. Then place it in a bucket of distilled water. The salts and other small molecules will diffuse out into the water, whereas the higher molecular weight materials like toxins will stay in the bag. You may need to change the water in the bucket a few times to get rid of all the salts.

You will still need some chemistry to figure out what substances remain in the bag. With seawater, it could be a very complex mix. However, looking at it with a fluorescent light might show something interesting. Good luck, and hope you get inspired to study chemistry!


How to demonstrate the safety of wearing leather shoes in the chemistry lab? Quite often (from my own experience anyway), one of the safety rules in a school (high school or university) laboratory is that the students must wear leather shoes.

I like to demonstrate some of the safety rules, so my question is, what is an effective and safe experiment to demonstrate the safety of wearing leather shoes while experimenting in the chemical lab?

*This is in comparison to say sneakers or canvas shoes

The simplest thing to do is to sacrifice two pairs of shoes, one with leather uppers and the other with canvas/cloth/whatever. Put something inside the shoes to represent "feet". Do something nasty to each pair, like pour concentrated acid over them (using secondary containment to avoid spills and splashes!). Then, extract your indicator and compare (assuming the different in damage to the shoes was not enough). You can also try lighting them on fire (though canvas shoes probably have fire-retardant compounds added), or seeing how long it takes water to soak through (leather will win here).

And, whatever you do, record it to video so that next time you only need to play the video and not buy more shoes. If this is for your students, I would make sure you appear in the video to increase "authenticity".


Difference between polarity and polarizability How do the terms polarity and polarizability differ from each other?

In my opinion, polarity is the degree of ionic character in covalent compounds and polarizability implies towards degree of covalent character in ionic compounds. Am I correct?

Polarity and polarisability are almost entirely unrelated.

Polarity refers to the distribution of partial electric charge in molecules (or even materials). Or, more technically, the degree to which a molecule has a dipole moment (which relates to its inherent electric field). Some molecules, like benzene, are non-polar as they have no net dipole moment; others, like nitromethane, have a strong dipole moment. Polarity is a fixed property of the molecule that doesn't depend on the external field.

Polarisability refers to the degree to which the electron clouds in a molecule or atom can be influenced by an external electric field. Everything, polar or not, has a polarisability. Crudely (i.e. simplifying a lot) you can think of it as the degree to which electrons are held tightly in the atom or molecule. So xenon atoms are fairly polarisable compared to helium atoms as their electron cloud is more spread out and less tightly bound. Water is very polar but a lot less polarisable than hexane which is non-polar. Bulk substances have complicated polarisability that depends on the crystal structure and may be inhomogeneous.

Polarity is about the ground state electron distribution; polarisability is about how that distribution changes when subject to an external electric field.


Number of diastereomer pairs of 1,3-dichloro-1,2,3-triphenylpropane How can I determine the number of possible pairs of diastereomers here?
My first guess was two: RR with RS and SS with RS. But what about the potential chirality center middle carbon? There are two identical branches attached to it; however, those two branches can change R/S. So, is it a chirality center?

I basically agree with Ron's answer, but had to draw all of the possible structures to confirm it. The complication is that carbon-2 is non-stereogenic but it may be chirotopic depending on the configuration of the neighboring atoms. The IUPAC Gold Book calls this a pseudo-asymmetric carbon atom. Carbon-2 is not a stereocenter, because it does not have four different "things" attached to it. However, its configuration is important relative to carbons-1 and -3. Below are all of the possible combinations.

Out of the eight possibilities, we see there are only four unique compounds (names generated in ChemDraw):

A: (1R,2s,3S)-1,3-dichloro-1,2,3-triphenylpropane

B: (1S,3S)-1,3-dichloro-1,2,3-triphenylpropane

C: (1R,2r,3S)-1,3-dichloro-1,2,3-triphenylpropane

F: (1R,3R)-1,3-dichloro-1,2,3-triphenylpropane

The small r and s are essentially used to indicate the stereochemistry at carbon-2 relative to carbons-1 and -3. Note that in compounds B and F, the stereochemistry at carbon-2 is inconsequential as in both cases the phenyl group is syn to one chloride and anti to another chloride. I haven't been able to find an adequate explanation of assigning the r/s designations, but I believe that the R configured substituent receives higher priority than the S configured substituent, and Cahn–Ingold–Prelog system is followed from there.


When can we add multiple reactions to compute yield of products in an equilibrium? It is widely known that we can add multiple equilibrium reactions to obtain a net reaction and a corresponding equilibrium constant for the net reaction. However, I am not sure of exactly when this method can be applied. For instance consider a diiprotic acid $\ce{H_2A}$

$$\ce{H_2A <=> HA^{-} + H^{+}} $$
$$\ce{HA^{-} <=> A^{2-} + H^{+}} $$

with hypothetical first and second ionization constants $10^{-5}$ and $10^{-10}$ respectively. Note that $K_{a_1} \gg K_{a_2}$ as is the case for real acids. If we start with $0.1~\rm M $ solution of the acid then the pH is straightforward to calculate since $\ce{H^{+}}$ from the second ionization is negligible we can write

$$\frac{[\ce{H^{+}}][\ce{HA^{-}}]}{[\ce{H_2A}]} = 10^{-5}$$

Since the concentration of the terms in the numerator are equal and the concentration of the un-dissociated acid remains approx constant we get

$$[\ce{H^{+}}] = 10^{-3} ~\mathrm M$$

If instead we take the net reaction with equilibrium constant $K_{a_1}\cdot K_{a_2} = 10^{-15}$

$$\ce{H_2A <=> A^{2-} + 2H^{+}}$$

and write the equation ($x$ is amount of acid dissociated):

$$\frac{(x)(2x)^2}{0.1-x} = 10^{-15}$$

This gives

$$2x = [\ce{H^{+}}] \approx 5.8 \times 10^{-6}$$

which is quite different from the result obtained using the first method.

Note that this is just one of the many cases I have encountered in which adding equations does not work.

Why does adding the equations not work in this case? Also, in general, when is adding equations to figure out equilibrium concentrations justified?

When you combine the reactions and the equilibrium constants like you do, you're calculating the amount of reaction which goes all the way to completion (the fraction which reacts in both steps).

The complication that you're seeing is that the protons are not just coming from the $\ce{H_2A}$ which completely dissociates, but also a (rather substantial) portion of the molecule which undergoes only partial dissociation.

