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Oxyanionic Series and Chemical names
If there is only 1 Oxyanion in a series, what is it’s prefix/suffix? Also, if I have Copper(I) Iodate, how do I know how many oxygen atoms are in the compound?

Also, does a specific amount of oxygen correspond to the prefix/suffix? Example: compound with 1 oxygen= Hypo-X-ite, 2 oxygens= X-ite, 3 oxygens= X-are, and 4 oxygens = Per-X-ate

Like this matters, really. On one hand people want some complicated IUPAC names, on the other still use naming from what XIX century?

I agree, but this is also for homework XD

If there is only 1 oxyanion in a series, what is it’s prefix/suffix?

This anion gets the "-ate" suffix and no prefix. Example: Carbonate $\ce{CO3^2-}$. There is no "carbonite" to freeze Han Solo in; the $\ce{CO2^2-}$ anion does not exist. Carbon does form other oxyanions, like oxalate, $\ce{C2O4^2-}$, but these anions also vary in the number of carbon atoms.

Also, if I have Copper(I) Iodate, how do I know how many oxygen atoms are in the compound?

Iodine can form multiple oxyanions, although hypoiodite and iodite are not particularly stable:

$\ce{IO-}$ - hypoiodite
$\ce{IO2^-}$ - iodite
$\ce{IO3^-}$ - iodate
$\ce{IO4^-}$ - metaperiodate
$\ce{IO6^5-}$ - orthoperiodate

With the exception of the "meta" and "ortho" periodates, this same trend is seen for two other halogens, chlorine and bromine (chlorine shown):

$\ce{ClO-}$ - hypochlorite
$\ce{ClO2^-}$ - chlorite
$\ce{ClO3^-}$ - chlorate
$\ce{ClO4^-}$ - perchlorate

Thank you, but let’s say I didn’t know Iodine could form multiple oxyanions, how would I know how many oxygen atoms are in Iodate?

And vice versa. If I have BaIO4, how do I know it’s name if I don’t have the other oxyanions with iodine?

@Jondue - You have two choices: Look it up or memorize it.

Actually carbonite was detected! https://chemistry.stackexchange.com/questions/64416/does-the-carbon-dioxide-dianion-exist

Or, as a surface species: https://doi.org/10.1007/BF02069193


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How would you determine the RDS from the rate and order of a reaction
I know what a rate determining step is, and also have knowledge about zero order and first order reactions. But, in one of my tests, I was unable to solve the following question:

A hypothetical reaction A2 + B2 --> 2AB has a rate law, Rate = k[A2]1 [B2]0 and mechanism of reaction is
Step-I: B2 --> 2B
Step-II: A2 --> 2A
Step-III: 2A + 2B --> 2AB
The correct statement is
A) Step-I of mechanism is rate determining step of the reaction
B) Step-II of mechanism is rate determining step of the reaction
C) Step-III of mechanism is rate determining step of the reaction if step I and II are reversible
D) Step-I and Step-III both are rate determining step of the reaction if step II are reversible

The correct answer is option B, but how would you approach the problem and solve it?
EDIT: What if rate was given as
1) Rate = k[A2]1 [B2]1
2) Rate = k[A2]2 [B2]1

The reaction is 1st-order in A2, so for starters we look for a step with A2 on the left, with a coefficient of 1. In fact, only step II has A2 on the left at all, so really we don't have to look any further.
Comments which may (or may not) make it clearer:
The rate equation says that greater the A2 concentration, the faster the reaction goes, in a linear relationship, and the rate doesn't depend on [B2] at all, since [B2] has an exponent of 0, and anything to the zeroth power is 1. (I suspect that was just put there to make it hard.)
If the whole reaction can be speeded up by adding A2 alone, [A2] must be rate-determining; i.e., [A2] determines the rate, and A2 reacts only in step II.

The answer is dependent on the assumption that the reactions are elementary reactions. In other words that the rate equation has reaction orders equal to the stoichiometric coefficients for each reactant. In chemical kinetics that doesn't necessarily have to be true.
So for the irreversible reactions only the forward rate, $r_f$, needs to be considered.

Step-I: $\quad\ce{B2 ->[r_f] 2B}$

$\ce{r_f = k_1[B2]}$

Step-II: $\quad\ce{A2 ->[r_f] 2A}$

$\ce{r_f = k_2[A_2]}$

Step-III: $\quad\ce{2A + 2B ->[r_f] 2AB}$

$\ce{r_f = k_3[A]^2[B]^2}$
For the reversible reactions both the forward rate, $r_f$, and the backwards rate, $r_b$ need to be considered to get the overall rate, $r$.
$r = r_f - r_b$

Step-I: $\quad\ce{B2 <=>[r_f][r_b] 2B}$

$\ce{r= r_f - r_b = (k_{1f}[B2]) - (k_{1b}[B]^2)}$

Step-II: $\quad\ce{A2 ->[r_f] 2A}$

$\ce{r= r_f - r_b = (k_{2f}[A2]) - (k_{2b}[A]^2)}$

Step-III: $\quad\ce{2A + 2B ->[r_f] 2AB}$

$\ce{r= r_f - r_b = (k_{3f}[A]^2[B]^2) - (k_{3b}[AB]^2)}$
So for the hypothetical reaction $\ce{A2 + B2 -> 2AB}$ with a rate law, $\ce{Rate = k[A2]^1 [B2]^0} = k[A2]$ and we just match to the various steps to see that Step II is the rate determining step.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Claisen-like condensation with acid chloride
In the Claisen condensation, an ester enolate attacks an ester, with an alkoxide anion leaving subsequently to give a $\beta$-keto ester. Is it possible for a typical enolate of an aldehyde or ketone to react with an acid chloride (acting as the electrophile) to give a $\beta$-keto aldehyde or ketone, with the chloride ion leaving in a similar manner to how the alkoxide leaves in Claisen condensation?
Similarly, could I also extend it to the anhydride as well? Would we observe a similar Claisen-like mechanism?

It seems that an enamine is able to react in this fashion with an acid chloride: http://chemtube3d.com/Enamine_acylation.html

I think it is possible.

Certainly possible if the enolate is generated with a non-nucleophilic base

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Stability comparison on the basis of violation of Bredt's rule between bicyclic carbocations

Compare stability between the two bicyclic compounds.

Both compounds I and II seem to be unstable because of violation of Bredt’s rule: they are sp2-hybridized.
Then how to compare the stability between I and II?

The carbocation's energy would be effectively minimised when it adopts an $\ce {sp^2}$ hybridisation scheme, that is, when it adopts a trigonal planar molecular shape. This would ensure that all the precious s orbital character of the carbon atom would be used for bonding with the other three surrounding atoms, leaving a vacant p orbital. However, this bicyclic structure prevents the carbocation from adopting this scheme, destabilising the carbocation formed. This is the basis of Bredt's rule.
On p. 435, Carey & Sundberg (2007) wrote the following:

... bridgehead carbocations become accessible in more flexible structures. The relative solvolysis rates of the bridgehead bromides 1-bromoadamantane, 1-bromobicyclo[2.2.2]octane, and 1-bromobicyclo[2.2.1]heptane illustrate this trend. The relative rates for solvolysis in 80% ethanol at 25 degrees Celsius are shown.

When there is greater flexibility in the structure, the carbocation is better able to adopt a structure that is closer to the preferred planar structure, thus it is better able to be minimise its energy. This results in a lower activation energy and an increased rate of reaction, explaining the data above.
Reference
Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


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I smelled an organic chemical in a lab. (Hexamethylenediamine) Will it cause brain damage?
Our teacher told us to keep a distance while smelling organic chemicals. I forgot and smelled from the opening of the tiny container. Can this substance be corrosive and may that have neurological implications? Could it have had an effect on my brain (i.e death of brain cells)?

The Material Safety Data Sheet for hexamethylenediamine is here MSDS here.
The main danger is associated with skin or eye contact. If you did not feel any immediate effects from smelling the sample then it is highly unlikely that any lasting harm was done. Take this as a lesson to be very careful when handling chemicals.


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What makes dimethylmercury ((CH3)2Hg) so penetrative through latex and human skin?
I've read the story of Karen Wetterhahn and how she was poisoned and died from two drops of dimethylmercury that accidentally fell on her latex glove, and I was wondering what is the chemical explanation for such easy penetration of the compound through latex and skin.
I do not have a vast knowledge of heavy metal chemistry, but here are the points that gave rise to this question:

There are many dimethyl compounds with a similar structure to dimethylmercury. Some are very common, such as dimethyl ether (CH3-O-CH3), propane (CH3-CH2-CH3). Oxygen is a lot smaller than mercury, and methylene is larger but is (presumably, I might be wrong here) more similar to the methyl groups than mercury in hydrophobicity etc., so why are they not as penetrative?
According to this answer by AMM:

Actually metallic mercury is very hard to absorb by ingestion or through the skin so if ingested one will most likely survive (according to studies on animals). However, if one ingests the same amount in a soluble salt form (or any other absorbable form, eg HgMe2) he is in serious trouble (see here: Karen Wetterhahn).

Is it because the most common form of mercury is the cation $Hg^{2+}$, whereas the methyl groups make the compound uncharged and thus soluble in hydrophobic matter, such as skin and latex?

Yes, that's pretty much the size of it.

I'm not sure rate of skin absorption is the problem. For one thing, I can't find skin absorption values for HgMe2 to make a comparison. Nevertheless, a crucial difference is that if your skin comes into contact with 0.1 mL of HgMe2, your life is in serious danger, whereas 0.1 mL of dimethyl ether or propane on the skin (well these are gases but still), no matter how efficiently these penetrate the skin, you aren't going to feel it—your body takes care of it. As for why HgMe2 kills you so easily? Try reading sections 1 and 3 here: https://en.wikipedia.org/wiki/Methylmercury


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How to calculate the magnetic exchange coupling in a charged radical complex?
I have a neutral complex with two magnetic centers. For this case it is easy to calculate the exchange coupling as the broken symmetry state is converged properly. But when I am making the complex charged, there is radical that is introduced. So now I have charged radical species and this radical is not localized on just one site but spread over a group of atoms. In this case I am not able to reach the proper broken symmetry state using Noodleman's formalism employed in ORCA. What are the possible ways to achieve the proper broken symmetry state and calculate the exchange coupling.


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If copper becomes copper oxides because of oxygen in the air, is it a competing reaction?
If copper becomes copper oxides because of oxygen in the air, is it a competing reaction?
My science class made pure copper with a simple displacement reaction. We let it set on the window to let it dry then 2 days later it became green.

Green isn't due to an copper oxide but rather copper(II) carbonate-hydroxide, $\ce{Cu2CO3(OH)2}$. The mineral is known as malachite.


