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Does a neutral dimercury molecule exist?
Is a neutral $\ce{Hg2}$ molecule possible, as a gas under extremely low (partial) pressures? What is the enthalpy of formation or similar?
It is my impression that usual mercury vapors are monatomic? I think it is well-known that a dimer ion $\ce{Hg2^{2+}}$ exists under some conditions.
Somewhat related threads: Is it possible to have a diatomic molecule of sodium in gaseous state? and Which elements can be diatomic?

TL:DR: It does exist, but it's a vdW molecule. Has much stronger bond then He2, but still...

Does a neutral dimercury molecule ($\ce{Hg2}$) exist? Simple answer is yes. But when I look at the answers given so far, a simple answer might not be good enough without giving some evidence. So, I decided to dig into the question a little deeper. I'm not a physical chemist, but can't resist the data I have seen in review and research articles for this subject so far. Thus, I’d like to share it with the community:
It is a fact as M. D. Morse (Ref.1) pointed out that, dimercury, $\ce{Hg2}$, was the first of the transition-metal molecules to be investigated, albeit $\ce{Hg2}$ is amongst the most weakly bound of the transition-metal dimers (In truth, most physical chemists found the weak attractive potential of the $\ce{Hg2}$ dimer was found to be a much harder problem in different computational approaches, e.g., Ref.2). The dissociation energy $\left(\mathrm {D^0_0} (\ce{Hg2})\right)$ estimations were first obtained by Franck and Grotian in 1922 (Ref.3) by monitoring the temperature dependence of the absorptions associated with $\ce{Hg2}$. This measurement has been repeated by numerous workers throughout the years, and an agreeable range of $\left(\mathrm {D^0_0} (\ce{Hg2})\right)$, $\pu {0.055 \lt D^0_{0} \lt 0.07 eV}$, has been obtained by 1977 (Ref.1). Then, using a third-law expression, Kuhn (Ref.4 & 5) has calculated the range of $\left(\mathrm {D^0_0} (\ce{Hg2})\right)$ as $\pu {0.065 \le D^0_{0} \le 0.091 eV}$. It is also noteworthy that some have investigated the interatomic potential in $\ce{Hg2}$ by gas-phase viscosity studies and all of them have provided $D_e$ in the range of $\pu{0.068-0.134 eV}$. In 1995, Fry and coworkers have proposed an experimental realization of Bohm’s spin-$\frac12$ particle version of the Einstein-Podolsky-Rosen experiment (Ref.6). They have used $\pu{350 cm^{-1}}$ ($\pu{0.0434 eV}$) for $D_e$ to calculate other parameters such as bond distance $r_e$ (3.63 Å), harmonic frequency $\omega_e$ ($\pu{18.5 cm^{-1}}$), anharmonicity constant $\omega_ex_e$ ($\pu{0.27 cm^{-1}}$), and rotational constant $B_e$ ($\pu{0.0127 cm^{−1}}$), using measured spectroscopic data for the $\mathrm{X0^+_g \leftarrow ^1\Sigma^+_g}$ transition. Also, they have calculated the same set of parameters in addition to the vibrational-rotational coupling constant $\alpha_e$ (in $\pu{cm^{−1}}$), for the $\mathrm{D1_u \leftarrow ^{3}\Sigma^+_u}$ transition as well.
Recently, a systematic analysis of the rotational structure in the (57, 0) band of the $\mathrm{D1_u \leftarrow X0^+_g}$ transition in $\ce{^{A_1}Hg^{A_2}Hg}$  molecule (where $A_1$ and $A_2$ are mass numbers of seven stable isotopes of natural abundance $\ce{Hg}$) is presented by Koperski and coworkers (Ref.7). It is very notable work because their goal was to examine the structure without any limitations and simplifications that were assumed in previous works of others. They also wanted to improve the rotational constants and bond lengths of the two electronic states involved in the transition.
In this work, a free-jet expansion beam of $\ce{Hg2}$ was produced in a vacuum chamber evacuated by an oil-diffusion pump backed by a mechanical pump. The molecules in the supersonic beam were then irradiated with a laser beam and optical date were recorded. The abundance of the seven stable $\ce{^{A}Hg}$ isotopes present in natural mercury are $\ce{^{196}Hg}$ (0.15%), $\ce{^{198}Hg}$ (9.97%), $\ce{^{199}Hg}$ (16.87%), $\ce{^{200}Hg}$ (23.10%), $\ce{^{201}Hg}$ (13.18%), $\ce{^{202}Hg}$ (29.86%), and $\ce{^{204}Hg}$ (6.87%) Wikipedia. If all of these isotope have contributed to make dimers, then the resulting dimers would consist of 28 isotopomers with 15 distinct total mass numbers. The following Figure presents details of the (57, 0) vibrational band recorded with a low laser resolution (i.e., without the intracavity etalon). The total isotopic composition of the (57, 0) band consists of only twelve components of theoretical 15, which were sufficiently abundant to be observed in the experiment (Ref.7):

It is also noteworthy that this experiment can explain antisymmetric rotational levels, which do not exist for homonuclear isotopomers such as $\ce{^{199}Hg^{199}Hg}$ with zero nuclear spin ($I_1 = I_2 = 0$) that has been proposed previously (Ref.6). Following schematic energy level diagram shows rotational levels in the $\mathrm{X0^+_g}$ and $\mathrm{D1_u }$  states in $\ce{^{A_1}Hg^{A_2}Hg}$  molecule, including their symmetries and the transitions between them forming the R, Q and P branches. Broken lines represent antisymmetric rotational levels, which do not exist for homonuclear isotopomers (where $A_1 = A_2$) with zero nuclear spin (vide supra):

Finally, Table 1 below summarizes the spectroscopic constants for the $\ce{Hg}$ dimer, as compared to known experimental data and most recent theoretical calculations obtained at different theoretical levels. Most of theoretical calculations are done in Ref.8 and reliable experimental data so far published in literature are as indicated in the Table foot notes, which are through 2010 (Ref.8):

References:

Michael D. Morse, “Clusters of transition-metal atoms,” Chem. Rev. 1986, 86(6), 1049–1109 (DOI: 10.1021/cr00076a005).
P. Schwerdtfeger, J. Li, P. Pyykkö, “The polarisability of Hg and the ground-state interaction potential of $\ce{Hg2}$*,” Theoretica Chimica Acta 1994, 87(4-5), 313-320 (https://doi.org/10.1007/BF01113387).
J. Franck, W. Grotian, Z. Techn. Phys. 1922, 3, 194 (Reference 323 in Ref.1).
H. Kuhn, “Pressure Broadening of Spectral Lines and van der Waals Forces I—Influence of Argon on the Mercury Resonance Line,” Proc. R. Soc. London. Ser. A – Math. Phys. Sci. 1937, 158(893), 212-229 (https://doi.org/10.1098/rspa.1937.0015).
H. Kuhn, “Pressure Broadening of Spectral Lines and van der Waals Forces II—Continuous Broadening and Discrete Bands in Pure Mercury Vapour,” Proc. R. Soc. London. Ser. A – Math. Phys. Sci. 1937, 158(893), 230-241 (https://doi.org/10.1098/rspa.1937.0016).
E. S. Fry, T. Walther, S. Li “Proposal for a loophole-free test of the Bell inequalities,” Phys. Rev. A 1995, 52(6), 4381-4395 (https://journals.aps.org/pra/pdf/10.1103/PhysRevA.52.4381).
J. Koperski, X. Qu, H. Meng, R. Kenefick, E. S. Fry, “Rotational analysis of the (57, 0) band of the $\mathrm{D1_u \leftarrow X0^+_g}$ triplet–singlet transition in $\ce{Hg2}$ produced in a free-jet expansion beam,” Chem. Phys. 2008, 348(1-3), 103-112 (https://doi.org/10.1016/j.chemphys.2008.02.035).
E. Pahl, D. Figgen, C. Thierfelder, K. A. Peterson, F. Calvo, P. Schwerdtfeger, “A highly accurate potential energy curve for the mercury dimer,” J. Chem. Phys. 2010, 132(11), 114301 (6 pages) (https://doi.org/10.1063/1.3354976).
J. Koperski, J. B. Atkinson, L. Krause, “Molecular beam spectroscopy of the  $1_u(6^3P_2)~- X0^+_g$ and $1_u(6^3P_1)~- X0^+_g$ transitions in $\ce{Hg2}$,” Canadian Journal of Physics 1994, 72(11-12), 1070-1077 (https://doi.org/10.1139/p94-139).
J. Koperski, J. B. Atkinson, L. Krause, “The  $G0^+_u (6^1P_1)~- X0^+_g$ Excitation and Fluorescence Spectra of $\ce{Hg2}$ Excited in a Supersonic Jet,” Journal of Molecular Spectroscopy 1997, 184(2), 300-308 (https://doi.org/10.1006/jmsp.1997.7345).
R. D. van Zee, S. C. Blankespoor, T. S. Zwier, “Direct spectroscopic determination of the $\ce{Hg2}$ bond length and an analysis of the $\pu{2540 Å}$ band,” J. Chem. Phys. 1988, 88(8), 4650-4654 (https://doi.org/10.1063/1.453777).
J. N. Greif-Wüstenbecker, Ph.D. thesis, Philipps University, Marburg, Germany, 2000 (Reference 28 in Ref.8).
P. Schwerdtfeger, R. Wesendrup, G. E. Moyano, A. J. Sadlej, J. Greif, F. Hensel, “The potential energy curve and dipole polarizability tensor of mercury dimer,” J. Chem. Phys. 2001, 115(16), 7401-7412 (https://doi.org/10.1063/1.1402163).
P. Schwerdtfeger, R. Wesendrup, G. E. Moyano, “Erratum: ‘The potential energy curve and dipole polarizability tensor of mercury dimer’ [J. Chem. Phys. 115, 7401 (2001)],” J. Chem. Phys. 2002, 117(14), 6881 (https://doi.org/10.1063/1.1507115).
V. Lukeš, M. Ilčin, V. Laurinc, S. Biskupič, “On the structure and physical origin of van der Waals interaction in zinc, cadmium and mercury dimers,” Chem. Phys. Lett. 2006, 424(1-3), 199-203 (https://doi.org/10.1016/j.cplett.2006.04.002).
K. A. Peterson, C. Puzzarini, “Systematically convergent basis sets for transition metals. II. Pseudopotential-based correlation consistent basis sets for the group 11 $(\ce{Cu, Ag, Au})$ and 12 $(\ce{Zn, Cd, Hg})$ elements,” Theoretical Chemistry Accounts 2005, 114(4-54), 283-296 (https://link.springer.com/article/10.1007/s00214-005-0681-9).

