2- or -2 as superscript? When writing the formula for an ion, should the operator come before or after the number that indicates the ion's charge?

Here are some examples:

$\ce{SO_4^{-2}}$ or $\ce{SO_4^{2-}}$
$\ce{Fe^{+2}}$ or $\ce{Fe^{2+}}$

I'm not great at chemistry, so I'm sorry if it's a stupid question.

I just haven't been able to find many answers.

As @MaxW has said in comments, the correct form is:

$\ce{SO_4^{2-}}$ and $\ce{Fe^{2+}}$

It is just as in math.
If you have $a+a$ it results in $2a$, not $a2$, and thus:

$\ce{SO_4^{--}=SO_4^{2-}}$ and $\ce{Fe^{++}=Fe^{2+}}$

It is not a stupid question.
IMHO it is a perfectly legitimate question.
I apologize for those who are downvoting your question.


How to identify hydrogen bonds and other non-covalent interactions from structure considerations? Chemistry is governed by a wide range of interactions, from ionic and covalent bonding, or other types of strong interactions, towards weaker types of bonding, attraction, or repulsion, that typically occur between molecules. The latter are often referred to as chemical, molecular, or non-covalent interactions. Popular examples include ion-dipole interactions, hydrogen bonds, and London dispersion forces.

These non-covalent interactions play very important roles in DNA, protein-drug interactions, catalyst-substrate interactions, self-assembly, chemical reactions, among others.

How can I identify and analyse hydrogen bonds and other non-covalent interactions, possibly from experimental and calculated structures?

It is safe to say that there will always be intermolecular forces at play. At the time where you will consider these you should already have a good idea about the molecules involved in your system.
Based on the composition and molecular structures you can make certain assumptions. In a molecule it is straight forward to estimate (bond) polarities based on electronegativities, then infer from these how they might arrange. I will work an example based on the interactions between adenine and thymine later.[1]

With the advent of the information age, tools that every chemist has at her or his disposal have become more sophisticated. We have access to many digital resources like databases,[2] or publication servers[3] to retrieve a vast amount of information. Molecular modelling,[4] or even more sophisticated quantum chemical calculations,[5] have become more important; and free tools are available for everyone to use.

With that being said, here are some points that might help you identifying hydrogen bonds and other non-covalent interactions. For that purpose, let's have a look at our example molecules:

Immediately we can formulate a couple of assumptions based on the schematic representation. In adenine there are five nitrogen atoms, which have a higher electronegativity than carbon. Negative partial charges will therefore be located at mostly at these. Hydrogens in organic compounds usually carry positive partial charges; albeit $\ce{C-H}$ tend to be a lot less polarised than $\ce{N-H}$ bonds. Whenever a hydrogen is involved in a bond, that bond can potentially act as a hydrogen-bond donor (see below).
Similar observations can be made for thymine. Here we have two terminal oxygen atoms, which will carry a negative partial charge, since they have one of the highest electronegativities. These are often able to accept hydrogen bonds. On the other hand we also have $\ce{N-H}$ bonds, which can act as hydrogen bond donors.

Charges & Electrostatic Potential Surfaces

For many molecules structures are readily available. If not, some molecular editors give you the possibility to use implemented force fields to optimise built (guessed) structures. Based on those you can already do a few analyses. One tool that is quite powerful for various tasks is Avogadro, it let's you read crystal structures, perform basic calculations and much more. If you are just playing around, this is a really good choice.
For example, I have imported the crystal structure of adenine into Avogadro, optimised it, and calculated the electrostatic potential. Or after extracting Cartesian coordinates, Molden let's you easily calculate the charges.[6]

Hydrogen Bonds

Many molecular editors try to guess hydrogen bonds based on their implemented cutoff values. That certainly is very helpful, but not everything can be automatised in this way. And especially weaker interactions won't be found. One has to go a bit deeper then.
As a nice and concise example I have picked an intermolecular 2:1 complex between adenine and thymine, for which the crystal structure is available.[1]

There are two principle structural parameters to decide about hydrogen bonds:  (a) The distance of the hydrogen $\ce{H}$ and the hydrogen-bond acceptor $\ce{Y}$ is significantly shorter than the sum of their respective van-der-Waals radii, $d(\ce{XH\bond{...}Y})<r_\mathrm{vdW}(\ce{H})+r_\mathrm{vdW}(\ce{Y})$.[7]
(b) The angle around the hydrogen is nearly linear, $\angle(\ce{XH\bond{...}Y}) \approx 180^\circ$. For weakly polarised $\ce{XH}$ bonds, isotropic dispersion forces become more important (while the directional electrostatic and covalent contributions decrease), therefore the angle becomes more flexible.

We easily see that the bond angles are close to what we expect for hydrogen bonds. I have reproduced a few values from Batsanov's paper below, with the caveat that the value for hydrogen strongly varies depending on the chemical environment from $\pu{110 - 161 pm}$, so I used the classic from Bondi.[7c] Since all the distance are around $\pu{200 pm}$, they are well below the threshold we set earlier.

$$
\begin{array}{lr}
\text{Element }\ce{Y} & r_\mathrm{vdW}(\ce{Y})/\pu{pm}\\\hline
\ce{H}   & \approx 120\\
\ce{C}   &  196 \\
\ce{N}   &  179 \\
\ce{O}   &  171 \\\hline
\end{array}\hspace{2ex}
\begin{array}{lr}\\
\text{H-Bond }\ce{XH\bond{...}Y} & \sum r_\mathrm{vdW}(\ce{Y},\ce{H})/\pu{pm}\\\hline
\ce{CH}   &  316 \\
\ce{NH}   &  299 \\
\ce{OH}   &  291 \\\hline
\end{array}
$$

A quite interesting approach of revealing non-covalent interactions was presented by Johnson et. al., and the corresponding program is easy to use and only requires Cartesian coordinates.[8]

The surfaces between the molecules represent these interactions, where green represents weak interactions, typically found for dispersion. Blue represents stronger interactions, typically found for hydrogen-bonds. Red displays repulsive forces, typically found within ring or cage systems.

If you have access to quantum chemical software, then you can obtain this plot also for wave function files .wfn.

Another possibility is to analyse the electron density in terms of the quantum theory of atoms in molecules (QTAIM).[9] For this you do need a wave function file. The analysis, however, is straight forward and will yield a bond path or not. If there is a bond path, we can estimate the strengths of these bond with the methodology developed by Espinosa et. al.. According to this the bondstrength is approximately half the value of the potential energy density at the bond critical point.
$$E_\mathrm{H-Bond} = \frac{1}{2}V(r_{\mathrm{BCP}[\ce{XH\bond{...}Y}]})$$

I have performed such a calculation on the DF-B97D3/def2-TZVPP level of theory with Gaussian 09. The optimised geometry will be at the end.

\begin{array}{lr}
\text{H-Bond} & E_\mathrm{H-Bond}/\pu{kJ mol-1}\\\hline
\mathrm{N(37)H \cdots  O(1)}   &  46.6\\
\mathrm{N(5)H  \cdots  N(36)}  &  38.5\\
\mathrm{C(41)H \cdots  O(2)}   &   3.2\\
\mathrm{N(20)H \cdots  O(2)}   &  50.5\\
\mathrm{N(3)H  \cdots  N(24)}  &  29.9\\
\end{array}

A general warning shall be applied to the above. Absolute values of these are only approximate, but fall within the range of what is expected. A very nice side effect of this methodology is, that it can be applied to intramolecular hydrogen bonds, too.

Concluding remarks

Dispersive interactions and hydrogen bonds become more and more important in rational reaction design. Be it for understanding of molecular structure of biomolecules, or as a guiding principle for catalyst-substrate interactions.
With further development of computer technology, it should become more accessible to everyone. I hope this post demonstrates that gaining more insight can actually be quite easy (and free).

Notes and References

(a) Based on the structure from S. Chandrasekhar, T. R. Naik, S. K. Nayak, T. N. Row, Bioorg. Med. Chem. Lett. 2010, 20 (12), 3530-3533. DOI: 10.1016/j.bmcl.2010.04.131 PMID:
20493694 CSD: 739016
(b) Adenine, CSID: 185
(c) S. Mahapatra, S. K. Nayak, S. J. Prathapa, T. N. Guru Row, Cryst. Growth Des. 2008, 8 (4), 1223–1225. DOI: 10.1021/cg700743w, CSD: 652573
(d) Thymine, CSID: 1103
(e) G. Portalone, L. Bencivenni, M. Colapietro, A. Pieretti, F. Ramondo, Acta Chemica Scand. 1999, 53, 57-68. DOI: 10.3891/acta.chem.scand.53-0057, CSD: 136916
(a) The Cambridge Structural Database (CSD), https://www.ccdc.cam.ac.uk/
(b) Crystallography Open Database (COD), http://www.crystallography.net/cod/
(c) Computational Chemistry Comparison and Benchmark DataBase, http://cccbdb.nist.gov/ (Only for 1799 small molecules and atoms)
(d) Handbook of Chemistry and Physics, http://hbcponline.com/faces/contents/ContentsSearch.xhtml
(e) ...
(a) SciFinder, https://www.cas.org/products/scifinder
(b) Google Scholar, https://scholar.google.de/
(c) Web of Science, (formerly known as Web of Knowledge) http://www.webofknowledge.com/
(d) ...
(a) MolCalc, http://molcalc.org/
(b) Pitt Quantum Repository, https://pqr.pitt.edu/ (At the time of writing it was dead.) Github: pittquantum
(c) Many open source molecular editors include the possibility to use force field calculations. For example: Avogadro, molden
(d) For more on molecular modelling in the open source domain see S. Pirhadi, J. Sunseri, D. R. Koes, J. Mol. Graph. Model. 2016, 69, 127-143. An updated online version of this catalog can be found at https://opensourcemolecularmodeling.github.io.
(a) For an extensive, but not necessarily complete, list of quantum chemistry software see Wikipedia.
(b) For the purpose of this demonstration I will be using the proprietary software Gaussian.
(c) To view crystal structures, Mercury can be obtained (for free) from the Cambridge Crystallographic Data Centre (CCDC), which also hosts CSD.
https://www.ccdc.cam.ac.uk/solutions/csd-system/components/mercury/
(a) Tutorial for Avogadro (b) Tutorial for molden
(a) A concise (and as far as I can tell newest) list of van-der-Waals radii of many elements can be found in S. S, Batsanov, Inorg. Mat. 2001, 37, 871-885. DOI: 10.1023/A:1011625728803 (mirrored pdf)
(b) A list of van-der-Waals radii can also be found on Wikipedia
(c) A. Bondi, J. Phys. Chem. 1964, 68, 441-451. doi: 10.1021/j100785a001
(a) The original publication: E. R. Johnson, S Keinan, P. Mori-Sánche§, J. Contreras-García, A. J. Cohen, W. Yang, J. Am. Chem. Soc. 2010, 132, 6498-6506. DOI: 10.1021/ja100936w
(b) The presentation of the program: J. Contreras-Garcia, E. R. Johnson, S. Keinan, R. Chaudret, J-P. Piquemal, D. N. Beratan, W. Yang, J. Chem. Theory Comput. 2011, 7, 625-632. DOI: 10.1021/ct100641a
(c) Download the code: http://www.lct.jussieu.fr/pagesperso/contrera/nciplot.html
(d) You'll also need VMD (Visual Molecular Dynamics) from the University of Illinois
(a) A very brief introduction can be found on Wikipedia. The corresponding book:
Bader, Richard (1994). Atoms in Molecules: A Quantum Theory. USA: Oxford University Press. ISBN 978-0-19-855865-1. (publisher)
(b) Multiwfn - A Multifunctional Wavefunction Analyzer; http://sobereva.com/multiwfn/ corresponding paper: T. Lu, F. Chen, J. Comput. Chem. 2012, 33, 580-592. DOI: 10.1002/jcc.22885
(c) Startup script (and examples) for Linux version: https://github.com/polyluxus/runMultiwfn.bash (shameless self-plug)
(d) E. Espinosa, E. Molins and C. Lecomte, Chem. Phys. Lett., 1998, 285, 170–173.

