Does 1-bromo-4-chlorobenzene have a permanent dipole moment? My first thought was there would be a dipole moment as chlorine and bromine are unequal in electronegativity. I have since looked up the electronegativity of both chlorine and bromine through the Pauling scale. The difference between the electronegativity is 0.20 according to the table. For reference, the difference in electronegativity between carbon and hydrogen is 0.25 on the same table.

Since this is the case, I am very confused about the dipole moment of this molecule. Since the carbon-hydrogen bond is so negligible, why should this difference be significant?

To add to my confusion I found a source stating Dipole moment = 0.
And Wikipedia only states: Dipole moment = D.

However, my textbook questions lists it as having a dipole moment with the simple explanation that Br and Cl have differing electronegativities (question 1.61).

As others have stated before in the comments, the molecule itself has a dipole moment, but it is indeed very, very small, to the point where you would say it is negligible. To illustrate this, I have performed a few quick calculations, rounded to two decimal places.
Remember that the dipole is defined from negative to positive pole, i.e. $\displaystyle \ominus \xrightarrow{~\mathbf{q}~}\oplus$.
\begin{array}{lr}\hline
\text{Method}  & \mathbf{q}(\ce{Br-C4H4-Cl})~/~\mathrm{D}\\\hline
\text{BP86/def2-SVP}           &     0.07\\
\text{BP86/def2-TZVPP}         &     0.11\\
\text{B3LYP/def2-TZVPP}        &     0.09\\
\text{B3LYP+D3(BJ)/def2-TZVPP} &     0.09\\
\text{PBE0/def2-TZVPP}         &     0.10\\
\text{M11/def2-TZVPP}          &     0.10\\
\text{B2PLYP/def2-TZVPP}       &     0.09\\
\text{MP2/def2-TZVPP}          &     0.06\\
%\text{CCD//MP2/def2-TZVPP} &     ----\\
\hline\end{array}

In this case the dipole is directed towards the chlorine. Here are the NBO (natural bond orbital) charges for the CCSD(T)//MP2/def-TZVPP level of theory:

We can see that the chlorine (light green) is essentially neutral, while the bromine (dark green) is slightly positive polarised. From this we would assume the dipole moment to be directed towards the bromine. This is also what you would expect from the electronegativity differences.
Slightly different values are obtained from an QTAIM (quantum theory of atoms in molecules) analysis on the MP2/def2-TZVPP level of theory, where the volume is cut off at an electron density smaller than 0.001.

 Normalization factor of the integral of electron density is    1.000001
 The atomic charges after normalization and atomic volumes:
      1 (C )    Charge:    0.071729     Volume:    68.316 Bohr^3
      2 (C )    Charge:    0.078474     Volume:    79.386 Bohr^3
      3 (C )    Charge:    0.088508     Volume:    78.816 Bohr^3
      4 (C )    Charge:   -0.122884     Volume:    73.483 Bohr^3
      5 (C )    Charge:    0.088513     Volume:    78.816 Bohr^3
      6 (C )    Charge:    0.078474     Volume:    79.386 Bohr^3
      7 (H )    Charge:    0.026701     Volume:    46.501 Bohr^3
      8 (H )    Charge:    0.025890     Volume:    46.475 Bohr^3
      9 (Br)    Charge:   -0.097472     Volume:   260.461 Bohr^3
     10 (H )    Charge:    0.025886     Volume:    46.475 Bohr^3
     11 (H )    Charge:    0.026702     Volume:    46.501 Bohr^3
     12 (Cl)    Charge:   -0.290521     Volume:   214.364 Bohr^3

However, the calculated dipole moment points in the opposite direction. Reasons for this are similar to the case of carbon monoxide (see here). In short: it is basically because bromine is larger and more diffuse than chlorine. The σ-lone pairs, which are mainly responsible for the direction of the dipole moment, are further away from the atomic centres of the molecule for bromine compared to chlorine.
This causes a larger charge difference than the polarisation of the bonding structure.

Above graphic corresponds to the QTAIM analysis, the bold blue line represents the electron density contour of 0.001.


Were gold(II) complexes synthesized for the first time in 2017? Recently in the Google news feed, I discovered this news:

  Gold stabilized in very rare oxidation state +II

  Missing golden link found: Divalent gold complex isolated for the
  first time in a pure form; new fundamental insight also provides
  possible explanation of the mechanism of action of cytostatic
  gold(+III) porphyrins

  Date: August 8, 2017

  Source: Johannes Gutenberg Universitaet Mainz

  Summary: A team of chemists has been able to isolate and analyze gold
  in the very rare oxidation state +II. This provides the missing links
  in the homologous series of the coinage metal ions copper(+II),
  silver(+II), gold(+II), and in the 'relativistic' triad of
  platinum(+II), gold(+II), and mercury(+II).

And there are loads of similar news- Here, here and here.

Now, I started digging for more information and found out that gold(II) complexes were known previously and has been synthesized but they are unstable and prone to disproportionation.

  Coordination chemistry of gold(II) complexes (October 1999)

  The number of complexes with gold in a formal oxidation state of
  two have increased considerably and nowadays this oxidation state can
  be considered nearly as a common state in gold chemistry. This
  review covers the evolution of the knowledge of this oxidation state
  from the first dithiocarbamate derivatives $\ce{[Au2(S2CNR2)2X2]}$,
  stable only at very low temperature, to the room temperature stable
  dinuclear gold complexes $\ce{[Au2(L–L)2X2](L–L=CH2PPh2CH2, CH2PPh2S,
 C5H4PPh2-2; X=halogen}$). Although a dinuclear structure with a
  gold(II)–gold(II) bond supported by two equal bridging ligands were
  the first and most common, this review shows the present diversity as
  mononuclear, $\ce{[Au([9]aneS3)2](BF4)2}$, dinuclear with two
  different bridging ligands, $\ce{[Au2(S2CNR2)(CH2PPh2CH2)X2]}$, with a
  non-supported metal–metal bond, $\ce{[Au2(Ph2PC8H6PPh2)2Cl2](PF6)2}$,
  and polynuclear gold(II) complexes either with chains of only gold
  centers or with other metals, such as
  $\ce{[{(C6F3H2)Au(CH2PPh2CH2)2Au}2Au2(CH2PPh2CH2)2]ClO4}$ or
  $\ce{[Au2Pt(CH2PPh2S)4X2]}$
  Disproportionation of Gold(II) Complexes. A Density Functional Study of Ligand and Solvent Effects (July 13 2006)

  For the disproportionation of $\ce{Au(II)L3}$ complexes with neutral
  ligands, disproportionation is highly endergonic in the gas phase.
  Calculations imply that for synthesis of a monometallic $\ce{Au(II)}$
  complex, a nonpolar solvent is preferred. With the exception of
  $\ce{[Au(CO)3]^2+}$, disproportionation of $\ce{Au(II)L3}$ complexes
  to $\ce{Au(I)L}$ and $\ce{Au(III)L3}$ is exergonic in solution phase
  for the ligands investigated.
  Structural Studies of Gold(I, II, and III) Compounds with Pentafluorophenyl and Tetrahydrothiophene Ligands(February 28
  2007)

  A golden opportunity was seized to prepare an unbridged, dinucleur
  gold(II) compound, $\ce{[Au2(C6F5)4(tht)2]}$
  (tht=tetrahydrothiophene), without any stabilizing chelating ligands
  (see picture; Au orange, S yellow, F green, C black) and to
  crystallographically characterize the gold(I) and gold(III) complexes
  participating in this conversion. A unique ligand scrambling of the
  latter gold(III) compound occurs in solution.

Gold(II) complexes has been intensly studied previously. They have been trying to synthesize a stable complex but failed miserably. The stable ones are only present in low temperature and only one was found to be stable in room temperature. So, although gold(II) complexes were previously discovered and (synthesized?), why scientists at University of Mainz and those who discovered it in 2017 claims to have synthesized gold(II) complexes for the first time?

You are correct in noting that there are a number of Au (II) small molecule species synthesized prior to the report of Professor Katja Heinze in 2017.
The heart of the Au (II) claim by Professor Katja Heinze and team as stated in the article you referenced, see reference 1 given below, has to do with the a) relative stability of the mono-nuclear gold-porphyrin complexes compared to other Au (II) systems and b) that the molecule does not have a Au-Au bond.  The latter point being the more important.
Au (II) is a very stable oxidation state for molecules containing Au-Au bonds.  We synthesized and characterized a number of such species years ago, see references 2 and 3 below for two papers describing the work.
For those working in the field, the nature of the claim by Heinze et al. is understandable.  For those outside the niche field of Au (II) chemistry I can see how the distinction is less than clear.
References
https://www.sciencedaily.com/releases/2017/08/170808145949.htm
https://pubs.rsc.org/en/content/articlelanding/1985/c3/c39850001278#!divAbstract
https://pubs.rsc.org/en/content/articlelanding/1986/C3/c39860000321#!divAbstract


The H NMR analysis of the hydrogens in salicylic acid I am currently working with an unknown and from the IR spectrum I was able to figure out that the compound I am working with is salicylic acid.

I am looking at the H NMR, but I am a bit confused on the deshielding, integration, and upsheilding of the hydrogens that are near the $\ce{OH}$ and $\ce{COOH}$ groups on the compound. Can someone explain this to me:

Attached is a picture

Since you have identified your unknown as salicylic acid (2-hydroxybenzoic acid), I'd help you understand how easily understand the substituents of aromatic nucleus induce shielding-deshielding effects. The following diagram illustrate the $\mathrm{^1H}$-$\mathrm{NMR}$ assignments of salicylic aromatic protons:

A lot of experts do not like explaining shielding-deshielding effects using electron densities around the sought atom. I agree with them, but a student of your capacity could easily related to that explanation. A good rule of thumb is o,p-directing electron donating groups (coursing positive mesomeric or resonance effect) usually shield the relevant protons at o,p-positions (inducing $\delta^-$-charge at the positions). Similarly, m-directing electron withdrawing groups (coursing negative mesomeric or resonance effect) usually deshield the relevant protons at o,p-positions (inducing $\delta^+$-charge at the positions). Accordingly, $\ce{OH}$-group shields $\ce{H_a}$ and $\ce{H_c}$ while $\ce{COOH}$-group deshields $\ce{H_b}$ and $\ce{H_d}$.

Thus, it is safe to say that the most up-field aromatic signals (sheilded) at $\delta \ 6.97$ (apparent $triplet$) and  $\delta \ 7.03$ (apparent $doublet$) are due to $\ce{H_a}$ and $\ce{H_c}$. The assignments of $\delta \ 6.97$ is belong to $\ce{H_c}$ (two vicinal protons) and  $\delta \ 7.03$ is belong to $\ce{H_a}$ (one vicinal proton) are solely due to the number of neighboring (vicinal) protons to each. Keep in mind that chemical shifts of these two signals can be moved to up- or down-field positions depending on the using locking solvent. For example, the locking solvent used to get present $\mathrm{^1H}$-$\mathrm{NMR}$ spectrum here is apparently $\ce{CDCl3}$ (based on solvent residue at $\delta \ 7.28$). However, two signals have been moved to more up-field positions (compared to that in $\ce{CDCl3}$) when the locking solvent is changed to $d_6$-$\ce{DMSO}$ (Ref.1). In that spectrum, two signals at $\delta \ 6.92$ ($t,\: ^3\!\!J=\pu{7.4 Hz}$) and  $\delta \ 6.96$ ($d,\: ^3\!\!J=\pu{7.5 Hz}$) are assigned for $\ce{H_c}$ and $\ce{H_a}$, respectively.