There has been recent research on mercury dimers in its ground state. Bond energies, transition energies, band spectrum and other spectroscopic parameters has been calculated. Here are the abstracts of two papers on mercury dimer:

The potential energy curve of the ground electronic state of the $\ce{Hg}$ dimer has been calculated using the CCSD(T) procedure and relativistic effective core potentials. The calculated binding energy ($\pu{0.047 eV}$) and equilibrium separation ($\pu{3.72 Å}$) are in excellent agreement with experiment. A variety of properties, including the second virial coefficient, rotational and vibrational spectroscopic constants, and vibrational energy levels, have been calculated using this interatomic potential and agreement with experiment is good overall.(Source)

The mercury dimer is among the most weakly bound metal dimers and has been extensively studied. The ground state $\ce{O^+_g}$ dissociation energy has been considered to lie between $\pu{0.55 eV}$ ($\pu{440 cm−1}$) and $\pu{0.091 eV}$ ($\pu{730 cm−1}$). We report here a spectroscopic study of $\ce{Hg2}$ in a supersonic jet. The first optical transition, $\ce{1_u←O^+_g}$, was characterized by its fluorescence excitation spectrum and the binding energy of the ground state has been measured precisely through the threshold of collision induced dissociation of $\ce{Hg2 1_u}$ to $\ce{Hg(^1S_0)+Hg(^3P_0)}$.(Source)

Yes, $\ce{Hg2}$ has a bond length of $\pu{0.334nm}$  and a dissociation energy of $\pu{7.5 kJ/mol}$.
See Mercury Handbook: Chemistry, Applications and Environmental Impact at page 10.
and
Mass spectrometric equilibrium study of the molecule $\ce{Hg2}$ J. Chem. Phys. 1982, 77(3), 1425-1427 (https://doi.org/10.1063/1.443968).

Respectfully, this isn't a molecule.

@permeakra is the potential well enough to have at least one vibrational state?

Doesn't matter. Water dimer is not considered a molecule, even though the bonding it fairly strong.

@permeakra it matters to IUPAC https://goldbook.iupac.org/html/M/M04002.html

The answer by Nilay Ghosh quotes $\pu{0.047 eV}$ (per molecule, I suppose) as the binding energy. That corresponds to $\pu{4.5 kJ/mol}$. Your answer and source say $\pu{7.5 kJ/mol}$. So are your data outdated?

@JeppeStigNielsen The book in my answer is 12 years more recent, but cites to older references.  When I have access to full text of Nilay articles, I'll try to figure out the discrepancy.

The nature of mercury associates was studied. Quick googling gave paper with words "$\ce{Hg_x}$ cluster transition from VdW to metallic bahavior between 20 and 70 atoms", suggesting that $\ce{Hg_2}$ associate, if exists, have VdW bonding (i.e. not a chemical, but physical bond) only.

According to IUPAC, the type of binding isn't relevant, as long as there is "a depression on the potential energy surface that is deep enough to confine at least one vibrational state" https://goldbook.iupac.org/html/M/M04002.html

@DavePhD What IUPAC says and what is accepted in the community are two entirely different things.

permeakra, I agree, and I cited a J. Chem. Phys. article that says "molecule Hg2" in the title. https://aip.scitation.org/doi/10.1063/1.443968  Here's one that says "Hg2 molecule". https://aip.scitation.org/doi/pdf/10.1063/1.449261

@DavePhD Again, what is accepted in physics community (J. of Chemical Physics) and chemical community (like chemistry stackexchange) also are two entirely different things. $\ce{He_2}$ associate is a thing and from physical perspective might be a molecule. But from chemical perspective He is not capable to form chemical bonds (except in conditions that are domain of physics) and thus $\ce{He_2}$ is not a molecule. Same for $\ce{Hg_2}$

The Journal of Physical Chemistry, published by the American Chemical Society, says "the Hg2 molecule".  https://pubs.acs.org/doi/abs/10.1021/jp952808p?journalCode=jpchax  The Canadian Journal of Chemistry says "Hg2 molecule".  http://www.nrcresearchpress.com/doi/pdf/10.1139/v68-031 It's just a word.  It doesn't change the science.

@DavePhD shrug Then they live in a different world then I live. Because Hg2 doesn't have covalent bonds, hence it isn't a molecule.

"formally van der Waals bonded systems that exhibit significant covalent bonding contributions such as Hg2 "  and "intermediate cases Hg2 and Be2 exhibiting significant covalent contributions" https://books.google.com/books?id=K2y5BgAAQBAJ&pg=PA445&dq=neutral+%22hg2%22+covalent&hl=en&sa=X&ved=0ahUKEwj1mNfY3frfAhUJKqwKHbuMDJcQ6AEIKDAA#v=onepage&q=neutral%20%22hg2%22%20covalent&f=false


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How many sig figs should go after this calculation?

How many significant figures should be reported in the answer to the calculation (Assume all numbers are experimentally determined).
$\frac{12.501 \times 3.52}{0.0042}+6.044$

I tried to do this problem as follows:
$12.501 \times 3.52 = 44.0$ (three sig figs because $3.52$)
Then, $\frac{44.0}{0.0042} = 1.0 \times 10^4$ (two sig figs because $0.0042$)
Finally, $1.0 \times 10^4 + 6.044 = 10006$ (no points past the decimal)
I think the answer is 5 because $10006$ has 5 sig figs. However, the correct answer is 2. Can someone please help me find out where I'm wrong?

I'm voting to close this question as off-topic because it's not about chemistry.

@Mithoron - Really? Significant figures are discussed in every beginning chemistry class.

@MaxW How said only chemistry is discussed in such classes? Chemistry students have classes about physics, mathematics, even some statistics. Sig figs may be useful in chem. but so is standard deviation and integrals and have as much sense to be asked here.

I'm totally with @MaxW here. We're way too trigger happy with marking stuff as not chemistry. Significant figures aren't even the correct way to measure errors. I've only ever had the chemists insist on using them.

@Mithoron I fully disagree. Insofar as the integral is about chemistry, it should be asked here and be expected to be answered here. What's next? We're going to insist quantum mechanics and statistical mechanics are just physics and applied math? Are those off-topic? Part of learning chemistry is learning to use the correct tools in the correct way. Please reconsider your position.

Addition follows the rule of keeping the greater of the least significant digits.
You have $\underline{10}000 + \underline{6.044}$ (underlines are significant). The largest least significant digit is the thousands place. Therefore, your final answer should be rounded to $1.0 \times 10^{4}$.
Another way to look at it is to consider an error of $\pm 1$ in the least significant digit of each addend:
$1.0\pm 0.1 \times 10^{4} + 6.044 \pm 0.001$
The largest answer you can obtain is $11006.045$ and the smallest number you can obtain is $9006.043$.
These correspond to the raw sum plus the summed uncertainties: $10006.044\pm 1000.001$. The most significant digit in the error (thousands) tells you which digit you are uncertain about, so that's the last significant figure in your answer.

how is the zero after the one a significant figure?

Because $1.0\times 10^{4}$.

I noticed the question had a significant zero.

Ok, given the expression:
$$\dfrac{12.501 \times 3.52}{0.0042}+6.044$$
the desired is to find the answer with the correct number of significant figures.
(1) Be careful when rounding intermediate calculations. That can lead to a greater overall roundoff error.
(2) Let's look at the first part
$$\dfrac{12.501 \times 3.52}{0.0042}$$

12.5011 has 5 significant figures
3.52 has 3 significant figures
0.0042 has 2 significant figures

Since the term only has multiplication and division, the result of this part of this term should not have more than two significant figures. Generally I like to carry two extra significant figures in the intermediate calculation. No problem with modern calculators, but you can only get 3 digits with a slide rule, and 4 digits are pretty much the limit for even a good set of log tables. In this case, I'd use 5 digits for the intermediate result.
$$\dfrac{12.501 \times 3.52}{0.0042} = 10477.$$
(3) 6.044 has four significant figures.
(4) Now the question is what to do with the addition. 10477 is only really good to the thousands place, thus 6.044 is insignificant and the final result is $1.0\times10^4$
(5) Notice however if the first term had evaluated to $0.010477$, then the result would have been $6.054$ which has 4 significant figures.

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Degradation of anode in electrolysis
In my electrolysis experiment, I used a 9 volt battery, 1:100 mL baking soda:distilled water, and stainless steel electrodes. I noticed that the anode degraded faster than the cathode. Out of curiosity, is there a phenomena to explain this?

Can you explain what you mean by degraded?

Deposits and erosion on the anode that inhibit function: EG I use it once and then I have to replace it, due to the fact that it stops producing as much gas.

Gosh! You would think that electrolyzing NaHCO3 with stainless electrodes would be a clean reaction to produce H2 and O2.
Perhaps OH. is powerful enough to attach to some atoms at the surface and make it more insulating. So, significant degradation could occur at the positive electrode because of oxidation.
Perhaps lower voltage and currents would be more gentle with the electrodes. And perhaps a more resistant grade of stainless would show less degradation (316 is more resistant than 304, and knife blades are typically 400 series, harder but less corrosion-resistant).
At the cathode, hydrogen would be produced in a nascent (atomic) state, which can reduce the oxide coating on stainless steel, making it active. This activity would not result in corrosion while the electrolysis is going on - it is being cathodically protected. This electrode would be a good conductor even as its protective oxide coat is reduced, but the anode will have its oxide coat thickened till it becomes somewhat insulated - or degraded.
A similar effect can be seen with aluminum.

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Is electrolysis with AC current possible?
Optimizing everything else, would efficient electrolysis be possible with AC current? Most sources suggest to use DC, but can the water still be split/disassociated with AC?

It would work, but you'd alternately be generating hydrogen and then oxygen at each electrode. You really want to mix hydrogen and oxygen? It goes BOOM really easily.

Especially when electricity is involved... but it could be a sort of water filled AC charged power cell, or something. The gas is stored until needed and then is exploded. However, you would have to be careful about energy fluctuation and electrical surges.

In addition to what MaxW wrote you would not expect electrolysis throughout the wave, only when the voltage exceeds the redox potential, so it would be less efficient. Also, AC can mean a lot of things. It would also depend on the frequency.

I found a similar question with the answer I was looking for: https://chemistry.stackexchange.com/questions/62980/high-voltage-ac-water-electrolysis-not-working?rq=1


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What is the Mechanism of hydrogenation of alkenes by Wilkinson's catalyst?
I've learnt that the Wilkinson's catalyst $[\ce{Rh(PPh3)3}]$ is used for hydrogenation of alkenes into alkanes. But I'm curious to know the mechanism. Could someone help me?

https://chemistry.stackexchange.com/questions/99344/rate-of-catalytic-hydrogenation-of-alkenes/99594?r=SearchResults#99594

Have you seen the wiki page?

Yes, I tried looking at the wiki page.. but I didn't quite understand it

Catalytic hydrogenation or any catalytic mechanism of any transition metal is normally not taught in high school as it is a more difficult topic to study. But the fact is that the mechanism remains same. There is no easy way to interpret. What happens is what you have to study. So i would suggest to go through mechanism of basic catalysts and then go through transition metal as catalyst.

Roger that! The teacher shrugged me off saying that it is out of the high school scope. But my curious mind wouldn't take that as an answer

You can't draw arrow pushing mechanisms exactly for some of the steps, but as with any other reaction mechanism, you want to know the basic steps. These are: oxidative addition (specifically, concerted), $\beta$-migratory insertion, and reductive elimination. There's also a regular ligation in the middle, which is just a Lewis acid/Lewis base interaction.


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Please explain plane of symmetry

i am getting option (a) but answer is (b) where i am going wrong please help
SO IT IS OPTION A ONLY FOR ME BUT HOW B HAS PLANE OF SYMMETRY PLEASE HELP

Please edit in your reasoning as to how did you obtain A and why's B wrong. It is far better to understand a concept than knowing an answer.

What's to explain? That's right, (a) has a plane of symmetry, and (b) doesn't.

The meso cmp. must be member of a set with at least two optically active stereoisomers.

Answer (a) is not optically active to begin with so it's not compound "A"!.