Appendix

Optimised Structure of the adenine-thymine 2:1 complex calculated at DF-B97D3/def2-TZVPP in Gaussian 09 Rev. E.01

  45
E(RB97D3/def2TZVPP/W06) = -1388.51095169
O         19.03780       11.79565        1.63996
O         14.74303       13.36808        1.71115
N         15.07940       11.09762        1.64043
H         14.04669       10.93403        1.64128
N         16.88238       12.55083        1.67499
H         17.23771       13.53695        1.70257
C         17.83067       11.52964        1.63864
C         17.29332       10.17534        1.60048
C         15.51719       12.40225        1.67775
C         15.94232       10.03610        1.60357
H         15.46489        9.06076        1.57654
C         18.24710        9.02139        1.56030
H         18.89656        9.08263        0.68030
H         18.90669        9.02997        2.43483
H         17.71085        8.06866        1.53480
N          8.28060       10.09397        1.65692
H          7.68055        9.28386        1.63653
N          8.91986       12.24943        1.71837
N         10.48101        9.01368        1.60709
N         11.91467       12.94472        1.71757
H         12.92612       13.11883        1.71578
H         11.26613       13.71385        1.74609
C         10.03193       11.42408        1.68464
N         12.26612       10.63929        1.64378
C         11.41986       11.70069        1.68284
C          9.65939       10.07380        1.64587
C          7.89700       11.42267        1.70074
H          6.85459       11.71169        1.71748
C         11.75955        9.39245        1.60917
H         12.49648        8.59107        1.57896
N         21.22700       16.46826        1.76840
N         17.31372       17.43807        1.81488
H         16.80939       18.31038        1.84233
N         19.61652       18.27309        1.82815
C         18.69253       17.30751        1.80467
N         17.71219       15.24141        1.74956
N         20.64990       14.21957        1.70686
H         19.98577       13.44513        1.68592
H         21.63800       14.02512        1.69520
C         20.27365       15.50797        1.74532
C         16.77892       16.17095        1.78070
H         15.71578       15.97320        1.77996
C         18.91965       15.92447        1.76371
C         20.85479       17.75435        1.80728
H         21.67003       18.47621        1.82413


Calculating the theoretical percent purity of a recrystallization The sample contains some percent A and some percent B with A being the majority. The solubility for both is given at two different temperatures. One where both are very soluble and one where both are much less soluble. From these five values I'm supposed to determine the theoretical purity of the recrystallized compound.

I'm surprised that this is possible and I have no idea how to even begin approaching it. I always thought that crystallization was structurally depended as well. Shouldn't the lattice favor the dominant compound? What if the impurity is less soluble at the low temperature than the majority compound? How can we mathematically represent such a (seemingly) complex phenomena with such little information?

Note: I don't mean percent yield, the question specifically stated purity.

I can give you a hint, imagine you have a mixture containing one consitutant $A$ only in your solution with a solvant.

Doing the balanced equations about the mass of $A$ in each phases you have :

$$\begin{cases}
m_A^\mathrm{t}=m_A^\mathrm{c}+m_A^\mathrm{liq} \\
m_A^\mathrm{i}=x_A^\mathrm{c}\cdot m^\mathrm{c}+x_A^\mathrm{liq} \cdot m^\mathrm{liq}
\end{cases}$$

Where $m^\mathrm{t}$, $m^\mathrm{c}$, $m^\mathrm{liq}$ and $m^\mathrm{i}$ denote respectively the mass of $A$ total, in the cristal form, dissolute in the solvant (liquid), and at the begining (initial) and $x$ the corespondant mass-fractions.

Then you must obtain, $$m^\mathrm{c}=m^\mathrm{t}\left(\frac{\frac{m_A^\mathrm{i}}{m^\mathrm{t}}-x_A^\mathrm{c}}{x_A^\mathrm{liq}-x_A^\mathrm{c}}\right)$$

Notice that $\frac{m_A^\mathrm{i}}{m^\mathrm{t}}=x_{A,\mathrm{i}}^\mathrm{liq}$ which is the mass-fraction at the begining in the liquid and $x_A^\mathrm{liq}$ is the mass-fraction in the liquid at the end.

If you know how what you put at the begining, you can determine the yield. Now if you want to know the purity, you'll need to do the same with both of your component to see how much of them have cristalised at the end and then determine the purity of your marjoritory compound.

I hope it can help, have a good day ! :)

Be careful when you do the balanced equations if there is also a gas phase which can happend sometime.


Why is the overall change in entropy of Photosynthesis positive? Net Reaction:
$$\ce{6CO2(g) +6H2O(l) -> 6O2(g) + C6H12O6(aq)}$$
The net reaction shows that 12 moles become 7 moles; 6 of gas,and 1 complex molecule.
This implies a decrease in entropy over the net reaction.
However, from reading further sources, it seems that the change in entropy is positive, but they don't justify it. Why is the overall change in entropy positive, over the series of reactions that make up this process? Could this be linked to the role of UV Light in this reaction?

To answer this question, we need to be clear whether we are talking about the change in entropy of the system, the surroundings and the Universe.
With regards to the entropy of the Universe, the answer is direct: photosynthesis is a process we observe to happen on a macroscopic scale. Therefore, by the second law of thermodymamics, photosynthesis must cause the entropy of the Universe to increase, period.
With regards to the entropy of just the chemical reagents, if we consider those as our system, then indeed the entropy of formation of glucose is negative, equal to approximately $\mathrm{-40\ cal\ mol^{-1}\ K^{-1}}$.
The way to reconcile these facts is to note that there is a significant increase in entropy which comes from the destruction of high-energy visible photons from the Sun, whose energy is scattered into chemical bonds and multiple low-energy far infrared photons.
In a simplified analysis, if three photons of $\mathrm{700\ nm}$ light are absorbed to generate one molecule of glucose and excess energy is radiated away with black body spectrum at $\mathrm{300\ K}$, then photosynthesis is allowed by the second law of thermodynamics, so long as the photosynthetic efficiency is less than 88%. The observed efficiency of photosynthesis using red light is on the order of 59%, so this requirement is cleared with room to spare.
Depending on the level of detail you want, it gets complicated to work through all the checking and balancing of entropy changes at every individual microscopic step, but the overall results must hold as stated above.
References:
Negative entropy and photosynthesis. Wesley Brittin and George Gamow, Proceedings of the National Academy of Sciences, 1961, 47 (5), 724-727. DOI: 10.1073/pnas.47.5.724
Entropy balance in photosynthesis. Wolfgang Yourgrau and Alwyn Van Der Merwe, Proceedings of the National Academy of Sciences, 1968, 59 (3), 734-737. DOI: 10.1073/pnas.59.3.734
Entropy production and the Second Law in photosynthesis. Robert S. Knox and William W. Parson, Biochimica et Biophysica Acta (BBA) - Bioenergetics, 2007, 1767 (10), 1189-1193. DOI: 10.1016/j.bbabio.2007.07.004


Why does bond angle decrease in the order H2O, H2S, H2Se? I know that bond angle decreases in the order $\ce{H2O}$, $\ce{H2S}$ and $\ce{H2Se}$. I wish to know the reason for this. I think this is because of the lone pair repulsion but how?

Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths:
\begin{array}{lcc}
\text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\
\hline
\ce{H2O} & 104.5 &  96 \\
\ce{H2S} &  92.3 & 134 \\
\ce{H2Se}&  91.0 & 146 \\
\hline
\end{array}

The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the $\ce{H-X-H}$ angle to contract slightly. So instead of the $\ce{H-O-H}$ angle being the perfect tetrahedral angle ($109.5^\circ$) it is slightly reduced to $104.5^\circ$.  On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ have no orbital hybridization.  That is, The $\ce{S-H}$ and $\ce{Se-H}$ bonds use pure $\ce{p}$-orbitals from sulfur and selenium respectively.  Two $\ce{p}$-orbitals are used, one for each of the two $\ce{X-H}$ bonds; this leaves another $\ce{p}$-orbital and an $\ce{s}$-orbital to hold the two lone pairs of electrons.  If the $\ce{S-H}$ and $\ce{Se-H}$ bonds used pure $\ce{p}$-orbitals we would expect an $\ce{H-X-H}$ interorbital angle of $90^\circ$.  We see from the above table that we are very close to the measured values.  We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two $\ce{X-H}$ bonds the angle opens up a bit wider.  This explanation would be consistent with the $\ce{H-S-H}$ angle being slightly larger than the corresponding $\ce{H-Se-H}$ angle. Since the $\ce{H-Se}$ bond is longer then the $\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum.  The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the orbitals in the two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.


Why does the rate of the reaction decrease with an increase in temperature? Why does the following reaction

$$\ce{2NO + O_2 -> 2NO_2}$$

have negative reaction coefficient, i.e., why does the rate of the reaction decrease with an increase in temperature ?

I have been taught that this is the only reaction that has this property.

It's strange, but there are mechanisms whereby this type of thing can happen.  Basically you have a rate determining step that involves an intermediate species, and that rate determining step must have the usual positive temperature dependence.  But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence that dominates.

Let's translate this into concrete terms for the reaction given.  A quite plausible mechanism would look like this:

$\ce{NO + O2 <=> ONOO}$ Fast, reversible

$\ce{ONOO + NO -> 2NO2}$ Rate determining step

The equilibrium in the first step is given by

$K_\mathrm{p1} =\dfrac{p^\mathrm{eq}(\ce{ONOO})}{p^\mathrm{eq}(\ce{NO})p^\mathrm{eq}(\ce{O2})}$

Assuming $\ce{ONOO}$ is a minor species, it's formation hardly affects the equilibrium partial pressures of the other species at any time, so we may just use $p(\ce{NO})$ as the nominal nitric oxide partial pressure and similarly for the molecular oxygen.  Then we can solve for the partial pressure of the $\ce{ONOO}$ intermediate:

${p^\mathrm{eq}(\ce{ONOO})}=K_\mathrm{p1}(p(\ce{NO}))(p(\ce{O2})$

Now put this result into the rate determining step with rate

$r_2=k_2{p^\mathrm{eq}(\ce{ONOO})p^\mathrm{eq}(\ce{NO})}=k_2K_\mathrm{p1}(p(\ce{NO})^2)(p(\ce{O2})$

Now for the eye-opener:  put in the temperature dependence of the equilibrium and rate constants

$K_\mathrm{p1}=K_\mathrm{p1}^0\exp(-\Delta H_1/RT)$

$k_2=k_2^0\exp(-E_{a2}/RT)$

So then

$r_2=k_2^0K_\mathrm{p1}^0\exp(-\dfrac{\color{blue}{E_{a2}+\Delta H_1}}{RT})p(\ce{NO})^2p(\ce{O2})$

The activation energy then contains two terms, one from the inherent rate in Step 2 and the other from the equilibrium in Step 1.  The latter is negative if Step 1 is exothermic, as it might well be if a bond is being (temporarily) formed between the reactants.  Then if this enthalpy term is the absolutely larger part of the sum ... we see a reversed temperature dependence!


why there is no simple rule to predict the possible ionic charges of element in middle of periodic table I know element in left two columns have 1+,2+... and right ones 1-,2-.