Similarly, the most down-field aromatic signals (desheilded) at $\delta \ 7.55$ (apparent $triplet \ of \ doublet$) and  $\delta \ 7.96$ (apparent $doublet \ of \ doublet$) are due to $\ce{H_b}$ and $\ce{H_d}$ (Note: The splitting of each signal with a small coupling constant is due to meta-or $\omega$-coupling generated by the presence of proton at meta to each; This phenomenon is generally called $^4\!\!J$-coupling). The assignments of $\delta \ 7.55$ is belong to $\ce{H_b}$ (two vicinal protons) and  $\delta \ 7.96$ is belong to $\ce{H_d}$ (one vicinal proton) are also due to the number of neighboring (vicinal) protons to each. Similar to $\ce{H_a}$ and $\ce{H_c}$, these two signals have also been moved to more some what up-field positions (compared to that in $\ce{CDCl3}$) when the locking solvent is changed to $d_6$-$\ce{DMSO}$ (Ref.1). In that spectrum, two signals at $\delta \ 7.52$ ($t,\: ^3\!\!J=\pu{7.7 Hz}$) and  $\delta \ 7.82$ ($d,\: ^3\!\!J=\pu{7.7 Hz}$) are assigned for $\ce{H_b}$ and $\ce{H_d}$, respectively.

References:

Milena Jadrijević-Mladar Takač, Dražen Vikić Topić, “FT-IR and NMR spectroscopic studies of salicylic acid derivatives. II. Comparison of 2-hydroxy- and 2,4- and 2,5-dihydroxy derivatives,” Acta Pharmaceutica 2004, 54(3), 177-191 (PDF).


Maximum metal-metal distance for antiferromagnetic coupling Imagine a transition metal complex with multiple open-shell transition metal cations. If the metals are relatively close in distance, it is possible that antiferromagnetic coupling will occur between the metals. However, at approximately what distance does ferromagnetic vs. antiferromagnetic alignment of spins become isoenergetic due to negligible coupling between the metals? To provide some context, I'm asking about this in the context of density functional theory calculations. To make sure we're on the same page, let's call two magnetic configurations "isoenergetic" if the energies are, say, within ~1 kJ/mol or so.

The annoying answer is that this will be system dependent. I'll try my best to explain with a quick recap of magnetism.

Unpaired electrons produce a magnetic moment
One of the factors that determines the strength of the magnetic moment is the number of unpaired electrons.

If you have a specific system in mind, I would suggest performing a series of single-point (no ionic relaxations) calculations where you increase the distance between the ions. Contrast the energetic between FM and AFM configurations. If you have a license, VASP can provide information on the magnetic moments.


Calculating the dissociation constant of one acid in a mixture A solution contains $\pu{0.09 M}~\ce{HCl}, \pu{0.09 M}~\ce{CHCl2COOH}$ and $\pu{0.1 M}~\ce{CH3COOH}$. The $\mathrm{pH}$ of the solution is $1$. Given that the $K_\mathrm{a}$ of acetic acid is $10^{-5}$, calculate the $K_\mathrm{a}$ of $\ce{CHCl2COOH}$.

My attempt is as follows. We have $[\ce{H+}] \text{ from } \ce{HCl} = [\ce{HCl}] = \pu{0.09 M}$.

Then, for the other two acids:

$$\begin{align}
[\ce{H+}] \text{ from } \ce{CH3COOH} &= x \\
\frac {10^{-2}\cdot x^{2}}{10^{-1}-x} &= K_\mathrm{a} \\
[\ce{H+}] \text{ from } \ce{CHCl2COOH} &= y \\
\mathrm{pH_\text{total}} &= 1 \\
-\log_{10}(0.09+x+y) &= 1 \\
0.09+x+y &= 10^{-1} \\
\end{align}$$

But the answer key says the answer is $\pu{1.25*10^{-2}}$.

I just wanted to add an answer based on "no" assumptions, except (and that is occasionally a big assumption) that concentration equals activity. At first it does not seem very practical, as of its size, but I couldn't stop me from adding it here as I like to see the whole picture. In the end, it is actually not that hard.

The exact equation for the proton concentration is:

$$[\ce{H+}] = [\ce{OH-}] + [\ce{Cl-}] + [\ce{CHCl2COO-}] + [\ce{CH3COO-}]$$

which is the sum of the hydroxide ions from the autodissociation of water and the sum of the resulting acid residue ions from the three acids. As is shown here [1], this system expands to the following (and can be reduced again, if appropriate simplifications are taken into account for each acid, as is shown here [2]):

$$
[\ce{H+}] = [\ce{OH-}]
+ \frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}
+ \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}}
+ \frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}
$$

where $K_{\mathrm a1}$, $K_{\mathrm a2}$, and $K_{\mathrm a3}$ are the $K_{\mathrm a}$'s of $\ce{HCl}$, $\ce{CHCl2COOH}$, and $\ce{CH3COOH}$ respectively.

Now what you need to find is $K_{\mathrm a2}$. Solving it at this stage for $K_{\mathrm a2}$ is probably not a good idea aka nothing nice to handle, as the system is quite large. But as you know all values except $K_{\mathrm a2}$, it's actually not that bad.

$$
\underbrace{[\ce{H+}]}_\pu{0.1 mol/l} = \underbrace{[\ce{OH-}]}_\pu{10^{-13} mol/l}
+ \underbrace{\frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}}_\pu{0.09 mol/l}
+ \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}}
+ \underbrace{\frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}}_\pu{10^{-5} mol/l}
$$

$\ldots$ I'll leave units from here on (everything is in $\pu{mol/l}$). This simplifies quite much into:
$$\begin{align}
0.1 &= 10^{-13} + 0.09 + \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}} + 10^{-5}\\
0.1 - 10^{-13} - 0.09 - 10^{-5} &= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\
0.01^* &= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\
0.01\,(0.1 + K_{\mathrm a2}) &= 0.09\,K_{\mathrm a2}\\
0.001 + 0.01 K_{\mathrm a2} &= 0.09 K_{\mathrm a2}\\
0.001 &= 0.09 K_{\mathrm a2} - 0.01 K_{\mathrm a2}\\
0.001 &= (0.09 - 0.01)\,K_{\mathrm a2}\\
0.001 &= 0.08\,K_{\mathrm a2}\\
K_{\mathrm a2} &= \frac{0.001}{0.08}\\
&= 0.0125
\end{align}$$

* I rounded to $0.01$; it was actually $0.00999\ldots$ which would lead to $\pu{0.0124861... mol/l}$

[1] Equations 2.1 and 2.2
[2] Assumption 1 for HCl, Assumption 3 for acetic acid


Can a protein complex be a machine? Discussing an article with friends we began discussing whether a protein complex that punches holes in cells (membrane attack complex) would be considered a machine.

Now, my friends are biochemistry students while I am a computer science student so this is far from my expertise. In my imagination this would be a mechanical action, hence the definition of a machine found on Wikipedia.
My friends counter argument is more of an emotional one, saying it feels wrong to call it a machine but also couldn't come up with a better word.

So:
Is it a machine? Is it something else? Or is a complex just a complex?
Does chemistry/biology have its very own definitions for things that do stuff in coordination without being "alive"?

According to the Oxford dictionary, a machine is:

  An apparatus using mechanical power and having several parts, each with a definite function and together performing a particular task.

Based on that definition, I think some proteins and enzymes could definitely be considered machines. A good example for that, in my opinion, would be ATP Synthase and all of the proteins that participate in the oxidative phosphorilation process.


What are the products formed when CH3CH2OH reacts with NaH? When $\ce{CH3CH2OH}$ reacts with $\ce{NaH}$, why are the products $\ce{CH3CH2ONa + H2}$? Why not $\ce{CH3CH3 + NaOH}$? Can you please explain in detail?

Consider firstly the ease of reaction: the hydrogen in $\ce{NaH}$ is basic $\ce{H-}$ and the hydrogen in ethanol is acidic $\ce{H+}$. The easiest reaction then, is a simple acid-base neutralization.

The reaction products you propose requires that a $\ce{C-O}$ bond is broken, or a situation wherein which the hydride attacks a carbon center which then ejects a hydroxide ion $\ce{OH-}$. In reality, however, you would have to eject $\ce{O^{2-}}$ which is much harder to remove.


Redox and acid-base reactions in planetary atmospheres I'm a chemistry layperson looking for chemistry expertise regarding atmosphere composition.

Earth's atmosphere is oxidizing, but it was somewhat reducing in the distant past before photosynthetic life developed. Venus has an acidic atmosphere primarily composed of carbon dioxide which includes clouds of sulfuric acid. Some planetary bodies have reducing atmospheres containing hydrogen sulfide or carbon monoxide.

How do redox and acid-base reactions interact in the context of planetary atmosphere compositions? Specifically, what combinations of reducing/oxidizing and acidic/basic atmospheres are possible? Are there other planetary atmosphere types that should be considered when categorizing them?

Quick terminology question: Does it even make sense to refer to an atmosphere as acidic or basic?

We most certainly can call atmospheres acidic or basic.  Put some quicklime, or even magnesia, in our air and you see that it undergoes an acid-base reaction with acidic materials (water vapor, carbon dioxide) in our atmosphere.  Over time, our weakly acidic atmosphere gradually dissolves in water and then reacts with alkali and alkaline earth metal silicates too, turning rock into clay.

Our atmosphere became acidic as it picked up oxygen, which is indeed an "acid former" through the volatile (thus atmospheric) compounds it forms with several nonmetals.  In contrast, the gas (or at least, clouds therein) on the giant planets contain some ammonia and would appear basic if we brought typical Earth chemicals there.


Volatile alkali metal compounds Most elements are or have compounds that are gases or have a significant vapor pressure at room temperature. Fluoride is a powerful tool for producing highly oxidized covalent and volatile compounds. Tungsten hexafluoride is gaseous at room temperature despite being made with the element with the highest melting and boiling point. Refractory elements can also have volatile oxides such as $\ce{CO2}.$

Organometallics such as iron pentacarbonyl and dimethyl zinc are liquids at room temperature but have a decent vapor pressure, boiling at 103 °C and 46 °C, respectively. Other examples of "volatilized elements" include diborane and heaver analogous of ammonia such as phosphine.

All of the compounds I found of alkali metals, however, have very low vapor pressures at room temperature. For example, organolithium compounds tend to form lithium clusters or ionic crystals and decompose into lithium hydride when heated (which isn't volatile), rather then the covalent examples previously. Alkaline earth organometalics seem to suffer a similar problem with dimethylmagnesium forming a solid lattice.

Are there any known examples of an alkali metal compounds that has a reasonable vapor pressure, say 100 Pa at 300 K?

