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Complete Oxidation of Methanol
I know that complete oxidation of primary alcohols produces acids:
$$\ce{R-CH2(OH) ->[\ce{[O]}] R-CH(OH)(OH) ->[][-\ce{H2O}] R-CHO ->[\ce{[O]}] RCOOH}$$
I wonder if that is the case of methanol. Why doesn't formic acid complete oxidation to $\ce{CO2}$?
$$\ce{HC(O)OH ->[\ce{[O]}] C(O)(OH)(OH) ->[][-\ce{H2O}] CO2}$$
Note: by [O] I mean an oxidizing agent as $\ce{KMnO4}$ or acidified potassium dichromate.

Permanganate and dichromate are not powerful enough under normal conditions to oxidize fully to carbon dioxide.

Although the direct complete oxidation of methanol (or formaldehyde or formic acid, on that matter) to carbon dioxide ($\ce{CO2}$) using  oxidizing reagents such as $\ce{KMnO4}$  or $\ce{K2Cr2O7}$  is hard to find in literature, electrochemical oxidation of methanol to $\ce{CO2}$ through formaldehyde and formic acid is one of the concepts used in fuel cell research (Ref.1). The extensive use of simple organic compounds such as methanol, formaldehyde, and formic acid as fuels is because they have several advantages, likes of being easy to store and handle, and having possess a high energy density of the order of $\pu{kWh/kg}$. Aside from that, one would also expect they should have the simplest and most straightforward reaction mechanisms of all the possible organic fuels because of their simple structure. Yet, in their review article (Ref.1), Parsons and VanderNoot revealed that the oxidation of methanol to $\ce{CO2}$, involves $\ce{6 e-}$ so the oxidation must occur in several steps with several products or intermediates. The product resulting from complete oxidation of methanol (and other individual organic fuels such as formaldehyde and formic acid) is of course $\ce{CO2}$. The reactions of all these three fuels have been studied on a variety of electrodes, and with a wide spectrum of experimental conditions to optimize the outcome (most of those references are sited in Ref.1). Keep in mind that the studies involving methanol are outweighed those of formaldehyde and formic acid, simply due to the fact that its complete oxidation to $\ce{CO2}$, generates 6 electrons compared to 4 and 2 for formaldehyde and formic acid, respectively. Electrode potentials of these reactions are still debatable, and Koppenol and Rush have collected and calculated some reduction potentials for $\ce{CO2/CO2^{.-}}$ couple and plotted them in a Frost diagram (Ref.2).
We noted above that the electrochemical oxidation of methanol to $\ce{CO2}$ involves $\ce{6 e-}$.  As expected, the oxidation occurs in several steps with several products or intermediates, and, for example, mass spectral measurements has showed that both $\ce{H2CO}$ and $\ce{HCO2H}$ were produced (Ref.3).  Their conclution was listed as:

(a) At room temperature, a stationary decomposition of methanol takes place on a platinum catalyst, with formation of formaldehyde and formic acid;
(b) $\ce{HCO2H}$ reacts with $\ce{CH3OH}$ to form $\ce{HCOOCH3}$ (methyl formate);
(c) $\ce{HCO2H}$ is also produced during electrolysis, and is an intermediate in the oxidation to $\ce{CO2}$;
(d) $\ce{H2CO}$ reacts with $\ce{CH3OH}$ to form $\ce{CH2(OCH3)2}$ (methylal), but, contrary to the results of other authors, $\ce{H2CO}$ is not an intermediate of electrolysis.

Accordingly, you’d see even under electrochemical conditions, this conversion is not straightforward. Ref.1 also proposed a step-wise mechanism to explain all those intermediate and product, involving main byproduct, carbon monoxide ($\ce{CO}$), which is believed to be poisoning the $\ce{Pt}$ electrode in most cases.
References:

R. Parsons, T. VanderNoot, “The oxidation of small organic molecules: A survey of recent fuel cell related research,” Journal of Electroanalytical Chemistry and Interfacial Electrochemistry 1988, 257(1–2), 9-45 (https://doi.org/10.1016/0022-0728(88)87028-1).
W. H. Koppenol, J. D. Rush, “Reduction Potential of the $\ce{CO2/CO2^{.-}}$ Couple. A Comparison with Other $\ce{C1}$ Radicals,” J. Phys. Chem. 1987, 91(16), 4429-4430 (DOI: 10.1021/j100300a045).
T. Iwasita, W. Vielstich, “On-line mass spectroscopy of volatile products during methanol oxidation at platinum in acid solutions,” Journal of Electroanalytical Chemistry and Interfacial Electrochemistry 1986, 201(2), 403-408 (https://doi.org/10.1016/0022-0728(86)80064-X).


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How does monoethanolamine (MEA) increase pH?
I know pH of a solution increases with increase concentration of $\ce{OH-}$.
I'm wondering: when adding MEA, is the $\ce{OH-}$ group from the MEA breaking off, or is it the whole MEA molecule taking a hydrogen from $\ce{H2O}$ leaving the $\ce{OH-}$?

Nitrogenous bases work as Brønsted bases because the nitrogen can pull a proton from water to produce hydroxide.

It is the latter case where "the whole MEA molecule is taking a hydrogen from $\ce{H2O}$ leaving the $\ce{OH-}$". Alcohols very seldom acty as arrhenius bases adding $\ce{OH-}$ to solution. One exception is triphenylmethanol but I cannot think of a second that isn't a variation of the given example. Instead the unbonded electron pair of the amine will abstract a hydrogen from water to form an ammonium ion leaving a hydroxide counter ion and thus raising the pH.

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Can an exothermic (endothermic) process increase (decrease) internal energy?
We classify exothermic and endothermic processes based on enthalpy change, $\Delta H < 0$ and $\Delta H > 0$ respectively.
I assume this translates to $\Delta U < 0$ and $\Delta U > 0$, respectively, for most processes. (where $\Delta H = \Delta U + \Delta (PV)$, in general).
But not all? In which case we might have processes where $\Delta H < 0$ (exothermic) but $\Delta U > 0$? Could you give an example?
Otherwise exo and endo could just be classified based on $\Delta U$ alone instead of $\Delta U + \Delta (PV)$.
Background: I'm just trying to find a motivation for using the concept of enthalpy. I know about the motivation due to additional $PV$ work needed for  a given process (hence $U$ not enough to account for all the energy involved). Just wondering if the exo/endo classification is another motivation for using enthalpy?
Side note: We know something like this for $\Delta H$ v $\Delta G$, for classifying spontaneous v non-spontaneous processes. In that case $\Delta H$ is not enough and we need $\Delta G$. Just wondering if a similar thing happens in case of $\Delta U$ v $\Delta H$ for exo/endothermic classification.

There are certainly examples (e.g. chemical lasers) where the output of the chemical reaction includes a large fraction of excited state molecules (which then lase). Is that kind of what you mean?

I have heard of exothermic and endothermic reactions, but never about exothermic and endothermic processes.  Are you using "processes" and "reactions" interchangeably?

this is the assumption I made Chester... it is a good question

A primary reason for using enthalpy is the fact that heat transferred ($q$) is a fairly straightforward quantity to measure, whereas change in internal energy ($\Delta U$) is often not.
Under conditions where volume is held constant and no form of work other than pressure-volume work is possible, $\Delta U = q$, so we can get direct information about the change in energy from the amount of heat transferred.
But under the much more common laboratory condition of constant pressure, $q \neq \Delta U$. For this case, it is useful to equate $q$ with the change in a state function, so we have defined enthalpy such that $\Delta H = q$ when pressure is constant. Because enthalpy is a state function, we can make general conclusions about other paths between states based on our measurements of $q$, even though $q$ is path-dependent.

From my perspective, this is easiest to tackle from a straight forward energy balance.
The energy balance determines how energy, heat and work are distributed when you go from State 1 to State 2.
Thermodynamic States
Just to emphasize this, a state is a fixed point where all properties are set in stone. So state 1 can be an unreacted system, state 2 a reacted system... State 1 can be the inlet of a compressor at a certain T and P and state 2 can be the outlet at a different T and P. State 1 can be a bomb calorimeter before combustion (at a T and V) and state 2 can be the calorimeter after combustion (same V, different T and consequently, different P).
You can do an energy balance between any two states, and it will tell you directly what is happening to the energy, heat and work of your system. It is a beautiful thing actually.
Energy Balance
For an unreacting and open system at steady state
$$ \left[ U + \frac{P}{\rho} + KE + PE + W + Q \right]^{\rm State_1} = \left[ U + \frac{P}{\rho} + KE + PE + W + Q \right]^{\rm State_2} $$
Where $U$ is internal energy, $\frac{P}{\rho}$ is flow energy, $KE$ is kinetic energy, $PE$ is potential energy, $Q$ is heat and $W$ is work. I should say each variable is really the summation of all sources contributing. There could be several sources producing heat, same with work. There could be several streams entering and several exiting. Traditionally $State_1$ and $State_2$ represent IN and OUT respectively.
If there is flow, then $\frac{P}{\rho}$ is present, and we typically denote it as $Pv$. Thus on both sides we have $U + Pv$ which we write as $H$. It is typical to say changes in $KE$ and $PE$ are negligible. I will also put IN and OUT in place of States Our energy balance is now
$$ \left[ H + W + Q \right]^{\rm IN} = \left[ H + W + Q \right]^{\rm OUT} $$
At this point the actual problem naturally falls out. If you want to know what the change in enthalpy is rearrange the sides
$$ \Delta H  =  W_{IN} - W_{OUT} + Q_{IN} - Q_{OUT} $$
It is entirely possible for $\Delta H$ to be positive or negative for either positive $\Delta Q$ or $-\Delta Q$ so I don't see how you can draw any conclusions about $\Delta H$ with respect to what heat is doing. If there are no work interactions, this simplifies the equation
$$ \Delta H  =   Q_{IN} - Q_{OUT} $$
With no work interactions the system would need to be exothermic i.e., lose heat for $\Delta H$ to be negative.
Let me know if I made a mistake in my derivations, It would not be the first time. But remember all of the assumptions that led us to this point!

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Metal catalysts used in C-H bond functionalisation
Why are the late transition metals used for C-H bond activation/functionalisation? What atomic properties of the late TM make them attractive for coordinating C-H sigma bonds?
Thanks.


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Oxirane and ring opening
I have a problem given as :-  I thought of a sn1 type mechanism in acidic medium and a sn2 pathway in basic medium .  Am I correct or not . Please also check the mechanism according to which I formed the answer for acidic conditions...... help would be appreciated and thanks in advance

Looks about right

Any improvements that could be made ?

Note that the selectivity in the SN2 addition is driven by steric factors


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Oxidation of diols using periodic acid

What are the orders of rates of oxidation with $\ce{HIO4}$ of the following diols? Explain with reasons.
(a) $\ce{Me2C(OH)C(OH)Me2}$
(b) $\ce{Me2C(OH)CH(OH)Me}$
(c) $\ce{CH2(OH)CH2(OH)}$
(d) $\ce{MeCH(OH)CH(OH)Me}$

I know that the reaction proceeds by forming a cyclic diester of periodic acid followed by cleavage of $\ce{C-C}$ bond and formation of $\ce{C=O}$ bond.  But how do we compare the rates of the given compounds?
My first thought was to think $\ce{C-C}$ cleavage as formation of a pseudo carbocation which would be stabilised by +I effect of the alkyl groups and also the hyperconjugation, but I don't think it is right.