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Is the relative humidity of pure water vapour 100%?
If there is pure water vapour in the gas phase, can the partial pressure be taken to be equal to the vapour pressure of water at that particular temperature?
edit: How is relative humidity calculated for pure water vapour?

The nature of your post seems to indicate that you've conflated some different things. Let me just sort them out and I think you answer will be clear.
Vapor pressure of water
Just the partial pressure of the water vapor, which may or may not be at equilibrium.
Equilibrium vapor pressure of water
For a closed isothermal system, with a gas phase and a liquid phase and between 0 C and 100 C, there is a particular vapor pressure of water that exists for the system at equilibrium. In general the higher the temperature in that range the higher the vapor pressure of water at equilibrium.
The vapor pressure of the water vapor does indeed depend on the purity of the water. So at a given temperature the vapor pressure of salt water would be less than that of pure water. Also if there is a solvent mixture of water the vapor pressure water will be less than that for pure water.
Relative humidity
For a given temperature there is a maximum amount of water (vapor pressure) that can exist in the vapor phase. Any excess will condense. So the relative humidity for a given temperature is expressed as the percentage of measured vapor pressure as compared to the maximum vapor pressure over pure water.

Thanks a lot. So is it right to say that even if a gas phase consists only of pure water vapour, the relative humidity can vary between some small value to 100%?

@Monisha The concept of relative humidity is irrelevant for pure water vapour. It is a measure of how much water vapour is in a mixture. If you have pure water vapour everything you need to know will be given by the temperature and pressure (and the major issue will be whether there is any liquid present at equilibrium).

For a temperature, T, 100% humidity is the equilibrium partial pressure of pure water at temperature T. Relative humidity at temperature T has to vary from 0% to 100%.

@matt_black Initially I was of the view that the concept of RH can only be applied to air-water mixtures. But a paper in a chemistry journal mentioned conducting their metal oxidation experiments with pure water vapour and 100% RH.


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Why do stronger intermolecular forces lead to a larger change in temperature?
We performed an experiment where tissue paper soaked in various chemicals was wrapped around the end of a temperature probe.
The chemicals with stronger intermolecular forces had a smaller drop in temperature (from room temperature to final, after most of the alcohol had evaporated off). Two of the chemicals were ethanol, with hydrogen bonding, and n-pentane, with only London dispersion forces.
My question is why?
Below are just my initial attempts to understand
My initial feeling was that molecules with stronger intermolecular forces would take more energy to evaporate and would have a larger temperature change than those with weaker bonds, but we see the opposite in the data.
I was also thinking that maybe went the other way: chemicals with stronger forces had fewer molecules evaporate, so the temperature drop was smaller. Is this more correct?

It's a matter of kinetics, not thermodynamics. If you measured heat not temp. you should get expected results.

The heat (enthalpy) needed vaporise 1 g of each solvent is approx 840 J for ethanol and approx 360 for the pentane. So unless you have equal or at least known quantities in your tissue your experiment can give misleading results. However, your initial idea about stronger forces needing more energy to evaporate makes sense.


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Gallium - Practical use to remove aluminium
Would gallium pass through aluminium into steel?
The situation is two joined pieces of metal.
Gallium could be applied to the aluminium to degrade it, for mechanical removal, but would it pass through to attack the steel beneath or remain within the aluminium?

Possible duplicate of Does Gallium (liquid or solid) corrode all forms of brass?

Not a duplicate of the brass corrosion thread, this is specifically about practical gallium/aluminium/steel interactions. Literature indicates that gallium will damage steel, but whether this is occurs much more slowly than in aluminium and whether the gallium would transfer out of the aluminium are unanswered.

Did you read the answer? "Gallium is corrosive to all metals except tungsten and tantalum, which have a high resistance to corrosion."

Hi MaxW, I did. Apologies if my questions are not well phrased, I think I see the confusion and will edit the question phrasing. The question is not whether gallium is corrosive, it is.

Think you're looking at this incorrectly. (1) I doubt that you could get the gallium only on the aluminum post. (2) When breaking apart the aluminum post you're going to get crud on the steel frame. // So what you propose isn't impossible, but what is the correct procedure to get it to work? Got some spare bikes to experiment with?

Perhaps 2) This is another way of looking at the crux of the question.... By crud I assume you mean the bits of brittle aluminium infiltrated by gallium. Would this crud itself damage the steel, or is the gallium 'happy' to stay within the aluminium doing no harm?


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Order of basicity

I tried using conjugate acid method to decide stability but I’m unable to differentiate between b,c,d based on inductive effect. There is no conjugation so no resonance.

In aprotic or very small polar solvents like chlorobenzene, solvated ions cannot form so there is no solvent effect. It follows a 'similar' order as that in gaseous phase.
Coming to the question, on protonation,  option A has 2
Alpha Hydrogens (with respect to N+). B has 1 alpha Hydrogens, C has 6 alpha hydrogens and D has 4 alpha hydrogens.
Number of hyperconjugative structures = No. Of alpha hydrogens. So we would have C>D>A>B based on hyperconjugative stability.
But actually, B>A. This is because in B, there is a possibility of hydride shift which shifts the + charge on nitrogen to the carbon connected to it. On doing so, B has 4 alpha hydrogens.
Hence C>D>B>A is the answer. (although now D and B have same no of alpha hydrogens (same no of hyperconjugative structures), D is more stable than B as it would require some amount of energy to perform Hydride shift).


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Gold/platinum + HF
I know that noble metals such as gold/platinum are unaffected by conc. mineral acids such as HCl or H2SO4 or HNO3 but Can be dissolved in aqua regia (mix of HCl and HNO3 in 3:1), but never heard about HF .Can conc. HF (100% pure) dissolved noble metals.If yes, what will be the products
NOTE:-I only mean 100% HF (superacid) not aqueous HF.

In short, no. Besides, HF is not much of a superacid.

Pure HF is a superacid due to H2F+ and have a H_0  of -15.1 . Refer Wikipedia

Yeah, sure, but it is not all that much when we talk about superacids.

It isn't the acidity of aqua regis that enables it to dissolve precious metals: it is its oxidising ability combined with the solubilising nature of the chloride in solution. HF doesn't oxidise and doesn't solubilise. Which is why it doesn't work here.

A look at the Nernst equation shows that increasing proton activity is of only limited impact in dissolving noble metals via hydrogen displacement.  Say you have a metal $M$ for which the standard reduction potential of $M^{n+}$ is $-1\text{ V}$ and an acid whose proton activity is $15$ orders of magnitude greater than $1\text{ M }\ce{HCl(aq)}$.  Then the equilibrium potential for generating one atmosphere hydrogen gas at $25°\text{C}$ with the metals in the standard states is given by:
$E=-1+\frac{(2.303)(15)(8.314)(298.15)}{96487}=-0.11\text{ V}$
and the reaction remains unfavorable even with a fairly good superacid.  And the most noble metals go well below $-1\text{ V}$ in that first term.
So we use a stronger oxidizing agent like nitric acid to help things along.
Hydrofluoric acid can be used to break passive oxide films on some substrates.  For example, a combination of hydrofluoric and nitric acid is one way to dissolve silicon, which would otherwise form a passive oxide.  See here for the reaction, and a Google search on "silicon etching" gives many references (most of the good ones, of course, behind paywalls).

Gold or platinum is more or less unaffected by hydrofluoric acid even at elevated temperatures. From the handbook of corrosion data:

The noble metals including gold, silver or platinum is more or less unaffected by anhydrous hydrogen fluoride or aqueous hydrofluoric acid of any concentration at ambient temperature.[...] The resistance of silver to hydrofluoric acid can only be affected in presence of sulfides and oxygen. Silver is not passivated by hydrofluoric acid as it is by other halogen acids. Platinum and gold are unaffected at high temperature.

For more information, see this yahoo answer or this reddit post.


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H0 value for fluoroantimonic acid
I am confused about $\mathrm{H_0}$ value of fluoroantimonic acid. Wikipedia says $\mathrm{-23<H_0<-21}$. Others say around $-31$. There's a huge gap. I won't worry for $3-4$ fluctuations but worry for $10$.
Can anyone provide confirmed data?

I think it's an example mistaken data copied from Wikipedia, later (now) fixed there (with refs).

In this question there was a similar confusion over how strongly basic tetramethylammonium hydroxide is.  Apparently Wikipedia, or more precisely the references it used, confused TMAH with trimethylamine.  The erroneous information is now on the cutting room floor.

Superacids consisting of a mixture are going to have a certain range for the Hammett acidity function $\mathrm{H_0}$. For example, $\mathrm{H_0}$ for both Magic Acid $\ce{HSO3F-SbF5}$ and fluoroantimonic acid $\ce{HF-SbF5}$ relies on concentrations defined by the equilibrium of the acid's formation, which, in turn, is affected by how much $\ce{SbF5}$ has been added [1]:

$$
\begin{align}
\ce{2 HF + 2 SbF5 &<=> H2F+ + Sb2F11-}\\
\ce{2 HSO3F + 2 SbF5 &<=> H2SO3F+ + Sb2F10(SO3F)-}
\end{align}
$$
The acidity function of $\ce{HSO3F}$ increases from $-15.6$ to $-21.0$ on addition of $25$ mole percent $\ce{SbF5}$ ... as shown in Fig. 1. Extrapolation of the $\ce{HSO3F-SbF5}$ curve in Fig. 1 would lead to an $\mathrm{H_0}$ value of about $-25$ for Magic Acid. Fluoroantimonic acid is even stronger. As shown in Fig. 1, with $4$ mole percent $\ce{SbF5}$ the $\mathrm{H_0}$ value for $\ce{HF-SbF5}$ is already $-21.0$, a thousand times stronger than the value for fluorosulfuric acid with the same $\ce{SbF5}$ concentration. At present it is difficult to estimate the acidity of $1:1$ $\ce{HF-SbF5}$, but a value of $-28$ can be predicted ... on the basis of isomerization kinetics data.

Fig. 1. Relative acidities of $\ce{HF}$ and $\ce{HSO3F}$ on addition of $\ce{SbF5}$.

That is why most textbooks (e.g. [2]) list $\mathrm{H_0}$ from $-21$ to $-25$ for Magic Acid and $\mathrm{H_0}$ from $-21$ to $-28$ for fluoroantimonic acid.
References

Olah, G. A.; Prakash, G. K. S.; Sommer, J. Superacids. Science 1979, 206 (4414), 13–20. https://doi.org/10.1126/science.206.4414.13.
Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, Fifth edition.; Pearson: Boston, 2014. ISBN 978-0-321-81105-9.


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what is the issue in evaluating two electron integral?
What sort of issue is there in two electron integral in quantum chemistry ? Is it a problem of convergence when we do numerically? Can someone suggest any review article or reference articles which has the numerical algorithm to evaluate and to understand it fully ?