You are right that after (a) is hydrogenated the result is a meso compound.  But, (a) is also a meso compound before hydrogenation, you can draw a mirror through the double bond; and so (a) misses the first part of the question.  (b) satisfies both parts.


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How is "the surrounding" and its temperature defined to calculate entropy change during a reaction?
Lets say supercooled liquid water at $263\ \mathrm K$ isobarically changes to solid ice at the same temperature. I wish to calculate the change in entropy of the surroundings and I happen to know the $\Delta H$ for the reaction.
This should be simple enough. Since the process is isobaric, $q_\text{rev} = \Delta H$ and $q_\text{surr}=-q_\text{reac}$
$\Delta S_\text{surr} = \frac{-\Delta H}{T} $
But what should T be here? Is it $263\ \mathrm K$? Or the temperature of the surroundings? Or are they both the same, ie;
Do we consider the temperature at which the reaction is taking place to be the temperature of the surroundings as well?
If so, why?

The phase transition no reaction. Yes. That would be the minum for that process

Correction: Unless its an adiabatic process the surroundings have to be at the same T as the system (isothermal), otherwise its not at equilibrium.

In thermodynamics, the surroundings are typically treated as an ideal constant temperature reservoir whose temperature T is specified.  And for such a reservoir, its entropy change is always $q_{surr}/T$, irrespective of any irreversibilities in the system.  Also, the temperature at the interface between the system and the surroundings is always the reservoir temperature, not the average temperature of the system.

You are not giving a realistic problem. The entropy of the surrounding is independent of the system unless you say otherwise. The surrounding will only be affected if there is heat transfer. You haven't said if this is the case, and it is unreasonable to assume either way.

Thermodynamics deals with changes between systems at equilibrium, which is to say that the initial and final states of a process are regarded as equilibrium states. As explained in the comments, the surroundings are usually defined as an ideal reservoir of infinite size and thus infinite heat capacity and constant temperature, clearly an approximation, but the key point remains that stated in the previous sentence.
It does not make sense to define an equilibrium diathermal state in which the surroundings and the system have a different temperature. Since  in the stated problem the system undergoes exchange of heat with its surroundings, and the system is at an initial and final T of 263 K after the transition, then the surroundings have to be at 263 K throughout.
One potential point of confusion regarding this type of problem concerns the metastable nature of the supercooled state. This is not a true thermodynamic equilibrium but is regarded as sufficiently stable to be regarded as such. An additional point of confusion may be that heat is transferred to the surroundings even though system and surroundings are at the same T. This is possible because the surroundings have infinite size and heat capacity. One way to envision such surroundings is as a reservoir of a substance with a melting point of 263 K.


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How do I write orders of magnitude in scientific notation when margin of error is involved?
My TA has me confused about how to use scientific notation when margin of error and orders of magnitude are involved.
The number I wrote: 3.78 ± 0.02 * 10^-3
My TA told me that this means: 3.78 ± 0.00002
I thought that this meant: 0.00378 ± 0.00002
Which one is correct? I've been looking online to find something that explains the proper notation when writing out margin of error with orders of magnitude but I haven't been able to find anything (I'm at a loss of what terms to search for). If what I wrote isn't the correct way to say 0.00378 ± 0.00002, then what is? Writing it as 3.78 * 10^-3 ± 0.02 or  3.78 * 10^-3 ± 0.02 * 10^-3 seems kind of foolish.

Maybe $(3.78\pm0.02)\cdot10^{-3}$?

Related: What does a number in brackets after another number mean? I.e. 21(1)cm-1

Writing $3.78\pm0.02\cdot10^{-3}$ instead of $(3.78\pm0.02)\cdot10^{-3}$ is clearly wrong for the same reason as writing $3.78\pm0.02\ \mathrm g$ instead of $(3.78\pm0.02)\ \mathrm g$ is wrong.

Sometimes you need to go the extra mile to be entirely unambiguous. Your original formulation (3.78 ± 0.02 * 10^-3) could be interpreted in two ways, the way you intended and the way your TA suggested. The majority of people, I expect, would use context to interpret it correct as you intended, but some (as your TA demonstrates) will not.
There is no clear cut widely agreed standard that will resolve this. But you can apply standard mathematical notation (which will look superfluous to some) to make what you write entirely unambiguous. As Ivan suggested in a comment the simple addition of brackets giving (3.78 ± 0.02) * 10^-3 makes your statement unambiguous in its interpretation. This is more compact than 3.78 * 10^-3 ± 0.02 * 10^-3 but just as unambiguous.
If you were using properly typeset formulae, you might well be able to avoid some of the apparent redundancy of the last formulation, but that would take more effort. Better to accept that redundancy is sometimes required to make things clear.

This is good to know, thank you! I wanted to make sure I hadn't missed some crucial part of how to use margin of error with scientific notation. I'm supposed to be learning how to write scientific papers that could be put in journals.  I'll keep in mind that redundancy is sometimes okay.

The answer Loong linked to the question mentions accepted standards of notation. It also notes that the general form discussed here should be avoided due to flawed interpretation. // Redundant is also the wrong word to use, because for unambiguity it is necessary to write it out, and therefore the exact opposite.

I think your TA is wrong, based on the following argument.
When you write a number derived from calculation or experiment, and following standard protocols for the propagation of uncertainties, you regard the last reported integer in the number as uncertain. Therefore, it simply does not make sense to report an uncertainty of ±0.00002 for 3.78, since the uncertainty in 3.78 must be at least ±0.01. I would say that although there may be cases in which explicit notation should be followed, it is generally safe to write

$3.78 ± 0.02 \times 10^{-3}$

as meaning

$0.00378 ± 0.00002$

A safe alternative (consult a standard analytical chemistry textbook such as Skoog, West and Holler's) is to write

$3.78 (±0.02) \times 10^{-3}$

The use of parentheses emphasizes that the uncertainty defines a range of possible values.
Postscript: Another similar argument can be made for the placement of units, for instance should one write

$3.78 ± 0.02 ~kcal/mol$

or

$3.78 ~kcal/mol ± 0.02 ~kcal/mol$

Clearly the second case borders on the absurd.
This is a recurrent problem during drafting of scientific text. Another similar situation is the following: should one write

A and B were determined to be 1.0 and 2.0 kcal/mol, respectively.

or

A and B were determined to be 1.0 kcal/mol and 2.0 kcal/mol, respectively.

or

We determined values of A=1.0 kcal/mol and B=2.0 kcal/mol.

Clearly the first case is acceptable, even if it could be construed as ambiguous, and is not necessarily the most concise. Scientific writing is subject to different rules than standard prose.
Clarity is very important, conciseness too, and there are, in general, recommended conventions. See this answer for instance, which suggests that using the "±" symbol is not recommended unless you mean to use the uncertainty to estimate the distribution of errors.

Thanks! I was thinking it would be along these lines. I feel like my TA might have graded too many papers by the time she got to mine because there is no reason I can see that an uncertainty of 0.00002 could be meaningfully attached 3.78.

@BradleyH to play TAs advocate here, they are trying to make sure you understand uncertainty and are grading accordingly. If they are interpreting what you wrote in a way where it couldn't "be meaningfully attached" to what you wrote, then that could mean (from their perspective) you don't understand uncertainties and so they mark it as incorrect

I think as long as the question is how to write a number in the format yyy ± xxx, and not "what is the uncertainty in yyy", then my answer holds.

@BradleyH There is no point in reporting an uncertainty, if you are sloppy with the notation. Your TA certainly is correct to point out your mistakes, after all that is how you learn. The is enough sloppy notation in well known textbooks, old and directed theories, it is important to teach to the new generation not to make the same mistakes again. // I think this answer also shows very bad style; a construct like 'the value is determined to be' should never be used in a scientific paper. Use active voice instead 'the value is'.

You're right @Martin-マーチン in general, but the literature is full of similar expressions, since humans write articles for other humans.

And despite what I just wrote, you should always make an effort to report an uncertainty, since a reported value is essentially useless without an associated reported uncertainty. Also, I would argue that @BradleyH was far from sloppy, and in fact followed a very logical line of argumentation coming to the conclusion that the TA was, effectvely, wrong. This is not to discredit TAs in general, but this question does not ask whether TAs should or should not have the final word.

@Tyberius It did occur to me that my TA might have been intentionally misunderstanding what I wrote to show how it might be perceived by the reader and I know they are just trying to help. Perhaps it wasn't very important or useful to bring up the fact that my TA was involved at all. I just wanted to make sure I was using the correct convention because I was having difficulty finding anything that showed the correct way to write this.


text:
What is the crystal structure of Calcium Acetate?
I tried conducting an experiment I saw at school with some materials I had at home. The experiment consisted of mixing a solution of Acetic Acid with Calcium Carbonate to yield Calcium Acetate: $$\ce{CaCO3 + CH3COOH -> Ca(CH3COOH)2 + CO3 + H2O}$$ I'm providing as much detailed in a summarized form to show what I did step-by-step if I did anything wrong. I didn't have any lab grade source of $\ce{CaCO3}$, but I found that chalk has a large concentration of $\ce{CaCO3}$, and I used that instead. I worked out a small calculator on Spreadsheets and found the appropriate quantities of each reactant. I then combined both of the ingredients and left the reaction to settle. After a couple days, I stirred and saw little to no fizzing, however, there was a lot of precipitate at the bottom of the beaker which I assumed to have been non-$\ce{CaCO3}$ residue from the chalk. I filtered the solution through a two coffee filters and let the solution evaporate. I should note that the solution was crystal clear after it was filtered. I came back a week later only to see a hard, styrofoam look-a-like on the sides of the beaker, and a layer of cloudy crystals at the base of the beaker. At the center of the base there's a crystal structure sprouting separately from the rest of the crystal layer.

I initially though perhaps my calculations or initial data gathered were incorrect but even after triple-checking them there was no error. Looking for possible structures of Calcium Acetate there was only one image that looked somewhat like the crystal sprouting at the center of the beaker, but still it doesn't explain what all the other substances are. Does Calcium Acetate have multiple crystal structures? Or has there been some sort of contamination from the chalk. Any help would really be appreciated:)

Too long for a comment, hence an answer:

What you are asking about is not crystal crystal structure, it's called crystal morphology. Crystal structure is a model describing atomic positions and overall symmetry within a unit cell, whereas crystal morphology is the appearance of the crystal on a macro-scale. And yes, crystal morphology is also affected by the conditions crystals are growing under (concentration and evaporation rate, in your case), so that the same substance can have crystals of different shape.
The formula for calcium acetate and hence the equation is wrong, the correct one would be $(\ce{OAc-} = \ce{CH3COO-})$
$$\ce{CaCO3 + 2HOAc → Ca(OAc)2 + H2O + CO2}$$
and, since your reaction's stoichiometry suggested 1:1 ratio between calcium carbonate and acetic acid, I suspect you didn't use enough acid leaving $\ce{CO3^2-}$ in the solution. Then, as the primary product – calcium acetate $\ce{Ca(OAc)2}$ – has crystallized (white edgy crystals on the bottom), the remaining calcium carbonate $\ce{CaCO3}$ started to precipitate or co-crystallize on the walls of the beaker (what you call a "styrofoam look-a-like" phase), resembling how calcite grows in nature (see this and other images of natural calcite on MineralMiner's FTP server):

Also note that there is no such thing as $\ce{CO3}$. It's either carbonate $\ce{CO3^2-}$ (solution) or carbon dioxide $\ce{CO2}$ (gas).
Regarding the the impurities from the chalk, usually there are parts of shells of marine animals and silicates (can be eliminated by filtering), also minor amounts of $\ce{MgCO3}$. However, if you took a chalk bar from the classroom, this is a different story as it's often nearly 50% $\ce{CaSO4}$ (I hope you didn't do that).


text:
What algorithm is GaussView using for the clean function?
A colleague showed me GAUSSIAN 09 which comes with GaussView 5
for drawing structures. What impressed me was the snappiness of the
Clean function.
What algorithm is used for the Clean function?
The GaussView 5 Reference can be downloaded from the publisher’s
web site. It doesn’t seem to contain the information that I’m looking
for.
Background: As part of a project called
 SAN – Self Aware Network, I have a similar problem, though with
 additional constraints. To find the distribution of points in space,
 I implemented a genetic algorithm. Unfortunately, with large
 structures, this algorithm tends to get stuck in local minima. I
 documented the algorithm in the report SAN: Location Optimizer.
 A mathematician friend implemented a
 solution using gradient descend. Yesterday, he showed me some
 nice results, but it’s nowhere as snappy as GaussView.
Update
I also contacted Gaussian technical support. Their response:
“Unfortunately, no. We do not have additional documentation about this
function, other than what is already present in the GV5 manual. This
function just uses a very simple MM expression with crude parameters
just to provide a simple way to "clean" a structure drawn "by hand" in
GaussView. It is certainly not intended to give accurate geometries
(it is expected that one is going to use some other model in Gaussian
to run a geometry optimization and get a good geometry). The purpose
is simply as another aid as part of building a structure in GaussView,
so one may "clean" a structure faster (not always) than otherwise
using "modify bond", "modify angle", "modify dihedral" tools.”