Could it possibly be related to stearic hinderance for the ring to form?

Could be but I am not sure

@SidharthGiri and Aditya One thing which you can be assured is that,whatever the differentiating factor is,it will certainly be before the formation of thr cyclic intermediate as after that,it is a pericyclic intramolecular rearrangement which is pretty easy to be done in all the cases with minimum requirement of any activation energy..

Also.. Is the answer a>b>d>c?

I don't have the answers and that is also a big problem


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Kinetic Shift of Simple Cleavage
What is the kinetic shift for the simple cleavage fragmentation if the ion source residence time is 100 microseconds, the time to reach the detector after leaving the ion source is 10 microseconds? (given that the simple cleavage is the green line below)

My work: Because kinetic shift is Ef-E0, E0 would be the energy at which the ion is in the ion source, or where the green line and 4 (from 100 microseconds, or 10^-4) intersect. Then Ef would be where the green line and 5 (from 10 microseconds, or 10^-5) intersect. Does this seem correct?


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Finding Pressure of M in a Charge Exchange
Ar (RE = 15.5 eV) is the reagent gas in a charge exchange experiment. If Ar+ reacts with anisole with a rate constant of 2.26x10-10cm3 molecules-1/s-1 and 65% of the Ar ions react in 25 milliseconds, what is the PRESSURE of anisole if the temperature of the ion source is 25 C? The IE of anisole is 7.78eV.
Not exactly what to do here. I know that -Hrxn=RE-IE, meaning that -Hrxn=7.72eV. Converting this would be -Hrxn=749.21kJ/mol. How would you go about using the rest of the information given along with this -Hrxn to find the pressure?

65% is about the amount consumed in one turnover, i.e. $1 - e^{-1} \approx 0.65$.  Therefore, maybe you could just multiply the rate constant times the 25 milliseconds to get a number for mL per molecule.  Multiplying by the gas constant $R$ converts that to a molar volume  $V$ with units of moles / volume.  Then the ideal gas law is $P = R T / V$.  The only assumption I think is that anisole is in vast excess relative to $\ce{Ar+}$.


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What is the largest coupled cluster calculation that has ever been done (as of March 2019)?
I am doing some research and trying to make sure the numbers I am reporting are accurate.
I found that the paper by Sylvetsky et. al. says:[1]

We  were,  however,  able  to  complete  a CCSDT(Q)/cc-pVTZ  calculation  on  benzene,  which entailed  2.2 trillion (Q)  contributions  and  3.1  billion  iterative  CCSDT  amplitudes

But there is no indication that this was the largest number of amplitudes (3.1 billion), or largest number of perturbative contributions (2.2 trillion). Furthermore, the paper was done in 2016 so surely larger might have been possible since then.
Does anyone know what the largest couples cluster calculation done as of March 2019 was? By "largest" I mean containing the largest number of cluster amplitudes (of any excitation level).

Nitai Sylvetsky; Kirk A. Peterson; Amir Karton; Jan M.L. Martin; Toward a W4-F12 approach: Can explicitly correlated and orbital-based ab initio CCSD(T) limits be reconciled? J. Chem. Phys. 2016, 144, 214101. DOI: 10.1063/1.4952410 arXiv:1605.03398 [physics.chem-ph]

@CARNEGIE I've voted to reopen and I do think it is an interesting question (+1). My point was just to clarify what you wanted out of an answer. For example, I wanted to make sure it wouldn't make a difference if the largest was (nonreal examples to follow) ccsdtqp of water or ccsd of a protein.

@Tyberius, the CCSD for a protein (done by Frank Neese) was with DLPNO, which vastly reduces the size of the virtual space. In the end, does it still have more amplitudes than the benzene calculation which involved 3.1 billion amplitudes? If not, then the coupled cluster calculation was not bigger (although the size of the system they worked on, is certainly bigger). As for CCSDTQP on H2O, it seems only possible in a double-zeta basis set, which means it probably has fewer than a billion amplitudes involved!

You might get more answers over at Materials Modeling, it’s new and active!


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How to determine the greatest d orbital splitting?
This question comes specifically from an IB Chemistry HL Paper 1 in May 2018 TZ1, namely question 8.
Which complex has the greatest d orbital splitting?
It gives 4 Complexes $\ce{[Fe(H_2O)_6]^{2+}}$, $\ce{[Fe(H_2O)_6]^{3+}}$, $\ce{[Co(H_2O)_6]^{3+}}$, $\ce{[Cr(NH_3)_6]^{3+}}$ and it says that they give the colours green, orange, blue and violet respectively.
Initially I thought that the answer would be $\ce{[Cr(NH_3)_6]^{3+}}$ because it gives the highest energy light, being violet. However, the answer is given as $\ce{[Fe(H_2O)_6]^{3+}}$, why is this?

The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.

Please explain how is it 'complementary'.

From what I understand, $\ce{[Cr(NH_3)_6]^{3+}}$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $\ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights?

@AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series.


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Important Stoichiometry question, just need formula or help and answer checking
How many grams of Cobalt(II) chloride  $\ce{CoCl2}$ is needed to make $\pu{0.645 L}$ with a concentration of $\pu{1.25  mol\:L^{-1}}$? (the molar mass of anhydrous $\ce{CoCl2}$ is $\pu{129.839 g\: mol^{-1}}$, I believe)

This is definitely a homework question. You must need to show your effort to solve it.

well, could you show me the formula possibly? I don't know how to get help with this question and I don't know how to do it...

First, you need to find the amount of $\ce{CoCl2}$ in $\pu{mols}$ to make $\pu{0.645 L}$ of $\pu{1.25 M}$ solution.

I believe it is .807, what do I do after this?

Correct answer is $\pu{0.806 mol}$. So, do you remember, $g= \pu{mol} \times \text{molar mass}$?

yes, what do i do with the .806

Remember the formula $c = \frac{n}{v}$ where $c$ is the molarity, which is moles of solute per liter of solution. You can rearrange it to get $c \times v = n$. Thus in your case $\pu{1.25M} \times \pu{0.645L} = \pu{0.806 mol}$.
With this information, you can convert moles to grams by multiplying the moles by the molar mass. Thus, $\pu{0.806mol} \times \pu{129.839 \frac{g}{mol}}$ which equals $\pu{104.68 g}$. This is your answer.


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The pH scale/calculating pH level
How much water do you need to add to 10 mL of a solution of HCl with a pH of 4 to change the pH to 5?

Since the pH is 4 the concentration of H+ in the solution is 10^-4. Therefore no.of moles of H+ from hcl ( neglecting autoprotolysis of h2o) is 10^-3 millimoles which remains constant as hcl dissociates completely even when we add more h2o. To get pH=5, we must make conc. to 10^-5 thus we need this solution to have a volume of 100ml( final volume =number of moles/concentration needed), thus we add 90ml h2o.

Thank you, I just needed a little help with that question.


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Effect of solid in equilibrium
CaCO3 (s) = CaO (s) + CO2 (g)
We know that pure solids quantity doesn't affect the position of equilibrium. But the application of pressure in gas does.
If we add CaCO3 in the closed system, the position of equilibrium should change, shouldn't it?
If we add CaCO3 the concentration shouldn't increase but the pressure in closed system should increase. Then it should affect in CO2 because the pressure has been increased.
Then why this is told that that pure solids quantity doesn't affect the position of equilibrium?

You are saying that CaCO3 is added to a system of unchanged volume or mass/matter (except the given addition)?

Because we live in a world that is open, and most systems are under constant pressure, not constant volume.

I am saying about a system of unchanged volume.

I'm sorry I meant to say unchanged volume and unchanged mass/matter (except the given addition).

I am asking simply that what should happen when i add CaCO3?

The reaction will go forward or backward?

@Zhe how can one observe equilibrium in a system that is under constant pressure not constant volume?

Yes! I was also thinking about that.

I don't understand the question. You see it all the time. You live in a world that is constant pressure and not constant volume.

@Zhe can you occur a reversible reaction without holding the volume constant? According to my knowledge reversible reaction can only be occured in a closed system where the volume is constant. If my knowledge is wrong then you inform me the right one. I will be thankful to you.

Just ask yourself if reversible reactions happen in the real world because the real world is basically constant pressure.

I knew that reversible reaction can only be occured in a closed container, where the volume is constant. Now you are telling the other thing. If you give me some link about this or explain I will be pleased


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How to determine the number of water ligands attached to Pd in palladium chloride aquacomplex PdClₓ · yH₂O?
Normally this can be achieved using UV-vis: there is a strong absorbance around wavelength 220 nm which varies based on the number of ligands. If the palladium concentration is too high, the solution can be diluted.
However, in my case I do not know the concentration of solution as the solution is reactive. I need to measure the palladium ligands after these reactions have occurred. Thus dilution is not an option. Does anyone know if UV-vis can be tweaked or another technique for testing palladium-water speciation in solution?
FYI I get an absorbance value of 4 (the peak is still quite broad at this value) when testing the palladium chloride under the desired conditions.

I hope the following two references would help you find the way. I honestly think Ref.1 shows much more promise, abstract of which states that:

Monomeric palladium(II) chloro aqua complexes of the form $\ce{PdCl_r(H2O)_{4-r}^{2-r}}$ ($r = [0,4]$) were studied both experimentally and theoretically to gain insight on both the stabilities and the nature of palladium-chloride interactions.
The thermodynamic stabilities of these complexes were studied in aqueous solutions from $5$ to $\pu{125 ^{\circ}C}$ with UV-vis spectrophotometry using a quartz flow-through cell. Tentative measurements up to $\pu{200 ^{\circ}C}$  were also carried out in pressurized titanium and gold-lined optical cells but revealed important losses in soluble palladium. The strong ligand-to-metal charge transfer bands of the palladium complexes below $\pu{350 nm}$ were used to constrain the step-wise thermodynamic formation constants at each temperature, using results of singular value decomposition of the spectra over a broad range of palladium:chloride ratios and wavelengths. The temperature-dependent constants were used to obtain changes in enthalpy and in entropy for each reaction. The thermodynamic stabilities of $\ce{PdCl(H2O)3^+}$, $\ce{PdCl2(H2O)2^0}$, and $\ce{PdCl3(H2O)-}$ are larger at higher temperatures, whilst the one of $\ce{PdCl4^2-}$ is smaller. All changes in entropies are positive for the former three species, but negative for the latter, presumably due to a larger solvent reorganization around the doubly charged $\ce{PdCl4^2-}$ species. The room temperature thermodynamic values derived from this study are also in agreement with previously published calorimetric data.
Theoretical calculations on the intramolecular distributions of electrons in the different palladium(II) chloro aqua complexes, using the methods of atoms in molecules and of the electron localisation function, showed $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ interactions to be of largely closed-shell/ionic nature. These interactions induce an important distortion of the outer core shell electrons of $\ce{Pd}$, as well as stable accumulations of electrons between adjacent $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ bonds known as ligand opposed core charge concentrations.