There is no issue, we can calculate such an integral all right. It is just that we have an awful lot of them.

The main problem of 2-electron integrals is their amount. They are given by
\begin{equation}
  I_{ijkl} = \langle ij|r^{-1}_{12}|kl\rangle
\end{equation}
where $i$, $j$, $k$ and $l$ run over all $N$ atomic orbitals. So we need to calculate $\mathcal{O}(N^4)$ integrals, which makes this a bottleneck (in terms of memory and computing time).
For Gaussian Type Orbitals we can evaluate the integrals analytically, which is much faster than numerical approaches. But this does not help with the $N^4$ scaling.
Evaluation for Gaussian Type Orbitals is explained in Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory by Szabo and Ostlund for s orbitals ($l=0$). More general evaluation schemes for higher angular momentum are explained in detail in Molecular Electronic-Structure Theory by Helgaker, Jorgenson and Olsen.

Thanks a lot.Can you also please tell whether monte carlo integration technique useful in evaluating this numerically ?

Monte Carlo would be a variant of numerical integration, which has pros and cons. Anyway there is no point in applying numerical methods when you have an analytic one. Unless you want to use Slater Type Orbitals. There is software available using them, but I don't know how they evaluate their integrals.


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Solving chemical equilibrium equation by hand

I was trying to solve question 22 above.
I obtained the equation:
$$50=\dfrac{4\alpha^2}{(1-\alpha)(2-\alpha)}$$
Now, the resulting quadratic is too much of a pain to  solve.
How would one use approximations to compute this by hand?

To do that, you should be able to make an educated guess. Is $\alpha$ close to 0, or close to 1, or close to 10?

Let $\alpha$ = amount of $\ce{A_2}$ that reacts, so the amount of $\ce{A_2}$ that is left is 1-$\alpha$. The amount of $\ce{B_2}$ that is left is 2-$\alpha$. The amount of product that is produced is 2$\alpha$.
$$50 = \dfrac{(2\alpha)^2}{(1-\alpha)(2-\alpha)} = \dfrac{4\alpha^2}{(1-\alpha)(2-\alpha)} = \dfrac{4\alpha^2}{2-3\alpha+\alpha^2}$$
So you set the equation up correctly.
Normally I'd just bite the bullet and solve the quadratic. With K=50 that means that the products are greatly favored, and the most product that can be made is 2. So with that multiple choice the only answer that makes sense is (c) 1.866, which means that
$$\alpha = 1.866/2 = 0.933$$
Checking that is quicker than solving the quadratic and checking that...
$$\dfrac{(1.866)^2}{(1-0.933)(2-0.933)} = 48.7$$
Any other answer would yield an even smaller value for the equilibrium expression, so (c) must be correct.
Now solving the quadratic to determine the rounding error...
$$46\alpha^2 -150\alpha +100 = 0$$
Gives
$$\alpha = 0.93444$$
which means that amount of AB = $2(0.93444) = 1.869$

This is a great observation, but could you apply some sort of approximation if I remove the options?

I'd never trust myself to solve a quadratic by hand without checking it. // If you remove the options, then I'd assume that 50 has two significant figures (at most...). So checking 1.9 gives 37 and 1.8 gives 17, so I'd go with 1.9. // The original problem is wacky. 50 has 2 significant figures, answers (a) and (b) have 3, and answers (c) and (d) have 4.

Why would 50 have two significant figures at most? In the form it is written, it has 1 significant figure, while in the scientific notation it can have any number of significant digits until stated otherwise (My source: http://chemistry.bd.psu.edu/jircitano/sigfigs.html). Also, what have you calculated as 37 and 17?

50 is ambigious. Does it mean $5\cdot10^1$ ( 1 significant figure) or $5.0\cdot10^1$ ( 2 significant figures). // If AB=1.9 then K=37. If AB=1.8 then K=17.

Can 50 also mean 5.00×10¹ ?

No, 50 is not so ambiguous as to possibly mean $5.00\cdot10^1$.


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CO2 is attracted to what?
People are using filters to remove $\ce{CO2}$ from the air.
What are this filters made of and are there any other substances that attract $\ce{CO2}$?
How is $\ce{CO2}$ absorbed or attracted to these substances?
P.S: Sorry, I did not get any suitable Tags for this.

The following wikipedia article does summarize different used techniques: https://en.wikipedia.org/wiki/Carbon_dioxide_scrubber


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Analytics of a paramagnetic complexes
I am working with salophene iron[III] complexes. Unfortunately, my 5th ligand cannot be determined via MS. I performed a 1H-NMR in methanol, but it's not a suitable method for a paramagnetic compound. What is a reasonable approach. Should i work on optimizing the MS method or is there a suitable method for iron salophene complexes via NMR.


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Can ideal gas law be applied to water vapour?
How to accurately estimate the molar concentration of pure water vapour at 298 K? Can ideal gas law be applied to water vapour to calculate the molar concentration?

Good or bad, you don't really have any other approximation, do you?

Get yourself a generalized corresponding states plot (from the internet) of the compressibility factor z as a function of the reduced temperature and the reduced pressure.  Knowing the critical pressure and critical temperature of the gas, as well as the specified temperature and pressure, you can judge from the plot (i.e., from the value of z) how accurate the ideal gas law answer will be.  For the state that you have defined, @Loong's comparison shows that z will be about 0.998.

It depends on the range of pressure you want to apply the ideal gas law to, and your definition of accuracy.

Yes, for many practical purposes, you may use the ideal gas law also as an approximation for water vapour.
However, you need two independent quantities to describe the state of the gas, for example temperature and pressure. The given temperature of $T=298\ \mathrm K$ alone is not enough. The ideal gas law does not give you a value for the second quantity.
For example, assuming equilibrium conditions for saturated steam at a temperature of $T=298.00\ \mathrm K$, the corresponding pressure is $p=3141.7\ \mathrm{Pa}$.
Using the ideal gas law yields a concentration of
$$\frac nV=\frac p{RT}=\frac{3141.7\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298.00\ \mathrm K}=1.2680\ \mathrm{mol\ m^{-3}}$$
The actual value for water vapour at this temperature and pressure is
$$\frac nV=1.2701\ \mathrm{mol\ m^{-3}}$$
Thus, using the ideal gas law introduces an error of about $0.2\ \%$ in this case, which could be acceptable for many practical purposes.


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Does gravity , say of another planet, affect the atomic mass and therefore the molar mass of water?
Does gravity , say of another planet, affect the atomic mass of elements and therefore the molar mass of water?

"Weight" depends on gravity, mass does not.

So the concept that you're getting confused is the difference between weight and mass. Mass is a fundamental property of an object that is not affected by its position in the universe. Any electron proton and neutron has the EXACT same mass as any other electron proton or neutron in the whole universe.
Weight, on the other hand, is the force an object feels due to gravity. Remember Newton's Second Law of motion $F = ma$. On Earth, the acceleration due to gravity is roughly 9.8 $m /s^2$ so a 1 kg object feels a force of magnitude 9.8 Newtons pointed downward. If this same 1 kg object were on the Moon where the acceleration due to gravity is 1.6 $m/s^2$ it would feel a downward force of only 1.6 Newtons.
Now, I suspect your confusion arises because we measure mass using the force of gravity. However, imagine a classic balance. If we put a 1 kg object on one side and a 2kg object on the other side both will feel a downward force due to gravity; however, the ratio of the forces (which is what determines the position of the needle) will be independent of the force of gravity.


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Francium has isotopes so how can its atomic mass be a whole number?
Why do francium, radon, radium, and actinium have whole numbers for average atomic mass even though they have isotopes? But then elements like beryllium, fluorine, and sodium are listed as having no isotopes but yet they have a decimal for their average atomic mass?

The Wikipedia data for Francium, Radon, Radium, and Actinium has mass numbers (protons + neutrons) not atomic masses since all the isotopes of these elements are radioactive.

Elements' names aren't capitalised.

It's not what's going on here, but note that it's perfectly possible for the weighted average of a set of integers to be an integer.

For elements with no stable isotope (i.e. francium, radium, and actinium), the atomic mass is chosen to be that of the longest lived isotope.
More generally, the masses for stable elements are reflective of the natural abundance of each isotope in a sample of the element. Sodium has more than one isotope, so the statement is not really true, though only one of them is stable and included in the averaging. In this case, it's important to remember that the mass of protons and neutrons are only approximately 1 amu, so even if there is only one isotope there's no requirement that it have an integer atomic mass.

For highly radioactive elements that don't occur in large quantities in nature, the mass number listed is the mass number of the most stable isotope. When this is done, the value is often written in brackets. The wikipedia page for the Periodic table has some nice details on the organization of the table and the values included.


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Reduction using sodium borohydride
When we react a ketone with NaBH4, it reduces to alcohol. I coudn't understand the mechanism of this reaction. The first step in my text book states that the hydrogen from NaBH4 attacts the carbonyl carbon. I didnt't understand why this happens. Specifically, I coudn't understand why the Boron didnt attack oxygen instead.

For starters, the formal charges should provide a good hint as to how the reaction proceeds.

Sodium borohydride is a complex unit made from the let's call it molecule $\ce{BH3}$ and the salt $\ce{NaH}$. Group 3 compounds like boron only have what we call an electron sextett instead of an octett. That makes them good Lewis acids, hence they tend to react with Lewis bases to form adducts. A famous example is the $\ce{H3B-NH3}$ adduct. In your example the hydride anion $\ce{H^-}$ is the Lewis base that forms an adduct with $\ce{BH3}$ to give $\ce{Na[BH4]}$. This means that at least one $\ce{H}$ in $\ce{[BH4]^-}$ can be somewhat easily removed again. The others tend to react as well but let's just call the $\ce{BH3}$ unit a stable molecule and assume it's a combination of $\ce{BH3}$ and $\ce{NaH}$. It's just a good way to store this hydride-anion so that it can be easily removed again. So once it sees your carbonyl group that has a positive partial charge on the carbonyl carbon (as the more electronegative oxygen pulls the electrons closer to itself) it can attack there as nucleophile. As you drew it already the double bond will turn to a single bond giving a negative charge on the oxygen.
Next you have to assume that $\ce{B}$ and $\ce{O}$ form quite stable bonds, hence the formation of a $\ce{B-O}$ bond is somewhat favored. And of course once the hydride ion is at the carbon we end up with an electron sextett on the boron again. Therefore it's the same thing, the carbonyl oxygen with an extra negative charge behaves like a strong Lewis-base again forming an adduct with the $\ce{BH3}$ unit.
This is later treated with water or acidified water to get the alcohol. To the question on the final product. I once talked to a boron chemist about that. He doubts that all four hydrogens will act as reducing agent under normal conditions. It's more likely that they will get hydrolyzed during workup.
Edit: Thinking about it. If you aren't sure about what attacks first you could probably also consider a transition state. I drew in the arrows here but you shouldn't do that (although you see it too often). I guess this would just be an alternative view on the reaction.