As Mithoron says in the comments, tert-butyllithium is one choice.  Like pure organolithium compounds generally, this is pyrophoric and reactive to many common atmospheric gases or vapors. An article by Schwindeman et al. [1] recommends using a nitrogen or argon atmosphere.

In the case of alkaline earth metals, magnesocene provides a relatively volatile magnesium compound, subliming at $\pu{100 °C}$.  This is also air- and moisture-sensitive and again, should be handled under nitrogen or argon.

References

Schwindeman, J. A.; Woltermann, C. J.; Letchford, R. J. Safe Handling of Organolithium Compounds in the Laboratory. Chemical Health and Safety 2002, 9 (3), 6–11. https://doi.org/10/ctj2sj. (PDF mirror: chemistry.ucla.edu)


How to determine whether paper was dipped in acetone? I was wondering if there was a solution of some sort that would cause a reaction to paper that was previously dipped into acetone much like in a counterfeit pen the iodine causes a reaction when written on standard paper .

If significant amounts of acetone are left (according to my old lab manual, the analysis works with a 50 mg sample), you could use the reaction with 2-nitrobenzaldehyde in the presence of sodium hydroxide, which generates indigo dye.

See also: Baeyer–Drewson indigo synthesis


NMR population difference from a pulse of known flip angle I want to know how a radiofrequency pulse of known flip angle $\beta$ changes the population difference of a set of isolated spin 1/2 nuclei.

I know that the equilibrium magnetization is related to the population difference by:

  $M_0= \frac{1}{2}\gamma\hbar\Delta{n_{eq}}$

I would like to know how to calculate the x, y, z components of the magnetization after the pulse, and then calculate the population difference from these values.

immediately after the pulse:

$M_z = M_0cos(\beta )$

$M_x = M_0sin(\beta )$

$M_y = 0$

so for example after a 90 degree pulse, all the magnetization has been rotated into the x-y plane, and after a 180 pulse, the magnetization is in the -z direction.


How many delocalised electrons in gold? How can we determine how many delocalised electrons every atom of gold contributes to the 'sea of delocalised electrons'? More generally, how can we determine the number of electrons any metal contributes? Must we have thermodynamic data to determine this?

Briefly: valence orbitals on adjacent metal atoms interfere to form "bands" of closely spaced molecular orbitals that spread throughout the metal.

If you have N metal atoms with occupied s orbitals, they interfere to form an "s band" with N molecular orbitals. The lowest energy orbital is fully bonding; the orbital on the top of the band is fully antibonding. The spread of the band will be finite, and since N is a huge number, the molecular orbital energies will be very closely spaced, practically continuous.

If you have two valence s electrons in your metal, the s band will be full (each orbital can hold up to 2 electrons, you have 2N electrons, and N orbitals in the band). If you have only one valence s electron, the s band will be half full (the first N/2 orbitals will be filled, and the remaining orbitals will be empty).

Above the s band you'll have a similar p band. In gold, the p band is unoccupied, while the s band is half full (gold has 6s$^1$). However, IF the bottom of the p band overlaps with the top of the s band, the s electrons can easily move into the bonding MO's at the bottom of the p band. They can also move into any vacant orbitals in the s band that are just above the highest occupied orbital.  In either case, the s electron will be delocalized.

To figure out whether the valence electrons will delocalize, then, you'll have to decide what bands are present, where the highest occupied molecular orbital is, and what the width and the spacing of the bands are (does the unoccupied band overlap with the occupied band, or is there a gap between the two?)


Synthesis using Cationic Hetero-Cope Rearrangement By searching for the reference of the synthesis below I arrived at an example of Cationic Hetero-Cope Rearrangement, but I can't access the paper. How does it work here, and what happens in the first step of the synthesis (the addition of iBuOH and TsOH at 80 degrees)?


How does volume contraction in solvent mixing work? I recall seeing chemistry demonstrations where a significant amount of a solute is dissolved in a solvent and the solution volume barely changes. Questions regarding this:

1) How is this possible?

2) What solvent/solute combinations best demonstrate this effect?

3) Is it possible to wind up with less solution, in other words Is it possible to have Volume(solution) < Volume(solvent) + Volume(solute) ?

Yes this happens fairly regularly. It is a deviation from ideal behavior associated with the property of partial molar volume.  For an ideal solution, the total volume always equals the summed volumes of the pure components, but ideal behavior neglects several important facts about real molecules, to wit, that they may have different shapes, and may also experience stabilizing or destabilizing interactions with each other.

It's fairly easy to see how molecular shapes and sizes could affect this property. Macroscopic objects experience similar effects.  Suppose I have a bucket "full" of golf balls and a bucket full of sand, if I work carefully, I can pour quite a lot of sand into the bucket of golf balls because the smaller size of the sand grains allows them to fit into the gaps between the golf balls.  Another example: Suppose you have a 10 liter bucket of golf balls and a 10 liter bucket of tiny cubes packed tightly together. If I then try to pour those two buckets into a 20 liter container simultaneously, it almost certainly will not be able to fit all of the balls and cubes, because they will not pack as efficiently in the 20 L container, costing some space.  These are extreme examples, but illustrate one way that different shapes can play a role.

Differences in intermolecular interactions can play an even bigger role.  If you are dissolving salt in water, the individual ions making up the salt crystal can be solvated very efficiently by the polar water molecules, which experience a strong attractive interaction with the ions.  (Recall that the O and H-atoms in water have partial negative and positive charges, respectively.)  This results in the solution having a significantly smaller volume than expected based on the summed volumes of the pure salt and water used to make the solution.

Partial molar volume is not too hard to measure experimentally. For something like aqueous sodium chloride it can be determined in a fairly straightforward way from precise measurements of the densities of a series of solutions with increasing concentrations.  However it is worth noting that the volume and mass measurements used for the calculation must be very precise, yielding measured densities with 5 or more significant digits.  This requires careful use of volumetric glassware and precision balances, and the temperature of the solutions must be kept constant throughout.

For purposes of demonstrating the qualitative effect, you could just use water and ethanol (or isopropyl alcohol), which also have a significant negative excess volume. (see the first graph on the wikipedia page for partial molar volume for details).

For comparison you could use mixtures of benzene and toluene, which behave almost ideally with respect to volume effects.


What happens to the structure of tetracyclines when they expire? I need to explain (chemically) why expired tetracyclines should not be used, relating to the changes in tetracycline structure that occur.

All I can find is that they cause nephrosis, but not why.

Apparently the culprits are anhydro-tetracycline (ATC) and 4-epianhydrotetracycline (EATC) - see Determination of tetracycline and its major degradation products by liquid chromatography with fluorescence detection. J. Pharm. and Biomed. Anal. 18 (1998), 839-845 for structural information and references to toxicity of those byproducts as known at the time of publication.

It is worth noting that renal tubular acidosis (RTA) - the condition that was reported in 1963 to have been precipitated by ingestion of expired tetracycline - is not presently considered as much of an issue as it was at the time (see this review). Finally, this piece in the New York Times from 2014 (entitled Do some drugs become dangerous after expiration?) has more information and references concerning tetracycline and RTA.


Fermentation in the presence of UV light Does the presence of UV light, or any radiation for that matter, affect the process of fermentation? I'm not sure, but I seem to recall somebody telling me before that yeast will not function properly, but I can't seem to find it anywhere. Thanks for any help.

Depends on the wavelength. Fermentation is basically metabolic activity by yeast and bacteria. UV-C (<280 nm) is pretty good at killing a lot of micro-organisms. Some species are more resistant than others, but it's safe to say that exposure to this kind of light will probably interfere with fermentation. Depending on the medium and how the light is applied (thickness of material, power, etc.), it may have more or less of an effect depending on how much light reaches the yeast or bacteria.


How do we predict the properties of the materials we make through chemical reactions? I find chemistry fascinating. I've taken a couple of classes and so am familiar with the rudiments of orbital theory, equilibria, acid/base chemistry, and a small bit of organic chemistry. So I can attempt to make predictions about which reactions might happen, but I'd like to know what the properties of the materials made are ahead of time. Is there a way to do this? I know that there are materials composed of the same elements in different structures that have vastly different properties (melting point, color, conductivity, etc.), so what can we actually do to predict these things (i.e. figure out the structure we'll get, and the properties these structures produce)?

Well, you've described one of the goals of computational chemistry. It turns out that this problem is harder than intuition might predict. I recently read an article on the state of quantum computing that happens to summarize the issue for chemists nicely.

  We do not have enough computing power to do this well.

Computational quantum chemical models have become more and more sophisticated to take advantage of increasing computing power, but there still is not enough computing power. Part of the issue is that at the center of what we are trying to do is a multibody problem, which means that any attempts to optimize the electronic state for one electron in a molecule perturbs the state for other electrons. Then we go and adjust those other electrons, which perturbs our first electrons. This is a certified Really Hard ProblemTM.

At the end of the day, we currently cannot determine exact solutions to multielectron wavefunctions. So approximations get made, and the approximations are getting better. In fact for smallish molecules, we are getting quite close to a variety of known experimental properties. However, for larger molecules, the algorithms still need trained and calibrated on real experimental data. As the molecules get bigger, or the materials being modeled become more complex, our ability to throw quantum chemical calculations at them diminishes. Modeling of proteins mostly uses molecular mechanics and statistical dynamics.


Does the Bohr model violate the uncertainty principle? In a book, it is stated that one reason for the failure of Bohr's theory was the fact that it violates the uncertainty principle. Is this fact true? How so?

The Heisenberg uncertainty principle states that

$$\Delta x \Delta p \geq \frac{\hbar}{2}$$

When you think about it, the whole of classical mechanics violates this principle. Think of a particle with mass $m$, a defined position $x_0$ at time $0$, a defined velocity $v_0$ at time $0$, and a constant acceleration $a$. The laws of kinematics state that at any time $t$, the position $x_t = x_0 + v_0t + \frac{1}{2}at^2$, and the momentum $p_t = m(v_0 + at)$, with zero uncertainty in both quantities, i.e. $\Delta x \Delta p = 0$. That works in most real-life cases because on a macroscopic level, quantum effects are negligible.

The Bohr model was derived via the classical assumption that the electron orbits the nucleus, much like how the Earth orbits the Sun, except that the origin of the centripetal force was electromagnetic instead of gravitational. It then went on to derive the energy levels of the electron in a hydrogen atom using the laws of uniform circular motion and electrostatics. This means that it is theoretically possible to know the position and momentum of the electron simultaneously with no uncertainty - a violation of the uncertainty principle.


Reaction of aluminium trichloride with water Since $\ce{Al^3+}$ is an acidic cation, shouldn't it react with water like this?
\begin{align}
  \ce{AlCl3 + 6H2O &<=> [Al(H2O)6]^3+ + 3Cl^-} \tag1\\
  \ce{[Al(H2O)6]^3+ +H2O &<=> [Al(H2O)5(OH)]^2+ + H3O^+} \tag2\\
  \ce{[Al(H2O)5(OH)]^2+ + H2O &<=> [Al(H2O)4(OH)2]^+ + H3O^+} \tag3\\
  \ce{[Al(H2O)4(OH)2]^+ +H2O &<=> [Al(H2O)3(OH)3](s) + H3O^+} \tag4\\
\end{align}

The 3rd step of the equilibrium rarely happens in water. But if we add a base, strong or not ($\ce{NaOH}$, or $\ce{Na2S}$) the equilibrium weighs to the right and $\ce{Al(OH)3}$ is formed.

