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How ammonia can be in equilibrium if formation of ammonia is spontaneous in one way?
If a process is spontaneous under certain condition, the reverse process is non-spontaneous. Under standard conditions, it can be calculated that the formation of ammonia from nitrogen and hydrogen is spontaneous.
How equilibrium happens in this situation as formation of nitrogen and hydrogen from ammonia is non-spontaneous?
I mean why the reverse reaction occurs (formation of $\ce{N2}$ and $\ce{H2}$ from ammonia) as it is non-spontaneous?

I assume you are asking about formation of ammonia from hydrogen and nitrogen.
Then you can write a reaction like this:
$$\ce{3H2_{(g)} + N2_{(g)} -> 2NH3_{(g)}}$$
Where I am indicating that ammonia is being produced.
Using our knowledge of thermodynamics we can write:
$$\Delta G_{\mathrm r} = \Delta G_{\mathrm r}^\circ + RT \ln \left[\frac{(p_\ce{NH3})^2}{(p_\ce{N2})(p_\ce{H2})^3}\right]$$
Where those pressures are the ones you have initially, at some temperature $T$.
So, as you say, if Gibbs free energy of reaction ($\Delta G_\mathrm r$) is negative the formation of ammonia in that condition will be spontaneous, but that will be until $\Delta G_\mathrm r$ is zero, then the system reaches equilibrium (and there is no spontaneous process). Mathematically:
$$\Delta G_{\mathrm r} = \Delta G_{\mathrm r}^\circ + RT \ln \left[\frac{(p_\ce{NH3})^2}{(p_\ce{N2})(p_\ce{H2})^3}\right] = 0$$
and
$$ \Delta G_{\mathrm r}^\circ = -RT \ln \left[\frac{(p_\ce{NH3})^2}{(p_\ce{N2})(p_\ce{H2})^3}\right]$$
where the pressures here are equilibrium pressures. Now the reaction should be written as:
$$\ce{3H2_{(g)} + N2_{(g)} <=> 2NH3_{(g)}}$$

Correct, only issue is it's 2 NH3 in the equation, not just 1

good..thanks..If you want to edit there won't be any problem..

i understand this calculation . but , as we know that whenever NH3 produced , the reverse reaction also happens to occur .That is formation of H2 and N2 back. But this reverse reaction is not spontaneous (the forward reaction is). so, why this reverse reaction occurs?

@AsifAhmed spontaneity only refers to the net forward reaction being favoured

@orthocresol so it doesn't matter whether reverse reaction(formation of H2 and N2 from NH3) is present ?

@AsifAhmed What $\Delta G$ indicates is what direction of the reaction will be spontaneous in the initial conditions you specify (first equation)..anyway, it can be confusing..

@hernan Miraola .Thank you for your answers.under certain condition , change in free energy indicates that the reaction is forwardly favored (that is , the production of more NH3). But in chemical kinetics , we learn that both the forward and reverse reaction are present(though not same rate , but both still present). so is this situation like this- both reaction is present , but net reaction is forward?

yes, you are right..@AsifAhmed. As orthocresol said, we are talking about the net process..


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Non existence of B3+ in solution
I read that B3+ ions do not exist in aqueous solution, because hydration energy cannot compensate for the sum of first three ionisation energies.
This leads me to the following questions:

If boron (III) ion really doesn't exist, how was the hydration energy data calculated? and ionisation potentials?
Boron being small in size and carrying a high charge of +3 (if it exists) should be greatly hydrated. This high hydration energy should be able to compensate for the ionisation energies, but it is not so.

Could someone please explain?
Thanks!

The formation of an ordered hydration shell decreases the entropy of the system. The hydration of the ion may result in a huge decrease in entropy, which makes the overall process unfavourable.

Why should all hydration shells be ordered?

Also, if ∆S < 0, the Gibbs free energy would be negative for highly negative values of ∆H. How do we rule out or explain that possibility?

According to most chemical literature I have read, the hydration of any ion has its own associated ordered hydration shell necessarily. This is because of the attraction between the partial negative charge on the water molecule and the positive charge of the cation, in the case of positive ions. This attraction sort of fixes the position of the water molecules.

We can rule it out based on our observation that it does not exist. For these kind of thermodynamics things, we can only use data and empirical observations to substantiate.

It is observed that B3+ ions don't exist in aqueous solution, so how was it possible to calculate ∆H of a process that never happened in the first place?