That mechanism looks cool, but while addition of BH3 to alkene is considered concerted, here I highly doubt that it works like that.

I'd doubt it as well. I'd just say if you are confused about what happens first and how the reactants might find each other in solution you can always just imagine it to be like that for example to see how exchanging one partner for another gives perhaps a more stable product.


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Conceptual understanding of solution / solute in a question
6.853g of sodium cyanide was dissolved in water and brought to volume in a 100ml volumetric flask. The final mass of the solution was found to be 107.166g
My understanding -
A solution is understood when I put a solute into a solvent (water) and dissolves it.
So in this case, if I wanted to find the composition of solution expressed as moles,
I must take mass multiply by the molecular weight of Sodium Cyanide, do I use the mass of the solution or the mass of solute ?
With this then I can go on to find molarity, normality as well as weight percent (mass of solute/mass of total solution) and weight/volume percent (mass of solute/volume of total solution).

Mole fraction is the ratio of the number of mole of a component to that of the number of moles of the total mixture. In your case, 6.853g of NaCN equals 6.853/49.002 = 0.139 moles.
The rest is water i.e. 107.166 - 6.853 = 100.313g. this corresponds to (100.313/18.015) = 5.568 moles.
Now, total moles = 0.1398 + 5.568 = 5.708moles.
The mole fraction of NaCN = 0.139/5.708 = 0.024.

If we do it like this, how do we calculate composition of solution expressed as molarity / normality ? For example, molarity, 6.853g of NaCN has 0.139 moles. I do not know the volume of NaCN in the solution.

And the same problem for normality. Thanks ...

Use proper significant figures $6.853/49.002 = 0.1399$ moles. (100.313/18.015) = 5.5683 moles

@user69734 - Molarity = moles/liter = (0.1399 moles)/(0.1 liter) = 1.399 molar. // A bit undefined as to how exact the 100 ml volumetric flask is. But I'd expect at least 0.100 and but not quite 0.1000.

@user69734 - "% by weight" has nothing to do with the "volume" of the solute, in this case NaCN. // A "5% w/w" means weight of solute divided by weight of solvent. So 5 grams of solute per 100 grams of solvent. // A "5%" solution would be 5 grams of solute per 100 grams of solution (solvent+solute).


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Electrophilic substitution in meta-nitrophenol
Where will electrophilic substitution take place in meta-nitrophenol (3-nitrophenol) ? The two mesomeric effects seem to compete against each other.

Electrophilic substitution can happen at only that place where electrons are present which can be donated/shared with some other electron-deficient species. In the resonance structures of m-nitrophenol shown above , there is a negative charge on hydrogen or carbon(because charge keeps on shifting between carbon and Hydrogen) and they are not that electronegative to handle the negative charge . So they want to stabilize themselves by giving off their electrons to electron deficient species by letting electrophilic substitution happen. You can see lone pair (negative charge ) of OH group  is not delocalised by NO2 group if it is present in meta position . Although NO2 group is still in Conjugation with benzene ring but for OH group, NO2 can show nothing but only -Inductive Effect.
So , now after making changes in my answer , In my opinion , electrophilic substitution will happen on left side of the molecule at ortho position. On the right side at ortho position , electron density is less as compared to the left side because NO2 is showing -I effect making conditions less suitable for electrophilic substitution to happen.

"+M effect of nitro group"? NO2 shows -M effect

@Harsh Yes. Sorry my bad. I have made changes in the answer . Any suggestions to correct the answer are more than Welcome.

Left side of the molecule at Ortho position means para position for NO2 group, and this will cause a great reduction in electron density. Will electrophilic substitution still happen? Shouldn't there be a list where +M and -M substituents both are included so that we can compare them in case they compete against each other? If there is, someone please post it here.


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Electronic effects of substituents in the benzyne mechanism
On the topic of nucleophilic aromatic substitution (S$_N$Ar) reactions, the benzyne mechanism is discussed. When considering the position of substitution, electronic and steric factors are mentioned. When considering electronic factors, effects of substituents on the ring are viewed rather differently, compared to how they are viewed in the conventional electrophilic aromatic substitution reaction.
Pg. 525 from Organic Chemistry (2nd ed.) shows the following diagram:

In the discussion of the benzyne mechanism, conventionally electron-donating groups are said to not be able to exert their donating effects and are even viewed as electron-withdrawing via the inductive effect. However, I cannot see why they behave in this way. Why is their donating effect no longer significant? Certainly, their p orbitals are still in conjugation with the rest of the $\pi$ system right?
However, if their electron-donating effects were more important, then we would observe ortho and para attack by the nucleophile, so as to generate the negative charge at the meta position, which is apparently not the case as experimentally observed.
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc

the electron pair that is creating negative charge on ring is sp2 hybridized and its orbital direction is away from the pi orbital electronic cloud of benzene ring so not able to resonate along with pi electrons. OMe inductively stabilize this electron pair as it is electron with-drawing inductively. thats why NH2 group attaches at meta position rather than ortho position.

The electron-donating effect only affects orbitals which are parallel to the orbital of the lone pair on O. But the orbital, in which the negative charge on benzyne is located, is perpendicular to the other p orbitals. This means there is no overlap, and therefore no conjugation with the lone pair on O.


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why is delocalization less effective when there is a separation of charge?
I've been reading Peter Syke's book "A guidebook to mechanisms in organic chemistry" wherein he has frequently mentioned that delocatlization is less effective when there is a seperation of charge i.e instead of a double bond formation and a negative charge, there is a positive and a negative charge occurance eg. Formate ion and Formic acid

Due to generation of non-equivalent resonating structures. Symmetry basically lowers the energy resulting in stability of resonating structures.

It takes energy to separate the charges, even when the difference in electronegativity is in the same direction as the proposed charge separation.  The energy that must he put in becomes part of the energy in the charge-separated contributing structure, making that contribution less stable and less likely to contribute to the molecular ground state.


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Expression for the entropy of an ideal gas
I want to obtain the Sackur-Tetrode expression for the entropy of an ideal gas based on:
$$S = K_B ln \Omega$$
We have an isolated system with a fixed volume V and fixed number of particles
N and there is no energy exchange with the external world (what statistical mechanics calls micro-canonical ensemble).
We know that $\Omega$ is:
$$\Omega(N,V,E)=\frac 1 {N! h^{3N}} \int_{\mathcal H<E} d^N\mathbf p \ d^N \mathbf q \tag{1}\label{1} = \frac{V^N}{N!h^{3N}} \frac{(2mE \pi)^{3N/2}}{(3N/2)!}$$
When I plugged equation 1 into  $S = K_B ln \Omega$ I got:
$$S = K_B ln \Omega = Nk_B ln [\frac{V}{N}(\frac{4mE \pi}{3 h^2 N})^{3/2}]$$
Which is incorrect as I should have gotten:
$$S = K_B ln \Omega = Nk_B ln [\frac{V}{N}(\frac{4mE \pi}{3 h^2 N})^{3/2}] + \frac{5}{2}NK_B$$
Why I am not getting the term $\frac{5}{2}NK_B$?

See http://diposit.ub.edu/dspace/bitstream/2445/59903/1/TFG-Pa%C3%B1os-Exp%C3%B3sito-FranciscoJos%C3%A9.pdf

You will get the right answer if you correctly apply Stirling's formula $\ln n! \approx n \ln n - n$.


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What substance consumes oxygen from atmospheric air
I'd like to find an economical substance that will consume oxygen from a container. Steel wool comes to mind but I don't have an intuition how feasible it would be.
The chamber is 5 gallons in size and I'd like the oxygen to be consumed in 48 hours or less.

Some polyphenols absorb O2 from air.

Nitrogen tanks are relatively cheap. N2 is more dense then oxygen, so filling with nitrogen several times should remove most oxygen from your chamber. This in conjunction with a vacuum pump is a standard procedure for removing oxygen from a container in a chemistry setting.

@mcole N2 denser then O2 ?

oxygen to what level? 0.01%, ppm?, ppb?

0.1% range. Ideally we would like reduce the oxygen concentration w/o significantly changing the pressure.

As mcole's comment suggests, it may be easier to do this by physical rather than chemical means.
There are two common approaches:

Purge

In the purge method, the vessel is filled from a high pressure source of another gas. The air needs some way to escape. If the new gas is more dense, all the better, but even repetitive purges of less dense gases will eventually push all the air out.

Pump and backfill

In this method, the vessel is connected to a vacuum pump, and the air inside is pumped out. The vessel is then refilled with an inert gas. Depending on your needs, you can repeat this process several times.


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Why was lime used by the Nazi's?
I am reading a book (Into That Darkness) about the Nazis, where there is the following passage:

causing unimaginable suffering to the deportees who were either left waiting, naked and without food or water, in the open, sometimes for days, or else were crammed into railway cars the floors of which had been covered with lime and were left to suffocate on sidings

What is the meaning of “the floors of which had been covered with lime”?
Is “lime” short for something? Was the floor covered with this intentionally, or did this “lime” accumulate naturally? What would be the consequences of this to the people in the railway cars?

Is “lime” short for something?

In this case it is not a fruit Lime, but is a mixture of calcium hydroxide and calcium oxide produced from calcining limestone.

What is the meaning of “the floors of which had been covered with lime”?...Was the floor covered with this intentionally, or did this “lime” accumulate naturally?

Not really the place to ask, but the lime likely was from previously treating bodies. As people were left to die, lime would be put on their corpse to keep the smell of decaying flesh down. There is also some belief that lime can speed up decay and reduce bacteria growth, but this is disputed. Any way not all of the lime would land on the body or it would fall off when moved. This process had happened so many times prior that the floor was covered in it. lime tends to build into layers as it is chemically similar to cement and will harden over time the same.

What would be the consequences of this to the people in the railway cars?

Lime is very caustic and causes great respiratory distress when inhaled. It is used to capture the putrid scents of decaying flesh.
Today lime is still used at mass grave sites to capture the scent of decay and keep soil pH high. Low pH soil is an indicator for a mass grave as the decomposition products are acidic and lower the soil pH. Adding lime reduces this acidity, masking (trying to mask) the presence of a mass grave.

The purpose of the lime was toilet facilities. the secondary purpose was torture.


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Why does wood burn but not sugar?
Fundamentally, they're both carbohydrates, although the cellulose in wood is essentially polymerized glucose, which combined with its isomer fructose forms sucrose.
So why does wood readily burn while table sugar chars?