But I've seen people write it as this.
$$ \ce{AlCl3 + 3H2O -> Al(OH)3 + 3HCl}$$
Is this correct? I need a certain answer.

Aluminum chloride is very soluble in water, according to wikipedia 44-49g/100ml water depending on the temperature, so you shouldn't expect any preciptation. The reaction is very aggresive and exothermic.
Assuming you dont have any buffers or bases in solution when you add $AlCl_3$, the reaction will lower the pH. At a pH lower than 4-ish, $Al^{3+}$ is the most stable cation. Thus, the reaction
$$\ce{AlCl3 + 3H2O -> Al(OH)3 + 3HCl}$$
is pretty accurate. Aluminum hydroxides are decreasingly stable at low pH, as you can see below. Since HCl is formed upon dissolution of $AlCl_3$ you can safely assume that $Al^{3+}$ is the only Al-species present.

!1


Why polar aprotic solvents don't surround negatively charged nucleophiles? Protic solvents can form a tight shell around a nucleophile. But what's stopping any polar solvent from doing so? Shouldn't a polar solvent surround negatively charged $\ce{F-}$ with its partially positive side?

Found a good explanation in Oxford University presentations.

It appears that partially positive side of aprotic solvents is hindered in the middle of the molecule. Only negative charge is on the surface. I've looked at DMSO, DMF, acetone, acetonitrile - all of them share this feature. So in case of a negatively charged ion - such solvents are not good at surrounding it.

It looks like elements that end up partially positive (except for $\ce{H}$) are those that can have 3+ bonds ($\ce{C}$, $\ce{S}$), so they are mostly surrounded by other atoms. It's only $\ce{H}$ who has 1 bond and therefore ends up at the surface of the molecule and can interact with negatively charged nucleophiles.


Calculating the electronegativity of an element from bond dissociation energies Calculate the electronegativity of fluorine from the following data:
$$\begin{align}
\text{BDE}(\ce{H-H}) &= \pu{104.2 kcal/mol} \\
\text{BDE}(\ce{F-F}) &= \pu{36.6 kcal/mol} \\
\text{BDE}(\ce{H-F}) &= \pu{134.6 kcal/mol} \\
\chi_\ce{H} &= 2.1
\end{align}$$

I thought that electronegativity should be related to effective nuclear charge, as it is a measure of how strongly atoms "pull" on bond pairs.
Is there a formula to calculate the electronegativity from bond dissociation energies, and what is the rationale behind it?

Considering the data you have been given, the equation you need is the one that defines the Pauling electronegativities:

$$\left|\chi_\ce{A} - \chi_\ce{B}\right|=\left(\text{eV}\right)^{-1/2}\sqrt{E_\mathrm d(\ce{A-B})-\frac{E_\mathrm d(\ce{A-A})+E_\mathrm d{\ce{(B-B)}}}{2}}$$

In the equation, $\left(\text{eV}\right)^{-1/2}$ is in the equation to negate the units of the bond dissociation energies. This means that you need to convert all your bond dissociation energies from $\pu{kcal/mol}$ into $\pu{eV}$. You can use the BDEs in units of $\pu{kcal/mol}$, but then the formula will require a constant in front to adjust for the difference between units.

$\pu{1 eV}$ is equal to $\pu{1.602 \times 10^-19 J}$, so you need to be able to convert from $\pu{kcal/mol}$ to $\pu{J/mol}$, then divide through by Avogadro's constant $N_\mathrm A = \pu{6.022 \times 10^19 mol-1}$. There is a handy website which can help with this.

$$\begin{align}
E_\mathrm d(\ce{H-H}) &= \pu{4.519 eV} \\
E_\mathrm d(\ce{F-F}) &= \pu{1.59 eV} \\
E_\mathrm d(\ce{H-F}) &= \pu{5.837 eV}
\end{align}$$

Plugging the information you have been given, we find that

$$\begin{align}
\left|2.1 - \chi_\ce{F}\right| &= \sqrt{5.837 - \frac{4.519+1.59}{2}} \\
&= 1.668
\end{align}$$

This gives you two possibilities for $\chi_\ce{F}$, $0.4$ or $3.8$, depending on whether the quantity $(2.1 - \chi_\ce{F})$ is positive or negative. This is where we need to use some knowledge of chemistry: fluorine is more electronegative than hydrogen, so we should have $\chi_\ce{F} > \chi_\ce{H}$, and consequently we take the solution $\chi_\ce{F} = 3.8$.

Why should bond energies be related to electronegativity? The idea that Pauling had when he devised this equation was that in a covalent bond between two of the same atom, there is no electronegativity difference and hence no ionic contribution to the bond strength. Hence, the "average covalent contribution" of each atom to a bond could be measured by the mean of the two bond strengths, i.e. $[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$.

However, when there is a difference in electronegativity, the bond strength has an ionic contribution. The larger the difference in electronegativity $\left|\chi_\ce{A} - \chi_\ce{B}\right|$, the larger the ionic contribution. This ionic contribution could be roughly measured by subtracting the covalent contribution, $[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$, from the $\ce{A-B}$ bond strength.


Double Bond or Aldehyde priority? Pent-3-enal. In the structure of pent-3-enal, the third carbon atom has aldehyde group or double bond?  Also is the priority same for all other functional groups?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), locants are placed
immediately in front of the part of the name to which they refer.

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are
placed at the front of their names.

Therefore, the locant ‘3’ in ‘pent-3-enal’ refers to the ending ‘ene’ (not to the suffix ‘al’).
The class name ‘aldehyde’ refers to compounds containing the group $\ce{{}-CH=O}$
attached to another carbon atom. In systematic substitutive nomenclature, the suffix ‘al’ is used for $\ce{{}-(C)HO}$ (see Subsection P-66.6.1). Thus, the aldehyde carbon atom is always a terminal carbon atom.
The characteristic group cited as suffix (here: ‘-al’) receives the lowest locant possible (here: ‘1’). However, terminal locants are omitted in preferred IUPAC names (PINs) for such aldehydes.

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

Therefore, the PIN for the example given in the question is ‘pent-3-enal’ (not ‘pent-3-en-1-al’).


How many electrons can each shell in an atom hold? I know that the first shell can hold up to 2, the second and third can hold up to 8, and the fourth 18. But what about after that?

$$\text{Electron Capacity for a shell} = 2n^2$$

where, n represents the Principal Quantum Number, a.k.a. energy level.

Keep in mind that an energy level need not be completely filled before electrons begin to fill the next level because beyond the fourth energy level, you would get into theoretical subshells.

See this.


Why is the halogenation of an alcohol a multistep process? Refer to the mechanism posted on this site, if you will: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch08/ch8-10-3.html

When the alcohol attacks the sulfur center, the pi electrons in the oxygen sulfur bond are taken by the oxygen. Then pyridine sucks up the proton. Afterwards the oxygen donates those pi electrons back and the chlorine anion leaves.

Why doesn't the chlorine anion just leave? What's up with the oxygen shuffling electrons around?

If the chlorine left straight away I.e after step 1 in the mechanism you linked to then you would have three charges on the structure - charges on two oxygens and a charge on sulfur. Doesn't seem too stable to me.

Instead what happens after step 1 is a proton transfer. Proton transfers are generally rapid. This proton transfer is also thermodynamically favored as it relieves the oxygen of its positive formal charge.


Counting Nodal Planes in cyclopropane The energy of molecule orbitals increases with more nodal planes. W1 (in the attached picture) has no nodal plane. I'd like to know how to draw the nodal planes in cyclopropane molecule orbitals but do not know how. I found no picture in the web with the nodal planes of cyclopropane.

How many nodal planes have W3, W4 and W5?

The assessment of nodal planes is not as trivial as it might seem at first, because they are usually not planes at all.

Orbitals can be explained in terms of symmetry operations. There are three different operations and each of them has a unique element (in cartesian coordinates):

Rotation: Axis
Mirror: Plane
Inversion: Point

When we talk about hydrogen type atomic orbitals, we usually refer to their cartesian form. It is already fairly obvious for the $\ce{2s}$, that there is no nodal plane, while there are certainly nodes.

The representations of $\ce{s}$ orbitals are always trivial with respect to the total symmetry of the external field (= Point group). That means any symmetry operation, which is given through the external field will be matched by the $\ce{s}$ orbital. Another term for this is totally symmetric.

The representation of $\ce{p}$ orbitals are always antisymmetric with respect to inversion in any given external field.

The representation of $\ce{d}$ orbitals is always antisymmetric to two mirror planes. That excludes of course the $\ce{d_{z^2}}$ orbital, which is not a cartesian, but a spherical representation. (Note: In cartesian space there are 6 $\ce{d}$ orbitals. As quantum numbers only allow 5 spherical orbitals, they have to be transformed. See for example at chemissian.com or Schlegel and Frisch.) Early basis sets for quantum chemical calculations used the cartesian representation, because it wa easier to code and calculate.

Due to the geometric arrangement of nuclei in a molecule, the external field is given through the point group of the molecule. Every molecular orbital can be expressed through symmetry operations of that point group.

Cyclopropane has $D_\mathrm{3h}$ symmetry. And every molecular orbital has to respect that. I have already written about orbitals these orbitals in Why does cyclopropane give bromine water test?
However, I would like to shed some more light on this, because I believe the picture you found is not quite correct.
The following picture shows the depicted orbitals from your source, as well as the LUMO.

Where W1 corresponds to HOMO-2, W2 and W3 correspond to both HOMO, W4 corresponds to LUMO+3, finally W5 and W6 correspond to LUMO+2.
The ordering of these orbitals is taken from a BP86/cc-pVTZ calculation of Gaussian09. A methodology that I believe gives fairly accurate results for molecular orbitals. (It is however noteworthy, that unoccupied orbitals are constructed from occupied orbitals. That means, that their physical meaning is limited.)

Let's have a closer look at the orbitals. The HOMO-2 is of $A_1'$ symmetry and therefore behaves like an $\ce{s}$ orbital. There is no antibonding nodal surface in this orbital. That means at any bonding axis of the ring structure the sign of the wavefunction does not change. There is however a nodal surface that does not affect bonding, as was already pointed out in some comments.

The HOMOs are of $E'$ symmetry and therefore behave like $\ce{p_x}$ and $\ce{p_y}$ orbitals. These orbitals have one antibonding nodal surface. In a result of symmetry restrictions these nodal surfaces are perpendicular to each other.

This is also true for the LUMO+2s. In addition to this there is another antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.

The LUMO again is of $A_1'$ symmetry and behaves like a $\ce{s}$ orbital. There is again no antibonding nodal surface with respect to the $\ce{C-C}$ bonds, but the main electron density would be outside of any bond and there is an antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.

The LUMO+3 is of $A_2'$ symmetry and does therefore not behave like any other atomic orbital. Here we find three antibonding nodal surfaces with respect to the $\ce{C-C}$ bonds.