References:

J.-F. Boily, T. M. Seward, “Palladium(II) chloride complexation: Spectrophotometric investigation in aqueous solutions from $5$ to $\pu{125 ^{\circ}C}$ and theoretical insight into $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ interactions,” Geochimica et Cosmochimica Acta 2005, 69(15), 3773-3789 (https://doi.org/10.1016/j.gca.2005.03.015).
L. Espinosa-Alonso, K. P. de Jong, B. M. Weckhuysen, “A UV-Vis micro-spectroscopic study to rationalize the influence of $\ce{Cl-(aq)}$ on the formation of different $\ce{Pd}$ macro-distributions on $\gamma$-$\ce{Al2O3}$ catalyst bodies,” Phys. Chem. Chem. Phys. 2010, 12(15), 97-107 (DOI: 10.1039/b915753k).

Thank you Mathew! The studies themselves do not help with the problem, but they do provide good references for the regions that I should look. I might be able to avoid the dilution and just look at the secondary peaks that show up in the 400 - 500nm region.

@Nilay Ghosh: I included the abstract. My intention was to give some insight to OP and I think OP found some lights.:-)


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How do amphoteric compounds act as acids?
According to my chemistry textbook, an acid is a compound that gives hydronium ion on dissolving in water but amphoteric compounds like ZnO do not yeild hydronium ion on dissolving in water . I understand that they give salt and water on reacting with a base but still do not satisfy the definition then how can they said to be to also act like an acid?

This confusion better should have been resolved before turning to amphoteric compounds. Look at H2SiO3: it's an acid, and so should give H+ on dissolving in water, but wait, it doesn't dissolve in water at all! Ditto for bases like Mg(OH)2 giving OH-.

Technically, magnesium hydroxide does dissolve measurably in water.  Milk of magnesia will turn red litmus indicator blue.  The solubility of the hydroxide is sufficient to displace heavy metals in water and soil treatment.

The trouble is that "amphoteric", when applied to oxides and sulfides, usually means: "what does the substance dissolve in?". (In the case of $\ce{ZnO}$, it would form either some $\ce{Zn^{2+}}$ complex when dissolved in an acid or $[\ce{Zn(OH)4]^{2-}}$ in a strongly basic solution.) This use must mentally be distinguished from "amphoteric" as applied to e.g. $\ce{NaH2PO4}$, which can accept and donate $\ce{H}^+$ ions (and their compounds with one, two or more water molecules).


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What is the reason for more stable keto form to tautomerise to a less stable enol form?
In the mechanism of SeO2 I have seen that a more stable keto form first converts to a less stable enol form and then yields a diketone. I have seen many other reaction where the reactant converts to an unstable form first and then reacts. My doubt is why does the reactant converts itself rather than just simply not react ?

Well... If you take a look at the mechanism: https://en.wikipedia.org/wiki/Riley_oxidation. The reaction cannot occur by starting with a ketone.  It can only begin with the enol form. Also, it is not the case where the enol form doesn't already exists in the first place. Although the carbonyls exist predominantly in the keto form, some of them are in the enol form. These enols can take part in the reaction. Even if these enols were consumed, the equilibrium would shift towards forming more enols, gradually the keto form would also deplete.

We can think of it an an equilibrium that is disturbed by another, irreversible reaction.
Think of a more familiar reaction.  When you try to dissolve magnesium hydroxide in water you get only a little solubility; the dissolved magnesium and hydroxide ions are only a minor species at equilibrium.  But when hydrochloric acid is added the acid reacts selectively with the hydroxide ions, neutralizing them; in response more magnesium hydroxide keeps dissolving into the water (to form magnesium and hydroxide ions, the latter then reacting with more acid) until you have a salt solution containing the accumulated magnesium ions and the chloride from the acid.  Reaction continues until you've dissolved all of the magnesium hydroxide or neutralized all of the acid.
Similarly, the enol form is only a minor species in the equilibrium, but when you add the selenium dioxide it reacts selectively with the enol form.  This draws more of the carbonyl compound into the enol form in the same way that the acid in the example above draws more magnesium hydroxide into its dissolved form.  Thereby the reaction keeps going until you run out of one reactant or another.


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How to take a sample of a hot salt solution without it precipitating in the cooler glassware due to the difference in solubilities at different temps?
I'm doing an investigation to see the effect of temperature on the solubility product of strontium hydroxide.
I made a supersaturated solution left in the oven overnight at 85°C for equilibrium to be reached. Now, I want to titrate it against $\ce{HCl}$, but when I try to pipette a 25 mL sample of the (very hot) solution, it quickly crystallises in the pipette due to the large difference in solubility at 85°C compared to room temperature.
(NOTE: doing at lower temperature is not an option.... )
One idea I can think of is to keep the glassware in the oven overnight with the solution so that it's more or less at the same temperature. (Then, I'd have to handle with a cloth, of course). However, I do not have access to an oven that is big enough to fit a pipette.
Hence, the only option I have is to use a cylinder rather than a pipette, but that would increase my uncertainties by A LOT.
Can anyone think of anything else that I could do?
Would really appreciate any help.

You can get glass measuring cylinders for that volume range (~25 mL), which should be smaller than your pipette in length.

Yes but I'm worried about the uncertainties. Using a glass cylinder, whether its 25 mL or more, would make my random error a lot higher. So I was just wondering if there's another way around.

I expect the least counts to be the same for a 25 mL cylinder and a pipette, though.

No, the smallest value you can measure with a 25 mL cylinder is around 0.5 mL (basically there's one line per 0.5 mL). It's an analog instrument, so the uncertainty will be half of that, which is 0.25 mL. For a 25 mL pipette, the uncertainty is 0.05. There's a big difference.

@Mohamad cost of the line does not equal uncertainty. The uncertainty here is inversly proportional to the square of the diameter of the tube whether it is a measuring cilinder or a pippete.

@Mohamad Anyway, if the problem is in the measurement, you can move measurement to happen after titration, if you can tolerate/eliminate uncertainties due to evaporation. Yes, you won't be able to titrate exactly 25 ml of your solution, but that's acceptable for analytical titration as long as you can deduce the value from other means. Another option is to prepare solution in 25 ml batches to begin with.

Well I can't really prepare the solution in 25 mL batches from the start since I'm gonna need to filter it, so the volume will decrease because of the filter paper.. What exactly do you mean by measuring after the titration. Do you mean taking a batch of random volume then titrating it, then measuring the volume of the new solution using a burette and subtracting the volume of acid? Hmm that could actually work. Thanks permeakra!

You are concerned about about this one uncertainty (volume error when removing 25 mL of supernatant). You should also worry about volume changes with temperature, and systematic errors of using volumetrics at temperatures they are not designed for. For these reasons, it is common to switch from concentration to molality for studies where temperature is varied by a lot. The density of water changes by about 3% from your equilibration temperature (d = 0.96859 g/cm3) to room temperature.

@KarstenTheis I am planning to find the solubility product at the different temperatures. Aren't you supposed to use concentrations in order to calculate Ksp?

Oh but the volume is technically changing because the water is expanding... ok got it.

Yes, I am saying there will be some systematic errors. It would be interesting to hear how you plan to use your data (is this one of many temperatures, for example).

For what is necessary to handle a supersaturated solution first of all?

There is another aspect of this. The Ksp is used for "dilute" solutions. The solubility of $\ce{Sr(OH)2}$ is large enough that activities should be used not concentrations. So you could determine the solubility of $\ce{Sr(OH)2}$, but not the Ksp.

Just use your room temperature volumetric pipette, and wash out the precipitate with water. The exact density at 85°C is going to be tricky anyway. The expansion coefficient of borosilicate you can find, and you need to measure or calculate (heat capacity of borosilicate, and your solution?!? IR thermometer?) the temperature of the solution.
You might think about using a second volumitric pipette with dest water, and wash everything into a beaker with known empty weight.
Have fun with calculating the uncertainties! ;-)
Afterthought:
Fill the RT volumetric pipette from the hot solution. Wait until cooling (the filling should get lower, because water has a larger expansion coeff. than glass) and precipitation is complete. Now completely fill the pipette, the additional amount of solution will be rather small, and you don't need to wait again for quillibration.
Now weigh and titrate the solution&precipitate (after washing out with a known volume of water). You know the density and molarity of cold strontium hydroxide solution, you know the density of solid hydroxide. I think that is sufficient information to calculate the concentration at 85°C.

I think you misunderstood something here. The problem is not with having to wash the precipitate away.. It's that just when the solution gets into the bottom of the pipette, it crystallises clogging it up so no more can even go in. Your advice would work if I could get the solution up into the pipette in the first place...

OK, that's a problem. Didn't think it was that bad. ;-) Heat the lower end of the pipette in a hot water bath?


text:
Why aren't all bond lengths equal in cyclobutadiene?

even when there is a possibility of conjugation why aren't bond lengths equal?

Could you cite where you found that the bond lengths are not equal?

Heard of anti aromaticity  it destabilizes a cyclic compound so to prevent it no conjugation

Your title does not match your question. All double bonds are equal all right.

Possible duplicate of Resonance structure of cyclobutadiene?

Anti-aromatic compounds have no delocalization of electrons hence no resonance. To check aromatic character we use huckel's rule of (4n+2)π e- present in a conjugated closed loop. An anti aromatic compound would have only 4nπ electrons which is the case with above mentioned compound . Hence the bond lengths are different.


text:
How does chelation control result in rate acceleration?
In the chapter on diastereoselectivity in Clayden, Warren & Greeves (2012), the Cram-Chelate model is discussed. Most significantly, under chelation control, the diastereoselectivity of the addition to the carbonyl group is reversed. Additionally, chelation control achieves high levels of diastereoslectivity, often greater than $\ce {95:5}$. Lastly, it also mentions that chelation control leads to rate acceleration. The above are discussed on p. 864.
It is not surprising that Felkin-Ahn control leads to rate acceleration, especially when both the steric and orbital factors are both satisfied. More details of the model are explained in this post. However, chelation control does not seem to provide any cause for rate acceleration. In fact, as chelation control goes against the steric and orbital considerations, we may even expect it to slow down the addition reaction, rather than speeding it up.
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

I could go into my usual refrain of how organometallic reactions are complicated, but more helpfully, when you add a chelating metal ion which binds to a carbonyl, is that not effectively the same as a Lewis acid?

@orthocresol I understand that organometallic reactions are often not straightforward... However, in this case, the results of chelation control are very obvious and has been shown to give consistent rate accelerations with high diastereoselectivity, as mentioned in Clayden's. However, I understand that the effect at play for the rate acceleration may not be straightforward. If anyone has any idea, please feel free to make a suggestion.

I thought I made a suggestion.