To see this, think about what happens when the hydrogens in the molecule are distinguishable: for example, the dissociation of $\ce{HDA}$ (hemi-deuterated compound), where the reactions are

$$\ce{HDA <=> DA^{-} + H^{+}} $$
$$\ce{DA^{-} <=> A^{2-} + D^{+}} $$

(in reality this wouldn't be the obligate order, but just assume it is for a thought experiment).

The $10^{-15}$ equilibrium constant is only applicable for the formation of $\ce{D^{+}}$ - you still have the rest of the $\ce{H^{+}}$ from the first step, which will also contribute to the pH.

So your assumption in your last step that $\ce{[H^{+}] = 2 [A^{2-}]}$ is incorrect, thus resulting in an incorrect result.

Further explanation

Let's actually work through the $\ce{HDA}$ case, assuming (incorrectly) that the dissociation has an obligate order. You start off with a certain amount of $\ce{HDA}$ ($T = 0.1M$). Now, imagine that at equilibrium we get $x$ total turnovers for the first reaction. A certain amount of $\ce{DA^-}$ (call it $y$) is then going to go and dissociate further. So in the end at equilibrium, we have the following concentrations:

$$\ce{[HDA]} = T-x$$
$$\ce{[H^+]} = x$$
$$\ce{[DA^-]} = x-y$$
$$\ce{[D^+]} = y$$
$$\ce{[A^{2-}]} = y$$

Now we can set up our equilibrium equations:

$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} = \frac{(x)(x-y)}{T-x} = 10^{-5} M$$
$$\frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HD^{-}}]} = \frac{(y)(y)}{x-y}= 10^{-10} M$$

We have a system of two equations with two unknowns. We can thus solve for $x$ and $y$ and get $x=0.995\cdot10^{-3} M$ and $y = 3.15\cdot10^{-7} M$.

Now, let's do the combined equation. First off, we can see where it comes from:

$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} \cdot \frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{DA^{-}}]} = \frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = 10^{-5} M \cdot 10^{-10} M = 10^{-15} M^2$$

So we set up the equation:

$$\frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = \frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$

Um ... we have two unknowns, but only a single equation. We can't solve this without additional information. We can't assume that $[\ce{H^{+}}] = [\ce{D^{+}}]$, or that all the $\ce{HDA}$ which disappeared went completely to $\ce{A^{2-}}$ - some of it can (has to) form $\ce{DA^-}$ which doesn't react further. The equation is valid (if you substitute $x=0.995\cdot10^{-3} M$ into the equation, as determined above, you'll get the correct answer for $y$), but only if we have some way of knowing how much of the $\ce{HDA}$ reacted and didn't go completely to completion.

The picture is further complicated because we're not dealing with distinguishable $\ce{H^+}$ and $\ce{D^+}$ which dissociate in an obligate order, but a single species which is involved in both reactions. That is, instead of

$$\frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$

we would have

$$\frac{(x+y)(x+y)(y)}{(T-x)} = 10^{-15} M^2$$

Again, the combined equilibrium constant is perfectly correct, and accurately expresses the relationship between the concentrations at equilibrium. It's just that the additional assumptions about the relationships between the concentrations that you're making are not. Specifically, $[\ce{H^{+}}] \neq 2 \cdot [\ce{A^{2-}}]$. Note this isn't particularly specific to this sort of combined equilibrium. Any additional reaction you're not considering can change the concentrations and make your assumptions less than correct - not just formation of intermediates which don't go all the way to completion to side reactions that draw off substrates, reactants or products into other forms.

The equilibrium constant tells you the relationship between the species concentrations, but the concentrations as they actually are in solution, not what the reaction generates. The latter is only true when the reaction in question is the only one under consideration.


Precipitation of CaCl2 and FeCl3 in Na2CO3 Why does $\ce{CaCl2}$ gives precipitation of $\ce{CaCO3}$ when $\ce{Na2CO3}$ is added, but $\ce{FeCl3}$ does not give the precipitation of $\ce{Fe2(CO3)3}$ in $\ce{Na_2CO_3}$  ?

This must have been a very good question for a geochemist. They'd give you a complete answer with details of why $\ce{Fe2(CO3)3}$ does not but $\ce{Fe2O3}$ exist in our surrounding. Yet, I would like to give only a simple answer for the question: No, $\ce{Fe2(CO3)3}$ does not form (in sort) when you added $\ce{Na2(CO3)3}$ to an aqueous solution of $\ce{FeCl3}$. The reason?

An aqueous solution of $\ce{FeCl3}$ is acidic because $\ce{FeCl3}$ is a salt of a strong acid $(\ce{HCl})$ and a weak base $(\ce{Fe(OH)3})$. Thus an aqueous solution of $\ce{FeCl3}$ is in following equilibrium:
$$\ce {Fe^{3+} + 2H2O <=> [FeOH]^{2+} + H3O+ }$$
Thus, $\ce{H3O+}$ ions in solution can react with added $\ce{CO3^{2-}}$ ions to form carbonic acid $(\ce{H2(CO3)3})$,which can decomposed to form $\ce{CO2}$  and $\ce{H2O}$. Remaining $\ce{Fe^{3+}}$ will combine with left behind $\ce{OH-}$ ions to make $\ce{Fe(OH)3}$, which would decomposed to $\ce{Fe2O3}$. Thus, when $\ce{Na2CO3}$ is added to an aqueous solution of $\ce{FeCl3}$, one can expect the following sequence of reactions to be taken place:
$$\ce {FeCl3 (aq) + Na2CO3 (aq) -> [Fe2(CO3)]^{3+} (aq) -> Fe2O3 (s) + CO2 (g) + H2O (l)}$$


Why would hexacyanidocobaltate(II) be uncommon while pentacyanidoaquacobaltate(II) is common? In Huheey's Inorganic Chemistry it is written that hexacyanidocobaltate(II) is uncommon but pentacyanidoaquacobaltate(II) is common. Something is mentioned about Jahn-Teller distortion, but I don't understand how it applies here.