The bonding situation in cyclopropane is very complicated and can be explained in many different ways. It is not as obvious, that the molecule is a $\sigma$ aromate, having in plane $\pi$ orbitals (HOMO). The explanation of its molecular orbitals were first composed in terms of a Walsh like behaviour on the basis of an extended Hückel calculation with a minimum basis.

When you disregard the $\ce{C-H}$ bonds and only allow three carbon $\ce{sp^2}$ and three carbon $\ce{p}$ orbitals, then you can only arrive at the boning picture that was depicted in your source (and many other publications), as you can only have six molecular orbitals formed from these. There are, however, many more atomic orbitals that mix into the final wavefunction. First of all it is necessary to state, that the used $\ce{sp^2}$ orbitals are not of the ideal $\frac13\ce{s}+\frac23\ce{p}$ composition (see more about that in the linked question and the answer of ron therein).

Now finally for the discussion about the correlation of nodal planes and increasing energy. It is a very common statement, that is missing it's most important limitation. It is only completely true for orbitals of the same symmetry. (This is also a necessary requirement.)

For example, the energy of the $\ce{s}$ orbital series increase with the main quantum number, as you add one more nodal surface per one increment in $n$: $\ce{1s->0; 2s->1; 3s->2;...}$. The same applies to the $\ce{p}$ orbital series: $\ce{2p->1; 3p->2; ...}$.

Now this statement became popular, when talking about aromatic systems. In the framework of Hückel's molecular orbital method, this must be true, since all regarded orbitals are of the same symmetry.

However, it is usually true, that an orbital with fewer nodal planes is more stable than another with more, but this is more a gut feeling, than actual, factual science.


Comparison between van der Waals radius and metallic radius I was reading about different types of atomic radius. It was written that if any atom can possess all the three types of radius, then van der Waals radius will be greater than metallic radius. However from the above diagram, it appears that the van der Waals radius and metallic radius are equal.

Then how can the van der Waals radius be greater than metallic radius?

Metallic radius is half of the distance between the nuclei of two atoms when they are in crystalline structure and bond is formed between positively charged atoms, that means they are tightly packed in metallic lattice.

And van der Waals radius is half of the internuclear distance of two non-bonded atoms of the same element in the closest possible distance.

Don't compare both the situations using above diagrams.  The radii are calculated in two different situations.

As in case of metallic bond, the atoms are bonded tightly so their shells can overlap, but in case of van der wal  forces the bond is weak. Therefore, in case of van der waal force, the internuclear distance will be higher as compared to metallic bond.

Via:
https://clay6.com/qa/94625/covalent-radius-of-cl-is-more-than-its-van-der-waals-radius-true-or-false-

https://socratic.org/questions/why-is-van-der-waal-radius-greater-than-covalent-radius-for-covalent-compounds


What is the meaning of signs in electrode potentials? Question

Suppose you have a cell set up between a copper metal/copper(II) ion electrode and a reference electrode.

Under standard conditions, the emf of this cell was −0.07 V. The standard electrode potential of the copper metal / copper(II) ion electrode is +0.34 V. Hence the standard electrode potential of the reference electrode is:

A: -0.41 V

B: -0.27 V

C: +0.27 V

D: +0.41 V

My attempt:

I was taught this equation: $\mathscr{E} _{cell} = \mathscr{E} _{a} - \mathscr{E} _{b}$

where:

$\mathscr{E} _{a}$ is the emf of the more positive electrode potential

$\mathscr{E} _{b}$ is the emf of the less positive electrode potential

But using this equation, none of the combinations of electrode potentials give -0.07V as the emf.

Thanks

Chapter 14 of Harris' Quantitative Chemical Analysis has a helpful section entitled, an intuitive way to think about cell potentials.  Here's the corresponding graphic:

In this case, the author is describing a Galvanic cell with cadmium as the anode and silver as the cathode.  The potentials of the two half-cells are known and marked on the line-graph.  Electrons always flow to the more positive electrode, which in this figure is towards the right.  The difference between the two half-cells is the cell potential.

In your case, you know the separation (-0.07 V) and one of two points (+0.34 V).  If we ignore the sign for the moment, we can use the line-graph to narrow the possibilities down to C or D, since each of these differ from +0.34 V by 0.07.  The question now is, which direction are the electrons flowing?  When determining the cell potential using this line graph, we subtract the right-most number from the left-most number in order to obtain a positive potential.  Since the cell potential in your case is negative, we must be doing the opposite.  Therefore, the reference electrode potential is +0.41 V (D).


Why don't labile protons such as -OH and -NH have a characteristic chemical shift? Is it simply because the H instantly disassociates so it is a proton with zero shielding?

Many $\ce{-OH}$ and $\ce{-NH2}$ protons do have characteristic shifts.  However, their characteristic ranges tend to be much broader than those of $\ce{C-H}$ protons.

For example alcohol chemical shifts ($\delta$) range from roughly 1-5 ppm  while pheonlics range from 4-8ppm and carboxylic acid protons are from 10-13 ppm or so.  The ranges are 3-5 ppm in width, while similar ranges for $\ce{C-H}$ are 1-2ppm in width.  Amines and amides are somewhat in between these two in terms of width of the ranges.  So, saying they are not characteristic is an unhelpful description of the actual situation.

The next (implied) question is why would the ranges be broader for these groups?

Chemical shift is a measure of electron density near the proton being measured.  The magnetic moment of electrons 'shields' the proton from the external magnet, changing the amount of energy require for the quantum mechanical transition of nucleus aligned with field versus opposed.  Normally, we consider electron density to be a reflection of the bonding arrangements.  However, for  $\ce{-OH}$ and $\ce{-NH2}$ protons, hydrogen bonding is common.  As such, often times nearby functional groups or NMR solvent are also contributing electron density, increasing the variability of the observe chemical shift from case to case.

Finally, in some cases the hydrogen bonding is dynamically changing slowly compared to the NMR time scale.  In this cases you will observe very broad peaks.  For example, water in $\ce{DMSO-d6}$ (a high-viscosity solvent) occurs around 3.3ppm.  In very old bottles of the solvent (with lots of atmospherically absorbed water), I have observed this peak as much as 1.5 ppm wide baseline to baseline, always with a peak at 3.3ppm.  However, in $\ce{CDCl3}$ water tends to have a pretty sharp peak at 1.5ppm (presumably because of limited water solubility).


Why do spectra appear continuous despite their allowed transitions being discrete? So in my physical chemistry class we have been looking into radiation, selection rules, etc.

Taking into account the discrete levels of energy an electron can have, I cannot understand why fluorescence and absorption spectra of compounds have continuous lines instead of discrete values.

If electrons in a system can only have a select number of energies, why is it that these graphs show a continuum of absorption and emission?

I also have this problem with blackbody radiation, since it is a continuous line for any temperature.

You are right that absorption and emission are quantized processes, but UV-visible peaks represent a broad range of wavelengths because there are multiple vibrational states of their representative ground and excited electronic states.

In the case on my diagram, you have six different quantized electronic transitions with slightly different ΔE values. In reality, there are more energy levels than that, plus the rotations and vibrations are constantly changing the orbital energy levels as a result of interactions with the matrix, so that helps fill in the peaks. The peak shape can actually be changed by altering the temperature. This gives rise to high-resolution UV-Visible spectrophotometric techniques, which have a few specific applications.


Resonance structure of cyclobutadiene? I studied cyclobutadiene as an anti-aromatic compound. But I am unable to draw the resonance structures for cyclobutadiene. Can anyone help me?

What is the driving force for the resonance to start?

Is it possible that a compound is having conjugate double bond system but not satisfying resonance structures?

Is a compound called as anti-aromatic if it is not undergoing resonance?

Antiaromaticity is a concept describing an explicitly fictional situation, no molecule is antiaromatic, because all molecules will react to avoid the situation.

So the answer to the implied question is that cyclobutadiene has very little resonance because the molecule is prevented from being antiaromatic (by the nature of the universe and its physics).

Elaboration:
To avoid this situation, butadiene is very reactive; it will undergo an intermolecular Diels-Alder reaction avoid this antiaromatic state.  When reaction is prevented something much more interesting happens.  When frozen in an argon matrix, x-ray crystallography of cyclobutadiene indicates the molecule is highly rectangular, not square.  For dienyl resonance, there is some changing of bond lengths.  In butadiene, the 2,3 bond shortens.  If this were to occur in cyclobutadiene, then there would be degeneracy in the orbitals leading to a Frost diagram that would describe an antiaromatic state.  So if reaction is prevented, the molecule distorts its own geometry.

The fictional resonance structures you would draw are double bonds on the top and bottom, and double bonds on the sides.  However, generally, one might assume there is no resonance for this molecule.


Why don't dicarboxylic acids show hydrogen bonding in spite of the fact that carboxylic acids exist usually as dimers in polar aprotic solvent? I have seen several examples:

The fact that maleic acid is more soluble in water than fumaric acid in spite of the fact that it must be able to show intramolecular hydrogen bonding.
trans-Cyclopentane-1,2-dicarboxylic acid is more stable than the cis form (cis form should be able to show hydrogen bonding).
o-Nitrobenzoic acid is more acidic than the para isomer.

I tried linking point 1 to acidic strength (maleic acid > fumaric acid because of hydrogen bonding in conjugate base) and point 3 to acidic strength (o-nitrophenol < p-nitrophenol)
What am I missing? Something related to the size of the ring?

I would like to object the statement dicarboxylic acids would not engage hydrogen bonding.  Terephthalic acids ($\ce{HO2C-C6H4-CO2H}$) are a prominent example since early attempts to formalize the description of intra- and intermolecular hydrogen bonding in the solid state. Consider the pairing of benzoïc acid and $p$-terephthalic acid, sharing the same motif of hydrogen donors and acceptors, described as $R_2^2(8)$ in Etter's notation:

(Etter, Acc. Chem. Res. 1990, 23, 120-126,  doi 10.1021/ar00172a005)
But there is no limit to $p$-substitution to find this motif again, e.g.:

(Holy et al., Collect. Czech. Chem. Commun. 2006, 71, 139-154, doi 10.1135/cccc20060139, researchgate.net)


Canonical Restricted Open-shell Hartree–Fock equation In Frank Jensen's Introduction to Computational Chemistry ($\mathrm{2^{nd}}$ edition), on page 99, it says:

  For an ROHF wave function, it is not possible to choose a unitary
  transformation that makes the matrix of Lagrange multipliers in eq.
  (3.41) diagonal, and orbital energies from an ROHF (Restricted
  Open-shell Hartree–Fock) wave function are consequently not uniquely
  defined and cannot be equated to ionization potentials by a
  Koopmans-type argument.

Here ROHF eq. (3.41) is the non canonical Hartree-Fock equation:

$$F_i\phi_i=\sum\lambda_{ij}\phi_j$$

But in ATTILA SZABO, NEIL S. OSTLUND's Modern Quantum Chemestry, they prove that the Hartree-Fock equation can always be put into the canonical form:

$$F_i\phi_i=\lambda_{i}\phi_i$$

What am I misunderstanding here?