I think it is unlikely you will find a direct answer to this question if the software is owned by Gaussian. Gaussian is notoriously strict about who they share their source code with and does not like when people share details of their algorithms. As an aside, I am almost certain that this "clean" function is not optimizing the structure. It is essentially making a sophisticated guess at the structure, so you might see faster convergence with GD if you use the clean output as an initial guess. Also, finding the global minimum of large structures is notoriously difficult and often unnecessary.

This webpage gives general information about what clean is doing :http://flex.phys.tohoku.ac.jp/texi/gview/clean.htm. Basically, it's using van der waal's radii and various weights assigned to different internal coordinates to make a good guess. Again, you'll need to do a real optimization if you are interested in true minima.

@jheindel Thanks! Unfortunately nothing new: This seems to be simply the online version of the reference guide I linked to in my question. It gives a rough idea about how the force field function f is constructed. What I am most interested in is: What algorithm is used to find the minimum of f? (e.g. gradient descent algorithm or genetic algorithm)

I would be very willing to bet that they are using gradient descent since they will already have a code for this to do actual geometry optimizations. If not, then it would be a Newton-Raphson, but given that they recommend 150 steps, this seems unlikely.

First off, it's worth stating that GaussView is not developed by Gaussian, Inc. You might find that AMPAC GUI looks familiar.

I strongly suspect that GaussView is using steepest descent or conjugate gradients on a very simple force field (e.g., Lennard-Jones). Based on your comments, it looks like you want to read up on Distance Geometry embedding or some of the Stochastic Proximity Embedding and Self-Organizing papers by Agrafiotis.


text:
Largest alkane having a given alkane as its base name
What is the largest (most carbon atoms) alkane having heptane as its base name?
For example, 2,2,3,3-tetramethylbutane is the largest (most carbon atoms) alkane retaining butane as its base name.

You put as many side chains on it as possible without creating a new longest chain. Then you do the same with each of the side chains, and then you hope that the final molecule's center hasn't become so dense that it snaps in two.

This is a subset of a very interesting theoretical question about how many isomers are possible for simple hydrocarbons discussed in this question: https://chemistry.stackexchange.com/q/16135/81 . Answers there also address questions about whether potential molecules are physically possible.

Thanks for the upvotes to my initial (stupid :-) ) comment: The surface of the molecule is densely packed, and the bonds in the center are stretched to the point where they rip.

Any alkyl substituent of butane in position 2 or 3 cannot be longer than $\ce{CH3}$ since that would lead to a longer parent chain. And obviously, there cannot be any alkyl substituent at all in the first or the last position of the butane chain. Therefore, the largest structure based on a butane parent chain is 2,2,3,3-tetramethylbutane.
This principle can be expanded to a heptane parent chain. The maximum length for alkyl substituent chains are 0 for position 1 and 7, 1 for position 2 and 6, 2 for position 3 and 5, and 3 for position 4. Therefore, the largest theoretical structure based on a heptane parent chain is 3,3,5,5-tetra-tert-butyl-4,4-bis[3-(tert-butyl)-2,2,4,4-tetramethylpentan-3-yl]-2,2,6,6-tetramethylheptane ($\ce{C53H108}$).

I believe this thing is much too crowded to survive, but then again, that's not what the question was about. Other than that, you are right.

@IvanNeretin Agreed, the inner C–C bonds would probably be stretched to more than 2 Å. That’s why I wrote “largest theoretical structure”.

@Loong make that definitely. ;-)

My head hurts trying to parse that name. Could you draw it and give its simple overall formula? Then I can look it up on the list given in answers to https://chemistry.stackexchange.com/q/16135/81

@matt_black It replaces any of the 12 Me groups of tetra-tBu-methane by tBu, so it's $\ce{C5 + 12 * C4H9 = C53H108}$. Too many isomers to even think about. ;-)

Yes, i got it as C53H108!!!

Preferred IUPAC names is a good concept, but “tetrakis(tri-tert-butylmethyl)methane” is somewhat shorter and clearer ;)

It would seem I have found the structure for the maximal pentadecane in a completely different setting.  I wonder if there's anything interesting that pops up from that unusual connection... Actually, it seems that if you take the infinite fractal and assign a single carbon atom to be a particular region size, then circle all regions of that size, you can recover the whole infinite family of maximal alkanes!

@NicolauSakerNeto I also found big resemblance in the Cayley graph of the free group with 2 generators. As a non-chemist, but someone with interest in maths, I wonder whether there is any deeper connection between algebra and chemistry. If someone has any pointers for me, I am happy to take them.

@ComFreek I think this is not so much a deep connection between algebra and chemistry as between algebra and graph theory.  And then there is a (shallow) connection between chemistry and graph theory via diagrams for covalently-bonded molecules.

@ComFreek Careful, other people have tried to find such connections, and it didn't exactly prove fruitful ... https://en.wikipedia.org/wiki/Jan_C._A._Boeyens is a hilarious example. ;-)

That's hard to tell, because already the next one after tetramethylbutane, tetra-tert-butylmethane (3,3-di-tert-butyl-2,2,4,4-tetramethylpentane) is so unstable it cannot exist. You want a parent chain that is even two atoms longer.
Actually it seems also removing one methyl group from $\ce{tBu_4}$methane doesn't make it sufficiently stable, so one has to be a bit more generous in weeding out branches for larger hyperbranched alkanes.
I believe that any permethylated (or per-n-alkylated) linear chain is principally possible (search for Hellmann, Beckhaus, Rüchardt "Thermolabile Hydrocarbons", e.g. hexamethylpentane is stable at RT), but adding more side-side chains to adjacent side chains will make the molecule snap in two.
So the biggest possible sum formula would be something based on
3,5-di-tert-butyl-2,2,3,4,4,5,6,6-octaamethylheptane = $\ce{C23H48}$
. You can surely make the inner methyl substituents longer,
3,5-di-tert-butyl-3,5-diethyl-4,4-di-n-propyl-2,2,6,6-tetraamethylheptane = $\ce{C29H60}$
, and then you can probably add 4 methyl to the nPr and Et to give iBu and i-Pr that's $\ce{C33H68}$. Any more and I'll bet it breaks, quite a way from the theoretical $\ce{C53H108}$.

Speaking of hypothetical structures, you can get, with little cheating, infinitely large “alkane” like poly(heptane-1,1-diyl)[1]

where $n=\infty$, i.e. with infinite degree of polymerization (or substituted branched ones like those in other, correct, answers, with even bigger infinity sizes).

Notes:

Alternative structure-based polymer name based on older organic chemistry nomenclature would be poly(1-heptylidene). However the “longest chain” rule seems to applied differently in polymer nomenclature, so somewhat more “preferred name” could be poly(hexylmethylene).

But in that case won't the parent be 1-chloro-4-(3-chloropropyl)infinitane? rather than dichlorosubstituted heptane?

@NandakumarUK I don't know if you are serious and where did you get chlorine. My name uses polymer nomenclature (see e.g. here), and it's a more systematic structure-based name, like poly(methylene) for more known “polyethylene” (source-based name).

@NandakumarUK .. but if you want to have fun non-polymer name, I'd name it rather like −∞,−∞+1,−∞+2,…,∞−2,∞−1,∞-infinitaheptylinfinitane.

Sir, i was referring to a  choosable monomer before getting polymerized.; that is why i mentioned a chloro compound. If you consider the compound to be a polymer, there must be a monomer for sure; i wounder, which is that?

Again poly methylene can have odd or even number of carbon atoms in the chain, but poly ethylene can have only even number of carbon atoms in the chain; so could we consider them to be the same always?

In my post to which you had replied i wrongly used the word parent, instead of monomer, my regrets

This is not really a useful answer as, if polymers were named by the same rules as alkanes, this one would not be based on the monomer or the size of the branches, but the size of the polymer backbone (which is potentially as long as you want). So it is, kind of, cheating.

@matt_black Some seem to find it useful, some seem to find unuseful the question itself. I admit it's little cheating, see original version, I've removed this info, because thought it is obvious, as I'm operating with infinity.

The general answer is easy. First we compute the maximum number of carbon atoms $a_n$ in a side chain of length $n$ from bond to end. Its difference equation is
$$a_{n+1}=3a_n+1$$
The $1$ is there to count the carbon that has the bond and the $3a_n$ counts the $3$ maximal side chains radiating from it. The homogeneous equation is
$$a_{n+1,h}=3a_{n,h}$$
With general solution
$$a_{n,h}=C\cdot3^n$$
And if we seek a particular solution of the form
$$a_{n,p}=K=\text{constant}$$
We get the solution
$$a_{n,p}=-\frac12$$
Since the above is a linear difference equation, the general solution is
$$a_n=a_{n,p}+a_{n,h}=-\frac12+C\cdot3^n$$
We know that the solution for $n=1$ is a methyl group, so
$$a_1=1=-\frac12+3C$$
So $C=1/2$ and
$$a_n=\frac12\left(3^n-1\right)$$
Now consider $b_{2N+1}$, the maximum number of carbons in a molecule with maximum length $2N+1$ carbon chain. To make this we just substitute $4$ maximal side chains of length $N$ to a methane molecule to get
$$b_{2N+1}=1+4a_N=2\cdot3^N-1$$
And to get $b_{2N}$ we substitute $6$ maximal sides chains of length $N-1$ to an ethane molecule to get
$$b_{2N}=2+6a_{N-1}=3^N-1$$
Hmmm... I guess we could have arrived at the same result by just joining $2$ maximal side chains of length $N$ together :)

So for heptane as the longest chain N=3 (2N+1 carbon atoms), that gives 53, correct. Pentane 17, correct. Hexane 26, also right. Nice math!