Related: Sugar burning with the assistance of ash

Sugar burns very well. It does take more effort to ignite it, I would hazard guessing that is due to higher thermal conductivity and lower specific surface area. If you powder and disperse it, it can even explode.

Sugar burns easily, it flames then forms black lumps.

Ever heard of sugar rockets?

Sugar will burn fine if you give it enough oxygen. Put sugar in a deep saucepan (it will splash a bit), and add a tiny bit of water to make it a slurry. Add in KNO3 in a 13:7 ratio, and start heating. Slowly add in just enough water to completely dissolve both sugar and KNO3, bring to a boil and evaporate the water until you get a smooth creamy color and texture. Quickly pour it in an old can and let it cool. Put the pan back on the fire and add a lot of water and boil to clean it. Ignite the can with a few magnesium sparklers. All of this should be done outside with an extinguisher on hand.

oh, but it does burn

@StianYttervik Sugar burns, but NOT very well. Much worse than wood. That you can make a dust explosion with it proves exactly nothing, same as the fact that it burns well if you mix it with nitrate.

Neither of the claims in the question are strictly true: wood usually doesn't ignite easily and sugar does burn under the right conditions. To avoid stupid answers trying to overgeneralise, this show be made clear. And don't forget a large part of the answer might be not about the fundamental chemistry but about the bulk physical state of the wood or the sugar.

@matt_black Both sugar and wood are solids. What difference in physical state are you thinking of?

@karl sugar consists of relatively large uniform crystals, for example, wood consists of heterogeneous fibres. Large pieces of wood ar hard to set alight; small pieces or thin pieces are not. Powdered sugar will burn or explode on contact with a flame; sugar lumps will not.

@matt_black Please just stop. You have to do funny tricks like powdering it or mixing it with ash or nitrate to make sugar be able to burn by itself, with wood you just ignite a solid piece, and it usually keeps burning. You know that as well as I do, so what the heck are you trying to prove here?

And please note that OP explicitly wrote "table sugar", not "powdered" or "a dust cloud of sugar". You are wilfully misunderstanding the question. That's just unconstructive, you might call it trolling. That five more people here did the same doesn't make it much better. :-/

Combustion is a gas phase reaction. The heat of the flame vapourises the substrate and it's the vapour that reacts with the air. That's why heat is needed to get combustion started.
Anyhow, wood contains lots of relatively volatile compounds so it's not too hard to get combustion started. Once combustion has started the heat of the flame keeps the reaction going.
However sugar dehydrates and emits water when you heat it. Water isn't flammable (obviously) so there's no way to get combustion started. Dehydration leaves behind pure carbon and that is non-volatile so again there's no way to get this to burn. Carbon will burn of course, but you need a high temperature to get it going.

If there is no way to get a combustion reaction started for sugar, then why are the TGA thermograms for wood and sugar similar?

@Agriculturist of course sugar will burn. Some of the ways to make it burn are discussed in the comments to the question (I would be cautious about mixing it with oxidising agents though as it might be a wee bit more exothermic than you expect! :-). The point is that unlike wood it doesn't burn when you put a match to it.

I'm trying to lead you to the correct line of thinking.  If you put a match to an oak log does the oak log burn?  In the 2008 imperial sugar plant explosion, why did the sugar dust explode?

A few things about this answer bothers me.  (1) The answer leaves out the fact that wood  dehydrates and emits water when heated the same as sugar.  This phenomena will certainly cause the flame in sugar to extinguish, but why not the wood?  (2) Dehydration is the removal of water, but it is not necessarily the removal of all water.  It can leave behind dry sugar.

@Agriculturist Read again - it's not that the water extinguishes the flame. It just doesn't combust in air, unlike the volatiles in wood (some of which are very volatile indeed). To get a flame, you first need to get burnable volatiles out of the target material - that's really what a flame is, volatiles violently reacting with ambient air and heating to high temperatures. John didn't talk about water preventing combustion at all - just that it doesn't facilitate the combustion.

It might be useful to point out that water vapor carries away heat.  IIRC, the thermal dissociation of sugar is endothermic, and then you need the high temperature to burn pure carbon.

@Agriculturist dust suspended in air has a very high surface area. iron dust in the air will burn quickly, an iron bar wouldn't easily burn

When you burn wood, most of the heat is produced by the oxidation of solid carbon (charcoal). The carbon is not a gas. So, out of curiosity, if that's not called "combustion," what is it called?

This is wrong. Sugar does burn if the reaction can get started. Catalysts like cigarette ash can be used to ignite sugar cubes with simple flames. And the burning doesn't go via elimination of water. Pyrolysis can cause dehydration of sucrose but that is determined by the conditions not the inevitable reaction that happens on heating.

@matt_black Pyrolysis of sugar definitely and inevitably causes dehydration. Perhaps not quantitatively (which leads to pure charcoal remaining, which is not very volatile), but to a large extent.

With hydrocarbons a certain amount of oxygen (n) and a certain amount of heat energy (Q) are required for complete combustion.  In complete combustion the byproducts are carbon dioxide and water in the form of steam.
$$\ce{C_xH_yO_z + nO_2 + Q -> xCO2 + \frac{y}{2}H2O}$$
Hypothetically, if wood and sugar were assumed to have an identical chemical composition and supplied the exact same amount of heat energy, then any difference in the reaction would need to be derived from the amount of oxygen.
A lack of oxygen is why sugar does not readily burn. This reaction was demonstrated back in the day by Mr. Wizard. He added some oxygen in a chemical form which made table sugar go up in flames quickly.
If oxygen is not readily available then another reaction, called pyrolysis, occurs. This reaction often gives off some combination of methane, hydrogen gas, and carbon monoxide, all of which are flammable, and often leaves behind carbon. This carbon can be observed as char.
$$\ce{C_xH_yO_z + Q -> aH2 + bCO + cCH4 + dCO2 + eH2O + fC }$$
The answer to the question is oxygen cannot readily get to the pile of sugar, which is why it does not burn well and it displays the byproducts of pyrolysis.  In this situation the wood is undergoing pyrolysis as well; however, enough oxygen is available such that all of the pyrolytic byproducts burn leading to complete combustion.

Why does oxygen have better access to wood rather than sugar? Is it due to wood's porosity?

The matchstick vs. lump of sugar scenario is a specific case.  Matchsticks are indeed made from porous woods such as pine and aspen.   When these woods burn the cellulose in the wood burns before the lignin.  Lignin gives wood its structure as it burns because it burns at a much higher temperature than cellulose.  Because of this wood will hold its porous structure as it burns thus giving it access to oxygen.   A lump of sugar has little structure as it burns so it goes out due to lack of oxygen.

@SwagMcMuffins is it written around in answers and comments. It is about how much flammable volatiles are released as compared to non flammable and heat carriers are formed (carbon dioxide, water). Look at the reaction in gas phase,or at the interface, and what is happening there. Not in bulk. If sugar is finely suspended in air it will burn and even explode. Same should do wood, of course.

Looking back at this assumptions can be made (such a match being the ignition source) to generate some very plausible sounding answers through inductive reasoning.  Using deductive reasoning it is known burning things require oxygen availability, charring things requires reduced oxygen availability, therefore something is different about the oxygen availability.  The oxygen availability is affected by some mechanism which cannot be deduced from the question the way it is stated.

Mr. Wizard's point is that with enough oxygen even sugar burns. You are reasoning by making analogies from there, and come to wrong and irrelevant conclusions.

Just so you understand why I downvoted your answer: The primary pyrolysis product of sugar is just water. You have to heat the sugar quite a bit before it has turned into sth flameable. Wood however already gives off copious amounts of flameable organics (from the lignin) at the same temperature.

@Karl.  You have the right to downvote for any reason you want.  I choose not to downvote for mere academic disagreement because such disagreement is fundamental to advancing science and understanding. My view comes from having worked in a research setting pyrolyzing multiple types of woods and sugars. When wood and sugar samples are properly homogenized, reaction differences are not as big as textbooks or as some of these answers would lead you to believe.  This can be seen on a TGA thermogram or by observing the byproducts.

The byproducts of these two pyrolysis reactions are water and water with some gunk in it.  The gunk is written about extensively in the literature because of the usefulness of tar.  It is also fundamental to gasification.  Just because there is much written about these organics does not mean there is much actually there.. This sort of shocked me when I first observed this because I was expecting more.  It is easy to see how people can get confused by the literature.

You are reporting anecdotal evidence and own limited (TGA) observations that contradict the literature and obviously do not fit to the elementar composition (see table in my answer) of the substances in question. Sorry.

@Karl you are entitled to your opinion.      https://www.researchgate.net/publication/268372341_6_Thermal_Properties   https://www.researchgate.net/publication/264972637_Determining_thermo-kinetic_constants_in_order_to_classify_explosivity_of_foodstuffs

What should those papers teach me? I never doubted that powdered wood, sugar or cacao can be the source of dust explosions. And the other one is about wood and fire retardants. You want to point me to a specific page?

@Karl You are not worth my time.  This discussion is over!

It's a matter of physics more than of chemistry. If you apply a lit match to a log, it won't catch fire. Doing the same to a (headless) matchstick will set it alight immediately. This is because the matchstick is surrounded by flame and heats up quickly as it is so thin. If you apply a match similarly to a stick of sugar as thin as a matchstick, then it will burn too. However, the sugar will melt as it heats up, and the burning sugar will tend drip off the remaining stick so that the combustion will not propagate as well.
There is also a chemical reason why cellulose inflames more readily than sucrose (ignoring the lignin content of wood for the present). Because of the additional cross-linkages, cellulose is more stable, and will start to decompose at higher temperatures than does sucrose. At the lower temperature of sucrose decomposition, the main reaction is loss of water vapour, which is incombustible, leaving behind a carbon-rich char which does not ignite so readily at moderate temperatures. While the same process happens as cellulose heats up, at the higher temperature of decomposition, volatile organic compounds, which are combustible, are produced in addition. As these ignite, the flame is enhanced and propagates by positive feedback. The same effects apply a fortiori to lignin in comparison with sucrose.

Thanks for this answer.  I was trying to figure out a unique case of why a cotton ball (which is nearly all cellulose) ignites more readily than cotton candy (which is nearly all sucrose.)    In this particular case, the air supply is nearly identical, but there is a noticeable difference in the way they combust if the flame is applied slowly.