@orthocresol Oh yes... I didn't read your comment carefully. Yes, it does make sense.


text:
What is the platinum doing in a Zippo hand warmer?
My Zippo hand warmer requires lighter fluid, platinum catalytic burner, and a heat source to start the flameless catalytic process. What is the platinum doing in this reaction?

Well, it is catalyzing the reaction. The surface chemistry of platinum allows the oxidation to occur there, not requiring a ‘stand alone’ flame. Quite common for propane heaters.

It's actually amazing how this works. Must be finely tuned equillibrium between conductive heat loss through the housing, cooling by inflowing air of variable oxygen content, and the actual reaction.

Platinum is a catalyst and decreases the activation energy, so the process can start at lower temperatures.

I think you can add a little more to this answer to explain why the catalyst is necessary in this particular case, and roughly what reaction it is catalyzing. Right now your answer was probably down voted because it's just too short and vague to be a good Stack Exchange question.


text:
Where can I look for a short chain photopolymer?
I am by no means an expert in this field, so please be patient with my explanation. I am looking for a photopolymer with very specific properties. I am not sure how realistic it is to find such a material, but I will list the properties in the hope that someone can point me in the right direction, to help me narrow my search.
What I am looking for is a short chain monomer (hopefully no longer than $\pu{5 nm}$) that polymerizes in response to light (I understand a photoinitiator will also probably be needed), and has a hydrophilic head, to allow it to bond to my aluminum (hydrophilic) substrate. In the polymerized form, it has elastic properties, as well as a hopefully low mass to elasticity ratio.
A few things that I have thought of that could work include negative photoresists, such as those used for mask lithography, as they would demonstrate very small features (however, they may be too brittle), or 3D printing photopolymer (I am cautious to use the latter, as it is not made for such small dimensions, though it would probably work for the elasticity).
What I am trying to do is make a very thin membrane to rest on top of a nanoporous surface (where the pores are $\pu{20\pm5nm}$, that, upon being exposed to light, hardens to an elastic thin film.
I have more of a biochemistry background, but I would very much appreciate a nudge in the right direction, either via a repository where I could possibly find such a material, or through learning resources specific to this type of material, or, even better, chemicals that have similar properties to the one I am looking for.

Welcome to Chem SE! If you haven't yet, take the [tour] and visit the [help]. I'm have very little experience with polymers, but I guess an okay place to start would be the wikipedia page for elastomers if elasticity is important. Best of luck in finding an answer! @Jmeeks29ig

So you want to graft this polymer from/onto your surface? And actually you want no polymer, but a crosslinked polymeric surface? Perhaps a photocrosslinkable surfactant?

@Tyberius - Thanks I appreciate it, I will look into that!

@Karl - interesting! Is the reason you mention a surfactant because that is what would allow the material to graft to the aluminum, with its polar end, and then cross link on the non-polar end?

That's what I was thinking.


text:
Why does monoammonium phosphate (MAP) have 52% of phosphorus (P₂O₅)?
The formula of monoammonium phosphate (MAP) is $\ce{NH4H2PO4}$. Manufacturers say that MAP has 10% of nitrogen (N) and 52% of phosphorus (P). However, its molecular mass is 115 u and the atomic mass of P is 31 u $(31/115 × 100\% = 27\%)$. How is there 52% of P?

They don't say there is 52% of P. They say there is 52% of P2O5.

The gist is that phosphorus can be added to fertilizers in several different forms. To have a standard way of reporting the fertilizer analysis, phosphorus is reported as being $\ce{P2O5}$ rather than whatever form it actually is.

Thanks for answering, but there is only one phosphate anion ($H_2PO_{4}^{-}$) in the original formula. How can it be defined in terms of phosphorus pentoxide $P_2O_5$?

Sorry! I understood after reading this wikipedia article.
The molecular mass of $P_2O_5$ is 142 u so that P is 43% of the compound ($231/142 \approx 0.4366$).
When I need to represent P in terms of $P_2O_5$, I must multiply the molecular mass of P by the inverse of that ratio ($1/0.4366 \approx 2.29$). Then, in this case, I have $31/1152.29 = 0.61$, which is 61% of $P_2O_5$.
This value - 61% - is acceptable for purified MAP, but I don't know why "basic" MAP has only 52% of phosphorus pentoxide having the same formula.

Apparently because it is not pure. Look, the content of N is also lower than expected.

When it comes to fertilizers, usually what is advertised by general name monoammonium phosphate 10-52-0, in reality is ammophos, which is a mix of monoammonium and diammonium phosphates, $\ce{NH4H2PO4}$ and $\ce{(NH4)HPO4}$ with some water, so the brutto formula is
$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}$$
where $0 ≤ x ≤ 1$.
To find out the range for $y$, let's find the average molecular mass from the given nitrogen ($\ce{N}$) and phosphorous ($\ce{P2O5}$) content $α_i$:
$$\bar{M} = \frac{1}{2}\cdot\left(\frac{M(\ce{N})}{α(\ce{N})} + \frac{M(\ce{P2O5})}{2α(\ce{P2O5})}\right) = \frac{1}{2}\cdot\left(\frac{\pu{14 g mol-1}}{0.10} + \frac{\pu{142 g mol-1}}{2\cdot 0.52}\right) \approx \pu{138 g mol-1}$$
Using the expression for molecular mass of brutto formula and its average value:
$$x = 1.35 - 1.06y$$
Since $0 ≤ x ≤ 1$, the following holds true for $y$:
$$0.33 ≤ y ≤ 1.27$$
resulting in formula with the following ranges for the variables:
$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}, \quad x \in [0; 1], y\in [0.33; 1.27]$$
Since monoammonium phosphate usually prevails in this mix, there is a substantial amount of water; e.g. one of the possible brutto-formulas would be the following monohydrate:
$$\ce{(NH4)_{1.3}H_{1.7}PO4 * H2O}$$
Obviously, all the above is valid in the absence of third-party contaminations, e.g. if pure phosphoric acid and ammonia were used for the synthesis.


text:
Help with missing output from cubegen utility after density calculation in Gaussian16
I am busy trying to do density calculations in Gaussian16 as I need the .cube output for input to the PIXEL software. But my Gaussian output seems to be missing an entire chunk of data and I don't know why.
The output is supposed to look like:

However, my output is missing the entire chunk of electron density data, formatted as 6E13.5 (the block of floating point values indicating electron density at each point in the block).
My input in the route section is
#MP2/6-31G** guess=core NoSymm Pop=MK Density=MP2

And I use the following to generate the .cube file
Formchk NAME.chk NAME.fchk

Cubegen 0 NAME.fchk NAME.cube 0 h

Am I missing a keyword/command to include the electron density data?

I haven't used it before, but you seem to have missed the kind keyword in your cubegen command @thalsit

Thanks @Tyberius. Since I included it in the input file to perform the calculation I thought it would be redundant also having it as part of the cubegen command, since I thought the cubegen utility would read from my checkpoint file that there is MP2 density data. But it seems I was wrong, adding the kind keyword solved the issue

I think the reason they don't just scan for what kind you want is that there are different cubes you can generate besides the density, so its clearer to just let the user specify rather than just generating the first set or all of the cube data that it finds.

@Tyberius I completely agree, and would suggest you turning your two comments into an answer. (Also could you imagine it would produce all-the-cubes, that would produce awfully much data.)

You aren't getting the density data in your cube file because you haven't specified the kind keyword in your cubegen command. Gaussian allows a variety of different types of cube files to be generated, including electronic densities/spin-densities/gradients, MOs, and electrostatic potential, along with magnetic current and shielding densities. Without specifying the type of cube file you want, the program doesn't know what to generate.
The reason it can't determine what you want by parsing the chk file is that the file in general will contain information that could be used for multiple types of cubes. The program could hypothetically use the first valid piece of information or some specific type of cube data by default, but this would often produce an undesired output. It could also make all valid cubes files, but these take up a lot of memory so this would be wasteful. They presumably decided the default case should be to do nothing if a type of cube file isn't specified so as not to produce unnecessary or incorrect output.


text:
Gap in ionic radii

The above picture was shown in a presentation by my prof in inorganic chem as we were discussing ionic radii more in depth (all in pm). These are all isoelectronic radii. Notice the sudden leap in radii going from the 6th main group to the 5th main group. Its about 30 pm overall. Now how can that be? Is this a known thing?
You lose a proton in the nucleus, but so do you when you go from the halides to the 6th main group.
I unfortunately don't know the source of this picture, but my prof was sure that it came from a reliable source.

When you go from He to Be+2, Ne to Mg+2 and so on, then you're losing the whole s shell from the next $n$ quantum number, but adding two protons. Why shouldn't the ion be much smaller?

Actually for me, the surprise is difference between main groups VI and VII. For example, $\ce{S^2-}$ to $\ce{Cl-}$ is only $\pu{3 pm}$ while $\ce{O^2-}$ to $\ce{F-}$ is $\pu{4 pm}$. On the other hand, others have more than $\pu{20 pm}$ difference.

This  might be helpful, particularly table 8.2.3, data from R. D. Shannon, “Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides,” Acta Crystallographica 32, no. 5 (1976): 751–767.

@NightWriter yeah there seems to be newer evidence that doesnt show such a big gap...but I have to look a little deeper before making a final statement

@MathewMahindaratne well its not that surprising since group VI and VII have very similar radii without any charge if you look at the table Night Writer provided


text:
Is 99% Isopropyl Alcohol sold in stores really 99%?
My understanding is that an azeotropic mixture of isopropanol and water is 91%. This makes sense as to why there are so many brands of rubbing alcohol sold at 91%. There are also some sold as 99%, which as far as I know is impossible to reach through distillation. Is there another process that the manufacturers are using to reach 99%, or is it just marketing lies?

Most brands I find on google sell it at 99.9% even (for cleaning purposes). I checked a few data sheets, inconclusive. The density they give is very close to that on wikipedia (0.786g/ml), whatever that means. Common sense says they sell the azeotrope, but then this stuff is not made via fermentation and distillation. :-)

Any decent chemical supplier certainly has >91% isopropyl available. Not surprisingly, higher purity costs more, but at least one supplier (a common one, no reason to name) has 99.7% pure isopropyl listed.

A few different concentrations of isopropyl alcohol are sold as rubbing alcohol. The highest I have seen is 90%

There are plenty of ways other than distillation to make pure isopropanol (or other alcohols). And most azeotropes can be broken by adding a suitable third substance.

@matt_black , an entrainer

You certainly can get high purity isopropanol, e.g. 99.99% from Fisher Scientific. However, that is about US$50/l for high performance liquid chromatography (HPLC) grade alcohol.
The isopropanol can be synthesized without water by hydrogenation of acetone, or water could be removed from the azeotrope using a desiccant. Membrane separation is also efficient. At home, table salt ($\ce{NaCl}$) can be used to remove some water.
That said, there are a few reasons it's unlikely to find isopropanol at much greater strength than 91% at your local pharmacy:

It is more expensive.
It offers little advantage in home use over 91%. In fact, 70% ethanol or isopropanol is a more effective antiseptic than 91%. (Reagent grade alcohol is recommended for cleaning precision optics.)
Water from air would slowly infiltrate most thin-walled plastic bottles, causing swelling (and possibly rupture) of the container.
After opening, water would be more rapidly absorbed until it approached 91%.