But in middle is all over the place. And with multiple charges.

What defines that if this element is gonna have this charge. All i have been told is that elements want to be close to as much possible to noble gasses because they are stable. Which doesn't explain much as to why such behavior is present. What does it mean to be stable. And why elements in middle find it stable just after 3 +/- charge even though they are far from noble gases.

Thanks, there are few questions in here but all very related i think.

TL;DR - The upshot is that there just happens to be a simple approximation governing small atoms but as the atoms get bigger and more complex, those approximations become less good and there is not another simple rule that works.

In more depth, the issue is a combination of two factors: distance from the nucleus and full/half-full valence orbitals.

In the first place, valence (outermost) electrons that are further away from the nucleus are more weakly bound than valence electrons that are closer. This means that the energy required to add or remove an electron is less, and so the charge given by a given electron matters less to the stability of the atom. Because of this, the further down the periodic table you go, the less the "normal" charges matter. As an example, Germanium is two rows below Carbon and is naturally found to have charges of +4 and +2 (as well as sometimes having +1, +3 and -4!).

The second factor is that elements find a special stability when their valence orbitals are either full (preferable) or half-full (less preferable, but still better than neither). So an element like Cr, which has almost no hope of gaining or losing enough electrons to look like a noble gas, "wants" to gain lose the 4 3d electrons, getting it down to a full 4s orbital. The tricky bit is that unlike the p orbitals, not all of the d orbitals have the same energy, so there are some that are more likely to be removed than others. There just isn't a convenient rule of thumb about which is which (nature is governed by physics, not approximations).


Calculating the strength of an ionic bond that contains poly-atomic ions So the bond association enthalpy for ionic compounds like $\ce{NaCl}$ and $\ce{NaBr}$ can be easily calculated from a Born-Haber cycle. But the way a Born-Haber cycle is constructed it uses info that only really applies to mono-atomic ions like $\ce{Cl^-}$. So how would one calculate the strength of an ionic bond for ionic salts that contain poly-atomic ions like $\ce{NaOH}$ and $\ce{K_2SO_4}$. Is there a completely different method, or can a Born-Haber cycle be adapted to use a poly-atomic ion?

With enough effort, Born–Haber cycle can be extended to polyatomic ionic solids, however it's practically never done in practice due to the lack of experimental data or because it's impossible to obtain any.

From [1, p. 117–118] (emphasis mine):

  Lattice energies cannot be measured experimentally since they represent hypothetical processes:

  $$\ce{M^n+(g) + X^n-(g) → MX(s)}$$

  However, the following reaction sequence, relating the heat of formation, $ΔH_\mathrm{f}$ of a crystal $[\ce{M(s) + 1/2 X2(g) → MX(s)}]$ to $U[\ce{M+(g) → MX(s)}]$ is thermochemically equivalent (and $ΔH_\mathrm{f}$ can be measured).

  $$
\begin{array}{ccc}
\ce{&M^0(g) &+ &X^0(g) &→ &M+(g) &+ X-(g)} \\
&↑\small ΔH_\mathrm{s} &&↑\small D &&↓ & \\
\ce{&M(s) &+ &1/2 X2(g) &→ &MX(s)}
\end{array}
$$

  In this diagram, $ΔH_\mathrm{s}^\circ$ gives the enthalpy of sublimation of the metal $[\ce{M(s) → M^0(g)}],$ $D$ gives the dissociation energy, or bond energy of the diatomic gas $[\ce{1/2 X2(g) → X^0(g)}],$ $IE$ gives the ionization energy of the gaseous metal $[\ce{M^0(g) → M+(g)}],$ and $EA$ gives
  the electron afﬁnity for the formation of the gaseous anion $[\ce{X^0(g) → X-(g)}].$
  The lattice energy is obtained through the relation:

  $$U = ΔH_\mathrm{f} - \left(ΔH_\mathrm{s} + \frac{1}{2}D + IE + EA\right)$$

  One difﬁculty with using a Born–Haber cycle to ﬁnd values for $U$ is that heats of formation data are often unavailable.
  Perhaps the greatest limitation, however, is that electron afﬁnities for multiply-charged anions (e.g. $\ce{O^2-}$) or polyanions (e.g. $\ce{SiO4^4-}$) cannot be experimentally obtained.
  Such anions simply do not exist as gaseous species.
  No atom has a positive second electron affinity; energy must be added to a negatively charged gaseous species in order for it to accommodate additional electrons.
  In some cases, thermochemical estimates for second and third electron afﬁnities are available from ab initio calculations.
  Even so, if there are large covalent forces in the crystal, poor agreement between the values of $U$ obtained from a Born–Haber cycle and Madelung calculations can be expected.

Already for some simple species such as oxoanion it is impossible to determine their electron affinity $EA$ in gaseous phase:

$$
\begin{align}
\ce{O^0 (g) + e- &→ O^- (g)}\label{rxn:R1}\tag{R1} \\
\ce{O- (g) + e- &→ O^2- (g)}\label{rxn:R2}\tag{R2} \\
\hline
\ce{O^0 (g) + 2 e- &→ O^2- (g)}\label{rxn:R3}\tag{R3}
\end{align}
$$

Enthalpy of reaction \eqref{rxn:R2} cannot be determined experimentally and can only be derived from the lattice energy $U$, which causes Catch 22 situation[2, p. 61].

References

Lalena, J. N.; Cleary, D. A. Principles of Inorganic Materials Design, 2nd ed.; John Wiley: Hoboken, N.J, 2010. ISBN 978-0-470-40403-4.
Smart, L.; Moore, E. Solid State Chemistry: An Introduction, 4th ed.; CRC Press: Boca Raton, FL, 2012. ISBN 978-1-4398-4792-3.


What is the effect of Helium-based surroundings (99% Helium) on electrochemical gas sensors? I am currently working on a project measuring gas composition of car exhaust. In order to dilute the gas concentration down to something electrochemical sensors can read, I diluted the sample with pure helium.

The following is true for these sensors:

  Importance of Oxygen. The reactions at the sensing electrode (anode) for some gases are as follows:
  \begin{align}
\ce{CO + H2O &-> CO2 + 2H+ + 2e-} \\
\ce{H2S + 4H2O &-> H2SO4 + 8H+ + 8e-} \\
\ce{NO + 2H2O &-> HNO3 + 3H+ + 3e-} \\
\ce{H2 &-> 2H+ + 2e-} \\
\ce{2HCN + Au &-> HAu(CN)2 + H+ + e-}
\end{align}
  Simultaneously, the reactions at the counter electrode (cathode) need oxygen molecules to complete the process:
  $$\ce{O2 + 4H+ + 4e- -> 2H2O}$$
  An inadequate supply of oxygen to complete the reaction will shorten the life of the sensors, hence the sensors will not operate properly.

Does this mean that doing what I intend is going to affect their performance and damage them?

Aiming for dilution of the exhaust gases, you need a gas that is inert, i.e. does not react at these conditions with neither the gases present, nor with the metals present in in the catalytic converter (if there is such device installed) and electrodes of the sensors.

Under this restraint, using a nobel gas was a good idea.  Yet instead of helium, that you still may use to spot leaks in the tubings, I would opt for argon, which is much cheaper per litre.  Note that at ambient conditions, its density as gas is higher than the one of helium and of air and has a slightly lesser thermic conductance.


Quasi-steady state hypothesis application Consider the reaction scheme:

$$\ce{S + E ->[$k_1$] C1} \qquad \ce{C1 ->[$k_2$] E + P} \qquad \ce{S + C1 <=>[$k_3$][$k_4$]C2}$$

where $\ce{S}$ is the substrate, $\ce{E}$ is the enzyme, $\ce{P}$ is the product, $\ce{C1}$ and $\ce{C2}$ are enzyme substrate complexes. Let $[\ce{S}] = s$, $[\ce{E}] = e$, $[\ce{C1}] = c_1$, $[\ce{C2}] = c_2$ and $[\ce{P}] = p$ be the concentrations of each respective chemical. I have simplified this system down to

\begin{align}
\frac{\mathrm ds}{\mathrm dt} &= -k_1se_0 + (k_1-k_3)sc_1 + (k_1s+k_4)c_2 \\
\frac{\mathrm dc_1}{\mathrm dt} &= k_1se_0 - (k_1s+k_2+k_3s)c_1+(k_4-k_1s)c_2 \\
\frac{\mathrm dc_2}{\mathrm dt} &= k_3sc_1-k_4c_2 \\
\end{align}

using the conservation equation $e=e_0-c_1-c_2$. I have found that $p(t) = k_2\int c_1(t) \,\mathrm dt$. Now I need to use the quasi-steady state hypothesis to show that $$\frac{\mathrm ds}{\mathrm dt}= - f(s),\qquad f(s) = \frac{k_1e_0s}{1+\frac{k_1}{k_2}s + \frac{k_1k_3}{k_2k_4}s^2}.$$

Now I'm not really given much information on this hypothesis. I have been told it means

  We assume that the initial stage of complex formation is very fast. After which it is essentially at equilibrium.

So how do I apply the hypothesis to transform $\mathrm ds/\mathrm dt$?

Thanks for the interesting question.  Before we get into the details, let me note a few things:

Your system is a model of substrate inhibition, a phenomenon where higher-than-optimal concentrations of a substrate for an enzyme decrease its own rate of conversion to product.
It sounds like you have a good handle on the enzyme mass balance, i.e. that $e_0 = e + c_1 + c_2$ is a conserved quantity whose time derivative is everywhere zero (no approximations or hypotheses required) in this kinetic scheme.  This makes sense because enzymes are catalysts that accelerate reactions without any net participation in them.

OK, onward to your questions.

The assumption of fast equilibrium for $c_2$ is a different assumption than the PSSH hypothesis.  Assuming a fast equilibrium for $c_2$ means that reactions 3 and 4 happen much more quickly than reactions 1 or 2.  Thus the concentration of $c_2$ is determined solely by the thermodynamics of reactions 3 and 4: $$k_4 c_2 = k_3 s c_1 \;\;\; \Rightarrow  \;\;\; c_2 = \frac{k_3}{k_4} s c_1$$
The pseudo-steady-state hypothesis could be applied to $c_1$.  $$\frac{
dc_1}{
dt} = k_1se_0 - (k_1s+k_2+k_3s)c_1+(k_4-k_1s)c_2 = 0$$  Substituting in the result of the fast equilibrium assumption (point #3) gives $$k_1se_0 - (k_1s+k_2+k_3s)c_1+(k_4-k_1s)\frac{k_3}{k_4}s c_1 = 0$$  You can solve this equation for $c_1$ and plug the result into your $\frac{
ds}{
dt}$ equation.
If you are interested in understanding more about the mathematical validity of the PSSH, you should know that it is related to singular perturbation.  For a treatment of a "simple" Michaelis-Menten system by singular perturbation, see this book excerpt.  The PSSH solution is an "outer" singular perturbation solution.