Another question. According to the argument you proposed, it shouldn't be possible to hydrate any ion on basis of purely entropical factors. Clearly, that's not the case. How do we explain ∆H? I'm still not very sure

Of course. Entropy change of the system is not the only consideration. To compensate for a negative entropy change, the reaction would need to be highly exothermic, so as to ensure that the entropy of the surroundings would increase.

What are you really asking about, 'cause I think you didn't even think what you postulate or what you do not

Are you asking about something like this: https://chemistry.stackexchange.com/questions/74528/formation-of-mno2-vs-mnh2o64-in-aqueous-solution ?

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Langmuir parameters for fixed bed continous column
I have prepared a model of fixed bed adsorption column in MATLAB. I have to insert Langmuir parameter value there, so how can I calculate them in case of continuous fixed bed column. I have calculated them in previous batch study. How to calculate them for continuous column, what data should I get?

Welcome to Chemistry.SE. Could you edit your question to add more info?

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Calculating the theoretical Density of a Lithium oxide
I am trying to calculate the theoretical density for lithium oxide using the following formula
$$\rho= \frac{n^{'}\sum M_c+\sum M_A}{V_c N_{AV}}$$
where
$n^{'}$ is the number of formula units in unit cell
$M_c$ is the sum of atomic weights of all cations in unit cell.
$M_A$ is the sum of atomic weights of all anions within unit cell.
$V_c$ is the unit cell volume
$N_{AV}$ Avogadro's number.
Here is what I have so far:
The structure for lithium oxide is similar to that of FCC so the number of oxygem atoms in this unit cell is $8/8+6/2=4$.
If the unit cell is to have an electrically neutral charge, it must have a total of $8$ lithium anions.
The radius of a oxygen ion with a coordination number of 8 is about $128$pm and for lithium with a coordination number of 4 is about $73$ pm
Since the crystal structure is cubic, the volume is $a^3$ but we need to write $a$ in terms of the radius of the lithium and oxygen ions.
From the structure we can deduce that $a=2\sqrt{2}(r+R)$ (since lithium and oxygen ions touch along the diagonal of a square).
The atomic weight for oxygen is $16$g/mol and the atomic mass for lithium is $6.491$g/mol
$$\rho = \frac{4\times 16+8\times 6.491}{16\sqrt{2}(73\times10^{-10}+128\times10^{-10})^3\ (6.022 \times 10^{23})}=1.05 \text{ g/cm}^3$$
The theoretical density however is closer to $2$ am I missing something?

I'm not too sure about your relation $a = 2\sqrt{2}(r+R)$. If you check the structure of the antifluorite unit cell, I suspect it should be $r+R = \sqrt{3}a/4$. That would lead to a correction factor of $[2\sqrt{2}/(4/\sqrt{3})]^3$, and a final density of $\pu{1.929 g cm-3}$. I'll be happy to post this as an answer a bit later.

Interesting how did you come up with that relation? I would like to see the drawing if you can. Thanks!

Welcome to ChemistrySE! Feel free to take a [tour] of this site.

By the way, if you would like to get your accounts merged, please use this link and select "I need to merge user profiles" from the dropdown list. @adam

Here's the unit cell of $\ce{Li2O}$, which adopts an antifluorite structure. The image is adapted from this webpage; technically it shows the fluorite structure but it doesn't matter.

The cation-anion contacts occur along the diagonal of the cube (black dotted line). By Pythagoras' theorem, the length of the black dotted line is $a\sqrt{3}/2$, as I indicated on the diagram. The sum of the ionic radii is half the length of this line, so
$$r + R = \frac{a\sqrt{3}}{4}$$
Using this I obtained a value of $\pu{1.929 g cm-3}$ for the density. By way of comparison, Wikipedia cites a density of $\pu{2.013 g cm-3}$. By the way, regarding your final equation, it's a good idea to leave the units in. At first I was fairly confused why you were using $10^{-10}$ instead of $10^{-12}$, until I realised it was meant to be in centimetres.

Very good representation of the unit cell geometry. It's hard to see these relationships

It has been too long since I studied things like this, but that structure could be reduced to a diamond-like tetragonal cell with a cation in the middle, no? Is that the standard way of visualizing this structure?