But we do find 99% isopropanol at the pharmacy (in the US) so it is not unlikely.


text:
Saponification side reactions
When soap is made from triglycerides and NaOH reacting, is saponification the only reaction that occurs or are there other side reactions. Information concerning this issue seems slim online

Hmm, have you any ideas? I can think of few, but I'm not sure about their viability - esters can react using their alpha carbons, maybe even diallylic carbons.

This may assist you - https://en.wikipedia.org/wiki/Claisen_condensation


text:
Which molecule is the most acidic?
Wouldn't the most acidic be the H connected to F? However, my professor says it is answer A, the alpha Hydrogen in carboxylic acid.
Edit: Sorry, I did not phrase my question right. What I meant to say is I thought that the H connected to the same carbon as F would be the most acidic. Also, my professor just sent the answer and did not say why A is the most acidic. I just assumed she meant the alpha H in the carboxylic acid, or is that only for ketone and aldehyde?

I'm having trouble seeing an H connected to F. Which molecule are you referring to? In A, it is the hydrogen connected to the oxygen that would dissociate, not the alpha hydrogen.

It would all right, but it is not there.

"A" is a carboxylic acid. While all the other compounds can give away a proton, do not lose sight of that fact.

You can find the list of $\mathrm{p}K_\mathrm{a}$ values of some common organic compounds at Master Organic chemistry website.
The higher the value of the dissociation constant of acid $K_\mathrm{a}$, the smaller is the $\mathrm{p}K_\mathrm{a}$ value and the more acidic the compound is. To explain the $\mathrm{p}K_\mathrm{a}$ values we check for the stability of conjugate base which is obtained by removing the most acidic hydrogen and here in this case the conjugate base of the carboxylic acid will be most stable due to resonance.
Also in option B there is no hydrogen directly attached to fluorine. Option A is carboxylic acid, B is alcohol, C is ketone, D is ester and E is alkyne. You can find in the link that out of all these the $\mathrm{p}K_\mathrm{a}$ values of carboxylic acid is the least.

Even though there is nothing wrong with this answer, I suspect that OP asked about the acidity determination from the presented structures in the absence of external data sources.

The hydrogens attached to carbon atoms have the least chance to be removed as there is negligible electronegativity difference between carbon and hydrogen however the hydrogen atoms attached to more electronegative species can easily be removed due to the polar nature of the bond . Acidic nature of organic compounds can be commented on the basis of stability of the resulting conjugate base . If the conjugate base thus formed is stable then the compound is more acidic or the hydrogen is more acidic and vice versa.
 For option A we can see there is only one acidic hydrogen ie the  hydrogen connected to the oxygen atom. It's resulting conjugate base is also stable as the minus charge generated on oxygen atom is in resonance with the pie bond thus it is the most stable conjugate base among all other conjugate bases as no other compound ,containing acidic hydrogen has resonance in it .


text:
Is there any book or text where ionic radius of elements are given?
So my research has me looking for ionic radius characterization. If anyone could post any link or give the name of any book where ionic radius of all elements are given that would be really great?

There is "classic" work by Shannon and colleagues [1, 2], also available online as Database of Ionic Radii.
Extended ionic radii for lanthanides with higher coordination numbers were published by Jia [3].
The values from the references above are also compiled in recent editions of CRC handbook of chemistry and physics, e.g. [4, p. 12-12].
More recently, an updated series of calculated ionic radii has been published by Rahm et al. [4].
References

Shannon, R. D.; Prewitt, C. T. Effective Ionic Radii in Oxides and Fluorides. Acta Cryst B 1969, 25 (5), 925–946. https://doi.org/10.1107/S0567740869003220.
Shannon, R. D. Revised Effective Ionic Radii and Systematic Studies of Interatomic Distances in Halides and Chalcogenides. Acta Cryst A 1976, 32 (5), 751–767. https://doi.org/10.1107/S0567739476001551.
Jia, Y. Q. Crystal Radii and Effective Ionic Radii of the Rare Earth Ions. Journal of Solid State Chemistry 1991, 95 (1), 184–187. https://doi.org/10.1016/0022-4596(91)90388-X.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97.
Rahm, M.; Hoffmann, R.; Ashcroft, N. W. Atomic and Ionic Radii of Elements 1–96. Chem. Eur. J. 2016, 22 (41), 14625–14632. https://doi.org/10.1002/chem.201602949.


text:
Predicting the products obtained after addition of bromine and then elimination using alcoholic KOH
If we have an alkene reacting with bromine, then I know that the the reaction happens by a cyclic intermediate. But if we add alcoholic KOH to the products formed, then what will be the major products formed?
I don't understand the meaning of trans addition in case of only one π and σ bonds where the bond formed after addition is free to rotate as it is σ bond.  If bromine adds in an anti manner, then can it not convert to the cis form?
I am not sure if using the terms like 'trans' and 'cis' in case of σ bond is correct. I am not sure whether it's a molecule of bromine or HBr will be eliminated after addition of alcoholic KOH.

HBr gets elimianted. Depending on the molecule you may get 1 or 2 equivalents of HBr lost so diene formation is possible.

@Waylander I am afraid. It's not very clear to me.

@Waylander can you please explain why will HBr be eliminated and not Br2 ?

The mechanism of alkene formation by KOH is removal of a proton by OH- to give H2O and leaving a carbanion. The carbanion very quickly eliminates Br-, KBr is formed in solution but overall the components of HBr are lost from the moelcule.


text:
Optimized General Contractions using the EMSL Basis Set Exchange
When using the EMSL BSE Database and selecting a basis set, one can choose the "Optimized General Contractions" or not. When doing so for, e.g., Nitrogen and the cc-pVDZ some basis functions are not included anymore while the others are kept.
Without "Optimized General Contractions":
N     0
S   9   1.00
   9046.0000000              0.0007000
   1357.0000000              0.0053890
    309.3000000              0.0274060
     87.7300000              0.1032070
     28.5600000              0.2787230
     10.2100000              0.4485400
      3.8380000              0.2782380
      0.7466000              0.0154400
      0.2248000             -0.0028640        <=====
S   9   1.00
   9046.0000000             -0.0001530
   1357.0000000             -0.0012080
    309.3000000             -0.0059920
     87.7300000             -0.0245440
     28.5600000             -0.0674590
     10.2100000             -0.1580780
      3.8380000             -0.1218310
      0.7466000              0.5490030
      0.2248000              0.5788150        <=====
S   1   1.00
      0.2248000              1.0000000
P   4   1.00
     13.5500000              0.0399190
      2.9170000              0.2171690
      0.7973000              0.5103190
      0.2185000              0.4622140        <=====
P   1   1.00
      0.2185000              1.0000000
D   1   1.00
      0.8170000              1.0000000

With "Optimized General Contractions":
N     0
S   8   1.00
   9046.0000000              0.0007000
   1357.0000000              0.0053890
    309.3000000              0.0274060
     87.7300000              0.1032070
     28.5600000              0.2787230
     10.2100000              0.4485400
      3.8380000              0.2782380
      0.7466000              0.0154400
S   8   1.00
   9046.0000000             -0.0001530
   1357.0000000             -0.0012080
    309.3000000             -0.0059920
     87.7300000             -0.0245440
     28.5600000             -0.0674590
     10.2100000             -0.1580780
      3.8380000             -0.1218310
      0.7466000              0.5490030
S   1   1.00
      0.2248000              1.0000000
P   3   1.00
     13.5500000              0.0399190
      2.9170000              0.2171690
      0.7973000              0.5103190
P   1   1.00
      0.2185000              1.0000000
D   1   1.00
      0.8170000              1.0000000

What makes the basis set optimized if those three functions are simply omitted?

In any case, the option "optimized general contractions" appears to have been removed from the new beta site - at least I couldn't find it...

There it is in the advanced section.

I tried generating the basis from the beta site and got same sets with/wo optimized general flag. Tried it a few times just to convince myself. The older website's output may be a bug?

Those basis functions are removed because they are considered redundant. There is a completely uncontracted shell with the same exponent and angular momentum, so they can be removed without much loss of accuracy.
N     0
S   9   1.00
   9046.0000000              0.0007000
   1357.0000000              0.0053890
    309.3000000              0.0274060
     87.7300000              0.1032070
     28.5600000              0.2787230
     10.2100000              0.4485400
      3.8380000              0.2782380
      0.7466000              0.0154400
      0.2248000             -0.0028640        <=====
S   9   1.00
   9046.0000000             -0.0001530
   1357.0000000             -0.0012080
    309.3000000             -0.0059920
     87.7300000             -0.0245440
     28.5600000             -0.0674590
     10.2100000             -0.1580780
      3.8380000             -0.1218310
      0.7466000              0.5490030
      0.2248000              0.5788150        <=====
S   1   1.00
      0.2248000              1.0000000        ***Same exponent

Unfortunately, this is not documented on the old BSE. Here is the reference:
T. Hashimoto, K. Hirao, H. Tatewaki
'Comment on Dunning's correlation-consistent basis set'
Chemical Physics Letters v243, Issues 1-2, pp, 190-192 (1995)
https://doi.org/10.1016/0009-2614(95)00807-G

There was a period of time when the option was not working on the new BSE website (https://www.basissetexchange.org) but has been fixed (I hope).
Note: I am the maintainer of the new Basis Set Exchange :)


text:
Gaussian syntax error on the input line
I ran a Gold complex in Gaussian 09 with the following exact Route Section PBE0/LANL2DZ Opt Freq NoSymm and
it gave me this error message:
QPErr --- A syntax error was detected in the input line.
#T PBE0/LANL2DZ Opt Freq NoSymm

May you kindly please assist me.
This was the input file:
#T PBE0/LANL2DZ Opt Freq NoSymm

au1i2se2pc4h10o2

0 1
Au 1.360814 1.469089 4.400746
C 5.307642 1.389607 6.843985
H 5.507275 1.229404 5.910229
H 4.870358 2.257890 6.915128
C 6.510720 1.389607 7.612332
H 7.089684 2.096802 7.304637
H 6.959811 0.547877 7.509174
H 6.306354 1.532993 8.538974
C 1.456104 -1.956071 7.162351
H 0.690196 -1.468381 7.503839
H 1.363535 -2.024224 6.198359
C 1.488953 -3.274870 7.738612
H 0.666753 -3.731582 7.532296
H 1.584937 -3.205832 8.690153
H 2.226021 -3.766101 7.374002
I 1.201101 1.134698 1.820913
I -0.420347 3.370638 4.490387
O 4.402687 0.338108 7.352659
O 2.673550 -1.232944 7.482496
P 3.034067 0.042485 6.682134
Se 3.109165 -0.249598 4.533606
Se 1.591287 1.649826 6.831890

I think that PBE0 is defined in Gaussian as PBE1PBE. What is the 'T' at the start of the input line included for?