Non-trivial cases where the frozen core approximation is bad? In computational chemistry, it is extremely common to freeze the core electrons after an initial Hartree-Fock calculation, meaning that the core electrons are left out of any treatment of the electron correlation. Which orbitals are considered core depends on the atom, but for the second row, it usually just means the 1s electrons are not correlated.

This is an intuitive approximation, as one would expect that correlation of the valence electrons would have the greatest effect on any actual chemistry.

Now, clearly one can imagine situations where freezing the core electrons is a very bad idea. One such case is if you were interested in modelling x-ray spectroscopy. I am less interested in this type of example as this is almost just a user error, and hence I would consider this a somewhat trivial example.

So, to be concrete, are there any known examples in which treatment of correlation of the core electrons results in a qualitative change in the modelling of a process which is primarily mediated by the valence electrons?

My guess is if there is an example it occurs in the third row and beyond.

The frozen core approximation (FCA) falls apart whenever you need orbital relaxation of the core to accurately describe your system. From a theory-based argument, there are multiple electronic structure methods which rely on this (CCSD, orbital optimized methods) to better account for correlation effects. However, given that the core orbitals do not change appreciably during most processes (ionization, binding, standard chemical reactions), the error introduced by the FCA is all but completely cancelled (you can still get sub kcal/mol accuracy with the FCA).

However, one case where it rears its head is when you're trying to calculate the single-point energy of a system. In this case, the FCA will get in the way of recovering (for example) the full CI result in the full basis set limit.


Is Pinacol rearrangement the source for the ring expansion? I have no idea what's happening as in the product (the answer is (a)) there is a new ring being formed on the left benzene. But as I studied after $\ce{H2O}$ leaves, a carbocation is then formed on stabilized by the right ring.

Can someone please explain the rest of the steps of this reaction? The thing I don't get is how does the three-membered ring opening happens.
I attempted the question, but the answer that I came up with isn't the one matching in the book. Where did I go wrong with the mechanism?

Your first and third structures are good but the arrow pushing in the second structure doesn't get the job done. As you have noted, diol 1 forms carbocation 2a the more stable of the two options. Now the lefthand cyclopropane bond with its pair of electrons migrates to the cationic center to form resonance-stabilized species 2b and on to ketone 3. This type of structure is a candidate for a Nazarov cyclization. The arrows in 4 may be drawn in the other direction as well. Rearomatization of 5 by loss of a proton and tautomerization leads to the product 6. [Note: Structures 2b and 4 are identical. Just a reprotonation of 3.]ADDENDUM: To answer your original question, yes it is a pinacol rearrangement. The bond that migrates (2a --> 2b) just happens to be attached via the cyclopropane ring to the site to which it is migrating. So, 1 --> 3 is a pinacol rearrangement.


How appropriate/adaptable is HPLC for in situ sample analysis on Titan? I have a limited background in chemistry (nothing beyond general Chem 2), so I am asking for help with a group project. We have two chemists on our team, but communication is limited at the time and I am trying to answer some specific questions for myself so that we can move forward. We're designing a probe to study the chemical composition of Titan's lakes and sediment deposits in the lake beds, among other functions. We would like to adapt HPLC for the project using UV-Vis spectroscopy as the detection method. However, from what I understand, a UV-Vis spectrophotometer must be calibrated using all known solutes, which we can't know in this situation. I read that when the detector is showing a peak, some of what is passing through the detector can be diverted to a mass spectrometer and the fragmentation pattern can be used to identify unknown compounds, which is what we would like to do. However, I'm not sure how this would work or whether it would be wise given the constraints of our project. Our motivation for adopting HPLC is that we can keep all substances in their original phase for analysis. But if I'm not mistaken, mass spectrometry will require vaporization. My question then is, is there a way to use HPLC to identify unknown compounds suspended in the lake without disturbing the state of the sample? Any advice would be appreciated.

It is a very good question because this is an active area of research. Having heard talks from the actual people who are making analytical instruments for space analysis, I can mention the actual issues before coming to your actual question so that you are aware of the latest problems. As far as I know, HPLCs have not been sent, but GC systems have been successfully used.

Key problems

(i) Weight. It costs an enormous amount (in millions) of money per kg to send anything in space. Routine HPLCs are very heavy. We need two heavy duty persons to lift the stack of standard equipment.

(ii) Latest HPLCs can now be reduced to the size of a small brief case. Check capillary chromatography. Also check HPLC on a chip. It is the size of a hand palm.

(iii) Elecric power consumption: Think how much power is consumed by the motors in the pump (huge) and other analytical equipment including the mass spec. You don't have lot electrical power in space. This the biggest problem besides weight issues.

(iv) Think of excessive radiation. Can the plastics and equipment tolerate huge amount of radiation from the Sun and elsewhere.

(v) Temperatures: Think of temperature extremes with liquid as solvents.

(vi) There is no HPLC column (or GC column) in the world which can separate everything. May be you would attach at least 7-8 columns at once and see which one separates the sample (having a system with multiple columns at once is commercially available. This is done by pharma people)

  Our motivation for adopting HPLC is that we can keep all substances in
  their original phase for analysis.

Yes, HPLC is a non-destructive technique. Mass spectrometer is not. However, UV-detection is useless for unknowns because you have no clue about their absorption spectra. Even if you have an absorption spectrum, it provides very little information about the full structure. Only mass spectrometer can be used in case of complete unknowns. Keep in mind that a typical HPLC C-18 column can only separate hydrophobic compounds under ideal conditions. The sole job HPLC or GC is to simplify the mass spectrum of a sample. Imagine you inject a sample containing 50 analytes into a MS. Imagine the complicated mess. With an HPLC, only compound will reach the detector one at a time. Mass spectrum is very simple then.

  My question then is, is there a way to use HPLC to identify unknown compounds suspended in the lake without disturbing the state of the sample? Any advice would be appreciated.

Short answer no. HPLC cannot detect or separate suspended matter. If you meant substances dissolved in the lake, yes, then it is possible. Every sample has to be carefully filtered to remove any particles.

Since you have two chemists, why don't you just do a real experiment on lake water.


What chemicals cause the acrid smell of burning food? When roasting vegetables in a oven, after a while there is a nice caramel smell. After that, if they start burning, there is a smokey smell, and if they continue to burn, the smokey smell gets more and more acrid. The smell really sticks in the back of your throat/sinuses, and induces gagging if it's strong enough. What chemicals make up the gasses and particles that cause that acrid smell?

Take a look at this link about Maillard reactions:

http://www.landfood.ubc.ca/courses/fnh/301/brown/brown_prin.htm

And what occurs if continue burning: pyrolisis

http://www.healthy-food-site.com/burnt-food.html


Why is the formation of NaCl an exothermic process? My book says,

  When NaCl forms, what really happens is that $\ce{Na+}$ and $\ce{Cl-}$ come together and form an ionic bond and thus NaCl is formed and it is a quite exothermic reaction.

Why and from where is this energy coming from? Ok Ok, I read that every system wants to reduce energy, but where literally is this energy coming from, is it because of some electrons or from the nucleus? Or from where?

The negative standard enthalpy of formation of sodium chloride (i.e. ‘the formation is exothermic’) is primarily due to the formation of a crystal lattice. This crystal lattice is a macroscopic arrangement of ions in a cubic fashion with each sodium ion surrounded by six chlorides and each chloride ion surrounded by six sodium ions.

If it were not for the generation of this lattice and its inherent stabilising properties (a very good charge distribution, charges neatly alternate, contact between species of identical charge is strongly inhibited) the formation of sodium chloride would be much less exothermic.

Note most importantly, that it is not merely the generation of sodium cations (that process is endothermic) or the generation of chloride anions (that process is only weakly exothermic) that causes the overall exothermicity of the entire reaction.


How to bond polymer to polymer (TPU on PMMA) I wanted to bond TPU films to PMMA slab. I  have tried thermal bonding but it is was not practical(I do not know why?). Can anyone please give me a method to bound them? Thanks

Sorry to say but you question is a bit vague because without knowing the project, it's pretty hard to help. Anyway, here is some general information.
Some of the sophisticated methods of joining these together include solvent bonding, vibration welding, friction welding and induction welding or pressure sensitive adhesives [ (Meth)acrylic copolymers are one of the most widely used polymer classes for the production of pressure-sensitive adhesives]. Acrylic requires an acrylic solvent adhesive. Cyanoacrylate (super glue) may work, depending on the project. UHU glue is very useful with all kinds of plastics, so you might wanna try that.  I myself have tried superglue and it works with both materials. But I think for your project, non adhesive methods woll be an optimal choice.
All the best.


What is the maximum molarity of aqueous HCl? Here is a problem a made up:

  We have some aqueous $\ce{HCl}$. What is the maximum possible molarity of the $\ce{HCl}$?

How do I solve this? I don't have any good ideas.

The HCl molecule is, at standard conditions, a gas. The reagent used in laboratories is HCl dissolved in water, which is why you'll find in the label that it is around 37% HCl in weight. The other 63% is water.

Considering 37% as the maximum solubility of HCl, you can calculate the molarity using the solution density (1.2 g/mL) and HCl's molar mass (36.46 g/mol).

For 1000 mL of solution, you will have 1200g of weight, of which 37% is HCl: 444g.

In 444g of HCl you have 12.18 moles (444/36.46), which means the concentration of P.A. HCl is around 12.18 moles per liter.

Note that because P.A. HCl is a solution saturated with a gas, vapors are likely to be released and escape the solution (which is why P.A. HCl shouldn't be handled outside fume hoods), which makes this concentration an approximation, and it should never be used to prepare a solution which you need to know the exact acid concentration.


Determining the equilibrium constant This is freely translated, I hope it's not too bad.

The reaction to dehydrate methylcyclohexane to toluene in the gaseous phase:

Formulate the reaction equilibrium's constant.

Now, that's not a problem - the problem is that I don't know the aggregate state. I know that $\ce{H2}$ is a gas and that other thing left of it seems to be toluene, which is also a gas according to the text above. So how should I now know what aggregate state the methylcyclohexane is in? I also assume everythings at standard conditions.
The solution is apparently (but it might be wrong, since it's not an official solution)

$$K_1 = a_\ce{C7H14}^{-1}\cdot a_\ce{C7H8}^1\cdot a_\ce{H2}^3 = \left(\frac{c_\ce{C7H14}}{c^\circ}\right)^{-1}\left(\frac{c_\ce{C7H8}}{c^\circ}\right)^{1}\left(\frac{p_\ce{H2}}{p^\circ}\right)^{3}$$
$$K_2=a_\ce{C7H14}^{1}\cdot a_\ce{C7H8}^{-1}\cdot a_\ce{H2}^{-3} = \left(\frac{c_\ce{C7H14}}{c_\ce{C7H8}}\right)^1\left(\frac{p_\ce{H2}}{p^\circ}\right)^3$$

whereas $K_1$ is right and $K_2$ is left.
I mean that's clear and everything but how do I know the aggregate state of $\ce{C7H14}$?

The standard for equilibrium constants of gas phase substances is to use $K_p$, so in this case $p^\circ$ would be more appropriate than $c^\circ$ for the organics as well as the hydrogen. If you use $c^\circ$, you should use it for all gas phase components and indicate your $K$ as $K_c$. Using different standards for gas phase components in the same reaction is unorthodox.
And if you use your equilibrium constant for calculation of $\Delta G$ for example, the simple form of the equation $\Delta G^\circ = -RT\ln K$ requires that the $K$ be $K_p$ for gas phase reactions. Using $K_c$ will give an incorrect result.