In Inorganic Syntheses (1946), a synthesis of $\ce{K3[Co(CN)6]}$ is described, which boils down to the following three equations:[1]

  $$\begin{align}\ce{CoCl2 (aq) + 2 KCN (aq) &-> Co(CN)2 v + 2 KCl (aq)}\tag{1}\\
\ce{Co(CN)2 + 4 KCN (aq) &-> K4[Co(CN)6] (aq)}\tag{2}\\
\ce{2 K4[Co(CN)6] (aq) + H2O &-> 2 K3[Co(CN)6] (aq) + 2 KOH (aq) + H2}\tag{3}\end{align}$$

The procedure requires precipitation of the cobalt(II) cyanide by carefully adding a $\ce{KCN}$ solution to a cobalt(II) solution. This precipitate is filtrated and washed, then redissolved in excess cyanide solution giving a deep-red solution. This in turn is heated to reflux for some 15 minutes, whereupon the colour changes to yellow indicating the formation of the hexacyanidocobaltate(III).[1]

I personally doubt equation $(3)$. Instead, I would assume that the oxidation of hexacyanidocobaltate(II) to the corresponding cobaltate(III) proceeds unter oxygen consumption, not hydrogen liberation. That would give equation $(\text{3'})$:

$$\ce{4 K4[Co(CN)6] + O2 + 2 H2O -> 4 K3[Co(CN)6] + 4 KOH}\tag{3'}$$

In any case, however, the presence of $\ce{[Co(CN)6]^4-}$ or hexacyanidocobaltate(II) as a stable intermediate in solution suggests to me that your premise is false and thus the book you cite from incorrect.

Reference:

J. H. Bigelow and J. C. Bailar (1946) Potassium Hexacyanocobaltate(III), in: Inorganic Syntheses, Volume 2 (ed W. C. Fernelius), John Wiley & Sons, Inc., Hoboken, NJ, USA. DOI: 10.1002/9780470132333.ch72.


Jahn-Teller Distortions in Square Planar Complexes? A Jahn-Teller distortion is predicted whenever a non-linear symmetric molecule has degenerate orbitals and has unequal electron occupation in those degenerate orbitals. Of course, this most often is discussed in the context of octahedral complexes, but it also occurs in other species - it's a general effect.

When I was teaching inorganic chemistry this term, my students asked if this could occur in square planar complexes. My immediate answer was "yes because, of course, there are degenerate MOs:"

Later, I was trying to figure out what the exact geometric distortion would be. As Wikipedia notes:

  Jahn–Teller theorem does not predict the direction of the distortion, only the presence of an unstable geometry.

I consulted the Walsh diagram posted above from Orbital Interactions in Chemistry by Albright, Burdett, and Whangbo.

My initial inclination would be a "scissor" distortion towards a tetrahedral shape since I know $\ce{T_d}$ complexes undergo similar Jahn-Teller distortions.

Based on the Walsh diagram, depending on the electron count (e.g., $\ce{d^3}$ as suggested by the students) it seems as if you would always have a Jahn-Teller distortion predicted, because you would have at least 1 electron in the $\ce{2e}$ orbitals (the structure would be $\ce{D_{2d}}$ as indicated in the figure.

I understand based on LFSE, such complexes wouldn't be stable, but the question remains: Wouldn't it drop down to $\ce{C_{2V}}$ to break degeneracy? If so, how?

What is the specific geometric Jahn-Teller distortion for unstable (e.g., $\ce{d^3}$ square planar complexes?

Introduction

Thank you for prompting me to look for square planar complexes that are not $\mathrm{d^8}$; I learnt some valuable stuff while researching the answer to this question! Do take this answer with a grain of salt, because I cannot access the paper I’m referencing here on holidays (I could when I’m back at work) but the supplementaries confirm what I’m trying to prove.

Dataset

Scott et al. reported the crystal structure of a square planar vanadium(II) complex in 1990. The central vanadium ion is surrounded by the oxygens of four substituted phenolates. A set of two oxygens is also coordinating a lithium ion to give an approximate trigonal planar geometry on the oxygens, and said lithium ions are further coordinated by a THF solvent molecule (again giving a trigonal planar geometry around lithium) such that there is an approximately straight line going through the following atoms: THF-oxygen, lithium, vanadium, lithium, THF-oxygen. I shall define this as the $x$ axis for further discussion; the square planar geometry being in the $x,y$ plane and $z$ being vertical.

Experimental section

I took the atom coordinates of vanadium, the six oxygens, the four oxygen-bound phenol carbons and the two lithii from the supplementary information. My first step was to move the coordinate system’s origin into the vanadium ion. Then I transformed into spherical coordinates and adjusted the axes so that when retransforming back into cartesian coordinates my definitions from above are valid.

Results

The vanadium-coordinating oxygens are slightly displaced from the $x,y$ plane — in sets of two such that one is about as much above the plane as the other is below the plane. This distortion carries through onto their corresponding carbons. The lithium ions are then again almost spot on $z=0$ — also coordinated by one oxygen with positive $z$ and one with negative $z$.

The oxygens do not form a perfect square. Two oxygens are slightly displaced outwards in $x$ direction; two in $y$ direction. The $y$ displacement is almost twice as strong as the $x$ displacement.

Discussion

The distortion of the square planar geometry resembles the one you predicted: Two opposite oxygens above the plane and two below the plane. This is coherent with a reported $\angle(\ce{O-V-O}) = 162^\circ$ (diagonally across the square).

The authors also reported a large $\angle(\ce{O-V-O}) = 103^\circ$ and a small $\angle(\ce{O-V-O}) = 79^\circ$ all between two neighbouring oxygens and a set of longer and shorter $\ce{V-O}$ bonds differing by $2~\mathrm{pm}$ all in accordance with the slight displacement in the crystal coordinates.

Seen as a whole, the entire thing is indeed en route towards tetrahedral, but also features additional slight distortion that further reduces symmetry from a $D$ point group to a $C$ point group.

Conclusion

The $\mathrm{d^3}$ ion vanadium(II) features a distorted square planar geometry. The distortion seems strong enough to consider a point group of $C_\mathrm{2v}$ around vanadium rather than $D_\mathrm{2d}$.

Reference

M. J. Scott, W. C. A. Willish, W. H. Armstrong, J. Am. Chem. Soc. 1990, 112, 2429. DOI: 10.1021/ja00162a059.


How does iceskating work? I was willing to understand a little bit better the physics behind iceskating.

Messy thoughts

From New Scientist (1964) (probably not the most up-to-date reference $\ddot \smile$), I found a diagram showing the relationship between both the static and kinetic coefficients of friction and the temperature (x-axis is reversed) for ice.

When we ice skate, we put a big pressure on the ice. As the ice is less dense than liquid water, this pressure may eventually be high enough to turn ice into liquid water. Intuitively I would think that high pressure would yield to high temperature but I suppose that this would hold true exclusively for gazes ($PV=nRT$). In any case, I don't quite have a good intuition for why warmer (or more pressurized eventually) ice would decrease the coefficient of friction of ice.