And if anything, it would be pbeZERO and not pbeOH - that is, a number, not a letter

@LigninPauling The formatting was bad, because not as code, so it ate the route sign. The #T produces terse output (if you didn't know already), and you are right, definition wise. Related: PBE vs. PBEPBE functional Btw: I highly recommend not doing compound jobs.

@Martin-マーチン That makes more sense now. I've noticed before you recommending not to do compound jobs. I'm curious as to why this is?

@LigninPauling there are three principle issues: 1. If you specify nonstandard keywords, they will not be copied to the second part. That might be an issue. 2. Creating a fallback, if anything after the optimisation fails. 3. Some visualisation tools can't really cope with them (molden). 4. You don't have to, but that's just a question of taste. It's easier to locate errors though. By far most important is the first. But if you know what you're doing, and know what you're expecting, you'll be fine either way...

Thank you both Lignin and Martin, My calculation ran successfully with the keyword PBE1PBE. Much appreciated!

Bryan, I am probably late, but here is a couple of suggestions:
The error that you get most likely comes from the wrong keyword: in Gaussian, the PBE0 method is defined using the keyword PBE1PBE. Change it, and it will be solved.
As for the "Opt Freq" keywords, it is not a good computational routine. You should run an optimization job first, and then a frequency calculation. What I would do if you don't want to run two separate jobs, is the following. After the geometry, leave a blank line. Then, on the following line, type:
--Link1--
On the next line, the new job specification. Don't forget to use the same chk, and the keywords Geom=Check Guess=Read in the input route.
In this way, Gaussian starts again, and you can always save these two jobs in two separate files.
I hope this helps, feel free to ask more questions!


text:
Can any body Suggest me a set up that I can build it up to study the surface roughness of transparent polymeric fibres?
I need to study the surface of nylon-6 blank fibres with a thickness of nearly 40 micrometre, I work in optics lab with many optical components such as beam splitters, lenses, mirrors ... etc, can I build a system to study the topology of this fibre?

Probably on Physics SE you get more chance of an answer


text:
What chemical regularly transported by road in the UK would cause serious injury/loss of life if it escaped in a road accident?
I'm writing a story where a serious multi-vehicle pileup happens in the winter during foggy weather. I need the fallout from this to include the escape of a substance that will cause significant injury/loss of life to those immediately involved in the accident.
What chemical can I use that is regularly transported by road in the UK that could cause this to happen?

Liquified Petroleum Gas (Flo Gas or Calor) - if it spilled and ignited

Chlorine, particularly in still, foggy weather (the main contributing factor to the only fatalities in large bulk transport of chlorine in the US that I know of).

Likely one of your choice https://www.gov.uk/guidance/moving-dangerous-goods

Liquid sodium? I used to work near a plant (now closed) where 40 Ton tankers of liquid sodium were a key feedstock. And this was considered safer than manual loading of sodium ingots into the plant.

Liquid nitrogen/carbon dioxide could asphyxiate many nearby.

A leaking lorry valve poured a large amount of solvent (acetone, perhaps) along a major Massachusetts highway, dissolving the asphalt of a large section. The fumes also nauseated and intoxicated drivers going over that section. Luckily, it did not catch fire, but it took some time to dig up the roadbed and remetal it.

Organophosphates in general would be a great idea as you can go from potent insecticides to chemical warfare as it fits the drama of your story.


text:
Intuition for why catalyst affects both forward and reverse reactions equally?
I have looked for quite a long time on the site however all the answers involve mathematical explanations involving Arrhenius's equation. I would really like an intuitive reason (or one using collision theory?) as to why adding a catalyst should increase both reactions equally.
It seems that if you add a catalyst it should decrease the activation energy by a constant for both forward and reverse reactions. Thus it should decrease the forward activation energy proportionately more than the reverse reaction then the forward reaction should go forward, and shift equilibrium. But it doesn't. Why not?

Well, if the catalyst shifted equilibrium, you would be able to make a perpetual motion machine out of that. As to me, that's a pretty solid reason to review one's assumptions.

What do you mean by " increase both reactions equally": the rates? or the energy barriers?

The equilibrium constant is fundamentally set by the energies of the reactant  and product (and not rate constants) so the reaction  pathway cannot matter to the equilibrium, just how fast it is reached. A catalyst causes the reaction to follow a different and faster pathway. As shown in the answer by @Andrew the change in activation cause by the catalyst energy cancels out.

I don't know if this qualifies as an intuition answer or not, but it's very easy to show mathematically. The key is that the rate constant is not linearly proportional to the activation energy. By the Arrhenius equation, $k = Ae^{-E_\mathrm{A}/(RT)}$, so the rate constant is dependent on an exponential of the activation energy. The equilibrium constant $K$ is a function of the ratio of the forward and reverse rate constants. For a simple reaction $\ce{A <=>[k_1][k_{-1}] B}$, the equilibrium constant $K=\frac{k_1}{k_{-1}}$.
If we assign the activation energy $E_\mathrm{A,fwd}$ to the forward reaction and $E_\mathrm{A,rev}$ to the reverse, we have
$$k_1 = Ae^{-E_\mathrm{A,fwd}/(RT)} \quad \text{and} \quad k_{-1} = Ae^{-E_\mathrm{A,rev}/(RT)},$$
so
$$K=\frac{Ae^{-E_\mathrm{A,fwd}/(RT)}}{Ae^{-E_\mathrm{A,rev}/(RT)}}.$$
Using simple rules of exponents, this reduces to
$$K = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}}.$$
Now if we subtract a quantity $E_\mathrm{cat}$ from both activation energies, the same derivation gives
$$K = e^{E_\mathrm{A,rev} - E_\mathrm{cat} - (E_\mathrm{A,fwd} - E_\mathrm{cat})} = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}},$$
so $K$ is unchanged.
Intuitively, it may help to think about the fact that the reverse reaction is necessarily composed of exactly the same microscopic states as the forward reaction, just happening in reverse order (by the principle of microscopic reversibility). The catalyst can stabilize one of more of the states on that reaction pathway, but cannot influence what direction that stabilized state reacts in. So making a microscopic state more easily achievable in the forward direction necessarily makes it more easily achievable in the reverse direction.

You are asking a quantitative question, but you are hoping for an answer besides the "mathematical explanations" already on the site. That might be impossible, but here is an attempt to talk about it without explicitly using the Arrhenius relationship.

It seems that if you add a catalyst it should decrease the activation energy by a constant for both forward and reverse reactions.

If you clarify this a bit and say the difference in activation energy with or without the catalyst is the same for the forward and the reverse reaction, then this is correct.

Thus it should decrease the forward activation energy proportionately more than the reverse reaction then the forward reaction should go forward, and shift equilibrium.

This contains a correct statement and and incorrect conclusion. It is true that when I subtract some value from a larger value, and then subtract the same value from a smaller value, the smaller value changes by a larger factor. So the first part of that statement is correct.
However, that does not mean that the rate constant changes by a larger factor (which is what the second part of your statement implies). The change in rate constant depends on the difference between activation energies (catalyzed or not), not the quotient of activation energies. Stating the Arrhenius relation in words, the rate constant changes by a factor that is proportional to the difference in activation energies. So forward and reverse rate constants change by the same factor, meaning that if the forward and reverse rates were the same without the catalyst, they are the same with the catalyst as well.

I would really like an intuitive reason (or one using collision theory?) as to why adding a catalyst should increase both reactions equally.

Collision theory is good to explain the concentration dependence of rates. It is less useful for explaining what fraction of the collisions leads to a reaction. It is easy to explain why a catalyst would be helpful for reactions in both directions. It is hard to make quantitative predictions based on a non-quantitative model. Even in collision theory, you have to know the distribution of the energies of collisions to determine how the fraction of successful collisions changes when you lower the activation energy by a certain amount. If you know the distribution and do the analysis, you arrive at the Arrhenius relationship.

Why not?

In his comments, Ivan Neretin had an excellent argument about how our world would change if catalysts could change the equilibrium constant. His argument works like a mathematician's proof of contradiction: It has to be this way because if it were any different, it would contradict something else we know (well, if we are talking about math, contradict an axiom we would like to keep). So here is my version of the contradiction: We could stick in catalyst into the reaction mix, the reaction would go in one direction until it reaches equilibrium, we would pull it out again, the reaction would go in the other direction to reach equilibrium. Imagine these are reactions in a battery. Batteries go toward equilibrium when we use them as a source of energy, and to charge them going away from equilibrium, it takes work. In our hypothetical scenario where catalysts change equilibrium constants, we would never have to charge the battery again. That is not our collective experience.


text:
Can ethylene glycol enhance the acidity of orthoboric acid?
My textbook says:

Boric acid is a weak monobasic acid but on the addition of certain organic polyhydroxy compounds, such as mannitol, glycerol, dextrose or invert sugar, it is transformed into a relatively stronger acid. Ethylene glycol cannot give this test (the reason is uncertain).

I do know that cis-diols form a boron-ether compound with the $\ce{[B(OH)4]-}$, leading to an increase in acidity. But is it true that ethylene glycol is exceptional? And is there any possible reason for it?

Maybe it is about the molecular geometries. I can't run the necessary calculations until at least the weekend, though. Would be interesting to know how buta-2,3-diol would behave (but then again, maybe it does not dissolve in water).

But a question in iit jee advanced(2014) says that on addition of ethylene glycol too the acidity of orthoboric acid is increased.
ques- The correct statement(s) for orthoboric acid is/are
(A) It behaves as a weak acid in water due to self ionization.
(B) Acidity of its aqueous solution increases upon addition of ethylene glycol.
(C) It has a three dimensional structure due to hydrogen bonding.
(D) It is a weak electrolyte in water.
ans- B,D (official)
i am also confused now beacuse its opposite is written in j d lee as per your question

But in my institutes exercise book the question is marked as of jee 2014 but only D option is mentioned as an answer

I had also watched a video on you tube claiming both option B and D check ques. 30 of uploaded link https://youtu.be/9J-SPkoPTno.

@TanmaySingh I cannot find where it is specifically written about ethylene glycol in J D Lee. It is mentioned that cis-diol will increase the acidity.

@ankitkumarmisra Your quoted text is from which book?

Yes it has been written in jd Lee that ethylene glycol don't give this test

Will somebody please make a solution of boric acid, check its pH, then add some ethylene glycol to see if the pH decreases, and by how much? It would save time for so many people checking their textbooks.