Do quantum particles have mass? Quantum particles are never objects but are always waves. But do they have mass or can they carry mass?

welcome to the chemistry StackExchange!
I think it would be helpful for you to review the wikipedia page on particle-wave duality. If you have access to a textbook on the origins of quantum mechanics, that would help clear things up,too.

The quantum mechanical model for particles states they have both particle-like and wave-like behavior, so it is incorrect to say they are never particles, which is what I assume you mean by "objects".

At the most fundamental level, quantum mechanics is just another tool with which to view and understand the world around us. It's an especially useful tool when we are looking at things in very fine detail (let's say, usually < $10^{-7}$ m). So, we can look at a given particle from the framework of classical mechanics or from the framework of quantum mechanics. This is an important point, because it is not quantum or classical mechanics that attributes mass to the particle. Rather, the particle had that mass to begin with.

So, to answer your question: A quantum mechanical treatment neither prescribes nor removes mass from a particle or an object. A photon is massless, a proton has mass, and that's that!


Why is 1H-borepine aromatic? For a compound to be aromatic, each atom in the ring must be a part of conjugation. Boron here has no lone pair, so how can 1H-borepine be aromatic as I was told by someone?

1H-borepine is isoelectronic to the cycloheptatrienyl cation. In both cases, you have a planar ring of seven atoms, whose hybridisation is close enough to $\mathrm{sp^2}$ to allow for a third p-orbital to be perpendicular to the ring plane. And in both, you are then filling these seven π orbitals with 6 π electrons, resulting in a Hückel-aromatic system ($4n+2 = 6$ for $n=1$).

There is no requirement for each atom along the chain of conjugation to have a lone pair; for example, in benzene no carbon atom has a valence lone pair. The only requirement for Hückel aromaticity is through conjugation, i.e. each atom must have an accessable p-type orbital that can overlap with the neighbouring atoms’ p-orbitals in a favourable π manner. However, neutral aromatic systems with empty p-orbitals are rare if boron is not used as most non-metals do not have favourable electronic configurations.


Why does hydrogen burn with a pale blue flame while its emission spectral lines are red in colour? I was studying oxidising flames and realized that sodium burns with a bright yellow flame the wavelength of which is around 588 nm. Then I searched for emission spectra of sodium and found that the lines around 588 nm are the most intense ones, so it appears yellow. The same happened with potassium.
But when I arrived at hydrogen, I knew it burned with a pale blue flame, but its emission spectral lines were most intense in the red region. So why does it still gives a blue flame?

As you see, it emits blue light too, but the red one is the most intense and so electrical discharge of hydrogen emits red light. But its oxidising flame is pale blue.
Would it burn with a red flame at higher temperature or something like that or are spectral lines not related to flame colours?

It is a very interesting question, but comparing a combustion spectrum with an atomic emission one is like comparing apples and oranges. A flame is a luminous gas phase chemical reaction where the hydrogen atoms are combining with oxygen atoms. It is a chemiluminescence phenomenon. A discharge tube emission is an atomic emission.
You posted a picture of a hydrogen spectrum from a discharge tube which consists of discrete lines in the UV, visible, and IR regions. In the same way, the hydrogen-air flame has an emission in the ultraviolet, visible and in the IR. The blue flame is not a line spectrum, as you might speculate. It is a continuum (broad band), which indicates that this is due to molecular emission in the flames, not from hydrogen atoms! This is in contrast with the alkali metals whose compounds easily atomize in the flame.
Also note that what may appear to us as a single color, like blue, is rarely a single wavelength.
There is a beautiful article by R.W. Schefer , W.D. Kulatilaka, B.D. Patterson, and T.B. Settersten on Visible emission of hydrogen flames, Combustion and Flame 156 (2009) 1234–1241. Go to Google Scholar and find it; it is open access on Google.
Look at their experiment: They view the "blue" hydrogen flames by various optical filters which block or pass certain wavelengths. See many colors are present in the "blue" flame of hydrogen. In fact most of the emission is in the ultraviolet region.

Direct flame luminosity photographs in a laminar, diffusion H2-jet flame. (a) Unfiltered, f /2.4 aperture; (b) short-wavelength pass filter with 550 nm cutoff wavelength,
f /2.4 aperture; (c) long-wavelength pass filter with 530 nm cutoff wavelength. Images taken at f /2.4 aperture with 90-ms exposure time. Fuel jet exit velocity is 47 m/s
with a coflow air velocity of 0.57 m/s. Reynolds number = 837.
Most of the emission is in the UV and infrared as you can see from the spike near 300 nm and intense band after 700 nm. Most people cannot see beyond 700 nm. The infrared emission is due to excited water molecule or perhaps its fragments.

Note: The above spectrum consists of two parts, UV on the left, and visible extending from $\pu{350 nm}$ to $\pu{850 nm}$ on the right hand side, recorded separately. For this figure, the later was amplified by a factor of 6.5 with respect to the OH in the ultraviolet.


Protecting elemental Li from the atmosphere I would like to make a small decorative object out of lithium, and keep it in an ordinary household (indoors) atmosphere.

The making itself aside, what can I do to protect it long-term from reacting with the atmospheric water and gases? Is there something that could react with it to form a protective outer layer, like oxides do on some other metals? Is it perhaps possible to electroplate it with Mg or Al? Are there other solutions?
To be clear, I don't want to just keep it submerged in oil; I want it to stand on a shelf and hopefully even get handled occasionally by curious hands.

I have no access to advanced equipment or supplies - I'm only armed with a high-school level of chemical understanding, access to ordinary hardware/paint shops, and patience.

I do understand that the question is unusual and has little broad practical utility, but it's still a proper chemistry problem to which I would like to see a proper chemistry solution (or an exhaustive explanation why a search for a solution is futile). Please don't answer along the lines of "your idea is stupid and you should feel stupid".

Industrially, this has been done by coating with Parylene.

The Rubber and Plastics Age, Volume 46 says:

  Parylene coated lithium metal granules on the left exhibit no reaction with water on which they float

and  New linear polymers says:

  Encapsulation of reactive chemicals in thin parylene C coatings provides protection. Parylene-coated lithium metal granules at left exhibit no reaction with water on which they float Unprotected granules at right react instantly and violently with water.

However, I don't know on what timescale this protection lasts.  As others are saying, lithium metal is an extreme fire hazard.  What if the coating is scratched or peels? Be careful!


Relationship between Lewis Theory and Redox theory? First of all, thank you ahead of time for taking the time to read my question and answer!

Let us take the reaction $\ce{SnI4 + I^{-} -> SnI2 + I3^{-}}$ as an example.

Here we see one obvious redox reaction, $\ce{Sn(IV)}$ gets reduced to $\ce{Sn(II)}$.
Less obviously, the iodide gets oxidized to $\ce{I(I)}$ in the triiodide anion. The other 4 iodines maintain their negative 1 oxidation state.

Is there also a lewis theory acid/base reaction here as well?
The tin acts as a lewis acid and accepts an electron from each of the two of the iodines (lewis bases)? Those iodines, now electron deficient (act as lewis acids), bond with the iodide (acting as a lewis base) to create $\ce{I3^{-}}$ product.

In this case, Tin accepts one electron from two atoms rather than a "pair of electrons" from one atom like you see in traditional examples of lewis acids.
So that would be my first question, can Tin act as a "lewis acid" even though it is accepting one electron from two different sources rather than a pair from one source? Or in other words, can the iodines be considered "lewis bases" even though they only donate one electron?

Or is the whole paragraph above a moot point because there is no acid/base reaction here because the Tin is simply regaining "ownership" of the electrons it let iodine "borrow" to create a covalent bond and thus its not actually getting any "new" electrons.

My second question is: Can we make a general statement about the relationship between lewis acids and their roles in a redox reaction? For example, can I say that ALL lewis acids are oxidants and ALL lewis bases are reductants? Or are these concepts mutually exclusive? In other words, sometimes lewis acids can act oxidants, sometimes as reductants, and sometimes as neither, like in the example below?:

$\ce{Al(III)Cl3 + Cl- -> Al(III)Cl4-}$ (no redox here)

Under the Usanovich definition of acids and bases, any substance which loses electrons is a base while any substance which gains electrons is an acid. Here, any redox reaction is also an acid-base reaction, whereas Lewis theory has nothing to state in these cases.

Regarding your second question, blanket statements are dangerous. Depending on exactly what you mean, I can provide a counter-example. Here is a strong oxidant ($\ce{HNO3}$) reacting as a Lewis base:

$$\ce{HNO3(l) + H2SO4(l) -> H2NO3^{+} + HSO4^{-}}$$


How does molecular collision excites the electron? I have this question for a long time that in LASER when we do the pumping by heating how does molecular collision causes the atoms(molecules) to reach in excited state. E.g. in $\ce{He-Ne}$ laser when the excited $\ce{He}$ atom collides with the $\ce{Ne}$ atom the $\ce{Ne}$ atom gets excited and $\ce{He}$ atom reaches its ground state.

My question are

What happens at the molecular level that causes $\ce{Ne}$'s electron to jump into higher state? Does Ne atom absorb some kind of photon by some mechanism?
Another thing is if we take $\ce{He}$ atom as the frame of reference then the high energy $\ce{Ne}$ atom is striking $\ce{He}$ so the high enrgy $\ce{Ne}$ must give its energy to $\ce{He}$ not the other way round.Where did I go wrong here?

The general rule is that you do not need to absorb the photon to excite an electron. What you need is to provide some portion of energy.
It can be done by means of a photon which is absorbed by an atom, of course, but it is not the only way.

Alternatively, you could collide an atom with some other object and upon the collision some of the kinetic energy is converted into internal energy which is then redistributed by its components. One of the components of the internal energy is the electronic energy, so if enough kinetic energy is converted into this particular part of internal energy, an electron gets excited. This process is usually referred to as the collisional excitation. This is the process by which an electron in $\ce{He}$ atom is excited: $\ce{He}$ atom in its ground electronic state collides with an energetic electron which causes electron excitation.

A collision between $\ce{He}$ atom in an excited electronic state and ground state $\ce{Ne}$ atom also involves an energy transfer. But this time it is the excitation energy (possibly together with some amount of kinetic energy as well) which is transferred from $\ce{He}$ atom to $\ce{Ne}$ atom. Due to a near coincidence between the $\mathrm{2\,^3S_1}$ and $\mathrm{2\,^1S_0}$ energy levels of the $\ce{He}$ electronic states, and the $\mathrm{3s^2}$ and $\mathrm{2s^2}$ energy levels of $\ce{Ne}$, the transfer of excitation energy from $\ce{He}$ atom to $\ce{Ne}$ atom is quite efficient.

Image courtesy of Wikipedia.

Disclaimer: I'm a professional chemist, not a physicist, so I might be wrong in what follows...

... but as far as I understand the matter at the microscopic level the energy is transferred by means of electromagnetic radiation the quantum of which is indeed photon. But do not think about this process as a photon being somehow "produced" in a form of a "billiard ball" by one atom, then transferred to another atom and finally magically "disappearing" here. This classical way of thinking is the root of confusion.