Questions

What is the intuition behind the fact the coefficient of friction of ice decreases as temperature increases?
Does the coefficient of friction of ice decreases with higher pressure too? For the same reason as above? As a correlated-side effect of increasing temperature?
Why do ice skates slide so well? Because it puts a high pressure on the contact surface or because it puts a high pressure on the contact surface which in turn increases the temperature?

Calderon & Mohazzabi give an excellent summation of the various theories proposed through the years to explain why ice is so slippery in their 2018 paper "Premelting, Pressure Melting, and Regelation of ice revisited" in the Journal of Applied Mathematics  and Physics.
They  offer both theoretical and experimental evidence that neither pressure melting nor friction melting explain the phenomenon and conclude from atomic force microscopy, among other evidence, that there is a pre-melting Quasi-Liquid surface layer with special properties that make ice skating possible.
References

Calderon, C. and Mohazzabi, P. (2018) Premelting, Pressure Melting, and Regelation of Ice Revisited. Journal of Applied Mathematics and Physics, 6, 2181-2191. doi: [10.4236/jamp.2018.611183.][1]

Preview/read online at :  https://www.researchgate.net/publication/328766489_Premelting_Pressure_Melting_and_Regelation_of_Ice_Revisited


Is enthalpy affected by isotopes? Enthalpy is a sum of $U$ and $PV$, and I thought that for different isotopes, internal energy values may differ. Is it true? There are three stable isotopes of $\textrm{Mg}$, mass number 24, 25, and 26. If enthalpy is different, which one is a standard for enthalpy $0$? I read somewhere on the web that enthalpy of a covalent bond is affected by isotopes, but I'm confused when there are no covalent bonds.

Enthalpy is slightly affected by isotope composition, but mostly below enthalpy measurement errors.

Generally, the standard enthalpy is considered for natural mixture of stable isotopes, unless considered for particular isotope explicitly.


How is the structure of triiodide ion (I3-) possible? I have seen the structure of triiodide ion ($\ce{I3-}$) but I cannot understand why this structure is even possible. I have seen in my textbook that

  $\ce{I_3^-}$ is formed by combination of $\ce{I2}$ and $\ce{I^-}$ ion in which $\ce{I^-}$ ion acts as a donor and $\ce{I2}$ molecule act as acceptor. It accommodates electrons in empty d orbitals:

But I think $\ce{I-}$ is stable due to its complete octet. Then why it should combine with $\ce{I2}$ molecule to form another new compound? I am confused because the Lewis structure of $\ce{I3-}$ would not make any sense if we consider the octet of central iodine atom.

First things first: the bonding orbitals of $\ce{I3-}$ do not contain any significant d-orbital contributions. In fact, the $\ce{I-I}$ bond lengths are significantly longer than in $\ce{I2}$, suggesting a lower bond order.

In fact, the triiodide anion should rather be imagined as a 4-electron-3-centre bond (bond order approximately 0.5) which might be easier to understand using this mesomeric depiction:

$$\ce{\overset{-}{I}\bond{...}I-I <-> I-I\bond{...}\overset{-}{I}}$$

What happens orbital-wise in this? Simplistically speaking, a 4-electron-3-centre bond is made up of the following molecular orbitals stemming from $\mathrm p_z$ orbitals of the individual iodine atoms:

an all-bonding lowest level orbital ($\ce{{+}I-\bond{...}-I{+}\bond{...}{+}I-{}}$, where $-$ and $+$ denote the different phases of a p orbital)
A far-reaching bonding orbital to which the central atom does not contribute ($\ce{{+}I-\bond{...}OIO\bond{...}-I{+}}$, where O denotes no orbital lobes; there is no p-type orbital on the central iodine that could contribute. s-type contribution is possible.)
an all-antibonding highest level orbital ($\ce{{+}I-\bond{...}{+}I-\bond{...}{+}I-{}}$) which remains unoccupied.

Why does this happen although the starting compounds iodide and $\ce{I2}$ are stable on their own? Obviously, most reactions happen although both starting materials are full-octet compounds and inherently stable in some way or another. And most happen because the products are more stable. So we just need to determine why it is favourable for $\ce{I- + I2}$ to react to form $\ce{I3-}$.

The simplest answer is distribution of charge. Rather than have a single negative charge on one atom you end up with a structure that has approximately half a negative charge on two different atoms if you take the resonance structures at face value (hint: don’t). Whatever the final picture is, the negative charge will be more distributed over a greater number of atoms than before. This is, very generally speaking, a great stabilisation and we can stop here.


Why does the Wurtz reaction form such an unstable compound with torsional strain? What would be the product formed when 1-bromo-3-chlorocyclobutane with two equivalents of metallic sodium in ether?

I know that there would be a free radical formation and then it will form a compound whose bond line structure would look like a square with a diagonal.

But why is such an unstable compound being formed as there is a lot of torsional strain?

In the Wurtz coupling reaction, an alkyl halide first reacts with a metal in a one-electron transfer step to produce an alkyl radical which quickly reacts in a second electron transfer with the metal to form a carbanion.  In the final step of the reaction, the carbanion displaces a halogen from a carbon-halogen bond by a $\ce{S_{N}2}$ process to form a carbon-carbon bond.  In the case of 1-bromo-3-chlorocyclobutane two reaction pathways are possible, the carbanion could react intramolecularly to form bicyclobutane or intermolecularly to from a cyclobutane dimer.  Considering the equation $$\ce{\Delta G = \Delta H - T \Delta S}$$ we would say that even though the $\ce{\Delta H}$ term is less favorable for bicyclobutane formation due to the increased ring strain, the change in entropy required to bring two separate molecules together to form the cyclobutane dimer overwhelms the enthalpic term and drives the reaction to proceed by the entropically more favorable intramolecular process.

Look at the cyclobutyl carbanion drawn in the figure below.  The carbanion is ideally set up for an intramolecular $\ce{S_{N}2}$ reaction (backside displacement) on the carbon containing the chlorine diagonally situated across from it.  Entropically, this is extremely favorable.

In summary, because 1) the carbanion and the chloride to be displaced are in the same molecule, 2) because they are so well positioned for the $\ce{S_{N}2}$ displacement reaction to occur and 3) because entropy strongly favors the unimolecular process (bicyclobutane formation) over the bimolecular process (cyclobutane dimer formation), bicyclobutane formation readily occurs.


How good is aromaticity at stabilization? I've been taught that aromaticity is often a very good stabilization factor. Then I came across this question in which another compound was said to have more stability.

Upon deprotonation arrange in the order of stability.