@StianYttervik I think, for the purposes of calculating density, that would be ok and it would certainly be easier to visualise. But that wouldn't be a unit cell, as it cannot generate the lattice by appropriate translations. The above is the full unit cell for the structure (and how it's nearly always portrayed, in my own limited experience)


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Will NH3 decompose into N2 and H2?
In chemical kinetics , we learn that in a reversible reaction ,both the forward reaction and reverse reaction occur at the same time. At equilibrium , the rates are the same. For example:Production of ammonia NH3 .During the production of ammonia both the forward & reverse reaction is present .
$\ce{ 3H_2(g) + N_2(g) ⟶ 2NH_3(g)}$ forward reaction
$\ce{2NH_3(g) ⟶ 3H_2(g) + N_2(g)}$ reverse reaction
Now come to the Gibbs free energy.
Under standard condition(298K) , it can be calculated that the reaction is forwardly favored (forward reaction is spontaneous,and thus the decomposition of NH3 is non-spontaneous).
I understand H2 and N2 will produce more NH3.My question is at this situation will NH3 decompose into H2 and N2 to some extent(though it is a non-spontaneous process)?.But chemical kinetics tells that both reaction(not necessarily same rate) have to be present to create equilibrium in future .

It does not matter in this case that the forward reaction is energetically favourable because the activation energy (energy barrier) between reactants and product is so large that the forward reaction rate is effectively zero at room temperature. i.e. Transition state energy is huge wrt. RT. The barrier is also extremely large for the reverse reaction, therefore reaction rate is also effectively zero at room temperature. Equilibrium  exists but rates effectively zero. At high temperature and pressure, and most importantly with a catalyst, the reaction will occur, see Haber process.

Under standard conditions (100kPa, 0°C), nitrogen and hydrogen do not react with each other in any measurable way. It follows logically that also ammonia cannot decompose (as it would to a small extent in an equillibrium reaction), because otherwise it would decompose completely with time, although it is stable.
In equillibrium, the forward and backward reaction rates must be identical and finite.

Here standard condition is 25 degree Celcius

@thephysicist No. What you mean is NTP. https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure

okay . I edited my question

@thephysicist 298 or 273 make absolutely no difference.

In equilibrium, as you say, forward and backward reactions have same velocity of conversion (that is equilibrium). Out of equilibrium, you could probably see that the non-spontaneous direction of the process has a very small rate.
Microscopically, a molecule of product can collide another molecule absorbing its energy, and then -because of the high energy- be converted in a reactant. Following your example, a very small quantity of $\ce{NH3}$ produced will be converted into a reactant. In short: it is improbable but occurs.

It's so improbable it will not occur. Not in a measureable way, not in any reasonable amount of time.

@Karl well, I don't think so. $k_{inv}$ will depend on activation energy of the inverse process.., I mean, it will be produced..at least that's what our theories implies.. :)

You are misapplying the theories. To get to ammonia is a multistep reaction, and each of the preliminary steps  has a terribly low propability at STP, and they have to happen in very close proximity etc. This is not like winning the lottery, it's like winning the lottery on your birthday, three years in a row. And even then you only get a single molecule. Perhaps.

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Will the mass of ethene change if it is polymerized into poly ethene?
Choose the correct answer  and give a reason :
When the ethene is polymerized into poly ethene , all coming will change except (mass- molar mass- density )

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.This appears to be a homework question, please share your thoughts and attempts towards the solution. If you receive useful answers, consider accepting one.

Mass of reaction mixture doesn't change, because of law of conservation but overall density of polyethene is higher than that of ethene, molar mass too.

Why won't the mass of the ethene change ?? Poly ethene consists of thousands of ethene moleclues ??


text:
How to accurately abbreviate solvents?
IUPAC Red Book in section IR-4.4.4 Ligand abbreviations [1, p. 63] states the following:

N.B. Abbreviations for solvents that behave as ligands should also be in lower case letters [e.g. dmso for dimethyl sulfoxide {(methylsulﬁnyl)methane}, thf for tetrahydrofuran]; the practice of capitalizing the abbreviation of a solvent when it does not behave as a ligand is strongly discouraged as an unnecessary distinction.

To my understanding, applying this rule to, lets say, $\ce{[CrCl2(C2H6OS)4]Cl * C2H6OS}$ complex [2] as example, leads to the following (correct me if I'm wrong):
\begin{array}{ll}
\ce{[CrCl2(DMSO)4]Cl * DMSO} & \color{red}{\text{wrong}} \\
\ce{[CrCl2(DMSO)4]Cl * dmso} & \color{red}{\text{wrong}} \\
\ce{[CrCl2(dmso)4]Cl * DMSO} & \color{red}{\text{wrong}} \\
\ce{[CrCl2(dmso)4]Cl * dmso} & \color{green}{\text{right}} \\
\end{array}
In the other hand, we are used to solvent abbreviations being capitalized, e.g. ACS Style Guide suggests abbreviating dimethyl sulfoxide as DMSO and it seems like it doesn't make any distinction between the ligand form and the free form [3, p. 177]:

DMSO $\qquad$ dimethyl sulfoxide (also $\ce{Me2SO}$)

Does the rule from IUPAC Red Book also mean that when using in text (e.g. referring to a solvent as to reaction media) "dmso" is preferred over "DMSO", or this rule is only applicable when solvent acts as a ligand and/or solvate?
Bonus question: If both ways of capitalization are possible, then when creating list of abbreviations (e.g. using glossaries package with $\LaTeX$), do I need to separately list DMSO and dmso? If not, which one should I pick?
References

IUPAC “Red Book” Nomenclature of Inorganic Chemistry; Connelly, N. G., Royal Society of Chemistry (Great Britain), International Union of Pure and Applied Chemistry, Eds.; IUPAC Recommendations; Royal Society of Chemistry Publishing/IUPAC: Cambridge, UK, 2005.
Al-Najjar, N. A.; Al-Sudani, A.-R. H.; Shanshal, M. A.; Mousa, F. H.; Kariuki, B. M. Acta Cryst E 2013, 69 (7), m395–m396 DOI: 10.1107/S160053681301622X.
The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. DOI: 10.1021/bk-2006-STYG.fw001.

Ancient IUPAC literature
The IUPAC Interdivisional Committee on Nomenclature and Symbols states in recommendations from $1979$, Use of Abbreviations in the Chemical Literature,

Since the style in which abbreviations are given (e.g., capital letters or small letters,
  with or without full stops) often depends upon well-established practices that vary widely
  from one journal to another, IUPAC does not recommend a specific style for general use.
DOI: 10.1351/pac198052092229, archived version: here.

This means there is no one general guideline. You may compare this to units which are to be written, without exception, in upright Roman. (ACS Style guide does recommend Italic for molality $m = \ce{mol / kg solvent}$ but IUPAC disagrees.)

The IUPAC–IUB Combined Commission on Biochemical Nomenclature (CBN), in $1967$, set forth a document of revised tentative, i.e., provisional, rules. Abbreviations and Symbols for Chemical Names of Special Interest in Biological Chemistry does not list $\ce{DMSO}$ specifically, however does mention the similar $\ce{EDTA}$.

7.$\ $Miscellaneous Compounds
7.1 The following abbreviations are permitted; although they are fairly common, they should be defined in any paper if it is thought that readers might be unfamiliar with them. Some abbreviations are taken from the list published by Annual Review of Biochemistry.
[---]

$\ce{EDTA}$ – ethylenediaminetetraacetic acid (or acetate)

DOI: 10.1111/j.1432-1033.1967.tb00070.x, archived version: here.

The above article mentions $\ce{EDTA}$ only as a separate compound, and is written once, exclusively in upper-case upright Roman.

Summary
Disclaimer: What follows is mostly personal opinion, and should be treated accordingly. It is a summary of my present understanding.

There really is not a recent IUPAC document discussing this topic.
IUPAC does not enforce a ubiquitous principle upon which rules of abbreviations are formed. Abbreviations should be used as rarely as possible, defined where possible, and the specifics depend on the field, journal, and purpose at hand.
That said, some rules exist. For instance, most new organic compound abbreviations should derive directly from acceptable full names set forth in IUPAC's $2013$ Blue Book.
If an abbreviation of a compound, such as $\ce{DMSO}$ or $\ce{EDTA}$, is applied in chemical formulae, they should be written in lower-case upright Roman. A non-exhaustive list of exceptions is listed in IUPAC's $2005$ Red Book, section IR-4.4.4, paragraph $\mathrm{(vi)}$.
The following cited quotation from the same paragraph is very confusing.

Abbreviations for solvents that behave as ligands should also be in lower case letters [e.g. $\ce{dmso}$ for dimethyl sulfoxide {(methylsulfinyl)methane}, $\ce{thf}$ for tetrahydrofuran]; the practice of capitalizing the abbreviation of a solvent when it does not behave as a ligand is strongly discouraged as an unnecessary distinction.

I am fairly certain it means the following: If a collection of atoms attached to a central atom does not behave as a ligand, you should still use lower-case abbreviations when composing a chemical formula. Unless exception. This does not mean nor imply you ought to use lowercase when discussing a solvent on its own.
The formula $\ce{[CrCl2(dmso)4]Cl * dmso}$ seems correct to me. Not entirely sure if you could also use $\ce{[Cl2Cr(dmso)4]Cl * dmso}$.
I would include both options in the glossary but with a footnote. Mention that uppercase for $\ce{DMSO}$ is generally used the compound itself, lowercase is reserved for compiling a chemical formula. This might be overkill; if you prefer one, I would go with including only the upright version. But again, if space is not an issue, add both for maximum clarity.