This equation was the one i always dreamt, attempted and miserably failed every time ;)...Kudos!!!


text:
Overlap matrix (H2 molecule, STO-3G)
I'd want to implement SCF method using Modern Quantum Chemistry (Szabo). But when I want to calculate overlap matrix for H2 (STO-3G basis) my results aren't equal to true overlap matrix that there's in the book. However when I compare my results in other basis (STO-1G) I get exact overlap matrix. Below my code for calculation overlap values.
class gauss():

  def __init__(self, alpha, Ra, k = 1):

    self.alpha = alpha
    self.Ra = np.array(Ra)
    self.k = k

  def __mul__(self, other):
    'calculate integral'
    lenght = (self.Ra - other.Ra)
    scalar = np.dot(lenght, lenght)

    return np.exp(-self.alpha*other.alpha*scalar/(self.alpha + other.alpha))*np.pi/(self.alpha + other.alpha)**(3/2)

def overlap(first, second):

  s11 = 0
  for i in first:
    for j in first:

        s11 += i * j * i.k * j.k

  s12 = 0
  for i in first:
    for j in second:

        s12 += i * j * i.k * j.k

  return s11, s12

## gaussian primitives:
s1 = gauss(0.168855404, (0.0, 0.0, 0.0), 0.444634542)
s2 = gauss(0.623913730, (0.0, 0.0, 0.0), 0.535328142)
s3 = gauss(3.42525091, (0.0, 0.0, 0.0), 0.154328967)
s1b = gauss(0.168855404, (1.4, 0.0, 0.0), 0.444634542)
s2b = gauss(0.623913730, (1.4, 0.0, 0.0), 0.535328142)
s3b = gauss(3.42525091, (1.4, 0.0, 0.0), 0.154328967)

first = [s1, s2, s3]
second = [s1b, s2b, s3b]

s11, s12 = overlap(first, second)

print(s12 / s11) # print normalized s12

0.781180936263708

but exact value from book = 0.6593 (1.4 Bohr)
and if I choose basis as:
s = gauss(0.4166127, (0.0, 0.0, 0.0))
sb = gauss(0.4166127, (1.4, 0.0, 0.0))

I get exact value:

0.6647924143035879

from book: 0.6648

I have down-voted this question, because it has little to nothing to do with chemistry. I believe this is a programming issue, since all that is given here is code. It is not even explained what this code actually is supposed to do; it is not even stated in which language it is written. I am surprised that no-one has voted to close it as off-topic.

@Martin-マーチン after looking the question over more, I agree. While the issue the  OP had wound up being chemistry related (misunderstanding about how to normalize overlap integrals), the question as phrased is more about debugging this particular code than addressing the underlying math. The OP does say they are trying to develop an SCF code, but not specifying the language doesn't help. I think a version of this question exists that would be on-topic, but as is I agree that it should be closed for now.

I'm voting to close this question as off-topic because it is about code debugging and not an underlying chemical principle.

I'm voting to close this question as off-topic because stack exchange's chemistry area should be devoted to the discussion of chemistry and not how to write code for a bit of computational chemistry software. To my mind it is like going on a diesel engine design forum and asking how to change the oil filter on a specific car when the forum is devoted to the theroy of how to design an engine.

You have two errors in your code, one minor and one major.
The minor error is in your multiplication function. The value you return should have $\pi^{3/2}$, but the way it is arranged now the power is only being applied to the denominator of that expression.
The major error is how you are normalizing your overlap integrals. You can't simply divide $\langle1|2\rangle$ by $\langle1|1\rangle$ in order to normalize it. You have to explicitly normalize each of the Gaussian basis functions. To do this you can include a factor of ((4*self.alpha*other.alpha)/np.pi**2)**(3/4) in the value you are returning from your multiplication (in terms of readability of your code, it may be wiser to define variables within the function like proportionality_constant, gauss_integral, and normalization and then return the product of these). Or you could simplify and just multiply the exponential by ((4*self.alpha*other.alpha)/(self.alpha+other.alpha)**2)**(3/4) This also means you only need the second part of your overlap function, since the multiplication will take care of the normalization.
The reason your code works for STO-1G is that dividing by $\langle1|1\rangle$ accidentally gives the correct normalization since there is only one normalization constant associated with the one Gaussian function of STO-1G.


text:
Why does Oxygen act as Nucleophile over here?

We were taught in class that when a carbon atom and an oxygen atom both have a negative charge in a resonating structure, then carbon atom attacks the nucleophile rather than oxygen atom because it acts as a better nucleophile.
In the question I have attached, in the first step the hydroxide ion abstracts the hydrogen from the oxygen of o-cresol. The negative charge so formed is delocalised onto ortho and para position.
There are 4 possibilities:

Oxygen atom attacks $\ce{CH3I}$.
The carbon atom attached to $\ce{CH3}$ attacks.
The carbon atom adjacent to oxygen (not attached to $\ce{CH3}$) attacks.
Carbon at para position attacks.

Now, based on the statement that Carbon atom should attack, Option 1 should be ruled out.
Option 2 also is unfavourable because the carbon atom attached to $\ce{CH3}$ would have less negative charge due to inductive destabilisation by $\ce{CH3}$.
Option 3 is unfavourable because the product formed would be sterically crowded.
Option 4 seems correct.
But the answer given to the above question is Option 2.
I confirmed from my teacher and he says that the statement that Carbon atom is a better nucleophile than oxygen is true but here attack from oxygen would take place (he doesn't explain why).
Also, in the well known Riemer Tiemann Reaction and Kolbe Reaction of Phenol, the attack takes place from Carbon side.
Question- where is my reasoning wrong? According to my reasoning, answer should be Option 1. Why isn't it?

Phenols are fairly acidic, the -OH is completely deprotonated and the -ve charge stays on the oxygen - forget any potential resonance structures, they do not apply here.

Fair explanation is given in: https://chemistry.stackexchange.com/questions/38006/carbonyl-oxygen-as-nucleophile.

You may be seeing the impact of the aromatic ring.  Putting a negative charge on carbon in your structure disrupts the aromatic coupling.  So carbanion contributions are reduced and the oxyanion contribution, where the oxygen is nucleophilic, is favored.

Well, here's the catch. When C attacks, the reaction proceeds through a carbanion intermediate. This carbanion in case of phenyl systems is stable enough to react only when the reaction takes place in a polar protic solvent which stabilizes the intermediate. But in case of an aprotic solvent such as iodomethane, O- wins the nucleophilicity race and hence it attacks.
You must not forget when we're talking about the product, we mean the major product. Experimentally, we're bound to obtain both products in certain quantities.

-1 from me: There's no carbanion intermediate when carbon acts as a nucleophile. Sure, some of the resonance forms have a formal negative charge on carbon, but these are relatively small contributors and it's a serious stretch to call the entire thing a carbanion. Methyl iodide is a reagent not a solvent here. I'm afraid your answer seems like you are applying rules you've read without thinking about the underlying chemistry and context.


text:
How to Calculate Lennard-Jones Parameters from TDDFT?
I am trying to develop parameters for AMBER force fields and to calculate Lennard-Jones parameters for the $\ce{Fe^3+}$ ion. I did a deep search and found that the best way to do this with a high accuracy is to use TDDFT. I found many articles about this, for example (I can put more if there is need):

Marques, M. A. L.; Castro, A.; Malloci, G.; Mulas, G.; Botti, S. Efficient calculation of van der Waals dispersion coefficients with time-dependent density functional theory in real time: Application to polycyclic aromatic hydrocarbons. The Journal of Chemical Physics 2007, 127 (1), 014107 DOI: 10.1063/1.2746031.
Banerjee, A.; Autschbach, J.; Chakrabarti, A. Time-dependent density-functional-theory calculation of the van der Waals coefficient $C_6$ of alkali-metal atoms Li, Na, K; alkali-metal dimers $\ce{Li2}$, $\ce{Na2}$, $\ce{K2}$; sodium clusters $\ce{Na_n}$; and fullerene $\ce{C60}$. Phys. Rev. A 2008, 78 (3) DOI: 10.1103/PhysRevA.78.032704.

But, I couldn't understand the articles, because in some there are so many mathematical formulas, in others the method is not clear. So, to get some things clear before starting, I need to answer these questions (let's say I did my TDDFT calculation):

After performing TDDFT calculations, are there software or tutorials that I can use to calculate van der Waals parameters?
If there is no software or tutorial to calculate this, where can I start?

I don't have access to the papers mentioned (so I cannot judge their level of detail), but have another reference for you: http://dx.doi.org/10.1063/1.3382344

I'm basing my answer on T. Gould, T. Bucko, J. Chem, Theory Comput. 2016, 12, 3603-3613.
In your TDDFT calculation, what you want to compute is the frequency-dependent dipole polarizability, $\alpha_{X/Y}(i\omega)$, that determines the dispersion coefficient, $C_6$.
$$ C_{6,XY} = \frac{3}{\pi}\int d\omega \alpha_X(i\omega)\alpha_Y(i\omega)$$
They also propose some approximate relationships that depend only on the static dipole polarizabilities, $\alpha(0)$.
$$ C_{6,XY} \approx \Xi [\alpha_X(0)\alpha_Y(0)]^{0.73\pm 0.01} $$
where $\Xi = (1.5 \pm 0.1)[\text{hartree} \cdot a_0^{1.62} ]$.
I don't know any tutorials for this, but you could start trying to reproduce some of the numbers on these papers.


text:
How does pyrosulfate compare to pyrophosphate?
That is, in terms of stability in aqueous solution and the energy released by the hydrolysis of the S-O-S bond compared with pyrophosphate hydrolysis. More specifically, could sulfates substitute phosphates in some kind of hypothetical biochemistry? AFAIK sulfates are generally not found in terrestrial biochemistry.

Biochemical systems can form sulfate and some organisms can even reduce sulfate as anaerobic oxygen sources.  But sulfate oligimers do not last long in water due to their Lewis acidity.

Pyrosulfates hydrolyse quickly. https://pubs.acs.org/doi/abs/10.1021/ic50004a019


text:
Regarding Clemensen Reduction of the following compound
While I was doing problems on reduction of aldehydes and ketones, I encountered this particular question
Here, I know that Clemensen reduction reduces the ketone group to $-CH_2-$ group. But, the answer key has given the option (c). I also know that the mechanism for this reduction is still shaky for the senior high school kid. But, how to deal with substitution like these as in case of this question

In this question option (a) is the answer though this is wolf kishner reduction.
Any suggestions how to solve these reactions?

Pardon me if I don't know any particular fact about it as I am reading and practicing myself on this. Reading from David Klein's organic chemistry book. But, practising from M. S. Chouhan's Problem on oRganic chemistry chemistry.

Under the conditions quoted I'm surprised you don't get hydrazine displacing Cl in a SNAR


text:
Can superoxide ion act as a ligand in a similar fashion as peroxide?
There is a question in my chemistry book:

Find oxidation state of platinum ion in the complex $\ce{[Pt(O2)(en)2Br]+}$.

My books says that there is peroxide, but I want to ask why it is peroxide? Can it be superoxide?
My answer was that oxidation state of platinum is $+3$, but the answer in the book given is $+4$ with explanation that $\ce{O2}$ here is peroxide. How can I get to know from the formula that $\ce{O2}$ is peroxide and not superoxide?

You don't know that from the formula alone.

Peroxide ligands are typically bridging ligands, I'd expect it to be a dinuclear complex at least. It  would be interesting to find out whether this standalone complex exists at all; if not, the textbook's question doesn't make much sense.