Why is S more stable than P?
(the answer given is (a); from Himanshu Pandey Organic Chemistry chapter General Organic Chemistry)

The situation in S and P is very similar.
Essentially you have a negative charge $\alpha$ to two double bonds which will delocalize these charges.
In the case of cyclopentadiene, this will be stabilized because the anion is aromatic.
But still, there will be a charge which is supported by carbon atoms.

In the case of the 1,3-diketone, the charge will also be delocalized by two double bonds, but instead of an aromatic anion of five members, the charge will be supported by only three atoms.
However, two of them are oxygens which can hold a negative charge much better than a carbon.

The electronegativity of carbon is 2.55 while for oxygen it is 3.44.
Note that the electronegativity scale ranges from 0.7 for $\ce{Fr}$ to 4.0 for $\ce{F}$.
This ability of supporting negative charge is much bigger than the small contribution of aromaticity to the stabilization of a carbanion.


Finding mass of MnF3 is produced in a reaction The question is as follows:

  If $\pu{1.23 g}$ of $\ce{MnI2}$ reacts with $\pu{25.0 g}$ of $\ce{F2}$, what mass of $\ce{MnF3}$ is produced?

$$\ce{2 MnI2 + 13 F2 -> 2 MnF3 + 4 IF5}$$

I think this is a limiting reactant problem but I'm having difficulties with starting the question off. Do I find the mass of both $\ce{MnI2}$ $\pu{(308.74 g/mol)}$ and $\ce{F2}$ $\pu{(38g/mol)}$ and divide them both by their given masses $\pu{1.23 g}$ and $\pu{25.0 g}$ respectively?

Once I do that, do I find the limiting and excess reactant, take the limiting reactant and do stoichiometry to find the mass of $\ce{MnF3}$ ?

Here we have an already balanced equation given in the question-

$$\ce{2 MnI2 + 13 F2⟶ 2 MnF3 + 4 IF5}$$

We find the given number of moles of each reactant by dividing their given masses by their respective molar masses

Given mass of $\ce{MnI2}=\pu{1.23g}$
Molar mass of $\ce{MnI2}= \pu{308.74 g/mol}$

Amount of $\ce{MnI2} = \frac{1.23}{308.74}= \pu{0.0039mol}$

Given mass of $\ce{F2}=\pu{25g}$
Molar mass of $\ce{F2}= \pu{38 g/mol}$

Amount of $\ce{F2} = \frac{25}{38}= \pu{0.657mol}$

Now we find the mole ratio from the given information and compare the calculated ratio to the actual ratio.
If more than $\pu{13 moles}$ of $\ce{F2}$ are available per $\pu{2 moles}$ of $\ce{MnI2}$, the $\ce{F2}$ is in excess and $\ce{MnI2}$ is the limiting reactant. If less than $\pu{13 moles}$ of $\ce{F2}$ are available per $\pu{2 moles}$ of $\ce{MnI2}$, $\ce{F2}$ is the limiting reactant.

Let's say that all of the $\pu{0.657 moles}$ of $\ce{F2}$ were to be used up, there would need to be

$\frac{2}{13}\times0.657=\pu{0.101  moles} \text{ of } \ce{MnI2}$.

But there is only $\pu{0.0039 moles} \text{ of }\ce{ MnI2}$ available which makes it the limiting reactant.

If all of the $\pu{0.0039 moles} \text{ of }\ce{ MnI2}$ were used up, there would need to be

$0.0039\times \frac{13}{2} = \pu{0.0254 moles} \text{ of } \ce{F2}$.

Because there is an excess of $\ce{F2}$, the $\ce{ MnI2}$ amount is used to calculate the amount of the products in the reaction i.e. amount of $\ce{MnF3}$ produced.

From the equation given its easy to observe that the stoichiometric ratio of $\ce{MnI2}$ and $\ce{MnF3} = 1:1$

Thus amount of $\ce{MnF3}$ produced = $\pu{0.0039 mol}$
Molar mass of $\ce{MnF3} = \pu{111.94 g/mol}$

Mass of $\ce{MnF3} = 111.94 \times 0.00398 = \pu{0.445g}$


Chair conformation chirality I have two questions regarding this solution:

I can't see how the structure drawn is the enantiomer.
How do I rotate the ring inversion chair form to obtain the enantiomer?

1.) The mirror plane is orthogonal to the green line drawn between the respective center of the two enantiomers.

2.) The rotation is around the red axis. The red marked carbon atom shows the rotation angle.


Rusting on an iron material I was wondering if someone could tell me why do we write down 'x' in front of $\ce{H2O}$ in the reaction that takes place to form rust on an iron material. I searched for it and went through many websites but couldn't get a specific answer so please someone help me.

Rust consists of hydrated iron(III) oxides Fe2O3·nH2O, iron(III) oxide-hydroxide (FeO(OH), Fe(OH)3.

The rusting of iron is an electrochemical process that begins with the transfer of electrons from iron to oxygen. The rate of corrosion is affected by water and accelerated by electrolytes. The key reaction is the reduction of oxygen:

$$\ce{O2 + 4 e- + 2H2O -> 4OH-}$$

The following redox reaction also occurs in the  presence of water and is crucial to the formation of rust:

$$\ce{2Fe2+ + 0.5 O2 <=> 2 Fe3+ + O2−}$$

Additionally, the following multistep acid-base reactions affect the course of rust formation:

$$\ce{Fe2+ + 2 H2O <=> Fe(OH)2 + 2H+}$$

$$\ce{Fe3+ + 3 H2O <=> Fe(OH)3 + 3 H+}$$

as do the following dehydration equilibria:

$$\ce{Fe(OH)2 <=> FeO + H2O}$$

$$\ce{Fe(OH)3 <=> FeO(OH) + H2O}$$
$$\ce{2 FeO(OH)  <=> Fe2O3 + H2O}$$

From the above equations, it is also seen that the corrosion products are dictated by the availability of water and oxygen. With limited dissolved oxygen, iron(II)-containing materials are favoured, including FeO and black lodestone (Fe3O4). High oxygen concentrations favour ferric materials with the nominal formulae Fe(OH)3-xOx/2.

Thus you can actually calculate the value of x (water of hydration) in the compound. Hope this works

Credits: Chemical formula of rust


Product of the reaction between hexane‐2,5‐dione and ammonium carbonate Which of the following statement is INCORRECT?
(A) The product X has a five membered ring.
(B) The product X is aromatic.
(C) The degree of unsaturation of the product X is 3.
(D) The product X has a six membered ring.