As noted in other answers, commercial table sugar is typically crystalline and the bulk of it must undergo a phase change into a form accessible to oxygen in the air. Melting the sugar consumes energy and may slow down its reaction (the melting point is reported to be ~186 $^o$C). According to more recent studies sugar undergoes thermal decomposition rather than melting. This suggests that the process of combustion of sugar is complex and can involve both thermal decomposition and oxidation by air. Presumably the formation of the molten caramelized mass reduces access to oxygen unless very high temperatures encourage further reaction with air.
Cellulose within dried wood also contains crystalline domains, as well as amorphous ones, but their size is smaller and the energy required to induce phase changes that provide access to air is presumably less. The polymeric structure probably interferes with decomposition and the porous structure of wood aids access to air.
As an interesting corollary to this problem and some of the answers and comments, the generation of coke from raw coal (containing impurities such as sulfur compounds) via heating in the absence of oxygen generates a cleaner burning carbon energy source.
EDIT: added consideration of thermal decomposition of sugar.

The heat of melting is totally insignificant compared to the heat of combustion.

@Karl I think your answer addresses many aspects more convincingly, but I am a little surprised after reading your answer that you bring this point up, since you make a very similar one (that the structure of sugar -molten or not - impedes access to oxygen). Perhaps you just think it's of secondary importance?

Perhaps I was a bit unclear. Molten sugar has a compact, closed surface, very much like a large crystal. The surface of a heap of table sugar is much more open, but it melts long before it can ignite. It seems powdered sugar keeps so much air even when it melts that its heat conductivity stays low, and it can actually burn by itself. (never tried that myself)

This is an experimental answer to a slightly different question.  The way the question is asked, it is confounded by the substances wood and sugar.  If wood and sugar are removed from the question it will read...

Why does substance A burn but not substance B?
Substance A and B are both carbohydrates of similar composition.  So
  why does substance A readily burn but substance B chars?

The fact that A burns and B chars is the key piece of information to discovering the correct answer.  Imagine substance A and B are both granulated cellulose.  (Cellulose happens to be both sugar and the dominant component of wood.)

This is a picture of what is left when substance A is heated to 550°C in the presence of air.

This is a picture of what is left when substance B is heated to 550°C in the absence of air.

Because the question says substance A burned and substance B charred the right answer is most likely substance A had enough oxygen for combustion while substance B did not.
In the original question, too many unknowns have the potential to affect the right answer.  The heat energy of the ignition flame is an unknown.  The type and size of wood are unknown.  The granularity of the sugar is unknown.  The size and shape of the sugar pile are unknown.  The presence or absence of any controlled conditions is an unknown.  The presence or absence of a flame is an unknown.
Because wood burns it is clearly evident that both heat and oxygen are present.  Because the sugar chars it is clearly evident that heat was present but oxygen was deficient.  Any explanation beyond this involves some level of speculation as to the conditions implied by the question.
-- cautionary note about speculation --
Once speculation starts, weird answers can become true.  For example, If someone posted an answer says that the sugar was buried under a campfire while the wood was in a campfire.  The question mentions no campfire nor does it say anything about burying sugar, it cannot be said for certain that this answer is right or wrong.
If the explanation is embellished with interesting facts it could sound awesome, but be completely wrong if one thing is off.  For example, someone could point out sugar makes caramel at 186°C and that wood contains a very flammable group of compounds called terpenes.  The question mentions no formation of caramel so there is no telling if it is present or not.  It is not known the extent to which the wood is dried so terpenes may not be significant.  It is not possible to say for certain if these explanations are right or wrong.

"The sugar/cellulose is identical in both reactions; however, in one situation it burned and the other it did not burn." Just to clarify, you mean the crucibles after experiments with either sugar or cellulose were identical, but in the presence of oxygen both combust to near completion?

The crucibles were identical.  The temperature profile was identical. The initial amount of cellulose was identical.  The only difference was the presence of oxygen.  One burned, one charred.  The cellulose can be switched with table sugar or ground up wood and the same results will occur.  One will burn, the other will char.   This is presented as direct evidence that the correct answer to this question involves the presence of oxygen in the reaction.   The wood had sufficient oxygen to burn, the sugar did not so it charred.

Without oxygen, neither wood nor table salt burn. So, logically, with enough oxygen, table salt will burn.

@karl no that is flawed logic. A duck is not a stone. You are not a stone. Therefore, you are a duck.

@StianYttervik I have just one more thing to say: Quack! ;-)

The oxygen (chemically reduced oxygen; that is, essentially, water) content of sugar is very high. It's already highly oxidised.
$$
\begin{array}{llr}
\hline
\text{Compound} & \text{Formula} & \text{Oxygen}/\text{wt %} \\
\hline
\text{Glucose} & \ce{C6H12O6} & 53 \\
\text{Saccarose} & \ce{C12H22O11} & 51 \\
\text{Cellulose} & \ce{C12H20O10} & 49 \\
\text{Dry wood} & & \sim 35 \\
\hline
\end{array}
$$
The oligo- and polysaccharides have lost water after polymerisation, but not very much. Cotton is not very flammable. That's why labwear is required to be made of cotton (plus it doesn't melt and then stick to your skin).
Wood however is not just a carbohydrate, but has a high content of lignin, which contains much less oxygen and has a sum formula in the range of $\ce{C3H3O}$.
Very simply put, you have to boil a lot of water out of sugar before anything can actually burn. If you look at the sum formulas above, about two thirds of the mass and all the hydrogen are gone at that point. Pyrolysis of sugar gives water + carbon, and very little flameable volatiles. Which are what you see on burning wood. When all the volatiles are pyrolysed out of the wood, the remaining charcoal is hot enough that it keeps burning.
Sugar has the additional problem that it melts, which makes it burn even worse, because the grainy or powdery structure fuses into a closed surface with good internal heat conductivity. There is an old trick, mixing sugar with ash, which makes it burn a lot better. Cellulose doesn't melt, that's why paper burns much better than sugar. (And because it's thin, a complete book is not so easy to burn completely.)
Btw. the autoignition temperature of sugar (probably saccharose) is ~410°C, wood 280-340°C. (from de:wp, without reference unfortunately)

Cotton is flammable!  The reason cotton is a popular choice for lab coats is because cotton does not adhere to the skin when it burns.

@Agriculturist That too, because it doesn't melt, but I said not very flammable. The flame smokes and stinks and often goes out when the ignition source is removed.

The cotton in treated lab coats resist flames.  Traditional cotton lab coats are not recommended for use around flames.

This is completely wrong. Not least because sugar does burn well once the reaction starts (cigarette ash is a good catalyst to kick it off). The key issue is activation energy not whether the overall reaction is favourable. And that barrier is as much about the physical properties as about the inherent chemistry.

@matt_black Sorry, no. Cigarette ash is not a catalyst, it simply keeps the sugar from melting into a viscous block with too little surface. What you mistake for an "activation energy" problem is actually the heat conductivity of said sugar block, cooling its surface. Wood is however also solid, and still burns without getting ground into dust beforehand.

@karl Ash enhances the ability of a block of sugar crystals to burn. We could argue about the precise chemical role Ime using "catalyst" in a casual not a precise sense), but it isn't important. You claim the reason sugar doesn't burn is that it has too much water.  But sugar does burn. So it can't be right.

@matt_black I'm not saying that sugar does not burn (I made that clearer now). It burns much worse than wood, and if you manage to ignite it, it will usually not keep burning by itself, unless you do special tricks like adding ash or even nitrate.

downvote -  because this line of logic could get a lab worker seriously injured.

@Karl This is actually my first downvote on the chemistry exchange.  I see strong evidence 75% of my downvotes from you. If you want to downvote every answer I have ever written I will turn the other cheek and say go ahead! ..  If my SE chemistry reputation is the price to pay to reduce the chance that a high school lab student might seriously injure themselves because some nameless chemist thinks cotton lab coats are not flammable then so be it.   Downvote away!!!

@Agriculturist Ah, that's still your issue. You did notice the four additional letters between "not" and "flameable", right? (And no, lab coats are not treated with flame retardants. You'd have to throw them away after washing them a few times, and I have never, in dozens of mandatory safety lectures, been told to do that.)

Following this trail of thought leads to thinking methanol shouldn't burn since it has to "boil so much water". Which, if believed, will cost you at least your eyebrows and other facial hair...

@StianYttervik Methanol is a flameable volatile (above its flash point at 9°C!). And if you remove one H20 from methanol, you are left with CH2, NOT pure carbon. Your analogy is very weak.

@Karl that is just not how chemistry works. If you claim that the degree that something burns is limited if it's reaction products are water, then I don't know how to help you.

@StianYttervik Look at the oxidation numbers. That's tenth grade school chemistry.


text:
Role of substituents in radical initialized polymerization

Here, the answer given is (iv), and I don't see the reason why methoxyethene is not forming an addition polymer with radical initiators. Clearly, the first option will form polystyrene and the second, polyacrylonitrile. But after that, I feel lost. After looking at the answer key, I felt that there would be some role of the OCH3 here as it may react with the peroxide initiator, but I'm not quite sure what. As far as I know, the radical is generated at the double bond which then keeps polymerizing independent of the substituent. Can somebody tell me what is the mechanism of polymerization in the (iv) part?

The second option would not yield PVC.

@Loong Yep, corrected that. It should be polyvinyl cyanide, shouldn't it?

No,  Polyacrylonitrile.

The compound IV is a great compound for cationic polymerization , the explanation is the oxygen which is giving electron with resonance : The form where the oxygen has a plus charge and the end carbon a minus is the reason why this compound will be able to start a radical polymerization but with a really low constant of polymerization (kinetic constant), and also the reason why the cationic polymerization will have high constant.
Compounds with strong (electron donor and tiny) resonance will not be good for radical mechanism, but if you put a phenyl group behind the ether function it will be better because the resonance effect will be "diluted".


text:
Are 3-center-2 electron bonds and and banana bonds same?
While reading about diborane, I came to know that the two bridging hydrogen atoms in diborane form 3 center 2 electron bonds. On account of repulsion between the two hydrogen nuclei, the delocalised orbitals of bridges are drifted away from each other giving the shape of a banana. Such bonds are called banana bonds. In diborane 3 center 2 electron bonds and banana bonds are same.
What if a compound has only one bridging atom which forms 3 center 2 electron bond. As there is only one bridging atom, it would not face repulsion so it won't form a banana bond.

In that case, it won't dimerise/polymerise to form 3C-2e- bonds(such type of delocalizations of electrons won't be possible).

@thewitness not really, https://en.wikipedia.org/wiki/Trihydrogen_cation

The phrase "banana bond" is usually reserved for situations where steric conditions prevent the orbitals of two atoms from aligning properly so that the bond appears "bent" and is weaker than an optimally aligned bond would be.  Cyclopropane is the poster child for such bonds, diborane is not because this description of banana involves only two atomic centers.
It's a different concept from three-center two-electron bonds, where typically the bonds are well aligned but the three atoms bonded together are not always collinear.  This is what diborane does.  The atoms in the diborane bonds are not collinear because we need two such bonds, not because either bond suffers from imperfect overlap.


text:
What type of interactions occur when CH4 dissolves in CCl4
When ionic compounds dissolve in polar solvents many types of factors come into play determining their solubility. Some of them may be: ion- dipole interactions, lattice energy, high dielectric constant of the solvent etc.
But I was unable to understand what kind of interactions take place between non-polar solvent and non-polar solute. I was reading about the solubility of $\ce{CH4}$ in $\ce{CCl4}$ and since they were both covalently bonded I couldn't explain the interactions. Any references would help too.
So my question is:
What type of interactions occur when $\ce{CH4}$ dissolves in $\ce{CCl4}$

Maybe it's just me, but it seems weird you know about things like lattice energy or ion-dipole int. but not about vdW, new to organic chemistry?