The treatment found in Szabo-Ostlund, although perhaps it is not explicitly stated, is appropriate only for non-degenerate spin-singlet electronic states, which in the Hartree-Fock method correspond to closed-shell systems.
For open-shell systems one may use the restricted open-shell Hartree-Fock method (ROHF), which is somewhat more complicated. In particular, in the ROHF method six arbitrary constants have to be introduced. Many choices for these parameters have been suggested.
The total ROHF energy, as well as the overall ROHF wave function and therefore the expectation value of any operator, are independent of these six parameters, so that all these ROHF sub-variants are largely equivalent.
However, the six parameters affect the Hartree-Fock orbitals and their orbital energies. Also, approximate post-HF methods such as MP2 or CISD will depend on the particular choice of those six parameters.
Knowles et al suggested in 1991 a particular choice leading to called "semi-canonical orbitals" whose orbital energies do obey Koopmans theorem and which are also a good choice for MP2 calculations.
I suggest the following paper for an introduction to the problem (the paper by Knowles metioned above is ref. 18):
T. Tsuchimochi and G. E. Scuseria, ROHF Theory Made Simple, J. Chem. Phys. 133, 141102 (2010) https://arxiv.org/abs/1008.1607
The paper above shows that one can obtain semi-canonical ROHF orbitals by a minor modification of the unrestricted Hartree-Fock method and that this approach is numerically very stable.


le Chatelier's Principle I am really confused about how reducing partial pressure of a gas by a certain percentage would impact the behavior of an equilibrium system.

e.g $\ce{POCl3_{(g)} <=> POCl_{(g)} + Cl2_{(g)}}$ is at equilibrium.  How would this system behave if the partial pressure of $\ce{Cl2}$ is reduced by 75%?

I understand the partial pressure is related to moles but having a hard time applying it here.  My guess is that the reactant would be consumed to produce more of the $\ce{Cl2}$ since its pressure and thus moles are decreasing.  However, my instructor said the system would react by consuming $\ce{POCl}$, the reason for which was not explained too well.

If I could get an explanation, it would be much appreciated.

Partial pressure can be imagined to be the pressure of a gas in the absence of other gases.

Le Chatelier's principle states:

  When a system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

Hence when the pressure of $\ce{Cl2}$ is reduced by 75% in the $\ce{POCl3_{(g)} <=> POCl_{(g)} + Cl2_{(g)}}$ system, the system naturally acts to counteract the change: by increasing the pressure of $\ce{Cl2}$.

Using the ideal gas law, $pv = nRT$ so pressure is directly proportionate to the number of moles of $\ce{Cl2}$ present. The position of equilibrium therefore shifts to the right, favouring the forward reaction that produces more $\ce{Cl2}$ (though not as much as to increase partial pressure of $\ce{Cl2}$ to its former value). This would mean it also consumes more $\ce{POCl3}$.

As for more $\ce{POCl}$ being consumed, if all other factors remain constant I'm not sure why this would happen.


Can I determine the rate law order from stoichiometric coefficients? I've read in two books, and seen in three separate videos courses, that you cannot determine the rate law order from stoichiometric coefficients.

Then, I have seen homework and "practice" problems that ask you to do exactly that.

Needless to say, I am confused.

This is how I am presently making sense of the issue: Given a balanced equation, and absent any other information, you can (and must) use the stoichiometric coefficients to determine the rate law. So, for example, given the equation $2A+B \rightarrow C + 3D$, the rate law would be $rate = k[A]^2[B]$.

But, this would only hold true if this were an "elementary reaction" (i.e., no equilibrium).

So, I wonder why the books and videos emphasize that you can only determine order by experiment and not from the stoichiometric coefficients. I am guessing that they do so because, in practice, you do not know if the reaction is an elementary reaction or not and therefore experiment is the only way to make that determination. And, because you don't really know if it is an elementary reaction, you can only determine the rate law experimentally.

The bottom line:

as a practitioner, determine rate law experimentally;
as a student, given no information other than the balanced equation, determine the rate law from the stoichiometric coefficients;
as a student, given rate and concentration data, determine the order (and rate law) from the data.

That is a pragmatic, but unsatisfying, solution to the problem but I would prefer a clearer explanation of what is going on.

There are multiple issues here:

You should not assume that there is a simple rate law. Chemistry is complicated, and even simple reactions may proceed through multiple pathways in parallel.
More generally, you should not assume that you are given enough information to determine the solution. Just because the stoichiometric coefficients are there, does not mean that you can use them to determine the answer. So, if you just have a balanced reaction, you know nothing. If you want to become a scientist, you should learn to be comfortable with not having the answer.

So, I would say that your bottom line is incorrect.

The experimental rate law should be determined experimentally.

Given a putative mechanism with elementary steps and knowledge of rate constants for those steps, you can propose a rate law that for that mechanism. Do not assume it has any bearing on reality, since your putative mechanism may be wrong.

Frequently, the approach is to take the experimental rate law and see if the derived rate law for a proposed mechanism is consistent with the experimental result.


Produce CuCl2 from CuSO4 I have a large amount of copper sulfate available and I am interested in obtaining copper chloride from it. What is the easiest way to do so?

I have considered mixing an aqueous solution of copper sulfate with sodium chloride, but I think that it would be difficult to separate the resulting sodium sulfate from the copper chloride. My other idea is to first react the copper sulfate with sodium carbonate and then react the resulting copper carbonate with hydrochloric acid, which would produce copper chloride (the other products would be $\ce{CO2}$ gas and water).  I am trying to find a cost effective method for performing this reaction.

You cannot obtain $\ce{CuCl2}$ mixing $\ce{CuSO4}$ and $\ce{NaCl}$ because both are very soluble, the second strategy seems good because $\ce{CuCO3}$ is insoluble in water, and you can react the new salt with $\ce{HCl}$.

Good luck.


Production process of kevlar How easy is it to produce kevlar?

What are/is the chemical recation/-s to produce kevlar, and which chemical compounds are required?

From this Wikipedia page:

  Kevlar is synthesized in solution from the monomers
  1,4-phenylene-diamine (para-phenylenediamine) and terephthaloyl
  chloride in a condensation reaction yielding hydrochloric acid as a
  byproduct. The result has liquid-crystalline behavior, and mechanical
  drawing orients the polymer chains in the fiber's direction.
  Hexamethylphosphoramide (HMPA) was the solvent initially used for the
  polymerization, but for safety reasons, DuPont replaced it by a
  solution of N-methyl-pyrrolidone and calcium chloride.

The structure of Kevlar is given from the same page:

However, Kevlar production on a small laboratory scale will likely be an expensive and difficult endeavor compared to purchasing Kevlar products, which are available in about any form you can imaging, many of them patented. Also, as Kevlar is water insoluble, concentrated sulfuric acid is required to keep it in solution during synthesis, molding, spinning, etc.  Good luck ;)


How to determine number of structural isomers? I have come across many questions where I'm asked to give the number of possible structural isomers. For example number, structural isomers of hexane is 5, while the number structural isomers of decane is 75.

How can I determine the possible number of structural isomers of a given organic compound?

It isn't easy but it is an interesting research topic

Determining the number of possible structures for a given range of chemical formulae isn't simple even for saturated hydrocarbons. The number of possible structural isomers rises rapidly with the number of carbons and soon exceeds your ability to enumerate or identify the options by hand. Wikipedia, for example, lists the numbers of isomers and stereoisomers for molecules with up to 120 carbons. But the counts are getting silly even at 10 carbons where there are 75 isomers and 136 stereoisomers.

It has been an interesting research topic in computational chemistry and mathematics. This old paper (pdf), for example, list some formulae for simple hydrocarbons among other simple series. Part of the interest arises because of the relationship to the mathematics of graph theory (it seems that chemistry has inspired some new ideas in this field of mathematics partially because enumerating possible isomers of hydrocarbons is strongly related to drawing certain simple trees which is intuitively obvious if you use the standard chemical convention of drawing just the carbon backbone and ignoring hydrogens).

You can look up the answers on the fascinating mathematics site OEIS (the online encyclopaedia of integer sequences). The sequence for simple hydrocarbons is here.

But the mathematical approach oversimplifies things from the point of view of real-world chemistry. Mathematical trees are idealised abstract objects that ignore real-world chemical constraints like the fact that atoms take up space in three dimensions. This means that some structures that can be drawn cannot exist in the real world because the atoms are too crowded and cannot physically exist without enough strain to cause them to fall apart.

Luckily, computation chemists have also studied this. There is, unfortunately, no obvious shortcut other than trying to create models of the possible structures and testing them to see if they are too strained to exist. The first two isomers that are too crowded are for 16 and 17 carbons and have these structures:

If you have any intuition of the space filling view of these, you should be able to see why they are problematic. A research group at Cambridge University has produced an applet to enumerate the physically possible isomers for a given number of carbons which is available here if your Java settings allow it. The results are discussed in a paper available in the Journal of Chemical Information and Modelling.


Turns in beta-alpha-beta loops (protein motifs) In the context of protein secondary structure, do the loops between strand and helix in a beta-alpha-beta motif form into the conformation known as a beta turn?

So a beta-alpha-beta structure will have two loops that could be turns - between the first beta strand and the alpha helix, and the helix and the second strand. They might be beta turns, but they might not.

Although there are different definitions of beta-turns, one common one is the presence of a hydrogen bond between the carbonyl of residue (i) and the amino of residue (i+3). I'm not actually sure how frequent this is, but you can see this from a search on PDBeMotif. There is also Motivated Proteins, which gives nice 2D (and 3D) images of motifs.

However, you could get other conformations of these loops that are not beta-turns, especially if they are quite long.


What is the standard potential of carbon? What's the value of standard potential of carbon?

The hydrogen and Carbon are in the electrochemical series despite being non metals hydrogen has zero potential difference what about carbon?

To look up the position of carbon in the electrochemical series, you would have to guess what species carbon is oxidised to. This is most likely carbon dioxide; you would then have to look up the value of $E^0$ for the $\ce{CO2}/\ce{C}$ couple in a textbook or some other data source. The standard reduction potential for the $\ce{CO2}/\ce{C}$ couple is given in Shriver & Atkins 5th ed as +0.207 V. Loosely speaking, this means that carbon is harder to oxidise than hydrogen, or in other words, carbon is below hydrogen in the electrochemical series (if you take Li to be at the top).

Just a note: The standard reduction potential for the $\ce{H+} / \ce{H2}$ half-cell is 0 V solely because we decided to define it as such.


1.0 g of AlCl3 is added to 100.0 ml of water. What would be the resulting pH? $1.0\ \mathrm g$ of $\ce{AlCl3}$ is added to $100.0\ \mathrm{ml}$ of water. What would be the resulting pH?

I think the answer may be $4.022$ but I'm not certain.