Here the oxygen "molecule" in its complex with hemoglobin is rendered as a superoxide ligand.  But remember, electrons are shared between the metal and the ligand; so there may be some latitude in how you can apportion them to determine "oxidation states".


text:
Reaction energy of oxygen and hydrogen
From the bonding energy of following bonds that I found in a Wikipedia article, I calculated the reaction energy that would be released when oxygen and hydrogen are reacted to produce water. The phase of each component is unknown.
$$
\begin{array}{lc}
\hline
\text{Bond} & D_0/\pu{kJ mol-1} \\
\hline
\ce{H-H} & 436 \\
\ce{O-H} & 497 \\
\ce{O=O} & 498 \\
\hline
\end{array}
$$
Therefore,
$$\ce{2 H2 + O2 -> 2 H2O} + \pu{618 kJ}$$
However, this Wikipedia article suggests that $\pu{241.8 kJ}$ of energy is released for every mole of hydrogen, which is $\pu{483.6 kJ}$ per $\pu{2 mol}$ for the context of the equation above.
Even when I consider the vaporization of water, which is $\pu{40.65 kJ mol-1}$, the total energy will amount to $\pu{536.7 kJ}$ per $\pu{2 mol}$, which is still far away from $\pu{483.6 kJ}$ per $\pu{2 mol}$ that is suggested in the Wikipedia article.
Where could this extra energy be going?

Bond energies are a great deal less universal than you seem to be thinking.

The catch may be the cleavage of OH in water. Also, you need to be careful with signs, vaporization is endothermic.

I am aware that it is endothermic. Therefore I have performed the following calculation:
2H2 + O2 = 2H2O(liquid) + 618kJ = (2H2O(gas) - 40.65kJ*2) + 618kJ = 2H2O(gas) + 536.7kJ

Please tell me. If my way is unconventional, what method is normally used to solve thermochemical equations in chemistry?

@Y.Yoshii Maybe this is useful: https://pubs.acs.org/doi/10.1021/jp013909s

I cannot speak for the source of the data on the wikipedia page on oxyhydrogen which you linked. However it may matter to what T the data refers, from the wikipedia page on bond dissociation energies, "The enthalpy change is temperature dependent, and the bond-dissociation energy is often defined to be the enthalpy change of the homolysis at 0 K (absolute zero), although the enthalpy change at 298 K (standard conditions) is also a frequently encountered parameter."

Regarding temperature change: If you are to burn oxy-hydrogen and burn it with different temperatures, then does that mean you will have different energy output for those different temperatures of flames?
That sounds like to me is breaking the conservation of energy.

The enthalpy of dissociation can show a temperature dependence due to changes with temperature in the heat capacity of reagent and products. Different rovibrational states are occupied to different extents at different temperatures, altering the mean dissociation energy.

Part of the discrepancy appears to be due to the need for two $\ce{O-H}$ dissociation energy terms to compute the $\ce{OH}$ bond energy for water: one for the $\ce{OH}$ bond in water (i.e. $\ce{H-OH}$), and another for the $\ce{OH}$ bond in the $\ce{OH}$ radical (i.e. $\ce{^.O-H}$). Values for these are reported by Ruscic et al. [1]. Using the average value of $\ce{D_0}$ for $\ce{H2O}$ (the $\ce{OH}$ bond energy for water), namely $\pu{459 kJ/mol}$, gives
$$\ce{2H2 + O2 -> 2H2O} \quad \Delta H = \pu{-466 kJ/mol}$$
This point is addressed in the Wikipedia page for "bond dissociation energy", the reference cited therein providing a slightly different bond energy for water of $\pu{461.5 kJ/mol}$.
Given various uncertainties common to the measurement of bond dissociation energies, and the potential effect of temperature on the bond energy, the final discrepancy seems reasonable.
References

Ruscic, B.; Wagner, A. F.; Harding, L. B.; Asher, R. L.; Feller, D.; Dixon, D. A.; Peterson, K. A.; Song, Y.; Qian, X.; Ng, C.-Y.; et al. On the Enthalpy of Formation of Hydroxyl Radical and Gas-Phase Bond Dissociation Energies of Water and Hydroxyl. The Journal of Physical Chemistry A 2002, 106 (11), 2727–2747. https://doi.org/10.1021/jp013909s.

That is brilliant. Thank you for your information.


text:
Rule of Thumb for separatory funnel volumes
Is there a general rule of thumb in terms of how one chooses the volume of separatory funnel used for an extraction? It's my understanding that it is the amount of product that determines this as opposed to the volume of the solvent it is present in. If that is correct I was wondering if someone could post what the general ranges in grams of product would be appropriate for the following sep funnel sizes:
250 mL, 500 mL, 1 L

This is a good question, but it may be hard for it to have one definitive answer. For example, I have always chosen my funnel based on volume of solution. If my compound is dissolved in 100 mL of solvent A and I plan to extract into solvent B, I generally choose equal volumes solvent B (100 mL), and thus the 250 mL funnel.

If a general rule must be given than the volume of the separatory funnel is mostly dictated by that of the liquid to be treated.
Indeed is better to extract the product by repeating several steps each using small portions of extracting solvent. This ensures a more efficient extraction based on repartition, I.e. a more quantitative recovery of the product as well as a more convenient work-up (cost-effective, less time consuming, less waste).
I am not considering difficulties due to a slow phase separation which might prevent the use of a narrow funnel but in general use the smallest that allows an easy shaking and inter mixing. (When you get to know the actual separation you might even prefer a  specific funnel shape not just that V).
Example: 200 ml to be treated with three to four 25 or 50 ml portions : 500 ml funnel.


text:
Compound having highest standard reduction potential
Which compound has highest SRP (standard reduction potential) value, e.g. the strongest oxidising agent? In most tables $\ce{F2}$ is at top with $E^\circ = \pu{2.87 V}$, but I've found a table listing $\ce{H4XeO6}$ at top with $E^\circ = \pu{3.0 V}$.
Is there a more strongest oxidising agent on basis of $E^\circ$ value and is $E^\circ$ value defined for every compound? If yes, what is $E^\circ$ value for $\ce{ClF3}$?

SRP is not an attribute of a compound, rather of a redox pair. Probably it would be a good idea to provide such pairs for the compounds you mentioned to begin with. Also, note that SRP is not dimensionless and is measured in Volts (hence the edit).

@andselisk I think "standard" means not just STP but also paired with standard hydrogen electrode.

@TryHard Sure, but that's not what I meant. One must introduce a redox pair befire assigning an SRP: e.g. for $\ce{F2 (g) + 2  e− → 2 F−}$ $E^\circ=\pu{+2.87 V}$, whereas for $\ce{F2(g) + 2 H+ + 2 e- → 2 HF (aq)}$ $E^\circ=\pu{+3.05 V}$. Posting an SRP for $\ce{F2}$ alone doesn't make much sense. To OP: note that for $\ce{KrF2 (aq) + 2 e- → Kr (g) + 2 F- (aq)}$ $E^\circ=\pu{+3.27 V}$ (Wikipedia).

Top 5 oxidants from the CRC Handbook of Chemistry and Physics [1, p. 5-83]:

TABLE 2. Reduction Reactions Having $E^\circ$ Values More Positive Than That of the Standard Hydrogen Electrode
$$
\begin{array}{ll}
\hline
\text{Reaction} & E^\circ/\pu{V}\\
\hline
\ce{F2 + 2 H+ + 2 e- ⇌ 2 HF} & 3.053\\
\ce{Tb^4+ + e- ⇌ Tb^3+} & 3.1\\
\ce{Pr^4+ + e- ⇌ Pr^3+} & 3.2\\
\ce{Cf^4+ + e- ⇌ Cf^3+} & 3.3\\
\ce{XeF + e- ⇌ Xe + F-} & 3.4\\
\end{array}
$$

Note that standard reduction potential is not characteristic for a single compound. Instead, it is only makes sense for a given redox pair – of course, assuming standard conditions ($\pu{298.15 K}$, $\ce{101.325 kPa}$), activity of all soluble species $\pu{1 .000 mol L-1}$ and measured against the standard hydrogen electrode.
For example, iodate $\ce{IO3-}$ has very rich redox chemistry and many SRP values [1, p. 5-79]:

$$
\begin{array}{ll}
\hline
\text{Reaction} & E^\circ/\pu{V}\\
\hline
\ce{2 IO3– + 12 H+ + 10 e- ⇌ I2 + 6 H2O} &1.195\\
\ce{IO3– + 6 H+ + 6 e- ⇌ I– + 3 H2O} &1.085\\
\ce{IO3– + 2 H2O + 4 e- ⇌ IO– + 4 OH–} &0.15\\
\ce{IO3– + 3 H2O + 6 e- ⇌ IO– + 6 OH–}& 0.26\\
\end{array}
$$

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.


text:
Why is water instead of hydroxide ions used to balance oxygen atoms in redox half equations?
When balancing half equations of redox reactions, $\ce{H2O}$ is used to balance any oxygen atoms? For example, this half equation
$$\ce{NO3- → NH4+}$$
is balanced into:
$$\ce{NO3- + 8 e- + 10 H+ → NH4+ + 3 H2O}$$
Why is $\ce{OH-}$ not used instead of $\ce{H2O}$? E.g. the balanced equation becomes
$$\ce{NO3- + 8 e- + 7 H+ → NH4+ + 3 OH-}$$

Most of these redox reactions are done in aqueous medium, thus, there are plenty of $\ce{H2O}$ molecules around to use.

It's not that we really have a choice in this case between water or hydroxide or pick one or another arbitrarily to fit the stoichiometry of the whole redox reaction. The redox reaction is obviously taking place in quite a strong acidic medium (makes sense, e.g. if $\ce{Zn}$ is used as reducing agent) as you supply protons $\ce{H+}$ to the nitrate according to your reaction.
In fact, 10 protons for each reduced nitrate group is needed; and even if you propose hydroxide ion formation, you'll still need 7 protons (second reaction). Either way, there is no chance for $\ce{OH-}$ to coexist with $\ce{H+}$ in acidic medium and equilibrium will be shifted towards water formation instead.
Note that it doesn't mean that reduction of nitrate is forbidden if hydroxide ions are formed. For example, you can use another reducing agent such as $\ce{Al}$ and do perform the reduction in basic media instead. In this case you are going to end up with hydroxide ions all right:
$$
\begin{align}
\ce{Al + 6 OH- &→ [Al(OH)6]^3- + 3 e-} &|\cdot 8 \tag{ox}\\
\ce{NO3- + 8 e- + 6 H2O &→ NH3 + 9 OH-}&|\cdot 3 \tag{red}\\
\hline
\ce{8 Al + 3 NO3- + 21 OH- + 18 H2O &→ 8 [Al(OH)6]^3- + 3 NH3} \tag{redox}
\end{align}
$$
Just as with the formation of human personality, environment also plays gigantic role in chemistry.


text:
Gibbs free energy ΔG=ΔG°+RTln(Kp) different expressions for ΔG
ΔG=ΔG°+RTln(Kp)
If I understand it correctly, the ΔG in this equation gives the "distance" in Gibbs free energy of a chemical system at a particular moment in a reaction (encoded by the Kp) to the gibbs free energy of the chemical system at equilibrium.
So when we write this in symbols
for a general equilibrium reaction A + B <=> C + D:
ΣG(products at eq) + ΣG(reactants at eq) - (ΣG(products at k) + ΣG(reactants at k)) with k denoting a random point with certain concentrations in the reaction and eq denoting the equilibrium point.
However, ΔG is also ΣG(products at k) - ΣG(reactants at k).
So how are these two expressions for ΔG related? Basically how would you prove that
ΣG(products at eq) + ΣG(reactants at eq) - (ΣG(products at k) + ΣG(reactants at k))
= ΣG(products at k) - ΣG(reactants at k)

I can comprehend distance measured in seconds, but distance in Gibbs free energy is something new to me. Is it something related to the reaction coordinate? Also I'm not sure what exactly $ξ$ is here; I'm used to $ξ$ as extent of reaction when used alongside with Gibbs energy, but here it seems more like an arbitrary reaction coordinate.