@Mithoron oh yes.Totally. Thanks for the lead.Should I delete this question?

I think it can stay as duplicate.


text:
Angle of Light Source for Absorbance Spectroscopy
I apologize if this question is stupid, but have been puzzled by it for a few days. Say I have a spectrometer measuring in the visible light range and we're looking to measure the absorbance on a solid material. In lab examples I've seen and read, the light source is placed on one end of the sample and the spectrometer on the other end. so the light passing through goes into the spectrometer. But what if you have a sample around 3 feet away from you. Could you place the light source next to the spectrometer and focus it on the sample then use a collimating lens in an attempt to see what was absorbed or not?
Once again, I apologize if this is a stupid question, I feel like I am missing some fundamental points of this process here.

Let me clarify something. Do you just want to make the measurement on a solid, or do you need to make the measurement on a solid surface of some sort. More directly I guess is could the solid be in powder form?

I just want to make the measurement on a solid, it would not be in powdered form.

Ok, that rules out photoacoustic spectroscopy.

Is the sample something "small" that could be put inside an instrument?

Yes you could do what you propose provided that the light is properly collimated so that it all reaches the detector. It is quite common to place mirrors inside the sample housing of a spectrometer to  take the light out into the lab, through a sample (usually a weakly absorbing gas ) and then use a mirrors to get it back into the spectrometer. Multi-pass cells can also be used to do this.
In atmospherical studies very long path lengths (100's metres) are sometimes used to identify trace amounts of various species. Of course if you do something like this you have to make sure extraneous light does not reach your detector.


text:
Do bulky bases participate in substitution reactions?
I see how bulky bases influence the regioselectivity of E2 and E1 reactions. I do not see very much content on the involvement of bulky bases in substitution reactions.
I know that bulky bases like t-BuOK are generally not involved in substitution reactions because the t-BuO− ion would prefer to take a β-hydrogen instead. Are any  substitution reactions possible, and if so, what mechanism would be seen in substitution reactions involving bulky bases?

The clue is in your title - "bulky bases" . Steric hindrance is a very real effect

It is very tough, but cases of t-butoxide acting as a nucleophile are known, e.g. J. Org. Chem. 2011, 76 (24), 10129–10139:

n-Butyl-tert-butylether. To a stirred solution of 10 g of potassium tert-butoxide (89 mmol) in 100 mL of anhydrous dimethylsulfoxide was added 12.1 g of n-butylbromide (89 mmol) dropwise at 10 °C. The reaction mixture was allowed to warm and stand under stirring for 8 h at room temperature. The mixture was distilled under vacuum at −20 °C. The crude was redistilled at normal pressure to collect 6 g (52%) of the fraction distilled at 75−78 °C as a colorless liquid.

Mechanistically, I don't see any other real possibility by which this occurs apart from an SN2 reaction. Of course, there will be significant amounts of competing elimination, and that is probably why the yield is not amazing.


text:
Will UVC light/ozone affects color on fabrics?
I'm planning to put a low power UVC light bulb into my wardrobe to kill bacteria and prevent bad smell on clothes but I'm wondering will it affect color on the clothes? I've done some research but haven't got into a conclusion if UVC light will affect color faster than sunlight or not (because there's no UVC light reaches the earth surface).
I have 2 light bulbs, 1 able to produce ozone and the other doesn't, ozone can help reaching the dead spots which UV rays can't reach but I've read that ozone will cause cracking on some plastic do fabrics also get affected by that?

Ozone will attack any fabric. UV light tends to bleach dyes. All in all this is a very bad idea.

Ozone destroys bugs using exactly sort of chemical reactions that will destroy fabrics and bleach dyes. All UV light will cause some bleaching of dyes and other damage to fabrics. You will destroy your clothes faster than cleaning them.

I'm planning to put a low power UVC light bulb into my wardrobe to kill bacteria and prevent bad smell on clothes but I'm wondering will it affect color on the clothes?

If the UV light is strong enough to break bonds to kill bacteria it is certainly strong enough to bleach the organic dyes and degrade the clothing.

I've done some research but haven't got into a conclusion if UVC light will affect color faster than sunlight or not (because there's no UVC light reaches the earth surface).

This depends on the brightness of the bulb and the sun where you live

...but I've read that ozone will cause cracking on some plastic do fabrics also get affected by that?

Yes fabrics are also degraded by ozone.
Keep in mind that though the UV and ozone may cause degradation of your clothes, they may or may not do so faster than mold or normal use would too many factors to say with certainty which is best.


text:
Why heating up things before H2C2O4 &KMnO4 titration?
In my school, there is a redox titration between H2C2O4 being titrand and KMnO4 being titrant. After adding H2SO4 to the acid to give an acidic medium for redox, the whole mixture inside the conical flask is heated up to 80 degree Celsius before titration starts, what is the need of heating?

Chemicals reactions tend to happen faster in a hotter solution. The rule of thumb, for which there are exceptions, is that for every 10 C increase in temperature that the reaction rate doubles. // Also CO2 is less soluble in hot solution which drives the reaction to completion faster too.

In general, the higher the temperature the faster the reaction. This is because increasing the temperature increases the number of molecules with more kinetic energy, and hence more molecules would have energy greater than the activation energy. If the reaction between oxalic acid and potassium permanganate wasn't heated then it would take extremely long to see the self-indicating titration occurring.


text:
How did Zyklon hold and release HCN?
$\ce{HCN}$ is notoriously unstable: It boils at room temperature, and it tends to polymerize when concentrated in unbuffered liquid form.  Yet $\ce{HCN}$ was for decades widely used as a fumigant pesticide.  At least one commercial formulation called Zyklon became notorious after it was employed for genocide by Nazi Germany.
I was wondering what chemistry allowed large masses of $\ce{HCN}$ to be bound (presumably adsorbed?) to a solid (Zyklon was known to consist of pellets), put in cans, and then released simply by opening the cans?
The best research I have found concludes of the solid substrate:

Zyklon-B is composed of calcium sulfate but not the soluble anhydrite of calcium sulfate.  Due to its microcrystalline structure it is possible to determine that the Zyklon-B sample is either the natural form of anhydrous calcium sulfate, also known as the mineral anhydrite or, equally likely, the insoluble anhydrite resulting from heating gypsum at temperatures above 650 degrees centigrade. Both of these forms preserve the microcrystalline structure.

Is it known, or deducible, that calcium sulfate will adsorb large masses of $\ce{HCN}$, hold it in a "shelf-stable" (nonreactive) manner when enclosed, and then release it completely at STP?
(E.g., I don't know anything about adsorption of organics, but assuming that $\ce{CaSO4}$ has adsorbed $\ce{HCN}$ to saturation then, since calcium sulfate is a powerful desiccant, I can imagine that on exposure to a humid atmosphere it would preferentially absorb water and the resulting $\ce{CaSO4·2H2O}$ must not be capable of adsorbing $\ce{HCN}$?)

Discussing this borders on distasteful, but are you sure Zyklon-B was released simply by opening the cans in the Nazi death camps, or was water added at some point? Once unsealed the pellets could get water from the air, but that seems like a very slow process.

Also the Wikipedia article seems that Zyklon-B didn't have just one formulation.

Poked around some on internet and it seems that the solid pellets were just dumped into the gas chambers. I'd guess that the OP's last paragraph is probably the gist for CaSO4. However it does seem that other adsorbents might have been used also.

@MaxW – Indeed, it is hard to find answers to these simple chemistry questions because there is so much more material (and controversy) online covering the role of Zyklon in the Nazi genocide.  E.g., it does appear that adsorbed HCN will outgas at STP (i.e., "just opening the cans"), and that process could be accelerated by heating.  OTOH, p.234-238 of this reference suggest (unclearly) that humidity actually inhibits/slows the release of HCN as water binds to the substrate.

Not sure if it was accurate, but I remember old crime movies in US showing gas chamber to execute criminals. The notion was that a NaCN pellet was dropped into acid to create HCN. I'd agree That HCN would be somewhat soluble in pure, or unacidified, water. // In the Nazi gas chambers people sweating and breathing would raise humidity in the air in a sealed chamber quickly even in cold weather with low ambient humidity.

I should also emphasize that I'm not interested here in the production of HCN or its use as a toxin.  Rather, I am curious about the detailed mechanisms by which it was able to be stored and released in a stable fashion, and without resorting to a chemical/redox reaction.

I'll add that Cyanosil is the name of a very similar product that has been sold recently. Not sure if it is still being sold.


text:
Are any other orthonormalizations schemes besides Lowdin normally used when solving SCF equations?
When solving equations of the form:
$$F(C)C = SC\epsilon$$
(e.g. Hartree-Fock-Roothan or Kohn-Sham-Roothan) where $F$ is the fock matrix, $C$ is the coefficient matrix and $S$ the overlap matrix, the usual way to proceed is to first perform a change of basis and orthonormalize the AOs (atomic orbitals) by performing a Lowdin (Symmetric) orthogonalization. I know at least one case (The Psi4 program) where a different orthogonalization scheme is used (Canonical) but I haven't been able to find any information of other alternative orthogonalization schemes implemented in any programs. Are any other schemes used? If so, where and why?

The Wikipedia article for Gram-Schmidt orthogonalization is widely used, but the source is behind a paywall. Possibly modified Gram-Schmidt, as it is more numerically stable.

I was under the impression that Gram-Schmidt couldn't be written as a linear transformation and that a linear transformation was needed in order to transform the SCF equations into an eigenvalue problem. Is Gram-Schmidt used in this specific case or some other?

I know that modified Gram-Schmidt is used in the Davidson procedure for CIS for instance, but that is not what I was asking about.

As far as I know, regardless of how you make the orbitals orthogonal, doing so will put the SCF equations into standard eigenvalue form since it will turn the overlap matrix into the unit matrix. I don't have it with me at the moment, but I believe in Szabo and Ostlund they use Gram-Schmidt for one of the examples. @Ignacio

Gram-Schmidt is just a form of QR factorisation (https://en.wikipedia.org/wiki/QR_decomposition) and thus is very much a linear transformation. In fact all the orthogonalisation procedures can be viewed as matrix factorisations, where one of the resulting matrices is unitary.