The reaction wasn't provided but I think its this:
$$\ce{Al(H2O6)6^3+ <=> H^+ + Al(H2O)5(OH)^2+}$$
Assuming all $\ce{Al}$ dissolved in water:
Initial $[\ce{Al}]=7.49\times10^{-4}\ \mathrm M$

$$x^2/(7.49\times10^{-4}-x)=1.4\times10^{-5}$$

$$x=9.56\times10^{-5}$$

$$\mathrm{pH}=-\log(9.56\times10^{-5})=4.022$$

Not being given any other information assume the reaction goes to completion and that it is:

$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)5(OH)^{2+} + H+ +3Cl-}$$

molecular mass $\ce{AlCl3 = \pu{133.34 g/mol}}$. So:

$$\ce{\pu{mol AlCl_{3}} = mol [Al(H2O)_5OH^{2+}] = mol [H+]} = \frac{\pu{1 g}}{\pu{133.34 g/mol}} = \pu{7.496\cdot 10^{-3} mol}$$

$$\pu{Molarity} = \dfrac{\pu{moles}}{\pu{volume}} = \dfrac{\pu{7.496\cdot 10^{-2} mol}}{\pu{0.100 L}} = \pu{0.07496 M}$$

pH = - log(0.07496) = 1.12

Now the truth is a bit more complicated. You can assume the reaction happens in two steps. The first step will go to completion.

$$\ce{AlCl3 + 6H2O ->[excess H2O] Al(H2O)6^{3+}  +3Cl-}$$

The second step won't go to completion since $\ce{Al(H2O)6^{3+}}$ is a weak acid. There will be an equilibrium favoring the left side. But with no value for $K_\mathrm{eq}$ you are stymied.

$$\ce{Al(H2O)6^{3+} ->[excess H2O] Al(H2O)5(OH)^{2+} + H+}$$

$$K_\mathrm{eq} =\dfrac{\ce{[H+][Al(H2O)5(OH)^{2+}]}}{\ce{[Al(H2O)6^{3+}]}}$$


How to determine the work function and electron velocities from data on the photoelectric effect? A metal received light with wavelength $300~\mathrm{nm}$ in experiment A and light with wavelength $500~\mathrm{nm}$ in experiment B. In one experiment the electron speed was twice the speed of that in the other experiment. Calculate the velocities and the work function ($W$) of the metal.

Because it's the same metal in both experiments we have $W_\mathrm{A} = W_\mathrm{B}$, which means that $E_\mathrm{A} - K_\mathrm{A} = E_\mathrm{B} - K_\mathrm{B}$.

We know that $E=hc/\lambda$ and $K=mv^2/2$. Because $\lambda_\mathrm{A} < \lambda_\mathrm{B}$, we have $E_\mathrm{A} > E_\mathrm{B}$ and $K_\mathrm{A} < K_\mathrm{B}$ which means $v_\mathrm{A} < v_\mathrm{B}$ and $v_\mathrm{B} = 2v_\mathrm{A}$.

Then I made the substituions in my initial equality but when I try to calculate $v_\mathrm{A}$ I obtained a negative number inside my root.

Consider the difference between the kinetic energy in case A and B:
$$K_A-K_B=(E_A-W)-(E_B-W)=E_A-E_B$$ where the work function cancels out because it is the same metal in each experiment.
The right hand side of this is easy to evaluate using $hc/\lambda$, which gives $$E_A-E_B=hc\left(\frac{1}{\lambda_A}-\frac{1}{\lambda_B}\right)=(\pu{1.9878\times10^{-25}J\cdot m})\cdot(\pu{1.333\times10^6m^{-1}})=\pu{2.65\times10^{-19}J}$$

To solve for the left hand side, we simplify to a single variable by recognizing that $v_a=2v_b$ (as Wildcat said, the right hand side is positive and the only way the left hand side will be is if A has the larger velocity): $$K_A-K_B=\frac{m}{2}(v_A^2-v_B^2)=\frac{m}{2}(4v_B^2-v_B^2)=\frac{3}{2}mv_B^2$$
With an electron mass of $\pu{9.109\times10^{-31}kg}$, this gives $v_B=\pu{4.404\times10^5 \frac{m}{s}}$ and $v_A=\pu{8.809\times10^5 \frac{m}{s}}$. The work function can then be solved by evaluating either $K_A$ or $K_B$ and solving for the difference $E_x-K_x=W$ where $x$ is either A or B.


Why is the curve of an isothermal process above the that of a adiabatic process during compression? For an adiabatic process $pV^{\gamma}=k$ and after differentiating the equation we get that the slope of an adiabatic process is $$\frac{\mathrm dp}{\mathrm dV}=-{\gamma}\frac{\mathrm p}V$$
For an isothermal process $pV=k$ and after differentiating the equation we get that the slope of an isothermal process is $$\frac{\mathrm dp}{\mathrm dV}=-\frac pV$$
It says in my book that during compression adiabatic curve is above the isothermal curve and that during expansion the vice versa happens. My question is that since the slope of the adiabatic process is $\gamma$ times more than the slope of the isothermal process, shouldn't the curve of the former process always be above the latter.

The two cases you refer to seem to be  for the reversible adiabatic and reversible isothermal change in an ideal gas.

In an adiabatic expansion as there is no heat loss or gain to or from the surroundings ($q=0$) the energy needed for the work of expansion against the external pressure must come from the internal energy of the gas. This results in a lowering of the temperature.

A curve of reversible adiabatic p vs. T change must therfore be below that for a reversible isothermal change. In addition as $\gamma \gt 1$, for example $5/3$ for an monoatomic gas, the adiabatic curve becomes even further below the isothermal one. The slope of the curves is not relevent in determining which is greater.


Why does the dielectric constant of a solid increase due to a Frenkel defect? In crystalline solids, an atom may be displaced from its original position in order to create a vacancy in the original location. This is referred to as a Frenkel defect. Why does this defect result in an increase of the dielectric constant of the solid?

When Frenkel defect alone is present, there is no decrease in density, however
the closeness of the charge brought about by Frenkel defect tends to increase the dielectric constant of the crystal.
Compounds having such defect conduct electricity to a small extent.


A Chemdrawing software? I would like to know about a Chemdrawing sotware for 2-d molecular structures. it should be mainly linux based (but if it supports other OS's then that would be great)

I'd really reccomend the free Marvin suite from ChemAxon.

MarvinSketch (the actual drawing package) hasn't disappointed me in the past and has a wide feature set.  In addition to just being able to draw compounds in Lewis format, the software includes plugins to name what you have drawn (systematic/traditional), predict properties, change atom and bond properties, generate stereoisomers and much more that I haven't got the experience to do anything more than experiment with!

You can view what you have drawn in Lewis form and also as a 3D structure should you wish to.  It supports many chemical drawing filetypes.

Here is the product page, including brochure and technical details.  It is Java based so works on Mac, Windows, Linux and is even available on the web (but is quite slow to generate properties etc. - I'd reccomend doing this locally).


As you heat up a substance will it's bonds begin to dissipate? As you heat something up does it slowly start to dissipate? Water, for instance, boils at 100 degrees Celsius, but if you are heating it up at 90 degrees will its bonds be weaker?

Remember the energy of the bonds never increase or decrease. They remain the same. When things are heated what actually happens is the molecules begin to move around (in solids they just oscillate while in liquids and gases the movement becomes much more prominent) and their kinetic energy increases and the molecules bump into each other harder and quite often which makes it increasingly difficult for the bonds to hold the molecules together (intermolecular forces). Once a certain threshold is reached the bonds become inefficient at their job and hence small lumps of molecules move apart enough for us to perceive the action as a phase change.


How much ionising radiation is released during an atomic blast? I've been looking for how much ionising radiation is produced during an atomic blast (as a percentage).

I have seen a claim of 5% of the energy in an atomic blast is ionising energy (Wikipedia), but if it's so little why do we put so much emphasis on radioactivity protection when discussing a nuclear blast. It just doesn't add up to me.

The amount of energy in an atomic blast releases as radiation (gamma rays, beta particles, alpha particles and neutrons) depends on how the bond is designed but isn't usually a big proportion of the total energy released. But that's not why we worry about the radiation from a bomb.

The reason why we worry about radiation is the radioactive products created in the blast (eg from  fission of the large nuclei in a fission bomb or from the potentially radioactive nuclei created when the initial radiation blast interacts with other materials). Typically many dangerous isotopes are generated that continue to release radiation for days or years. Worse the explosive blast scatters many of these over a large area.

But not all bombs are the same. Neutron bombs are designed to release far more of their energy as radiation (in this case a large blast of high energy neutrons) with the express purpose of killing people in the immediate area while having a much smaller blast. The original idea was to be able to use them on invading Russians in Western Europe without doing too much damage to the infrastructure. Neutron bombs can release ~40% of their energy as neutron radiation which is far more than the 5% released in a bomb tuned to create a large physical explosion.


Is the rate of an SN1 reaction really independent of the type of nucleophile? Suppose we take two reactions

$\ce{(CH3)3CCl + H_2O \rightarrow (CH3)3COH +HCl}$

$\ce{(CH3)3CCl + CH3OH \rightarrow (CH3)3COCH3 +HCl}$

I know in $\mathrm{S_N1}$ reactions the RDS (Rate Determining Step) is the formation of a carbocation intermediate.

But I am quite certain that methanol is a more bulky and weaker nucleophile, as compared to water. Should that make the rate of the second reaction lower than the first reaction (after the RDS is crossed)?

Nope, the rates wouldn't be affected, even if the nucleophiles used are different (though if you're dealing with highly concentrated reactants, it might be a different story...).

Judging from the way you framed the question, you probably already know a fair deal about both the Unimolecular and Bimolecular Nucleophilic Substitution (SN) reactions. So I won't start from scratch.

In the SN1 case, as the name would suggest, the rate depends on only one concentration factor, namely the substrate. You already (correctly) mentioned that a carbocation intermediate is formed. At this stage, its just a matter of whether the nucleophile performs a 'frontal attack' or a 'rear attack', and doesn't really involve the nucleophilicity or the size (at least not much in your example) of the nucleophile. This is different from the SN2 reaction, which requires species with high nucleophilicity  and appropriate size to react (rear attack) with the substrate, leading to an unstable transition state.

How about an analogy?

Suppose you've got a lot of magnets stuck on your fridge door (at least I sure do..). Now suppose you're standing at a distance from the fridge, and you want to place another magnet on the fridge by throwing it. Since you've already got a lot of magnets stuck there, its going to be really hard to make your magnet land there. The weaker your magnet, the lower the chances of you landing it on the fridge door. However, if you have this really strong neodymium magnet, this wouldn't be much of a task; one throw, and it'll land so hard, it'll send the other magnets flying. This is a lot like the SN2 scenario. The attacking group must be a stronger nucleophile that the leaving group.

Now suppose your fridge door is devoid of magnets [which is precisely what my mother requires of our fridge.... :( ]. Then clearly throwing a magnet and hoping it'll land is a very, very good possibility. In this case it doesn't really matter whether the magnet is square or circular, large or small, strong or weak (not too weak). Apart from how large your fridge door is, there's no other factor that would keep the magnet from sticking to the fridge (albeit you aim). This is a lot like your SN1 scenario. The nucleophile's characteristics don't really matter (at least not in most typical cases). As long as there's space on the door, the magnet will stick.


Can I lock up CO2 into silicon carbide? Can it be possible to use atmospheric $\ce{CO2}$ in a reaction (or series) with silicon dioxide, both of which are plentiful, to sequester the carbon in a solid and more useful material, silicon carbide?

Yes, but not practically.

The reduction of silica and carbon dioxide to silicon carbide is a conducted at high temperatures (commonly higher than 1500 °C, reaching 2500 °C). The energy required to reach these temperatures is immense. Furthermore, this is done in a nearly pure carbon dioxide atmosphere. I remind you that our atmosphere has about 400 ppm of carbon dioxide.