Ammonium carbonate will probably lose a proton and one of the carbonyls may be protonated. But from here onwards I can't find any leads. Does the lone pair of other oxygen atom plays a role?
Usually the heat provided may lead to a cyclic structure, but I am not sure.

Heating $\ce{(NH4)2CO3}$ will generate $\ce{NH3 + CO2 + H2O}.$ Out of these, $\ce{NH3}$ will act as a nucleophile as it's the most nucleophilic of the products. Addition of $\ce{NH3}$ to the two ketones, followed by elimination of water molecules (due to the high temperature provided by ∆) will generate the final product as follows:

The product is a five-membered aromatic ring called 2,5-dimethylpyrrole.


Equilibrium constant and thermodynamic relationship Why is Equilibrium constant (Ratio of rate constants) considered as a thermodynamic quantity whereas the rate constant as kinetic quantity. Moreover why is extent of reaction also a thermodynamic quantity.?
Please explain briefly as I'm stuck.

Well first think of thermodynamics as determining what a system wants to do, and kinetics as what the system is allowed to do.  The rate  constant gives how fast something will react (kinetic rate).  The thermodynamic equilibrium constant gives how much of something will react.  Kinetics are dictated by the movements of atoms, Thermodynamics is dictated by the changes in energy and entropy.


Removing Rust with Coke I've seen a few videos on cola being used to remove rust, I've tried it myself too so I know that it works.

Now the thing that's bothering me is that this apparently contradicts what I've learnt about corrosion in class. If I'm not wrong, cola contains phosphoric acid, and therefore would most probably have free H+ ions floating around in there. I've been taught that H+ promotes (catalysis) the rusting of iron.

Then how come coke removes rust instead of promoting its formation? Also, why doesn't coke form rust on iron?

Rust is a hydrated iron oxide that has a volume greater than the iron it contains. When iron rusts, the oxide grows in layers and curls away from the iron, exposing it to more corrosion.

Phosphoric acid is a moderately strong acid, but it forms an iron phosphate that is insoluble but adherent to iron (and steel). In that sense, phosphoric acid deactivates corrosion on iron (not perfectly, but pretty good), and there are commercial products (like Naval Jelly) which are essentially gelled phosphoric acid, designed to rub on iron and steel to give some corrosion protection. It is a sort of passivation.

Phosphate ions adhere to rust, making rust particles negative and promoting dispersion in the liquid and disadhesion from the iron metal. Some elbow grease is generally needed.

Overall, the extra corrosion from Coke or other phosphoric acid is minimal, and the dispersive effect is helpful, and the iron is better protected after the treatment than before.


How does Carbon Dioxide have a degree of freedom of 6? Carbon Dioxide has a degree of freedom of 6, yet from what I understand, it is a linear molecular shape. Therefore, the number of axis of linear movement is 3 (x,y,z) but there are only two axis of rotational movement since the center of mass essentially does not rotate on one of the axis due to its linear shape. Yet I'm pretty sure my textbook is saying that carbon dioxide has a degree of freedom of 6 when that should be wrong. Can someone explain?

There are always 3N total independent degrees of freedom for a molecule, where N is the number of atoms. These come about because when each atom moves, it has three independent degrees of freedom: its position in each of the x, y, and z directions.

Now, having independent degrees of freedom for each atom isn't all that useful. Instead, we can make combinations of different degrees of freedom. The important thing when doing so is that the number of independent degrees of freedom are preserved: it's just that what a particular degree of freedom does to the atoms changes.

The standard breakdown of degrees of freedom subtracts out global movement in each of the three directions. So you have 3N total degrees of freedom, but you can set aside 3 of them as translation of the whole molecule in each of the x, y and z directions, leaving (3N-3) degrees of freedom.

Likewise, it's standard to subtract out the whole molecule rotation. For most larger molecules, there's three different degrees of rotational freedom: rotation around each of the x, y, and z directions. But for linear molecules like $\ce{CO2}$, one of those rotations (around the axis of the molecule) doesn't actually change the position of the atoms. Therefore it's not a "degree of freedom" which counts against the 3N total. So while for non-linear molecules there are (3N-3-3) = (3N-6) degrees of freedom which are independent from the global rotational and translational ones, for linear molecules there are (3N-3-2) = (3N-5) degrees of freedom which are independent from the global rotational and translational ones. -- A quick clarification. The reason why we ignore this rotation is not because the center of mass doesn't move. The center of mass doesn't move for any of the global rotations: in the typical assignment of degrees of freedom the axis of rotation goes through the center of mass. Instead, the reason the rotation is ignored is that none of the atoms move due to the "rotation".

So since $\ce{CO2}$ has three atoms and is linear, it has 3*3 - 5 = 4 degrees of freedom which are independent of the global rotation and translation. We call these the vibrational modes. There's different ways you can decompose them, but the most useful one is 1) the symmetric stretch (both oxygens going out and in at the same rate while the carbon sits still), 2) the asymmetric stretch (the carbon going back and forth while the oxygens sit more-or-less still), and 3&4) two different out-of plane bending modes. (If the molecule is aligned along the x-axis, one each where the carbon moves in the y and z directions.)

That's it for $\ce{CO2}$: any movement you make - any position you put the atoms in - can be thought of as a combination of those nine movements (three global translations, two global rotations, and four internal vibrations/bending).


Why doesn't 1,3-dichloro-2,4-dimethylcyclobutane have a plane of symmetry? In (1R,2R,3S,4S)-1,3-dichloro-2,4-dimethylcyclobutane, are the blue lines not planes of symmetry?

My book says it has a centre of symmetry, but not a plane of symmetry.  But if I took a cut through any of the blue lines wouldn't both sides be symmetrical?

Ivan already stated quite briefly in the comments, that the planes you drew would transform an upward facing chlorine to a downward facing chlorine and vice versa, and that the same holds for the methyl groups. This is of course true and the first and most obvious point, why there is no symmetry plane.

A more close investigation will, however, point out another important feature of the compound itself. The cyclobutane ring itself is not planar. In the (electronically) lowest conformation of the molecule it has no symmetry whatsoever (technically $C_1$ symmetry), not even inversion symmetry like your book states. This is due to the fact, that the four-membered ring is puckered. This is best visualised by looking at the 3D structure of the compound. There are multiple programs that will allow you to view such a structure and I will xyz coordinates at the end of this posting.

The following image will hopefully already show you what I am talking about. It was optimised on the DF-BP86/def2-SVP level of theory.