Before the energy transfer we have two systems each described by its own wave function. In order for the energy to be transferred the systems should start to interact with each other, but once they do so they become entangled and we do not even longer have separate systems. Rather, we have a big composite system described by its own wave function. After the interaction we again can speak about the individual systems each of which is again in own quantum state, though, the final state of each system is different from the starting one due to energy transfer. And in order to predict the final state of both systems (to be more precise, the probabilities of all possible final states) one could use the rules of quantum mechanics.


How should the hydrated proton be represented in chemical equations? I learned the equation

$$\ce{H2O + H+ -> H3O+}$$

And I know $\ce{H+}$ and $\ce{H3O+}$ really mean the same thing. But I am confused as to when I should use $\ce{H+}$ and when I should use $\ce{H3O+}$? Is any one of them a more accurate representation?

In addition to entropid's answer, let's remember why we invoke the hydronium ion $\ce{H3O+}$ in the first place.

We use $\ce{H3O+}$ as a shorthand for $\ce{H+(aq)}$, which looks more like protonated water clusters of the generic formula $\ce{H+.(H2O)_{$n$}}\equiv \ce{H_{$2n+1$}O_{$n$}+}$.

Almost ten years ago, a very interesting paper appeared in Science that examined the structure of these clusters ($n=2,3,4,5,6,7,8$). The linked article has been made freely available by Science, but you need to register for a free account at sciencemag.org.

One of the key points of the article was the determination of the inner structure of these clusters. For $n=1$, you have $\ce{H3O+}$, the hydronium ion, which we are reasonably familiar with. When $n=2$, you have $\ce{H5O2+}$, which has a different structure: the proton is evenly shared between two water molecules: $\ce{[H2O\bond{...}H\bond{...}OH2]+}$. Higher order clusters have these two structures, termed the "Eigen" and "Zundel" ions, respectively, at their cores.

The study used a combination of theory (MP2/aug-cc-pDVZ) and experiment (photodissociation vibrational spectroscopy). They found the for $n=2,6,7,8$, the clusters have a Zundel core, while the Eigen core exists for $n=3,4,5$. Whoa! And it gets more complicated as you increase the number of water molecules in the clusters (might be behind a pay wall) - though it turns out the Eigen ion is slightly more common in larger clusters.

In aqueous solutions, we can have an extensive dynamic network of clusters always in flux. In which case, we have a mix of solvated Eigen $\ce{H3O+}$ ions representing protons closely associated with water molecules and Zundel $\ce{H5O2+}$ ions representing protons in transit.

So, we use $\ce{H3O+}$ to keep things simple, and also to satisfy the needs of Brønsted-Lowry acid-base theory, in which every acid-base reaction needs a proton-donor and a proton-acceptor. Thus, the "dissociation" of acids in water is not really a dissociation, but an ionization reaction - an acid-base reaction in which water is the base and hydronium is the conjugate acid.

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

Of course, it is only appropriate to use $\ce{H3O+}$ in aqueous solution. In other solvents, the "proton" has a different structure. In general, the proton is attached to a basic site on the solvent molecule. For example, in methanol:

$$\ce{HCl + CH3OH <=> Cl- + CH3OH2+}$$


Understanding Gibbs free energy and enthalpy Let's say that a reaction releases -50kJ/mol of Gibbs free energy (delta G), and -75kJ/mol of enthalpy (we know that -25kJ/mol went to make sure that the total entropy of universe increased).

Now let's say we reversed the reaction.

Book says that the minimum energy required to make the reaction occur spontaneously in the reverse direction is the negative of the original change in Gibbs free energy (50kJ/mol).

What bothers me here is that I would think that you would need 75kJ/mol to reverse the reaction - the total energy the original reaction emitted needs to be put back in order to bring the system back to it's original potential energy level. It seems as though enthalpy shows that one amount of energy is needed to reverse the reaction, while Gibbs potential energy indicates a different amount of required energy.

I have no problem solving any of the computational problems relating to these concepts, but I'm curious about how to more fully understand them conceptually. I've noticed that a lot of people are great at solving technical problems related to these concepts, but are not too worried about how to interpret them conceptually. I guess for me, on the other hand, it's more important to understand than to compute...

Thank you for your help!

The equation for Gibbs Free Energy is:

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

The change in Gibbs Free Energy already includes the amount of energy associated with the change in energy. So to reverse the reaction, the total change in energy is given by $-\Delta G$.

Now the question is how much energy do you NEED to get the reaction to go backwards. The minimum amount of energy for it to go backwards spontaneously is $-\Delta G$ because that puts the product on the same energy level as the reactants so in principle they are energetically equivalent. However, what you should really be concerned about was the activation energy of the forward reaction, $E_a$, because this is the mountain that the reaction needs to climb. There is a diagram below, but it is based on the $\Delta H$ values. You can convert this to a graph with $\Delta G$ as well.


How do I name this compound I think I numbered it the correct way but I do not know which is the correct way to name it. Also please give the correct reason.

My answer was:

cyclohexane 1-carboxylic acid 2-oxo 4-carbaldehyde

OR

4-carbaldehyde 2-oxocyclohexane - carboxylic acid

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  15. aldehydes
  16. ketones
  (…)

Thus, the principal characteristic group of the compound that is given in the question is the carboxylic acid group. This principal characteristic group is expressed at the end of a name by means of a suffix or a class name, i.e. ‘cyclohexanecarboxylic acid’.

The other groups of the given compound cannot be expressed as a suffix. The cyclohexanecarboxylic acid is substituted using the usual principles of substitutive nomenclature.

For the aldehyde group, the suffix ‘carbaldehyde’, which generally can be used when the $\ce{-CHO}$ group is attached to a ring as in cyclohexanecarbaldehyde, cannot be used in this case. Instead, the prefix ‘formyl’ can be used in the presence of a characteristic group having priority to be cited as a suffix or when the $\ce{-CHO}$ group is attached to a side chain.

Likewise, the suffix ‘one’, which generally can be used to form the name of ketones as in cyclohexanone, cannot be used in this case. Instead, the prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix.

Therefore, the correct name for the given compound is 4-formyl-2-oxocyclohexane-1-carboxylic acid.


Why does sunlight cause colors to fade? If you leave something outside, its colors seem to inevitably fade or bleach due to exposure. Is this due to UV absorption? What sort of mechanism causes this - is it that man-made dyes deform on a molecular level? Are there notable materials that are exceptions?

I have a full undergraduate background in physics but not so much training in chemistry. A technical explanation would be cool, though.

To turn this string of questions into a focused query, let me refer back to the title:

Why does sunlight cause colors to fade?

Dyes and pigments work by absorbing certain wavelengths of light and reflecting or transmitting the rest. When a dye molecule absorbs a photon, an electron is excited to a higher energy state. Most of the time (neglecting fluorescence), the molecule de-excites by giving off heat and returns to the ground state intact. However, because the excited state is a high energy state, it has the potential to undergo a chemical reaction, breaking a covalent bond or otherwise irreversibly reacting with another molecule. This changes the electronic structure of the molecule which changes its absorption properties: e.g. many dyes that absorb visible light have large systems of conjugated double bonds and if these are broken, the absorbance can shift to much shorter wavelengths. How likely this kind of destructive chemistry is depends on the nature of the dye. Organic dyes tend to be more susceptible to photobleaching than things like quantum dots and inorganic pigments.

Touching on Nicolau's comment, red paint can often degrade faster because it absorbs higher energy (shorter wavelength) light. The more energy that is available, the more likely it is that reactions will occur. This is more pronounced as we enter UV wavelengths, which is why UV-B can cause thymine dimers in DNA, which can lead to skin cancer. For outdoor signs and such, UV-blocking coatings are often applied to extend the life of the dyes.

It's also not necessary for the light to interact directly with the dye molecules to cause bleaching: depending on what other molecules are present with the dye (solvents, etc), these other molecules may produce reactive species that then react with the dye.


How many ether metamers are possible for C4H10O? How many ether metamers are possible for $\ce{C4H10O}?$

According to me, the answer should be 2 but it was given 3. I know 3 structures can be made:

$$\ce{CH3—O—CH2—CH2—CH3}\tag{1}$$

$$\ce{CH3—CH2—O—CH2—CH3}\tag{2}$$

$$\ce{CH3—O—CH(CH3)2}\tag{3}$$

Structure 2 and 3 are metamers, but 1st and 3rd structures are positional isomers(?), then how the answer can be 3?

"Metamerism" is an extremely old, obsolete, and ill-defined property. It is also not educationally or synthetically useful. An additional, and pointless, way to categorise chemical compounds.

See this answer to the question "What is metamerism": https://chemistry.stackexchange.com/a/104576/17952

If you have to answer this in a test, do your best with whatever definition your teacher has provided. And tell him that he should stop teaching such nonsense.


Why does gold react with cyanide and sodium hydroxide Why does gold react with cyanide and sodium hydroxide and what is the chemical reaction's impact on gold's atomic structure and bonding?

What I think is since in the 'flotation' process, when the water, chemicals and other substances are added, gold isn't affected and neither is the pyrite, but when the cyanide and caustic acid is mixed with the slurry, it separated both substances, then the oxygen or 'air' is added into the mix, so then the gold rises to the top and frothing over the slurry. I'm not sure if i'm correct, but I don't understand how it's related to the atomic structure and bonding of gold? This process is from (Super Pit - Kalgoorlie exploration mine)

Can someone please explain how the separation of gold from the mixed compounds is related to its atomic structure and bonding? Is it to do with the 'Elution' process? Thank you in advance!

Flotation and treatment of gold with $\ce{NaCN}$ in the presence of air are two distictive steps!

Flotation yields fine particles of gold (and pyrite) at the surface of a flotation tank, attached to foamy bubbles while the rest of the minerals sink to the bottom of the tank.

Flotation does not rely on differences in specific density, but on hydrophobicity!

In the cyanide treatment, the following process - similar to what Uncle Al described for the seawater case in his answer - occurs:

Solid gold is oxidized, the cations are bound by cyanide anions and form a water-soluble complex:

$$\ce{4Au + 8NaCN + O2 + H2O -> 4Na[Au(CN)2] + 4NaOH}$$

In order to reobtain solid gold, the $\ce{Au(I)}$ in the cyano complex $\ce{[Au(CN)2]^-}$ has to be reduced to $\ce{Au(0)}$ again. This can be done either by adding zinc dust, according to

$$\ce{2Na[Au(CN)2] + Zn -> Na2[Zn(CN)4] + 2Au}$$

or electrochemically.


Will mixing hydrogen and oxygen without a spark result in combustion? I plan to do the electrolysis of dilute sulphuric acid. This will produce hydrogen gas and oxygen gas at the anodes. I do not plan to collect these gases, but to simply let it form and release into the air without separating them. Would this carry any risk whatsoever, considering that an electric setup is nearby?

For hydrogen to burn or explode in air, its concentration needs to be at least 4%.

Unless you're using very high power levels, you're unlikely to reach that concentration in the room's air. You could ignite bubbles of hydrogen as soon as they pop at the surface, but if you don't, the hydrogen (with about 1/8 the density of air) will rise quickly, diffusing into the air as it goes.

If you're doing this using a battery, or a small line-operated power supply, there's very little danger of explosion.


Why does rice become sticky after cooking? Rice when not wet isn't sticky at all. But when you soak it in water for some time it gets sticky.

What's the chemical change that occurs in rice that makes it sticky?