I think all are reading the Indian version of JD Lee the Sudarshan Guha one the original does have the ethylene glycol debate

Boric acid is a very weak acid with a $\mathrm{p}K_\mathrm{a}$ value of 9.2 and does not dissociate in aqueous solution as a Brønsted acid. Yet, in aqueous solutions, it acts as a Lewis acid by accepting a hydroxyl ion to form the tetrahydroxyborate ion $(\ce{B(OH)4-})$, as confirmed by Raman spectroscopy (Ref.1):
$$\ce{B(OH)3 + 2H2O <=> B(OH)4- + H3O+} \tag1$$
Thus, the dominant forms of inorganic boron in natural aqueous systems are mononuclear species such as boric acid $(\ce{B(OH)3})$ and borate ion $(\ce{B(OH)4-})$. The distribution of these two components depends on the first dissociation constant $K_\mathrm{a}$ of boric acid, which is equal to $\pu{5.80 \times 10^{−10} mol L-1}$ in fresh water at a temperature of $\pu{25 ^\circ C}$ (Ref.2).
Note that in solutions that are more concentrated than $\pu{0.1 M}$, boric acid acts as a much stronger acid than in diluted solutions, and becomes comparable to acetic acid. As shown in the equation $(1)$, the apparent ratio of the concentration of borate ions to that of boric acid molecules in the solution progressively increases from $1:1000$ at $\pu{0.2 M}$ to $1:5$ at $\pu{3.5 M}$ (Ref.1).
In an aqueous environment, boric acid and borates react with alcohols and multiple hydroxyl-containing compounds (polyols) forming boron esters. For example:
$$\ce{B(OH)3 + 3 R-OH <=> B(OR)3 + 3 H2O} \tag2$$

As shown in the above figure, complexation with polyols increases the acidity of boric acid due to the formation of cyclic borate esters. For example, the dissociation constant of boric acid becomes about $7 \times 10^{−6}$ in the presence of mannitol, i.e. a $10$- to $1000$-fold increase compared to that without the polyol (Ref.1). Such behavior is not shown on the addition of monofunctional alcohols, nor of glycols, nor of the trans-form of cyclopentane diols, but it is shown upon the addition of the cis-form of the latter compounds. AS described in the Ref.1:

The phenomenon depends on the formation of monocyclic or dicyclic compounds with the polyol groups, which are more highly dissociated than the boric acid itself, and it follows that the only compounds with two hydroxyl groups suitably placed on the same side of the C–C link can react in this way. Thus, the stability of the borate complex formed is strongly dependent on the type of diol used. If the diol involves the OH groups oriented in such a way that they accurately match the structural parameters required by a tetrahedrally coordinated boron, a strong complex is formed. For instance, erythritan (3,4-cis-dihydroxytetrahydrofuran) has produced a much lower $\mathrm{p}K_\mathrm{a} \ (\approx 4.8)$ with boric acid (more acidity, stronger complex; Ref.3).

The equilibrium constants of borate complexes have been investigated in several studies, and some of the data are listed in the following Table (Ref.3):
$$\begin{array}{l|c|r}
\hline
\text{Polyol} & K_1 & K_2\\
\hline
\text{1,2-Ethanediol} & 2.15 & 1.15\\
\text{1,3-Propanediol} & 1.27 & 0.11\\
\text{Glycerol} & 16.0 & 41.2\\
\text{Catechol} & 7.8 \times 10^3 & 1.42 \times 10^4\\
\text{D-Mannitol} & 1.1 \times 10^2 & 1.37 \times 10^5\\
\text{D-Glucose} & 1.50 \times 10^3 & 7.60 \times 10^2\\
\hline
\end{array}
$$
It has been shown that the amount of acidification produced upon the addition of polyol is proportional to the extent of ester formation (Ref.3). Basically, 1,2-Ethanediol is ethylene glycol and it also has a $K_1$ and $K_2$ values as well although they are not pronounce as those of D-Mannitol or even Catechol. Therefore, the increase in acidity may not significant, but it sure does the trick. The lower values of $K_1$ and $K_2$ are probably due to restrictions in structural parameters required by a tetrahedrally coordinated boron.
References:

Victor Kochkodan, Nawaf Bin Darwish, and Nidal Hilal, "Chapter 2: The Chemistry of Boron in Water," In Boron Separation Processes; Nalan Kabay, Marek Bryjak, and Nidal Hilal, Editors; Elsevier B.V.: Amsterdam, The Netherlands, 2015, Pages 35-63 (DOI: https://doi.org/10.1016/B978-0-444-63454-2.00002-2)(ISBN of the Book: 978-0-444-63454-2).
Benton Brooks Owen, "The Dissociation Constant of Boric Acid from 10 to 50°," J. Am. Chem. Soc. 1934, 56(8), 1695–1697 (DOI: https://doi.org/10.1021/ja01323a014).
Philip P. Power and William G. Woods, "The chemistry of boron and its speciation in plants," Plant and Soil 1997, 193, 1–13 (DOI: https://doi.org/10.1023/A:1004231922434).

In the figure, a particular polyol, reacts with both: boric acid and borate ions to increase the acidity. Can we say that the value of $K_1$ and $K_2$ (boric acid) should be less than $K_3$ and $K_4$ (borate ion) to increase the acidity more? Thus, the polyol preferably reacts to $\ce{[B(OH)4]-}$ to remove it and the equilibrium of reaction 1 also shifts in forward direction.

Does the vale of $K_1$ and $K_2$ in the table below that figure represents the same $K_1$ and $K_2$? And I think, in the balanced reaction for $K_2$, there will be $\ce{1H2O}$ and that for $K_4$, there won't be any $\ce{H+}$.

@ Apurvium 20: The reference did not report $K_3$ and $K_4$ so I won't able to discuss the difference.

I do not see any reason why the following reaction cannot take place, which will consume $\ce{[B(OH)4]-}$ and the concerned reaction will move in forward direction to increase the acidity of orthoboric acid$^1$.

According to wikipedia, "With polyols containing cis-vicinal diols, such as glycerol and mannitol, the acidity of the boric acid solution is increased."
If someone has aceess to this or this journal, then please add.

Reference

Page 375-376, Concise Inorganic Chemistry, 5th Edition, J. D. Lee, Wiley (ISBN 978-81-265-1554-7)

I think JD Lee writes "In effect, it acts as a strong acid in the presence of the cis-diol except ethylene glycol" which might be the source of confusion for others as it was for me as well. It probably means that all cis-diols enhance the acidity to an extent so as to make it a strong acid except ethylene glycol which does enhance it but not to the extent of others.

I do not see any reason either. The question is whether ethylene glycol is exceptional. Or, rather HOW exceptional. I predict that free-rotating ethylene glycol, like the structurally tight cis-diols, will increase the acidity of boric acid, but probably not as much as the sugars and other tight cis-diols. So if you do the acidity test with paper, you will see no definite pH drop; however, if you do the test with a two-place pH meter, you will see a small drop in pH. Can I offer a bounty for someone to do this test and report the actual numbers so we can all discuss it?

@Apurvium The reaction looks similar on paper to the formation of acetals in organic chemistry. These are equilibrium reactions, synthetically useful (i.e., at good yield within reasonable time) if run under acid catalysis, elevated temperature, and active removal of water (e.g., Dean-Stark apparatus).  Does your source details out such conditions in addition to an balanced reaction equation?

@Buttonwood No, my reference book does not provide more details but I added few more sources to my answer.


text:
Why is the principal energy of an electron lower for excited electrons in a higher energy state?
Several places state the 'principal energy of an electron' can be calculated as such:
$$E = \frac{2π^2mZ^2e^4}{n^2h^2}$$
Another equation I found was:
$$E = -\frac{E_0}{n^2},$$
where $$E_0 = \pu{13.6 eV}~(\pu{1 eV} = \pu{1.602e-19 J})$$
As seen in these equations, the greater the principal number ($n$) of the electron, the lower the principal energy $E$ of this electron.
However, the principal number $n$ is associated with higher energy states. The farther from the electron, the higher the energy state of this electron.
I seem to be fumbling the concept of 'principal energy of an electron.' What is the difference between the 'energy state' associated with the principal number, and the 'principal energy' of an electron? What exactly do the 'principal energy' equations mean? I read somewhere that it would be the energy it would take to "unbind" or ionize the electron, which would make sense, but I have not seen 'principal energy' explained as the ionization energy of an electron anywhere else.

Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~\mathrm{eV}
$$
which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom.
If we increase $n$ to say $n=2$, then we have,
$$
E=-E_0/4=-3.4~\mathrm{eV}
$$
which is a larger number than for $n=1$. Don't let the minus sign confuse you.
This is a very common source of confusion when these equations are seen for the first time. The confusion often stems from the fact that we are free to choose the zero of energy wherever we would like. So, in this case, zero energy corresponds to the case where the electron and nucleus are infinitely separated which is the $n\rightarrow\infty$ limit. So, more negative numbers correspond to lower energies and more tightly bound electrons.

My first exposure to this equation was the first, and read the relation between energy and quantum number was inverse. I don't quite see where the minus would come in the 1st equation, but do see it in others. Thank you

@jheindel your current formulation implies that you need negative amount of energy (i.e. $-E_0$) to ionize a hydrogen atom.

There's a similar thing for gravitational potential energy. If we set the GPE at infinity to zero, then the GPE for finite $r$ is proportional to $\frac {-1} r$

@Ruslan currently it says the energy of an electron in a hydrogen atom is $-E_0$ which means the energy required to ionize the hydrogen atom is $+E_0$.

Citing your first sentence: "we have, $E=-E_0$ which is the energy required to remove an electron...". So, $E$ is the energy required, and, since $E_0>0$, this energy required is $E<0$.

@Ruslan Ok, sure, that's the magnitude of energy required to remove an electron. Edited.

It maybe helpful if you think of the energy as interaction energy between the electron and a nuclei. In an hydrogen atom the interación potential is electrostatic, and we only have to account for the nuclei-electron atraction. Using the expression of the Coulomb potential you will obtain that this contribution is always negative. The zero of energies correspond to the proton and electron at $\infty$ distance.

I find the concept of "principal energy" quite confusing and prone to create misconceptions. Where did you find it?.


text:
Is cobalt (III) hexaaqua paramagnetic or diamagnetic?
The complex ion $\ce{[Co(H2O)6]^3+}$ has $\ce{Co}$ in the $+3$ oxidation state, meaning it has an electron configuration of $[Ar] 4s^0 3d^6$. Therefore it has 4 unpaired electrons and would be paramagnetic.
However, in the octahedral complex ion, the d orbitals split into two levels, with three lower-energy orbitals and two higher-energy ones. The six d electrons would therefore be in the lower set, and all paired. Following this logic, the $\ce{Co}$ atom would be diamagnetic.
Which is correct?

Quoting Housecroft and Sharpe "Inorganic Chemistry" (second edition):

The blue, low-spin $\ce{[Co(H2O)6]^3+}$ ion can be prepared in situ by ....

Greenwood and Earnshaw says about $\ce {Co}^{3+}$ complexes:

these are virtually all low-spin and octahedral
[...]
Even $\ce{[Co(H2O)6]^3+}$ is low spin.

Thus $\ce{[Co(H2O)6]^3+}$ is $d^6$, octahedral and low spin. It is thus diamagnetic.


text:
Grignard reaction on electron-poor benzyl bromides
I am trying to prepare a Grignard reagent from 4-nitrobenzyl bromide

but the reaction just doesn't work. My conditions are:

10 mL of THF (anhydrous);
7.04 mmol of 4-nitrobenzyl bromide;
7.04 mmol of Mg;
Argon atmosphere.