In order for the the molecule to have inversion symmetry, the shown angle needs to be exactly $0^\circ$. The following animation shows the molecule rotating hopefully further clarifying this point.

You can clearly see the puckering of the cyclobutane ring.

One final comment: For a molecule like this it really is necessary to specify the stereochemistry with the name. There are other stereoisomers, which do indeed have a symmetry plane. I leave it to you to figure out which of them do.

   Except the last one, all of them have at least one plane of symmetry.

Addendum: Transition state structure

The symmetric ($C_\mathrm{i}$) transition state is on the DF-BP86/def2-SVP level of theory only about $\Delta G (298.15~\mathrm{K}) = 8~\mathrm{kJ/mol}$ higher in energy than the puckered asymmetric forms. It is therefore worthwhile to have a quick look at its structure. In the following graphics the red ball indicates the inversion centre.
Below that is an animation of the transitional mode. It briefly pauses at the puckered structures (these are extrapolated, not optimised structures) and the transition state to make it easier to see the movement in between them.
I would expect this process to be readily available at room temperature and therefore almost impossible to distinguish the different lower lying structures.

Coordinates of the lowest structure:

18
G09D.01; E(RB-P86/def2-SVP/W06) = -1154.66760674 A.U.
C        0.051334000      1.246599000      0.089432000
C       -0.768087000      0.117382000      0.789824000
C        0.002884000     -0.950933000     -0.020424000
C        1.086640000      0.142749000     -0.293350000
H        0.449081000      2.008047000      0.790972000
H       -0.535742000      0.060683000      1.871003000
H       -0.532992000     -1.105843000     -0.982729000
H        1.517474000      0.183811000     -1.312785000
C        0.343625000     -2.291820000      0.610089000
H        0.939420000     -2.156389000      1.535300000
H       -0.586821000     -2.839917000      0.866159000
H        0.932752000     -2.926935000     -0.083624000
C       -0.606103000      1.926898000     -1.110977000
H       -1.475939000      2.532182000     -0.784546000
H        0.116584000      2.600709000     -1.616245000
H       -0.977177000      1.195914000     -1.858831000
Cl       2.502295000      0.083512000      0.845211000
Cl      -2.570729000      0.126790000      0.662589000

Coordinates of the transition state (X is a dummy atom indicating the inversion centre)

19
G09D.01; E(RB-P86/def2-SVP/W06) = -1154.6659061600 A.U.
C       -0.188954000      1.025076554      0.415795307
C       -0.873066584     -0.383572154      0.501227861
C        0.188954000     -0.991627446     -0.457606693
C        0.898153416      0.404477846     -0.509590139
H        0.242239584      1.289691385      1.404629307
H       -0.864241753     -0.820445015      1.518694431
H       -0.267326416     -1.206068615     -1.446440693
H        0.939502247      0.849712985     -1.522875569
C        0.953473154     -2.212350139      0.042720768
H        1.426842139     -2.018382400      1.026791199
H        0.267219307     -3.079692431      0.145048491
H        1.760843738     -2.494108569     -0.663022216
C       -0.974378846      2.203987861     -0.147249232
H       -1.806836262      2.489927431      0.529225784
H       -0.309030693      3.083873569     -0.266301509
H       -1.418479861      1.964027600     -1.135500801
Cl       2.591747000      0.489781723      0.126797539
Cl      -2.591747000     -0.506506277     -0.064080461
X        0.006271708      0.013588700     -0.012543416


Are the reaction rates of these equations equal? $$\ce{CH3CH2CH2Br + OH- -> CH3CH2CH2OH + Br-}$$

$$\mathrm{rate} = k[\ce{CH3CH2CH2Br}][\ce{OH-}]$$

If I change the $\ce{Br}$ with any element from halogens (without changing concentration, volume, mass etc.), does the rate of reaction change? If so, how does it change? What factors change it?

For the reaction kinetic rate constants, there is the well known Arrhenius equation:

$$k=A \cdot \exp{\left(-\frac{E_\mathrm{a}}{RT}\right)}$$

$A$ is sometimes called frequency factor, interpreted as the rate of collisions with the proper orientation of molecules.

It has 2 terms:

The rate of general collisions, that is function of temperature and molecular masses, which determine speed of molecular motion (close relation to the kinetic theory of gases). Note that the temperature dependence is much smaller than for the exponential Boltzmann term.
The probability of the proper orientation of molecules, what depends on the molecular geometry. For molecules of  otherwise the same geometry, it depends on covalent atom radii.

The exponential term follows the Boltzmann statistical distribution, determining the probability molecules would have enough kinetic energy to overcome the reaction activation energy barrier.

All 3 terms (2 for $A$ and the exponential term) Are different for different halogen atoms.

The mass molecules increases fluorine < iodine, so collision frequency as the rate constant term is the lowest for iodine.
As the reaction mechanism I suppose SN2. The geometrical aspects of $\ce{-CH2X}$ is tricky to determine from basic principles. Bigger halogen is more sterically blocking, but is also farther from the central carbon. More polar bond of the smaller halogen should cause stronger  repelling of the other 3 bonds, so the other side is more open for sn2 reaction.
Activation energy would decrease ( and the Boltzmann term for the rate constant increase) in order F ... I

That about the principles. To compare the particular rates, it is matter of experimental data.


Is carbon tetrachloride an electrolyte? Is $\ce{CCl_4}$ an electrolyte all by itself, or does something need to be added to it so that it becomes an electrolyte?

Also, is it weak or strong?

No, $\ce{CCl4}$ is not an electrolye.  An electrolye is a material that causes ions (charged entities) to form in the solvent.  Carbon tetrachloride neither dissociates into ions nor induces ion formation in the solvent.

The reason is that the $\ce{C-Cl}$ bond is rather strong and won't break under
normal solution conditions.  Nor does the molecule have any net dipole moment.


What is the common denominator of all the lipids? I've read the lipids article on Wikipedia, but I couldn't understand the answer for my question and I'll explain why.

In biochemistry we have monomers and polymers while the monomer is the basic unit molecule for the polymere. In protein for example, we have amino acid that this is the basic unit of the proteins. Amino acid is characterized as structure of 4 groups: Alpha carbon, amine group, carboxylic group, hydrogen and R chain. this is the basic of all the amino acids.

But when we are talking about lipid as a main family of molecules (like what we have we have with protein) then I can not understand what is the "amino acids" (parallel) of them, i.e. what is the basic unit of the lipids and what is the structure.