Starch is what makes rice sticky. Starch surfaces, when dry, are closely packed together and they do not stick to each other. If they get moist, especially if left to soak, or boiled, the starch will "unwind" as it absorbs water and become sticky. Depending on the source of starch and the type, you can have cleavage of starch to sugar and it will become even more sticky. You can read this pdf for more information

Funfact, you can make your own glue with flour and water as an example.

I do not take responsibility for any mess or any unwanted results if for instance you decide to glue the broken baby seat with home made glue. =)


order of basic strength 2 is the least basic as the lone pair is involved in aromatizing the ring
1 is the maximum basic as one of the lone pairs is used to aromatize the ring and the other lone pair is available
but how do we distinguish between 3 and 4
the options are:
a.1>3>4>2
b.3>2>4>1
c.4>3>2>1
d.3>4>2>1
by the above discussion we can make out that the answer is a.
but how to distinguish between option 3 and 4. how can we say that 3 is more basic than 4.

I think 4 should me more basic than 3 since $ -CH_3 $ has $ +I $ effect due to which electron density on $N$ increases.


What components of safety should be included in a chemistry laboratory experiment conclusion? The focus of my question here is this: In a laboratory there is a Bunsen burner, a hot plate, hydrochloric acid, and concentrated ammonia. What would you mention about safety precautions?

As students of chemistry and science, we often need to write detailed conclusions about our laboratory experiments. This eventually becomes second-nature, but to have an an idea of which of the safety measures taken to include are often useful to ensure ourselves that we have not left anything out.

I believe that the following are some of the the most important components of any well written conclusion in a lab entry, something that your lab instructor will read and grade you on.

Purpose: Explain the goal and purpose of the experiment in a clear and concise manner.
Findings: Presents a reasonable interpretation of, and logical explanation for all findings pertaining to the problem and stated purpose.
Discussion: Discusses possible sources of error in detail, including their effect on the results and ways of avoiding them in the future.
Referencing experimental findings and explaining the known/expected results we were looking for. Mentioning and discussing reasons for trends, if any.

In particular my instructors last year were often interested in:

Safety: This is often what I always got marked down for. I would explain that hydrochloric acid is a caustic substance and should be treated with care to not get on your tissue by wearing clothing that does not expose skin and closed toe shoes, and to always carry out the experiment with safety goggles securely fastened. It never seems to be enough for them, even if I mention eye flush and chemical shower in case of emergencies. Should I mention to not snort or freebase it? /end sarcasm

Have I missed any safety components that should be included in a "complete" conclusion for the average experiment? What else might I have left out that instructors tend to look for?

Perhaps your instructors want more specific information about the hazard itself, rather than generic safety advice.  Some examples:

1) HCl: Strong acid, highly corrosive (nitpick: caustic is usually reserved for strong bases, corrosive for acids), emits corrosive and toxic fumes (not all acids do)

2) 29% NH4OH: Caustic, emits strong fumes (lachrymator--makes your eyes water), work in a fume hood, outdoors or in a well-ventilated area

3) Bunsen burner: No open flames in the presence of flammable vapors or liquids with high vapor pressure (much bigger hazard than just burning yourself)

4) Hotplate: Hotplate looks the same whether hot or cold, so assume it is hot, take care with flammable liquids with low flash points

Again, these are just examples, but I tried to tailor the answer to the item in question.


Why is strontium(II) ion bigger than krypton atom? $\ce{Sr^2+}$ is exactly the same as $\ce{Kr}$, in terms of electrons and orbitals.
The only difference between the two, is that $\ce{Sr^2+}$ has a couple of extra protons in the nucleus (and probably a couple of extra neutrons too, but these don't influence ionic/atomic size).
Considering the two extra protons in the nucleus of $\ce{Sr^2+}$, and the overall 2+ charge on the ion,
shouldn't this draw the valence electrons in closer to the nucleus in $\ce{Sr^2+}$ in comparison to $\ce{Kr}$?
(Krypton size= 88 pm, while Sr2+ is 132 pm.)

Sr2+ is exactly the same as Kr, in terms of electrons and orbitals.

I think the appropriate word for this relation is isoelectronic.
According to Ptable and LumenLearning, $\ce{Kr}$ has an atomic radius of 88 pm, while $\ce{Sr^2+}$ has an atomic radius of 132 pm. Although it is true that $\ce{Kr}$ and $\ce{Sr^2+}$ have the same electron orbital configuration and $\ce{Sr^2+}$ has two more protons than $\ce{Kr}$, $\ce{Sr^2+}$'s greater nuclear attraction is likely countered by the nature of metal ions being more willingly to lose electrons, which implies the electrons are more loosely held by the nucleus than the noble gas $\ce{Kr}$.
Another reason I conjecture is that $\ce{Sr^2+}$ has unfilled 5s orbitals to readily accept promoted electrons, while $\ce{Kr}$'s electrons have less tendency to fill the 5s orbitals because the property of being a noble gas.


What happens to minerals once you boil off liquid in a water and fat mixture? Let us say you are cooking a curry.  As water evaporates the minerals will become more and more concentrated in the curry's water but they will not deposit in the fat which would be at the top layer of the liquid. Regarding the minerals not depositing in the fat whilst water exists, is this correct?

Once you have evaporated all of the water which I imagine happens before any fat starts evaporating, would the minerals then deposit in the fat?

Is there a way to know by observation that water has evaporated and you are left with minerals and fat?

Because dietary minerals are predominantly water soluble and not significantly soluble in fats, your first statement is correct; the minerals will not dissolve in the fats as you boil away the water.  This is true even after all the water has boiled off, so the answer to your second question is no. As the water boils away, the minerals will precipitate out until no water remains. At that point, you will predominantly have the minerals at the bottom of the pan, covered in the fats. It is unlikely any of the minerals will have a low enough density to float on top of the fats. With mixing, you may be able to keep a suspension of the minerals within the fats even though they don't technically dissolve.

It will probably be difficult to tell exactly when all of the water is gone. There should be a difference in the appearance of the way the stuff boils once the water is gone and it's just fats boiling, but this could be a subtle change. You might be able to tell when the water is gone by monitoring the temperature of the mixture. It should stay close to the boiling point of water until the water is gone and then increase when you just have the fats boiling.

I'm far from a chef, so I hope I haven't made any improper assumptions on that front. Don't hesitate to ask for clarifications in the comments.


What does "the natural bond angle" mean in the context of Universal Force Field (UFF) potential energy calculation and how are bond angles calculated? As the title says, what does "the natural bond angle θ0" mean in the context of Universal Force Field (UFF) potential energy calculation? The reference appears in section D. Angular Distortions, 1. Angle Bend). The article also talks about "natural bond length" (Section C. Bond Stretch).
I am curious about how molecular dynamics (MD) simulations estimate/calculate bond angles using the Universal Force Field.
At several points in the original UFF paper they mention a "natural angle" but it is not overwhelmingly clear to me what this means. Is it the empirically observed bond angle(s) in particular molecules? How do MD simulations, then, produce bond angles for molecules that there is no published data for?
I understand that the UFF potential energy calculation contains several terms that account for various properties like hybridization, van der Waals interactions, etc. Which of the terms, which are outlined in the first pages, account(s) for steric interactions between molecules and determine the final bond angles?
I am admittedly not very well versed in the types of maths that are involved, but I would appreciate any explanation possible.
Thank you very much in advance!

The original article on UFF is from 1992, so it is somewhat dated. The selection of "natural bond" values is explained on page 10028:

a. Standard Bond Angles. The natural angles for the group 15, 17, and 18 main group elements are obtained from standard reference structures of the parent hydrides. Thus 0-3 has Bo = 104.5' from HzO, while S3 has Bo = 92.2' from HzS. Exceptions include 0-32, OR, and N-2. The bond angles for 0-32,O-R, and N-2 are fit to (C13Si)z0, methyl vinyl ether, and dimethyl- diazene angles of 146O, 118.3', and 112.3', respectively. Where structural data are unavailable, the natural angles are extrapolated from the element above it in the periodic table. The remaining elements are all assumed to have regular octahedral, tetrahedral, trigonal, or linear structures. The natural angles are collected in Table I.

So yes, the angles are selected from experimental geometries where available, or where not by extrapolation from geometries for compounds in the same group.
References

A. K. Rappe et al.. UFF, a Full Periodic Table Force Field  for  Molecular Mechanics and  Molecular Dynamics Simulations. J. Am. Chem. Soc. 1992, 114, 10024-10039.


Mechanism for acylation of benzene with succinic anhydride I've just started with EAS. I wanted to write down a mechanism for acylation of a cyclic anhydride (for some reason the formation of the carbocation is not given in my textbook). So I wish to know whether this mechanism is correct or now. Since this is my first attempt at writing one, I don't know whether it's correct.
The reaction:

What I came up with:

The mechanism shown in the question seems reasonable enough to me. I would note that ring opening before the attack on the $\pi$-system is likely, but not absolutely necessary. Loss of $\ce{AlCl3}$ may not occur until workup. Direct proton transfer seems unlikely, but is a inconsequential detail. Note however, that the electron pushing arrow is drawn opposite to what it needs to be for that transfer.


What is the purpose of Gaussian, itself? I have been doing some research into computational chemistry, having never looked into the area before, and have read about a popular, paid-for software "Gaussian", and free alternatives such as "GAMESS". From what I understand they are both capable of employing methodologies such as PM3, but I am struggling to understand the point of the software itself; how is a Gaussian PM3 calculation any different to a GAMESS calculation? There must be a reason one would want to pay for it.

There are several interrelated reasons (disclaimer: I am a GAMESS developer).

Gaussian has a lot of features and interoperability.  To take your PM3 example. Yes, the results of a PM3 calculation are identical in GAMESS and Gaussian. But until recently you couldn't do a PM3 calculation in solution using the PCM method in GAMESS.  Also, what if you want to do a PM6 calculation instead. In GAMESS this is currently only possible only for elements up to F.

Because of this Gaussian has been the program of choice for many people, which also means they have trained a lot of people to use Gaussian.  Learning to use a new (and perhaps free) program well can be very time consuming, so it might be more cost effective to pay for a program you know how to use well.  And chances are you probably have to learn how to use several new programs to get all the functionality of Gaussian.


nomenclature of some organoselenide moieties what are the structures of these selenide moieties : selenenylate, selenolate ?

Can anyone tell the difference between the two in the terms of if they are anions or cations , electrophiles of nucleophiles.

Based on usage:

Selenenylate is a verb. It means to attach a selenium group to an organic compound - See this passage from Comprehensive Organic Synthesis: Selectivity, Strategy, and Efficiency in Modern Organic Chemistry, Volume VII

  The analogous morphinoselenamide has been shown to selenenylate the $\alpha$-ketoester...

Selenolate is a noun. It is the $\ce{RSe-}$ anion, especially in ionic or coordination compounds. See various usage at the links in this Google search


Why does sweetening milk need less sugar as compared to tea? If I have a cup of milk and a cup of tea; to get them to a sweetness I like, I usually add less sugar to the milk as compared to the tea.

Why is it that milk needs less sugar to be just as sweet as tea?

It's all in your taste buds. Lactose in milk is a form of sugar. It is naturally a little sweet. Tea is slightly bitter so it might seem you need to add more sugar to make it sweet.


Temperature and pressure for three phase equilibrium of aqueous sodium chloride solutions I am currently investigating the effect of adding Salt(NaCl) to water on its triple point. Let us assume the mass ratio of water to NaCl is $1:1$. I asked a similar question previously. But, I still haven't been able to find the temperature(s) of the triple point(s).