Molcas developer here, currently implementing symmetric and canonical orthonormalization for the RASSCF module in addition to Gram-Schmidt.
As a rule of thumb you should treat those Orthonormalization schemes (ON-schemes) as different tools for different jobs.
If there are $n$ linear independent atomic orbitals (AOs), of which $n$ molecular orbitals (MOs) are to be constructed, then symmetric ON is probably the best, because the squared deviation of the MOs to the AOs is minimized. This means that your MOs are as close to the AOs as possible, while still being orthonormal.
If there is linear dependency, symmetric orthogonalization can detect, but not cure it.
Canonical and Gram-Schmidt orthonormalization have the advantage, that they can be used to construct only $m$ MOs from $n$ AOs ($n > m$) and to cure linear dependence.
If you compare Gram-Schmidt with Canonical ON, Gram-Schmidt has the disadvantage, that it is not as numerically stable. Besides the AOs that are orthonormalized last, can look completely different after the procedure. If you manually constructed your basis with chemical knowledge, this is not what you want. Modified Gram-Schmidt cures the numerical instability, but still has the disadvantage, that MOs might look completely different to the initial AOs.
PS: Just an addition to the comments: Grahm-Schmidt can be written as linear transformation. Although one would rarely do this.

Interesting! Thank you for the clarification on Gram-Schmidt, (also pointed out by Ian Bush in the comments). Are there any advantages or use cases for modified Gram-Schmidt you can think of? Any reason why it was implemented before symmetric/canonical in molcas?

It is extremely easy to implement. You just need a Fortran compiler. For canonical or symmetric ON you need to diagonalize and hence need to link against BLAS libraries.


text:
Fishes and Heater
I had an aquarium for about an year and this week my mum just installed a heater in it. On asking, she told me that fishes would die in cold water.
But from Henry's Law, isn't it that more oxygen would dissolve in cold water and enabling fish to get more oxygen for Metabolism.
My research on this told that fishes are cold blooded animals, meaning they can change the temperature of their body to match the environment. It says, that warmer water can be deadly for most fishes, the reason being Henry's law.
So what could be the reason, for putting heater in aquarium.

An aquarium should also have a bubbler,  so dissolved oxygen shouldn't be a problem. The temperature of the aquarium should match the preference of the fish species. So tropical fish would  like between 75 and 80 °F year round. In the northern hemisphere we're approaching winter, so indoor temperature may fall below that. Freshwater goldfish don't need a heater.

Ah.. so the heater is to make the metabolism work in stable manner, so if you have multiple fishes, then what??

I'm voting to close this question as off-topic because it's a matter of biology not chemistry.

While this question does have a significant amount of biology content. I think it has a sufficient amount of chemistry question to be on-topic.

While some fish are very tolerant of water conditions, they are not infinitely adaptable, they are typically tuned to a limited window of conditions that includes salt concentrations (ionic strength), specific ion and organic compound concentrations (some heavy metals and ions are very toxic to fish), pH, and oxygen concentration, not to mention temperature and pressure. So make sure to consult with an aquarium specialist (perhaps a book or internet site will do) to find the right conditions for your fish.
Cold blooded animals such as fish cannot regulate their body temperature to any significant extent, and as metabolic processes tend to be tuned to a narrow temperature window, fish do best within that window, and tend to die outside of it.
FYI I killed my first pet fish (a goldfish) within 24h after placing it in untreated tap water. I got very upset and blamed my mom.
Back to topic: gas solubility typically decreases at higher temperature, as you rightly point out, and this can be a problem for instance in hot summers, often also due to a combination with other temperature-related factors. As you have had an aquarium for over a year, and presumably your fish have survived that long, they seem to be reasonably hardy cold water species (depending where you live) and I don't see a good reason to regulate the temperature unless something is expected to change such as a significant drop in the room temperature. However I'd prefer not to get involved in a dispute with your mom!


text:
How to identify the most stable Lewis adduct of boron trichloride?

Which of the following is the most stable Lewis acid–base adduct?
(a) $\ce{H2O\bond{->}BCl3}$
(b) $\ce{H2S\bond{->}BCl3}$
(c) $\ce{H3N\bond{->}BCl3}$
(d) $\ce{H3P\bond{->}BCl3}$

This is a question from KVPY 2016 examination. In my first attempt, I focused on the positive charge that is going to appear on the donating atom. So the less electronegative the donor, the better it is at donating its lone pair to the vacant orbital on boron. However, the answer key states that the answer is option (c).
I am unable to rationalize why (c) is the most stable adduct.

Since this is a Lewis acid-base interaction, the simplest way to go about it would have been to see the strongest Lewis base for this acid. As we move down a group, basicity decreases and as we move across a period,basicity again decreases. So ammonia is the strongest base here, and hence, answer would be (c)

Perhaps, what is expected of the question is for the student to use Pearson's Hard Soft Acid Base (HSAB) theory, which essentially states that hard acids have a preference to bind to hard bases and soft acids have a preference to bind to soft ligands. $\ce {BCl3}$ can be seen as a hard acid and would thus preferably interact with hard bases. Wikipedia classifies boranes as hard acids. Water and ammonia are much harder bases than phosphine and hydrogen sulfide since $\ce {O}$ and $\ce {N}$ are markedly more electronegative and possess higher electron density. Thus, we can likely expect either water or ammonia to be the correct option. Next, we need to compare between the two hard bases. Since they are both relatively hard, we need to compare now a different parameter, that is, their ability to donate the electron pair. In this aspect, nitrogen is clearly superior to oxygen due to the former's lower electronegativity. Thus, $\ce {NH3}$ is the correct option.

You have to consider also the inherent strength of the covalent bond.  A strong covalent bond can entice an electronegative atom to accept some positive charge or give up some negative charge.
Roughly speaking, hydrogen and period 2 nonmetals form much stronger bonds among one another than with other elements, and this tends to override the electronegativity difference.  For instance, fluorine is much more electronegative than chlorine, but fluorine bonds so much more strongly with hydrogen that hydrofluoric acid is only a weak acid in water whereas hydrochloric acid is strong.  Note that the dissociation of hydrogen chloride in water creates hydrogen-oxygen covalent bonds.
Here, boron binds especially strongly with the period 2 nonmetals nitrogen and oxygen.  So (a) and (c) have this advantage, and after that (c) wins between these two by electronegativity difference.

See b and d are out due to weak 3p-2p overlap as compared to 2p-2p...also that as oxygen is more electronegative than nitrogen, so even if it does share lone pair, it will develop a positive charge which won't make it happy and it will pull its lone pair back to a greater extent than nitrogen...hence the deficiency of electrons in BCl3 will be most compensated on adding ammonia, and this goes with the facts too


text:
Hartree product and the Slater determinant
I am very new to theoretical chemistry so my question may sound very basic to most. It is known that the Hartree product does not satisfy the antisymmetry principle for electrons while the Slater determinant is able to do so. Thus, the electronic wavefunction is often written as a Slater determinant, rather than a Hartree product. However, I am just wondering if the Hartree product can still be used to approximate the electronic wavefunction or if it is still being use to do so in some theoretical methods? And if the Hartree product is used, what other measures are implemented to ensure that the antisymmetry principle is not violated? Also, in Density Functional Theory, which is used to approximate the wavefunction, the Hartree product or the Slater determinant?

The Hartree product will always produce a physically unsound wave function guess; it cannot obey the antisymmetry principle. The Slater determinant is the easiest formalism to guarantee it. It is therefore used on most applications.

@Martin-マーチン But are there any cases where it is used instead? Because I was just reading this introductory DFT text, which made no mention of Slater determinants under the DFT section and only talked about Slater determinants under the section on HF... Anyway, is Slater determinant even relevant to DFT since the wavefunction isn' really computed?

@TanYongBoon strictly speaking, you don't need Slater determinants for DFT because it solves for the energy using the density and so you can do orbital free/pure DFT that doesn't use Slater determinants or orbitals at all. In practice, we often formulate DFT using the Kohn-Sham equations, which computes orbitals/Slater determinants for a fictitious 1 particle potential.

@TanYongBoon basically, the Hartree product can't be used unless it is antisymmetrized and the most intuitive way to do that is to form a Slater determinant from the same set of orbitals.

@Tyberius If I may enquire further, how is the Hartree product different from the Slater determinant in terms of approximating the wavefunction? Yes, we know that only the latter satisfies antisymmetry principle and that it is also more complicated to compute for large N. But are there any other advantages of using a Slater determinant over Hartree product as an approximation of the wavefunction?

@TanYongBoon A Slater determinant is essentially a linear combination of a bunch ($N!$) of Hartree products with the electron labels permuted to. By including all of these instead of one individual Hartree product, you account for the fact that exchanging the labels should lead the overall wavefunction to change sign. I would say the functional form of the two is very similar and that the key difference is that Slater dets properly describe the antisymmetry of fermions. This may not seem like much, but this antisymmetry is hugely important to properly describing electron behavior.

@Tyberius I think that answers my question. Maybe u can expand it into an answer.

The Hartree product can definitely be used for approximate calculations; as you might know Hartree developed his method for atoms in 1927 (for reference, the Schrödinger equation was discovered in 1926) and he implicitely used a Hartree product for the wave function, although the connection to the exact solution of the many-body Schrödinger equation via the variational principle wasn't know. It was Slater in 1930 (and, independently, Fock) to introduce the determinant form which bears his name. When using the Hartree product one has to account for the Pauli principle "by hand", by allowing only two electrons to occupy the same spatial orbital.
By starting with a Hartree product, enforcing the Pauli principle and applying the variational principle one obtains the Hartree equations, which are similar to the Hartree-Fock ones but are missing the exchange term, usually indicated by $K$; this term is, however, smaller than the  electrostatic term $J$ so Hartree calculations are meaningful and give useful results, although of course very approximate ones.
Kohn-Sham DFT calculations use in principle the Slater product form, but in practice, depending on the exchange-correlation functional used, they lead to equations which may be more similar to the Hartree ones (i.e., without the exchange term) than to the Hartree-Fock ones. In the original LDA (local density approximation) functional the exchange term is dropped and replaced by a functional of the electron density which is much simpler to compute. As a result, DFT-LDA is similar in complexity to the Hartree method, not to the Hartree-Fock (a rather significant simplification). However, most modern functionals for molecules (as opposed to solid state calculations), e.g. B3LYP, re-introduce the exchange term (with a scaling coefficient).
The Hartree form of the wave function can be also used to compute expectation values (e.g., to compute dipole moments); you can check by looking a the Slater-Condon rules  that the results you obtain with Hartree functions are, for one-body operators, the same you'd get using the proper Slater-determinant form.