You're right, that's confusing so I edited it. I meant "distance" in the sense that this site uses it https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Chemical_Energetics/Free_Energy_and_Equilibrium

Oh. Then maybe you should've also quoted the word "distance" as your source did; it's more like a metaphor rather than a physical quantity.

$K_p$ is normally reserved for equilibrium thus only apples when $\Delta G=0$ and conventionally $Q$ is used otherwise. In a plot of free energy $G$ vs extent of reaction $\zeta$, the slope is $(\partial G/\partial \zeta)_{T,p} =\Delta G$, and equilibrium occurs when the slope is zero then $\Delta G^\mathrm{o} =-RT\ln(K_p)$.  (The extent of reaction is zero with only reactants present, and 1 when one mole of reactants has become product.)
Thus to speak of 'distance' does not mean the energy difference from equilibrium as $\Delta G$ is the slope not the value of the energy $G$ vs $\zeta$.
In the sums you ask about, always calculate products - reactants. The only experimental quantities you have are the concentrations or pressures so it would be possible to calculate $\Delta G$, using your first equation, if the values are known at equilibrium and at some other point. This seems to be what you ask about.


text:
Finding mole fraction from molality

Calculate the mole fraction of ammonia in a $\pu{2.00 m}$ solution of $\ce{NH3}$ in water.

What I know is that the formula for mole fraction is
$$X = \frac{\text{no.-of-moles-of-solute}}{\text{(no.-of-moles-of-solute)} + \text{(no.-of-moles-of-solvent)}}$$
The solute is ammonia which is $\ce{NH3}$ with a molar mass (MM) of $\pu{17 g mol-1}$, whereas the solvent is water or $\ce{H2O}$ which has a molar mass of $\pu{18 g mol-1}$.
The $\pu{2.00 m}$ from the problem signifies molality (because of the small $\pu{m}$), and molality is
$$\frac{\text{no.-of-moles-of-solute}}{\text{mass-of-solvent-in-kg}}$$
With no. of moles
$$n = \frac{m}{\text{MM}}$$
Despite knowing the formulas, I can't seem to solve for the answer. The answer should be $0.0347$, but I can't seem to get the right solution.
Any help would be appreciated.

Please note: 1. The quantity ‘amount of substance’ shall not be called ‘number of moles’, just as the quantity ‘mass’ shall not be called ‘number of kilograms’. 2. Descriptive terms or names of quantities shall not be arranged in the form of an equation. 3. Multiletter abbreviated terms (such as ‘MM’) shall not be used in the place of symbols.

You don't have to memorize some weird formula like andselisk has proposed.
You have sufficient information to solve the problem:

Calculate the mole fraction of ammonia in a $\pu{2.00 molal}$ solution of $\ce{NH3}$ in water.

We can assume any quantity of solution, so let's assume 1.00 kg of solvent. So the mass of solvent (water) is $\pu{1 kilogram} = \pu{1000 g}$ in a molal solution by definition.
moles of water = $\dfrac{1000}{18.015} = 55.402$
For 1.00 kg of solvent there are 2 moles of $\ce{NH3}$ which has a mass of $\pu{2 moles}\times \pu{17.031 g/mole} = \pu{34.062 g}$
From the Op's formula:
$X = \frac{\text{no.-of-moles-of-solute}}{\text{(no.-of-moles-of-solute)} + \text{(no.-of-moles-of-solvent)}} = \dfrac{2}{2 + 55.402} \approx 0.0348$
Now I'll confess that the significant figures in this problem bother me. To have three significant figures the molality should have been given as 2.00 molal, not 2 molal.

Thank you. To be honest, I avoid memorizing too much formula. What confuses me though (until now) is the line "2.00m solution of NH3 in water". How did you know that there are 2 "moles" of NH3? Since that "2" from the question is the molal solution or molality of ammonia = 2 and its unit is mol/kg which is not the same with the number of moles (n), which is just "mol". Sorry for such question, I'm new to this.

@Jayce - The problem is open ended, so one can assume as much solution as desired. Frankly I tried solving the problem as 2 molar ( ie 1 liter of solution) which gave the "wrong" answer. Then I tried 2 molal (ie 1 kg of solvent) and got the "right" answer. An old convention is to use M for molar and m for molal. But without knowing which convention the particular book is using, that is somewhat of a guess. I think the newer convention is to be more explicit and use mol/L and mol/kg .

@Jayce - I edited the solution and moved things around a bit. Does that make the line of thought clearer?

Despite the unconventional notations, your formula is generally correct; however, you should've express mole fraction via molality explicitly and only then plug in the numbers. By definition, mole fraction of $i$-th component $x_i$ is
$$x_i = \frac{n_i}{n_\mathrm{tot}}$$
where $n_i$ – amount of $i$-th component; $n_i$ – total amount of all mixture constituents. For a simple solution of a single component the following hold true:
$$x_i = \frac{n_i}{n_i + n_\mathrm{solv}}$$
where $n_\mathrm{solv}$ – amount of the solvent that can also be found via its  molecular mass $M_\mathrm{solv}$ and mass $m_\mathrm{solv}$, which, in turn, appears in the expression for molarity $b_i$:
$$b_i = \frac{n_i}{m_\mathrm{solv}} \quad\implies\quad m_\mathrm{solv} = \frac{n_i}{b_i}$$
$$n_\mathrm{solv} = \frac{m_\mathrm{solv}}{M_\mathrm{solv}} = \frac{n_i}{b_iM_\mathrm{solv}}$$
Finally, mole fraction can be expressed via molality as follows:
$$
\require{cancel}
x_i = \frac{n_i}{n_i + n_\mathrm{solv}} = \frac{n_i}{n_i + \frac{n_i}{b_iM_\mathrm{solv}}} = \frac{\cancel{n_i}}{\cancel{n_i}\left(1 + (b_iM_\mathrm{solv})^{-1}\right)} = \frac{1}{1 + (b_iM_\mathrm{solv})^{-1}}
$$
Time to plug in the numbers:
$$
\begin{align}
x_i &= \frac{1}{1 + (b_iM_\mathrm{solv})^{-1}}\\
  &= \frac{1}{1 + (\pu{2.00e-3 mol g-1}\cdot\pu{18.02 g mol-1})^{-1}}\\
  &\approx 0.0347
\end{align}
$$
Few key points:

Note that you have to convert molality expressed in $\pu{mol \color{red}{kg}-1}$ before plugging in the value:
$$\pu{1 m} = \pu{1 mol kg-1} = \pu{1e-3 mol g-1}$$
In general, never omit units in your calculations and use standardized notations.
Mind significant figures. Since molality is given with two decimal points, you also should've taken molecular mass with higher precision.

Thank you. I would like to ask some questions though.

Xi stands for mole fraction of i-th component, so if for example I was asked to find the mole fraction of the solvent, instead of the solute, will the formula be the same?

The reason for expressing the molality in mol kg^-1 is so that it will have the same unit as the molar mass of the solvent?

This is too much to ask but can you answer the problem by using the formulas I've written above (if it's possible). Or at least how to transform / derive it into your kinda shortcut formula.

Again, thank you~

Yes, with respect to molar mass of solute, or just use $x_\mathrm{solv}=1-x_i$ for a single dissolved component; 2. No, 1 molal solution is a solution of 1 mol of the given compound in 1 kg of solvent by definition (not related to molar mass at all); 3. Since you used non-standard notations (or none at all) I'd rather prefer not to do that as it's going to bring a lot of confusion on both sides; I'll try to post an updated answer with the derivation later this day.

@Jayce The answer is updated with the derivation of the formula linking molality with mole fraction

Thank you again. It is clear now how the formula was derived. One of the reason I got too confused in answering the problem was due to the line in question: "2.00m solution of NH3". I assumed that the 2 molal is the molality of ammonia and not the solvent / water. Another reason was I kept on figuring out how can I insert the molar mass of NH3 into the formula and also how can I find the mass of water and ammonia given the limited givens. Again, thank you. I learned a new formula, thanks to you~

@Jayce No prob, and good luck with chemistry:)


text:
Waste water treatment
This is a very simple question: Why are oxidized compounds generally thought to be less toxic/polluting than reduced species? This is used in waste water treatment where oxidizing agents (such as chlorine) are added to remove organic AND inorganic pollutants.

I don't see the trend of reducing toxicity when pollutant is oxidized. In fact, it's often the opposite and may lead to more polluted water: chlorinated water might contain organochlorides which are more dangerous than its precursors; chlorinated water is also more corrosive towards metal water supply pipes, possibly introducing increased content of copper, iron and lead in tap water and so on (same applies to ozone treatment). Oxiding compounds are never used exclusively as their primary goal is to kill bacteria, not to get rid of pollutants.

Not an expert in this but it sounds fishy. For instance, you add oxidants to a swimming pool to impede growth of unwanted micro-organisms, not because the water is in any way less "toxic" after treatment.

Hmm that is for sure correct. However, we study this in an engineering course where chlorination is also used as an oxidant of contaminants such as arsenite, iron and manganese, sulfide etc... But yeah I don't understand the lecturer focusing on these aspects considering that negative outcomes of oxidation seem similarly important.

Source water is often oxygen-free and contains dissolved Fe(II) and Mn(II). They are removed by bubbling air through it, they are rapidly oxidised and precipitate on a sand filter. Chlorination sounds like total overkill for that purpose.

Generally, there are rather few (by mass) unpolar compounds in a biologic system. They are often bioactive (steroids, terpenes, etc.) and/or toxic. Coming from nonbiologic source, they are very often toxic.
Once they've been partially oxidised (as in "added oxygen", "made more polar", not as in "chlorinated"), there are much more possibilities for further metabolic breakdown. Polar substances don't usually (exceptions prove the rule) survive terribly long in nature, because some thing or other finds them nutritious. ;-)
Of course the intermediate metabolites can be even more toxic (it's often a metabolism that was "meant" by nature to do sth else), and it doesn't work out all the time, but the general trend is clear: keep oxidising, and you end up with CO2 and water. The smaller and more polar the fragments become, the quicker.
And one comment on chlorinating: Improperly done, esp. on dirty water, it produces organochlorides, which are practically all toxic. The modern way is to clean the water with excess ozone, and then add a tiny amount of hypochloride (if necessary, i.e. in warm countries or if the piping is not in good working order) to the pure water, to keep the water pipes from moulding. It's just to kill any microorganism that tries to sneak back in, not to actually oxidise him.
And for inorganic compounds: Underground water very often contains soluble iron and manganese (II), because the inside of our earth is generaly a bit reductive. They are removed by bubbling air through the water, the rapidly oxidised +3 ions then precipitate on a sand filter. Chlorination would do the same, but so does ozone, if you really need a very rigourous workup (e.g. surface water). I guess using chlorine is a 1960ies technology. We visited our local water station when I was in primary school in the 80ies, and I remember they explained it as "like in your municipal swimming pool".