Water is not a problem since I left all my glassware in the oven overnight and flame-dried the round bottom flask + stir bar before adding Mg and the bromide. I also did everything under argon. I know the reaction is not proceeding because virtually all of the Mg turnings I added are still there and the solution is still clear. I tried adding two crystals of iodine but it didn't help either.
Is there something about this particular bromide that prevents/slows down the formation of the Grignard? I was wondering if the fact that the nitro group is electron withdrawing could make it more difficult to form the carbanion-like structure of the Grignard.

Has this particular Grignard been made previously in the literature? AFAIK, Grignards react with nitro groups, e.g. Bartoli indole synthesis.

I could only find 4 references that use this compound, none of them describes its preparation (none of them are in very good journals also)... I took this as an red alert that it wouldn't work since this structure is not that complicated (someone would've tried doing it before). I decided to give it a shot though cause it would provide an important insight on the mechanism I am investigating. I will try again this with a milder EWG at this position.

I also know grignards react with the nitro group, but in this case some reaction would occur and the Mg would be consumed to some extent. But nothing is happening... I decided to abort the reaction already following my advisor's advice.

Yes, you are correct about that. I just wanted to mention it because it might not have been a productive use of time trying to troubleshoot this. So it was just a more general comment, not so much about the question itself. If you want an organometallic, Zn is compatible with nitros, but I think it's a real pain making them (I've no experience, though).

Might be worth trying Knochel exchange conditions with iPrMgBr

@ Waylander: May not work since there are two electrophilic centers in 4-nitrobenzyl bromide.

@Waylander this is drifting off topic for the comments section (!) but have you done Knochel's turbo Grignards on anything other than aryl halides? Last time I looked, Knochel's work was all on aryls, although I might have missed something.

I have not and I'm not keeping up with current literature. Thought it was worth throwing in as a possibililty before saying "this won't work"

This question has aged a bit with no other answers, so here I go.
The same nitro group that can oxidize Grignard reagents could be oxidizing the magnesium too, thus keeping it passivated.  What you need is not so much a milder EWG but one with less oxidizing potential.  Try your reaction with nonoxidizing EWGs like acetyl (watch for side reactions), a $\ce{CF3}$ group, or a chlorine atom attached to the aryl group.

CF3 is happily sfable under Grignard conditions and strongly electron withdrawing


text:
Is there a scale for stickiness of glue?
Non-specialist question stemming from standing in front of a dizzying aisle of consumer adhesives... These differentiate themselves via meaningless superlatives like "serious", "extra" and "ultra". I just wanted to know suitability-for-different-materials and some quantitative measure of "stickiness" :)
Update: Could there be a consumer rating system for glue - based on existing professional testing standards?
(Note a rating system doesn't have to be a single numeric measurement in scientific units so can include any scaling/grouping function and include material classfications etc)

I recall there being two factors for glue: adhesion (stickiness to another material) and cohesion (stickiness to itself). The former factor will vary wildly with the material (consider: "remove grease before applying glue"). The tests would be specific to the material combinations and probably measure the force at which failure of the glue happens.

ASTM certainly has a range of standards for adhesion and adhesive testing.

@JonCuster Well, ASTM offers a whole series of documents about glues if alone searching for "adhesives" (https://www.astm.org/Standards/adhesive-standards.html) including quantificative comparative assessments (e.g. https://www.astm.org/Standards/D5649.htm states "inch-pound units" in its preview as standard unit).  Given the multiple materials  (paper, wood, polymers, metals, biomaterials, etc. pp.) and applications possible, a rather broad question indeed.


text:
Why (SO4)^2- does not create 4 double bonds
I looked this question up and still couldn't understand.
Why, in $\ce{SO4^2-}$ don't the 4 oxygens create double bonds.
In that case the all the oxygens will have 0 formal charge while the sulfur will have -2.
In what I've seen only 2 oxygens create double bonds making the sulfur have no formal charge, 2 oxygens have -1 formal charge and 2 others no formal charge.
When comparing formal charge of -2 on the sulfur, it is indeed less stable than no formal charge at all, and that is why it should be the most common resonance structure.
But everywhere I looked that was not the case, $\ce{SO4^2-}$ only created 2 double bonds and I cant understand why.
Cant the oxygens create coordination bonds with the sulfur?

Having formal charges greater than 1 on a single atom will tend to be less stable than spreading it around. Also, oxygen is more electronegative than sulfur, so we would expect more of the negative formal charge to reside on the oxygens than the sulfur. @guesting

Actually, there are zero double bonds.

See this answer for a discussion about $\ce{SO3^2-}$ which is essentially identical but for a missing oxygen (and thus a sulphur lone pair).

What you have seen is not an accurate depiction of the bonding situation according to the currently accepted theory. The correct structure of sulphate, shown below, has exactly zero double bonds.  Martin performed a calculation on the closely related sulphite ion (in which there is one less oxygen leading to a lone pair on sulphur) which shows zero π-type bonding orbitals. Unfortunately, I found no calculation of sulphate’s structure in a quick search but rest assured that it will be the logical extension of sulphite.

In this structure, sulphur is surrounded by exactly eight valence electrons as predicted by the octet rule. If you come across any depiction of a main-group compound in which an atom has more electrons in its vicinity than the octet rule allows, chances are this depiction is either simplifying (for example: drawing a 4-electron-3-centre bond as if it were two single bonds), an unstable reaction intermediate or outright incorrect.
Sulphur cannot form more than four traditional 2-electron-2-centre bonds (double bonds counting as two bonds, triple bonds as three bonds) due to a lack of available orbitals. On a very basic, general and simplified level, such a (localised) 2e2c bond is formed when an orbital of each atom overlap, resulting in a bonding and an antibonding orbital. Sulphur only has four such orbitals available (one 3s and three 3p) for bonding, so any electron exceeding the first eight would have to be placed in an antibonding orbital – but that would cause a decrease in bond order rather than an increase.
Historically, the depiction of sulphate that you gave was ‘explained’ by sulphur using its 3d orbitals for bonding. These orbitals are virtual (unoccupied) but do exist mathematically. However, their energy is too high for any meaningful bond to be formed. It can be calculated (and has been, somewhere on this site that I cannot locate at the moment) that the participation of d orbitals in such compounds is very low – certainly far lower than an $\mathrm{sp^3d}$ or even $\mathrm{sp^3d^2}$ hybrid orbital would require. Thus, it is best that the idea of double bonds in sulphate be purged from the textbooks yesterday.

The sulfate ion is derived from the sulfuric acid molecule:

When it undergoes chemical reactions, it typically donates both hydrogens as $\ce{H+}$ ions. This leaves behind the sulfate ion:
$$\ce{H2SO4 -> 2H+ + SO4^2-}$$
When the $\ce{H^+}$ ion departs, it leaves its electron behind, so it has to go somewhere (it remains with the $\ce{O}$ atom).
Hypothetically if $\ce{SO4}$ existed, with all $\ce{O}$ atoms double-bonded to the $\ce{S}$, then the sulfur would have a total of 16 electrons in its valence shell, which would make it more unstable. But the main reason is that sulfur only has 6 valence electrons in the first place, so it can only form up to 6 covalent bonds. This gives it a total of 12 valence electrons.
In formal charge theory, the idea is indeed to try to keep individual FCs as close to zero as possible, but also to break the octet rule as little as possible. The sulfate ion is very stable: just because something is an ion does not mean it is unstable. In fact, it often is much more stable than uncharged molecules.


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When naming an organic molecule - which group has the highest priority: methyl or methylidene?
I have been tasked with naming Myrcene. (see image below)

Which of the following names is correct?
7-methyl-3-methylidene-oct-1,6-diene
or
3-methylidene-7-methyl-oct-1,6-diene
Thank you in advance :)

Where would the two terms appear in a dictionary?

Priorities of substitutents need to be considered only while numbering the chain. When naming it after you've numbered it properly, the different substitutents need to be arranged in alphabetical order. Thus the first name is correct, since methyl comes before methylidene in alphabetical order.

I wanted to figure it out for myself and specifically avoid looking up its name.
Why is it that methyl comes before methylidene in alphabetical order?- I mean, don't they consist of the same 6 first letters?

@Patrick — This is a convention so we can determine the alphabetical order for any two words; so "methyl" alone comes before any "methyl" followed by any character.

(Also note that no one is correct. There are 2 issues. Excessive dash between the last substituent and the parent chain name. And missing vowel beween oct- (locants) -diene.)


text:
What could cause a peak to split into 2 resonances as temperature increases in NMR spectroscopy
I understand that some peaks may coalesce upon an increase in temperature due to an increase in free rotation, this does occur with some peaks in my data. Why might you see a peak split into two other peaks? I have thought it may be to do with an equilibrium shift in a rapid conversion between species but really I would expect that to show as one peak shrinking and another one growing. The observed effect is one peak clearly splitting into two. Any ideas why this could happen?
Edit:
The sample is a diamagnetic complex with aromatic ligands with t-Bu groups in the meta and para positions. It is hard to assign peaks as I believe there may be an equilibrium between different conformations in solution. The peak that splits is in the alkyl region of the NMR (around 1.5 ppm) and probably corresponds to some part of the t-Bu groups. The 2 peaks produced do appear to get sharper with temperature and the two peaks are the same size, however the difference in chemical shift (between the 2 peaks) at higher temperature is much greater than the linewidth at the lower temperature (where it is one peak), so I'm not sure if this could be J-coupling being resolved.
If it was the case of 2 conformers being present, would you not expect the single peak to reduce in intensity as another separate peak appears and increases in temperature? Rather than a splitting into 2 new peaks?

Is the  peak (are the peaks) getting sharper as you increase T? Are the peaks symmetrical (same intensity)? You might be narrowing the lines, allowing you to resolve a J coupling.

May you add an example where you observe such splitting of one into two / multiple peaks while increasing the sample temperature?

What kind of sample are you measuring?

As others saud give an example.  Nevertheless, by intuition the T might be enough to access two conformers rather than provide free rotation and equalize the splitting. It seems reasonable.

That's pretty cool looking data. I don't know that this question needed to be put on hold. It helps that you provided the data. Would be nice with some additional info regarding T and particularly field strength (scale is ppm but what is it in Hz) etc, but it's quite clear my hunch was right, except that the coupling would be little big :), so actually prob not a J splitting.

I have a stupid question. Are you sure the higher temperature is top of that series?

I'd quite like to see the structure. This would really help eliminate some of the possibilities.

Sorry for not including too much, obviously this is unpublished so I wouldn't want to upset anyone in my research group by sharing too much! The top is definitely the higher temperature, even if it weren't there are two other examples of coalescence going in the expected direction so if if you reversed the temperature it would be even more surprising! Thanks for all the thoughts everyone, you've given me a lot to think about.

Since I do not know which compound it is, my guess is that you have two different  $\ce{H^1}$ that have almost the same chemical shift at room temperature. The chemical shifts change with temperature as for instance in sucrose:

Here  are represented mainly the  $\ce{H^1}$  of the $\ce{-OH}$ but the $\ce{C-H}$ (the peak of 1) shifts also.
In your case,  one is almost fixed, while the other one moves to higher $\delta$  and, therefore, they split into two.
More info on the sucrose spectrum here.