Well, I could understand that the fatty acids are the analogous for the amino acids, but the problem is that fatty acids according the article (and other schemes that I saw on google pictures), they are considered as subgroup of the lipids, and that says that not all of the lipids are made of them. So I don't have clue for the answer to my question.

As you note, most biomolecules — or more precisely primary metabolites — exist as polymers that can theoretically be infinitely long and consist of well-defined monomers. This is true for nucleic acids and the building block nucleotides, for proteins made up of amino acids and for carbohydrates made up of monosaccharides.

Lipids, on the other hand, do not have that kind of polymeric possibility. In essence, they are trimers or monomers by definition, depending on how you look at them. Furthermore, all lipids contain the same (with same I mean identical) connective molecule namely glycerin or propan-1,2,3-triol $\ce{CH2OH-CHOH-CH2OH}$. Glycerin has three ‘sockets’, if you wish, namely its three hydroxy groups.

The ‘plugs’ that connect to glycerin always contain an acid functionality: typically this is either a carboxy group (fatty acids) or a phosphate (e.g. phophatidyl choline). The bond between ‘plug’ and ‘socket’ is an ester bond.

Both the fatty acids and the substituted phosphates can be structurally diverse. If you wanted, you could label the different fatty acids ‘monomers’ and the overall lipid molecule a ‘trimer’. However, the most typical difference between fatty acid residues is simply their length ($\ce{C16}$, $\ce{C18}$ or $\ce{C20}$ — typically multiples of 2) and their degree of unsaturation (e.g. stearate, oleate, linoleate and linolenate have increasing degrees of unsaturation). Attempting to teach students more than a few exemplary fatty acid residues or phophate residues is rather pointless since the choice of fatty acid hardly has any effect on the properties of the overall molecule, and aside from generally knowing what constitutes cellular membranes the same can in principle be said for the phosphates. Hence, students are hardly ever bothered.


What are the main axis of research in Chemistry? I would like to know what are the main problems currently studied in Chemistry.

For some reason, it seems that there is far less vulgarisation in chemistry than the other fields, and it's hard to find an overview of the field accessible to a layman (compared to math or physics for instance).

I only found this wikipedia page, but I have no idea how relevant it is.

Even though I voted to close this question as not constructive, these are among the most important unsolved problems in chemistry:

Liquid Water - We still know so little about the molecular level interactions of water molecules with themselves and other species. The great extent of hydrogen bonding networks in liquid water make it hard to model. Just how many water molecules are solvating a proton in acid? Why are organic reactions between hydrophobic molecules accelerated in the presence of water?
Molecular origins of life, especially homochirality. The formose reaction is believed to be the origin of carbohydrates. Amino acids, lipids, and nucleic acids have similar hypothetical origins. However, how did these molecules come together to create living organisms? Why are most (not all) biomolecules present as one enantiomer and not the other?
Protein folding. Can we predict the three-dimensional shape of a protein simply from the order of the amino acids? If so, then we could design proteins to do whatever we wanted.
True designer proteins. Once we solve the folding problem, then we can design proteins with active sites of specific geometries and specific amino acids in those active sites to do whatever reaction we need.
High Temperature superconductors. Although room temperature superconductors would be nice, inexpensive superconductors that operate at liquid nitrogen temperature (77 K) instead of liquid helium (5 K) temperatures would make NMR, MRI, and a host of other important techniques much cheaper.
The Energy Crisis. We need a chemical solution to replace fossil fuels as the world's primary energy source. Hydrogen, fuel cells, biomass, and solar power all have serious drawbacks currently.
Replacement of petrochemical feedstocks. The petroleum industry remains the primary source of organic compounds. What will we do when the petroleum runs out?
Bonding theory. Modern bonding theories have developed into bloated mathematical formalisms that have removed any resemblance of intuition from the study of the structure of molecules. Can a unified modern bonding theory be developed that retains only the best features of valence bond and molecular orbital theories?
Improved ideality and efficiency of synthesis. Can the synthesis of compounds be improved so as to remove wasteful steps (like adding and removing protecting groups) or complex purifications to better adhere to the principles of Green Chemistry?
Structure-Property Relationship. All established relations between structure and activity is empirical, especially when the desired activity is biochemical. Can we design the best drug with the fewest side effects on the first try?

I find some of the unsolved problems on the to be too narrow (stability of norbonyl cation - this is a subset of the structure-activity problem) or trivial (origin of rotation barrier in ethane - the problem is important but the answer is useless).


Gabriel Phthalamide Reaction Can we use acid chloride in place of alkyl chloride in Gabriel Phthalamide Reaction?

Maybe it can be used as process to synthesize amides.

NO, this is not a viable way to make amides.

Phthamide anion will react with acyl halides, but the problem arises with the second step, the removal of the phthalimido group. The usual conditions described here of hydrolysis with hydroxide or reaction with hydrazine will result in removal of the acyl group to reform phthalimide plus the carboxylic acid, in the case of OH- hydrolysis, or the acyl hydrazide if using hydrazine. This is because the acyl group is the more electrophilic site for the nucleophile.


Why can't HS- participate in E1 reactions? In my textbook (Organic Chemistry by Klein), it says that certain molecules can only act as nucleophiles and not as bases; thus they can only undergo substitution reactions, not elimination reactions. This group of molecules includes: $\ce{Cl- , Br- , I- , HS- , H2S , RS- , RSH}$. I know that while E2 reactions require strong bases, E1 reactions do not. Even a base like $\ce{H2O}$, which is pretty weak, can undergo E1 reactions. Then why can't something like $\ce{HS-}$ undergo a E1 reaction? For example, if we have a tertiary alkyl halide and $\ce{HS-}$, I would expect a mix of $S_N1$ and E1 products. However, my book claims there are only $S_N1$ products. Why is this the case?

As has been pointed out in the comments, the use of the word only is a little misleading. What it should really say is that $E1$ is possible but $S_N1$ is so strongly favoured that to all intents and purposes the only product is the $S_N1$ product.

This can be explained in terms of the relative energies and sizes of the orbitals involved. $\ce{HS-}$ has a large, diffuse, high energy HOMO which can have much better overlap with the empty $2p$ orbital of the carbon than the small, low energy $\ce{C-H}~\sigma^*$. Therefore, the $S_N1$ reaction is strongly thermodynamically favoured and dominates.

If you compare this to something like $\ce{H2O}$, the HOMO is much more compact and lower in energy and so it has much better overlap with the $\ce{C-H}~\sigma^*$ and so $E1$ is comparatively much more favourable.