Silica is not abundant as you think. The main source for silica, sand, is actually quite rare and many countries are experiencing environmental damage due to illegal or legal mining of sand. Sand is commonly host to fragile and unique ecosystems. You can mine sandstone, but you will need to crush it to sand. Pulverising is the most energy intensive phase of any mining operation. For silicon carbide to form efficiently, it has to be crushed to a fine powder. Sand size is not enough. This is extremely energy intensive.

This is going to be an extremely expensive undertaking, and since the source of most of our energy is still fossil fuels, then this would be counter productive.


Understanding Scatchard Plots Im having trouble understanding Scatchard plots.
Y Axis = Bound/Free Ligand
X Axis = Bound Ligand
The graph has a negative slope.
Why when there is almost no Bound (Y axis = 0)  do we get a high positive number for Bound ligand on the X Axis corresponding to Receptor concentration? if there is no bound how is there a positive bound number?
it looks like the less bound ligand on the Y axis, the more bound on the X axis!

The Scatchard eqn can be written as $\displaystyle \frac{Y}{[L]}=\frac{n}{K}-\frac{Y}{K}$ with $[L]$ the free ligand concentration and $Y$ the fraction bound, (i.e. $[L]$ bound /total protein) so the x-axis intercept (when $Y/[L]=0$) is $n$ the number of ligands bound, the slope is $-1/K$ and the y-axis intercept is $n/K$. Your plot confirms this equation, the interpretation of intercepts and slope follow from the equation.
(Your plot seems to plot bound concentration not fraction bound so needs correcting with total protein to make $Y$ and this is presumably why $n$ comes out to be a small value)


Solid hydrophilic transparant materials that don't dissolve in water I am looking for hydrophilic transparant materials which are solid at room temperature and that don't dissolve in water. I need these for an experiment in which I condense droplets on a plate made out of/or coated with the material

Just thinking `chemically' I imagine that any polymer with a lot of hydroxyl groups will be hydrophilic (like Polyvinylalcohol), but PVA is water soluble. Polyethylenetereftalate (PET) is reasonable but still has a contact angle around 70 degrees (common polymer contact angles)

Does anybody know transparant materials with contact angles closer to 0 that don't dissolve in water? An additional `demand' is that the material does not break easily (like glass)

Several commercially available titanium dioxide coatings are available, commonly used to kill bacteria or aid UV stability (I think). When exposed to light, the titanium dioxide has a contact angle of almost 0 degrees.

You should be able to deposit a thin layer on something like a PVA or polycarbonate sheet. The commercial coatings are somewhat expensive, though, as far as I can tell, and the common process for doing it yourself involves heating the surface in a mid-temperature flame for 10-15 minutes which excludes pretty much all plastics I know of that would otherwise be suitable.


Bonds between metals and non-metals My teacher in college says that bonds between metals and nonmetals are ionic.

$ \ce{Metal - Metal} $ $\Rightarrow$ Metalic bond

$\ce{Non metal - Non metal}$ $\Rightarrow$ Covalent bond

I have to write about $\ce{CuCl2}$, and found in Wikipedia that if you subtract electronegativity you get what bond it is:

$$\chi (\ce{Cl}) = 3.16, \quad \chi (\ce{Cu}) = 1.90$$

$$\chi = 3.16 - 1.90 = 1.26$$

$\chi < 0.4 \Rightarrow$ covalent non-polar

$0.4 < \chi < 1.7 \Rightarrow$ covalent polar

$\chi > 1.7 \Rightarrow$ ionic

So it should be really a covalent bond. Which one is correct?

You should be careful with simple associations such as "metal + non-metal = ionic bond". These tend to throw out the idea of understanding the chemistry involved in favour of rote memorization. Note for example that mixing caesium metal with gold will produce a salt instead of an alloy, caesium auride ($\ce{Cs^+ Au^{-}}$). Mixing barium metal and platinum can also produce salts, though their structures are somewhat more complex. One can also argue that there is significant ionic character in solid xenon difluoride, even though both atoms are non-metals.

The idea of using electronegativity to determine covalent/ionic character is also meant as a helpful guide, not as a strict rule with black-and-white limits. Firstly, all bonds have both ionic and covalent character; both concepts are an oversimplification, and in reality it is more correct to say that a bond has a certain contribution from each type of bonding. This means there is a smooth transition from compounds with mostly ionic character and those with mostly covalent character. Also, the inequalities you mention rely on Pauling electronegativities. Electronegativity is surprisingly still a hotly debated topic, as we continue to search more general, more fundamental and more precise ways of defining it. Pauling electronegativities are based on empirical thermodynamic data regarding bond energies after applying a certain equation that was "picked", not derived from scratch. The values are particularly poorly defined for transition elements, such as the $\ce{Cu}$ in your problem. You get some not-so-easy to explain situations, like $\ce{HF}$ as a gas that is a borderline ionic compound.

Finally, in light of these comments, the answer to your question is that bonding in $\ce{CuCl_2}$ (I'm pretty sure that's what you actually meant to write) has intermediate characteristics between a purely ionic and a polar covalent bond, with similar contributions (though pinpointing which is highest sounds like an exercise in futility). A good way to study it more in depth is to analyze Fajans' rules. After a little self-calibration, you can get a good feel for the degree of ionicity and covalency of a compound. Some further but less certain evidence (lots of caveats!) for the intermediate character of $\ce{CuCl_2}$ can be found by looking at the substances' melting and boiling points ($\pu{498°C}$ and $\pu{993°C}$ [decomposition], respectively, according to Wikipedia). They are both quite high compared to substances with polar covalent bonds (dimethylformamide boils at around $\pu{150°C}$), but rather low compared to substances with very ionic bonds ($\ce{NaCl}$ boils over $\pu{1400°C}$).


How to approximate the pH of a box of water molecules with one proton? I have a cubic box of length $L = 12.42\ \mathrm{\overset{\circ}A}$, containing $60\ \ce{H2O}$ molecules and $1\ \ce{H+}$ ion. How can I approximate the $\mathrm{pH}$ of the system?

Following the basic definition, $\mathrm{pH}=-\log\left([\ce{H+}]\right)$, I tried to calculate it like the following two ways:

\begin{align}
  \mathrm{pH} &= -\log\left(1/60\right) = 1.78 \tag1\\
  \mathrm{pH} &=
    -\log\left(
      \frac{1/6.022\times 10^{23}}{12.42^3\times 10^{-30}\times 10^3}
    \right)
    = 0.0621 \tag2
\end{align}

However, I am not sure whether these approximations are valid or not in this regime.

For a rough, back of the envelope, sanity check sort of calculation, your second approach is more appropriate than the first. (Though check your value for Avogadro's number.) This is something that should jump out at you if you keep track of units in your calculation. The $\ce{[H+]}$ in the equation should be measured in moles of hydrogen ion per liter, so your calculation needs to result in an inserted value that's in moles of hydrogen ion per liter. You can also rule out the first equation by performing a thought experiment: if the size of the box doesn't factor into the equation, does it make sense that the pH wouldn't change if the box size changes? (No matter how much the size changes?)

However, the statement that pH is equal to the negative log of the hydrogen ion concentration is strictly speaking only true for dilute aqueous solutions near standard temperature and pressure. For concentrated acids/bases, solutions which are not primarily water, or for pressure and temperatures well outside standard, things get more contentious. There's a number of different interpretations of what "pH" means, and each have their advantages/disadvantages, depending on the situation.

One of the big considerations here is that "concentration" is not really really the thermodynamically relevant parameter when figuring out how chemical systems behave. Instead, it's the compound activity which is typically the more important quantity to keep track of. As it happens, the thermodynamic activity of a compound is normally proportional to the concentration in dilute solutions, so by convention we set unit activity to be equivalent to 1M solution.* This means the distinction between concentration and activity is normally ignored/glossed over in most introductory chemistry courses, but make no mistake, it's the activity of the compounds which should be used in things like the Henderson–Hasselbalch equation, not the concentration.

*(This normalization by 1 mol/L is the reason we can take the log and not run up against any strange results with dimensional analysis - it's technically a dimensionless value inside the log: the actual concentation, divided by the reference concentration of 1 M.)

The reason this matters for you is that you're dealing with a situation which isn't quite the same as a dilute aqueous solution. Nominally, it sounds like you're close. Your water density is around standard water density, and the amount of hydrogen ions are close to 1 mole per litre. But there are other considerations to take in to account. For example, what temperature and pressure are you using? Extreme values can change the hydrogen ion activity. Also, you didn't mention anything about the presence of a counter-ion. The unbalanced electrical charge might affect the predicted activity as well. The other issue is that just a single hydrogen ion will result in some pretty extensive fluctuations in the system as the system changes state. The "pH" (or hydrogen ion activity) of the system may actually exhibit time dependence, which might be important for the system you're modeling.

If you're truly interested in what the equivalent pH of the solution is, you should look into simulation techniques which attempt to actually gauge the chemical activity of the hydrogen ion under the conditions you're interested in. -- That is, figure out why you want to know the "pH" of the system, and look for more direct measures of that.


Oxygen Solubility in Water This is probably really obvious, but I really don't understand. Why does oxygen dissolve in water when oxygen is a nonpolar molecule and water is a polar molecule? It's obvious that oxygen does dissolve, or else fish could not survive, but I'm confused on how it does so.

Oxygen is relatively insoluble in water, its solubility being only 264 µM at 25 °C. That explains in part why you (and fish) require dedicated oxygen carriers in your blood to transfer sufficient oxygen around your body.
That oxygen is weakly soluble can be explained, as you point out, on the basis of the hydrophobic effect, which amounts to saying that water likes to hydrogen bond, and an oxygen molecule is not very good at that compared to another water molecule.
That oxygen is soluble at all can be explained on two grounds. First, statistics, just the random mixing of matter. Some oxygen molecules will happen to find their way into water and not find their way out for a while. The other explanation is that oxygen is not entirely nonpolar: it lacks a permanent dipole moment, but it is polarizable. Even molecules that we think of as highly water repellent can exist as liquids (oxygen is a gas at room temperature but can be cooled into a liquid state), which means something must be keeping them together, and that is dispersion forces due to the polarizability of all atoms and molecules.


Diffusion experiment with dialysis membrane bag Recently I have performed a diffusion experiment with a dialysis membrane containing methylene blue. A schematic picture of the experiment:

I want to calculate the flux (amount diffused/area*time) from the samples I have taken. However I am stuck with how to calculate the surface area of the dialysis bag so I can use Fick's law to estimate the flux. The dialysis bag is cylindrical, however when you put the 2 pins above to close it, it is not completely cylindrical anymore as you can see in the picture. Anyone that could help me out with this? If in need of membrane specifications I can post them.

You can double-check the area after you're done:

Poke a small hole in the membrane and drain it completely.
Stretch gently by pulling the clips (don't let them slip) until the area between them is rectangular.
Multiply that area (L x W) by two, because the membrane has two sides around the now-flattened middle.

There will always be some small error, because the wet membrane can be stretched somewhat, but that should be insignificant compared to other sources of inaccuracy, such as edge-effects where the clips narrow the membrane, impeding flow.