There is not such as general/ic rule but water treatment has used, waste water as inflow. It is not as we are discussing the oxidation of X to obtain Y, if you get the point.  The target is to get an outflow of water that can be reused or discharged in nature without significant impact. The lines below do not pertain to the treatment of industrial wastes nor the removal of a specific pollutant.
Normally no toxic compounds go to the sink, and even in the lab we are not supposed to flush chemicals down.
Oxidative processes in waste water treatment mainly consists of biological aerobic ones, basically the demolition of a wide range of chemicals as it would naturally occur in a natural environment by means of bacterial activity.
What is peculiar here is that the waste water is kept in a regulated physical and chemical environment able to sustains and ease the process (for instance the oxygen required by the microorganisms is guaranteed by injecting air, or the temperature is regulated, ecc.).
This aerobic biological oxidation is targeting biodegradable pollutants is not targeting any specific chemical. Again, keep in mind that we are speaking of waste water not of special industrial waste. Even a chemical toxic to the bacteria, or in other words interfering with the bacterial  metabolism, does not pose a limitation as far is getting degraded in the while.
Most of the organic pollutants are decomposed into carbon dioxide and water leaving a water with high content of nitrites, nitrates, phosphates that must be lowered in subsequent steps and is full of particulate and biomass that must be separated as well.
Let assume all the steps above have resulted in a sufficiently clear and relatively pure water, then yes a chemical oxidation is performed. This step might use chlorinated chemicals and other oxidizers, but it is just targeting the residual microorganisms and tailored just to do that, like in the case of the disinfection of a swimming pool. It is performed when virtually there is nothing to oxidize but the residual organisms that were at the start of the whole treatment.


text:
Opening the imidazoline ring in tolazoline

Hello.
I was studying about the alkaline ring opening or purines. While doing research, i came across this compound. This got me thinking if we can use the same procedure to open the imidazoline ring in tolazoline.
Please give me as much information as possible about the reaction and reagents that will be used in opening the imidazoline ring of tolazoline.

The compound you have drawn is an imidazoline not an imidazole - big difference.
Imidazolines are fairly easily to hydroyse in both acidic and basic media, there is a good review here.
Imidazoles on the other hand are much more stable aromatic compounds and the hydrolytic ring opening is far from straighforward.


text:
will the autoionization of water reduce due to common ion effect?
When a weak acid is added to water , will the dissociation of water reduce due to common ion effect caused by increased H+ concentration due to acid , my text book said common ion effect is caused by a strong electrolyte
Moreover this doubt is a part of a larger doubt - if multiple weak acids are added in water will they reduce each others and water's dissociation as compared to their dissociation in their individual solutions ( due to common ion effect caused by H+ ions ) They should ? Having this doubt because the textbook I'm using just adds up their normal dissociations irrespective of common ion.

Yes, if you have multiple equilibria with common species, they will influence each other. If you put an acid in neutral water, the $\ce{H+}$ will increase of course, but at the same time, the $\ce{OH-}$ concentration will decrease. For the case with multiple weak acids, adding the first acid will change the pH, so the other acids will dissociate less just like when you add strong acid to a weak acid.

Adding acid to the water will not stop the auto-ionization of the water, so: the auto-ionization will continue to give equal concentrations of hydroxide and hydronium. But the addition of acid to the water increases the concentration of the hydronium ions in the solution, pushing water balance with its ions to the left to resist the increase in the concentration of hydronium and reach a new equilibrium. The concentrations of hydronium and hydroxide produced by auto-ionization of water are reduced, where the concentrations of $\ce{H_3  O^+}$ or $\ce{OH^-}$ from water ionization is less than$\pu{ 10^{-7}}$.
$$\ce{{[H_3O^+]}_\text{(after acid  addition)}=\ce{[H_3O^+]}_\text{(produced by acid dissociation)} +\ce{[OH^-]}_\text{(produced by water auto-ionization )}}$$

so how do I know what is the new dissociation.

You have two equilibria: water auto-ionization and acid dissociation, and you can  use the following equation to calculate $\ce{[H_3O^+]}\text{(after acid addition)}$:$$\ce{{[H^+]}=\frac{K\mathrm{w}}{[H^+]}+\frac{K_\mathrm{a}\text{[Acid]}\mathrm{I}}{\ce{[H^+]}+K\mathrm{a}}}$$


text:
Why is are weak acid-strong base and strong acid-weak base reactions irreversible?
In the case of both are strong, I can see why it's irreversible, because the acid is very willing to give a proton and the base is very willing to take one.
Now here are my two main questions:

In the case of both are weak, we say the reaction is reversible. Does this mean the reaction happens then the products react again to give reactants, or that the reactants don't run out and the reaction doesn't happen at "full potential"?
In the case of a reaction between a strong base and a weak acid (or the other way around), why is it always irreversible? I mean since one of them is weak, shouldn't it give out an unstable product which is willing to react again, like
$$\ce{CH3COOH + OH- → CH3COO- + H2O}$$
The $\ce{CH3COOH}$ is a weak acid. The $\ce{CH3COO-}$ isn't stable so shouldn't it react again?

When is a reaction considered 'irreversible'? What's the equilibrium constant of the reaction between acetic acid and sodium hydroxide?

Yes the following reaction is reversible.
$$\ce{CH3COOH + OH-  <=>  CH3COO-  + H2O}$$
There is an equilibrium between the species $\ce{CH3COOH, OH-}$ and $\ce{CH3COO-}$ in aqueous solution.
Thus the gist is that particular $\ce{CH3COOH}$ and $\ce{CH3COO^-}$ molecules aren't stable, but rather continuously react.


text:
What is Delta H for dissolution?

An ionic compound has a solubility of $1\ \mathrm M$ in water at $25\ ^\circ \mathrm C$ and its solubility increases as the temperature is raised. What are the signs of $\Delta H^\circ$ and $\Delta S^\circ$ for the dissolving process?

Since the the solubility increases as the temperature is raised ($\Delta G^\circ$ becomes more negative), I know that $\Delta S^\circ$ is positive. However, I am unsure of how to determine the sign of $\Delta H^\circ$. The answer is $\Delta H^\circ > 0$.

Hint: What happens to the bonds in the substance as it dissolves...

I know the bonds break (delta H is positive), but I thought that some substances, like NaOH, dissolve exothermically (so delta H is negative)?

I am not sure the proposer of the exercise want all this but your doubt is legitimate. See here: https://chemistry.stackexchange.com/questions/51862/why-does-the-solubility-of-some-salts-decrease-with-temperature

See answer about Van't Hoff equation here.

You could estimate K from the equilibrium concentrations and use that to determine $\Delta G$. That might help figuring out the signs of the other quantities.

The statement about $\Delta G^\circ$ is not quite true. K becomes larger, and $\frac{\Delta G^\circ}{T}$ becomes more negative as the temperature increases.

@KarstenTheis That is a great point. The emphasis on $\Delta G$ makes one (me at least) forget to focus on $\Delta H$ rather than $\Delta S$ when looking at the T dependence.

@Mithoron The question has generated some interest, and while the question does not show "some effort to understand the underlying concepts", it shows some misconceptions about the underlying concepts, explaining why they got stuck. The comments address some of these misconceptions, but it would be great to eventually have an answer, maybe by opening the question again now, or when the OP has updated the question in light of the comments.

@Raphaël did the hard part. ΔH is larger than zero.

[OP] Since the the solubility increases as the temperature is raised (ΔG∘ becomes more negative), I know that ΔS∘ is positive

This argument linking solubility and Gibbs energy is incorrect. Le Chatelier or the Van't Hoff equation connects the temperature-dependence to the reaction enthalpy, not to the entropy. So how can we make a statement about the entropy? You might think the entropy of dissolution must be positive because the solid ionic compound is more ordered than the individual ions in solution. This is a good guess, but sometimes the solvent loses degrees of freedom, so it is not a sure bet.
Instead, let's consider the Gibbs energy of reaction. It is zero, because the reaction is at equilibrium. What about the standard Gibbs energy of reaction? Because all species are approximately at standard state, it is also zero (or near zero). With a positive change in enthalpy, the change in entropy has to be positive as well to get a Gibbs energy of zero.

I would solve this question rather simply.
Let's consider the general equation $\ce{ X-Y_{(s)} \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$ with a $\Delta H = C$. Intuitively, if the reaction is endothermic ($\Delta H > 0$), an increase in temperature should help the dissolution (more heat to be absorbed by the salt while dissolving). You can rationalize this further using the Le Châtelier's principle:
$\ce{ X-Y_{(s)} + C \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$
If you add more heat ($C$), the dissolution will be more favored.
More rigorously, you can use Van't Hoff equation (assuming a dissolution equilibrium): $\frac{d(ln(K))}{dT} = \frac{\Delta H}{RT^2}$. If $\Delta H > 0$, then $ln(K)$ increases with $T$, i.e. dissolution is more favored.


text:
High concentration of ferric ion catalyst decreases rate of decomposition of hydrogen peroxide in the Fenton reaction
I did a project on the Fenton reaction and rate of decomposition of $\ce{H2O2}$. I used colorimetry to measure the rate of decolourisation of methylene blue when I mixed $\pu{5e-6 mol}$ $\ce{FeCl3}$ and $\pu{0.2 mL}$ of different concentrations of $\ce{H2O2}$ (1, 3, 5, 10, 15, and 20%).
After processing the data, I found the following graph:

I was wondering why with higher concentrations of $\ce{H2O2}$ the rate decreases. I initially wanted to analyse this according to the Michaelis-Menten model (as $\ce{Fe^3+}$ ion can be regarded as a catalyst in this reaction). However, it seems my data goes in the opposite direction as the Michaelis-Menten plot. What surprised me most that it was decreasing.
I came up with a few hypotheses for this, and searched for papers, but I could not find any with results similar to mine. My main hypothesis was that because $\ce{Fe}$ ion and $\ce{H2O2}$ form $\ce{OH.}$ radicals and $\ce{OOH.}$ radicals (as reported here and here), and these radicals are what cause the decomposition of methylene blue, I thought that maybe if there was too high a concentration of $\ce{H2O2}$, there would be too many radicals, and this would in turn terminate the free radical reactions (when 2 free radicals meet). At lower concentrations, the radicals would continue propagating, hence decolourising methylene blue faster.
Moreover, I know that this is not a problem with the colorimeter, as I could see this decolourisation outside the colorimeter too. The following is a screenshot of a video I took; each reaction in the cuvette was started at the same time, and concentrations of $\ce{H2O2}$ used in order from left to right are 1, 3, 5, 10, 15, 20%.

I was wondering if anyone knew why this happened.

Do you have a negative control, i.e. no $\ce{H2O2}$ added? I don't quite understand the y-axis of your graph. Why isn't it change in MB concentration divided by time?

In this paper, a similar reduction in rate is observed at concentrations of H2O2 above 0.03 M final H2O2 concentration (which is about 0.1 wt %). They attribute it to $\ce{HO.}$ reacting with H2O2 (to yield $\ce{HOO.}$) more quickly than it reacts with MB. The result is that $\ce{HOO.}$ is much more abundant than $\ce{HO.}$ and its lower reactivity means a slower rate of reaction with MB.
Based on their data, I would recommend repeating the experiment with much lower H2O2 concentrations.