I understand that according to the Gibbs equation $f = C-P+2$ Therefore, in my case of Salt water, $f = 1$. Thus we have a range of temperatures/presures where the triple point exists.

But how can we find this range of temperature or pressures? IS there an experimental methods?

But how can we find this range of temperature or pressures? IS there an experimental methods?

The range of temperatures is fairly easy. Mix saturated aqueous NaCl (liquid solution, concentration about 5 mol/L) with ice (pure water, solid state), and measure the temperature. It will be about equal to the temperature of a household freezer (no coincident, freezers were using salt/water mixtures in the old days), i.e. -20 $^\circ$C. If you keep measuring while the ice melts and the NaCl concentration drops, you will get the temperature ranges, which will have the normal freezing point of water at the high end.

Partial pressure of water measured at atmospheric pressure

Using the same setup, you could measure the humidity above the ice/solution slush. In the experiment, water exists in the solid, liquid, and gas phase, just like pure water at the triple point. The gas phase contains mostly nitrogen and oxygen, so you would have to measure the partial pressure of water rather than the total pressure (which is about 1 atm and depends on the weather). The partial pressure will be a bit lower than that of pure water at the same temperature (the activity of water is less than one in a solution, so the partial pressure will be lower relative to pure water, Roult's law).

Pressure of water measured with the aid of a vacuum pump

In a slightly more involved setup, you would place the slush into a vessel that you evacuate with a vacuum pump. This removes all gas components (nitrogen, oxygen) from the gas phase, and replaces them with evaporating water until the water boils and there is pure water in the gas phase. At that point, you can close off the system and measure the pressure. The result should be the same as the first experiment.


Backbonding in phosphorous pentoxide I read this today in a book that $\ce{P=O}$ in $\ce{P4O10}$ consists of a coordinate bond and pπ-dπ backbonding, but why does this happen?
Can't phosphorus share its lone pair with one of the lone pairs available in oxygen to form the double bond?

Coordinate bonds are covalent bonds where both bond electrons stem from only one of the two bond partners. They are formed when a Lewis base donates two electrons into accepting orbitals of the Lewis acid. That's not what you're looking at in this case, where both $\ce{P}$ and $\ce{O}$ contribute one electron to the common $\sigma$ bond. The interesting part is about everything that is to be described beyond the bond order of $1$. For example, everything that contributes to PO bonding beyond that single $\sigma$ bond.
I will further regard to phosphane oxides $\ce{P(=O)R3}$, where the question was originally asking about $\ce{P=O}$ bonds in $\ce{P4O10}$. The problem might be reduced to PO bonding in phosphoric acid, as $\ce{P4O10}$ is nothing more than its anhydride. The general observations regarding the $\ce{PO}$ bonding situation are unaffected. Kutzelnigg noticed in $1977$ already:

[The PO] bond is somewhere between single and triple and [...] only by chance it may happen to be a double bond.

What we know about the '$\ce{P=O}$ bond':

Its length is remarkably shorter than the P-O bond ($\pu{1.43 Å}$ compared to $\pu{1.60 Å}$ for $\ce{P4O10}$ according to Wikipedia).

It is highly polar.

Its bond strength of about $\pu{540 kJ/mol}$ is higher than for a corresponding single bond.

A critical review regarding the bonding situation in phosphane oxides was given by Gilheany. If it comes to Lewis valence formula, one might find three reasonable structures:

None of those is suitable to represent all theoretical and experimental findings regarding the characteristics of the $\ce{PO}$ bond!
Some scientists prefer to use the zwitterionic representation A, where the predominant $\sigma$ character of the $\ce{PO}$ bond is highlighted, while its dipolar character is also taken into account. If one chooses to write a Lewis structure as shown in B, one implies (besides one  $\sigma$ bond) the contribution of a $\pi$ bond between oxygen p orbitals and phosphorus d orbitals. However, the $3$d orbitals of phosphorus are no suitable $\pi$ accepting orbitals for $2$p orbitals of oxygen, as they are quite high in energy. Besides compatible symmetry, atomic orbitals (AOs) of the atoms in question require to be somewhat similar in energy to allow for a combination to form molecular orbitals (MOs).
If one allows $\pi$ back bonding from oxygen into $\sigma ^*$($\ce{P-R}$) (anti-bonding orbitals between P and its substituents R), one might come up with Lewis valence formula C. However, it should be noted that the $\pi$ bond order for $\ce{Me3PO}$ is only $0.7$ - and not $2$, what is implied by structure C!


Hydrogen and deuterium exchange Is this a valid mechanism? I think it is. It is just a simple acid base mechanism.

Could I increase yield though by using a stronger base than heavy water? Perhaps heavy sodium hydroxide? Or is chloroform sufficiently acidic?

If you are interested in making deutero-chloroform, there are better ways to do it.

From a purely mechanistic standpoint, chloroform isn't a very strong acid ($\ce{pKa \sim 15}$) so $\ce{D2O}$ is unlikely to deprotonate it at a useful rate.  Further, chloroform isn't very soluble in water making your desired reaction even less likely (but you could engineer your way around this problem by using a phase transfer catalyst).

Your suggestion to use a stronger base is reasonable, but the problem you run into here is that a different reaction is brought into play.  Specifically, once a stronger base like potassium t-butoxide (in t-butanol) is used, dichlorocarbene formation occurs.  Once the trichloromethylcarbanion is formed, it quickly ejects $\ce{Cl^-}$ to form dichlorocarbene.  Dichlorocarbene is an extremely energetic species and would quickly react with whatever else is around - perhaps the t-butanol to produce dichloromethyl-t-butyl ether.


Why is vulcanization called "vulcanization"? I know it might sound a dumb question but whenever I come across a note on importance of vulcanization I get stuck as to the meaning of the term and why it was adopted.

According to Thomas Hancock's 1857 Personal narrative of the origin and progress of the caoutchouc or india-rubber manufacture in England

  In preparing to carry out the new process on a
  large scale, it appeared desirable to give the mate-
  rial a more definite name than " the change,"
  which I had adopted for the purposes of my speci-
  fication; and, whilst discussing the subject amongst
  my friends, Mr. Brockedon proposed the term
  " Vulcanization," and, as no better suggested
  itself, I determined to adopt it, and it is now
  known by that name both in Europe and America.
  It owes its derivation to the Vulcan of mythology,
  as in some degree representing the employment of
  sulphur and heat, with which that mythological personage was supposed to be familiar. I shall,
  therefore, of course use the term in the remainder
  of my narrative.


I suspended iodine over steel nails. What are these droplets that formed? I suspended about 15g of iodine crystals over a couple of nails to see how quickly they would begin to rust.

After a week, these droplets had formed. When prodded they broke apart like water drops.

What are these? My guess is that the iodine sublimated and then combined with water vapor.

The original question that prompted this experiment.

The setup (~one week prior)

Iodine vapors combine with iron to form iron(II) iodide:$$\ce{I2 + Fe->FeI2}$$ In an anhydrous form the compound is dark red-purple. But iron(II) iodide is hygroscopic and deliquescent which means that it dissolves in moisture it absorbs from the air, and forms those droplets. This complicates a bit the chemistry of the droplets - the $\ce{Fe^2+}$ ion exists most likely in a form of various aquo-, hydroxo- and iodo- complexes. Also, when an iodide anion and elementary iodine are present together they form triiodide anions:$$\ce{I- + I2->I3-}$$ Therefore, the droplets are a solution of a rather complicated mixture of iron and iodine compounds.


Isn't water a Lewis base in the ammonia–water reaction? Exactly one of these statements is false. Which one?
(a) When ammonia dissolves in water, the latter behaves as a Lewis base.
(b) The conjugate acid of $\ce{H-}$ is $\ce{H2}.$
(c) Boric acid is a monobasic acid.
(d) Ammonia behaves as an acid when it reacts with sodium metal to give sodamide and hydrogen.

The key says (d) is the false statements. How? I think (a) is the one.
(d) $\ce{2NH3 + 2Na -> 2NaNH2 + H2}$
The ammonia here loses a proton, which is characteristic for acids. In addition, it attaches to the electropositive sodium.
(a) On the other hand, when ammonia dissolves in water:
$\ce{NH3 + H2O <=> NH4+ + OH-}$
Ammonia gains a proton, which is characteristic for bases.

Your reasoning is correct. Perhaps the answer key is mistaken.
For part (a), in the reaction
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
water is donating a proton $(\ce{H+}$ ion) and hence is behaving as a Brønsted acid. Since all Brønsted  acids are Lewis acids, water is behaving as a Lewis acid.
For part (d) as well, the half reactions are:
$$
\begin{align}
\ce{2 NH3 &-> 2 NH2- + 2 H+} \\
\ce{2 Na &-> 2 Na+ + 2 e-}
\end{align}
$$
It's clear that $\ce{NH3}$ is acting as an acid and $\ce{Na}$ is acting as a Lewis base by donating electrons. The $\ce{NH2-}$ ion pairs up with the $\ce{Na+}$ ion to form $\ce{NaNH2}$ and hydrogen accepts the electrons and is liberated as $\ce{H2}.$
EDIT: As pointed out by Mithoron in the comments, the reduction of ammonia is not so simple. Sodium dissolves in ammonia to give a deep-blue solution containing ammoniated electrons (reference here):
$$\ce{Na + \text{(x+y)}NH3 <=> Na(NH3)_x^+ + e(NH3)_y^-}$$
The $\ce{e(NH3)_y^-}$ eventually loses a hydrogen atom to give $\ce{NH2-}$, which then combines with the $\ce{Na+}$ giving the product.


why does FeCl₃ have such a specific smell? When I use $\ce{FeCl3}$ solution for etching circuits it gives off an acidic smell.
I was wondering what is the cause of the smell.
I don't think it's caused by the $\ce{Cl2}$ from $\ce{2FeCl3->2FeCl2 + Cl2}$ because chlorine has a different smell.

The product used is the one in this picture.

The only clue I have been able to find is a one-off sentence in the Wikipedia article for $\ce{FeCl3}$, for which no reference is given:

  Iron(III) chloride undergoes hydrolysis to give an acidic solution.

If so, then the reactions are probably those along the conversion of $\ce{FeCl3}$ to $\ce{Fe2O3}$, producing hydrochloric acid $(\ce{HCl})$. The hydrogen chloride gas that escapes from the solution has a pungent acidic acrid smell.

Possible reactions include:

$$\ce{FeCl3 + H2O <=> FeOCl + 2HCl}$$
$$\ce{2FeCl3 + H2O <=> Cl2FeOFeCl2 + 2HCl}$$
$$\ce{2FeOCl + 2H2O <=> Fe2O3(s) + 2HCl}$$

Because these reactions are in equilibrium, solutions of $\ce{FeCl3}$ can be stabilized by addition of $\ce{HCl}$. Addition of $\ce{HCl}$ also promotes the formation of the tetrachloroferrate ion $\ce{FeCl4-}$, which is more resistant to hydrolysis.

$$\ce{FeCl3 + HCl -> FeCl4- + H+}$$

Ferric oxide (Iron(III) oxide $\ce{Fe2O3}$), is insoluble. Does your solution develop reddish precipitate as it ages? That would be  evidence in favor of hydrolysis as the source of $\ce{HCl}$.