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Scratching the inside of a flask to crystallize aspirin - what is it called and has it been scaled up?
I've just watched the Periodic Table of Videos on Aspirin and saw some "tricks" used to induce crystallization after the synthesis. Rapid heating with a heat gun followed by rapid cooling in ice was not sufficient, but scratching the bottom of the flask with a stirrer seems to have triggered rapid crystallization.
Warning: For those who are sensitive to the sound of chalk or fingernails on blackboards, better lower the volume during the actual scratching starting after 07:10, it's quite loud!
There is a little bit of an explanation for how the scratching helps:

When you scratch it you are actually forming a nice surface for the crystals to aggregate on.

but I am not sure I really understand what is really happening.
Question: Is the scratching just glass-on-glass? Is this crystallization happening on the (probably) invisibly small scratches only, or does it happen throughout the solution? Is there an "official" name for this technique? Has it ever been scaled up in some way?
For example, this answer says that the scratching produces a rough surface on the bottom of the glass for nucleation, but this answer says it is shards of glass in solution that are providing the nucleation sites. I'm trying to understand what is actually happening when chemists rub stirrers on the inside of flasks to induce crystallization, and once that's understood, to find out if the phenomenon is scaled up in manufacturing situations.

below: "Before" and "after" screenshots showing the clear liquid becomes thick with crystals after scratching the glass.

@andselisk there is no accepted answer there, and all three answers there would not even approach a complete answer to this question; glass-on-glass?  on the scratches only, or  throughout the solution? Has it ever been scaled up? so no, it is not possibly a duplicate. Not even close. Also, other answers here claim it is particles of glass in solution, not the scratches themselves, so I'm really looking for a more comprehensive understanding of what is happening.

Your question is not an exact duplicate, but it's very close to that question. If you seek an accepted answer, have a look at this one: https://chemistry.stackexchange.com/a/44486/41328

@andselisk The nucleation on shards explanation directly contradicts the nucleation on scratches explanation in your first link. Which is it? Where is the nucleation happening, on the scratches or in the solution? I'm looking for a definitive answers, not just a list of possibilities. Let's get something authoritative and conclusive, and then we can mark those questions as duplicate and point to the superior answer here. The goal in SE is good answers. In the mean time, I'll add links to both contradictory answers in my question. Thanks!

I see no contradiction as nucleation can occur both on the scratched glass surface and/or any particles present in the solution (shards, dust, density fluctuation etc.).

@andselisk why not leave the question open and see what other people think. It's quite possible someone more knowledgable will come along and know what is happening definitively, not just be able to make a list of what might be happening. SE is not best served by insta-closing questions that could produce more meaningful answers. Those answer could be considered pure opinion, as neither offers any way to confirm their hypothesis. No supporting links. Let's be scientific and see if we can find out if it is really known what is going on, or if those are just guesses.

I expressed my opinion, and you expressed yours. Now you claim I lack knowledge and that I should retract my close vote at the same time reminding in every comment how SE works. Wow. Bravo!

@andselisk I'm looking for someone who knows what's happening here, and who will answer my question definitively. Let's focus on getting an authoritative, non-contradictory answer. The insta-close prevents this from happening.

@uhoh I believe you are correct that this does not constitute a duplicate because the other similar questions do not have well founded answers. I would just recommend trying to not get frustrated. Don't let a misunderstanding devolve into a heated argument.

@Tyberius I don't see any of that. Insta-closes can snowball, so I've tried to encourage the user to reverse the close vote by explaining that I'm looking for someone who may have some knowledge of what's really happening, and that a quick-close would prevent that person from leaving an answer. It looks like andselisk has tried to make that look like it was something very different from what it is. If you read the comment again you'll see it's quite innocuous. Everyone has different knowledge, and SE lets us share.

I doubt anyone has scaled it up, as for large scale runs, you're likely to have seed crystals of the compound itself you can use for nucleation, and these would work much better than the random nucleation you get from scratched glass.

@R.M. If it is the introduction of "shards of glass" into the solution that makes scratching effective, then I think it's fair to call introduction of silica nanoparticles or other non-target compound particles as scaling-up, but as you suggest, introducing particles of the target compound itself would not qualify. Thanks for the comment.


text:
Validity of van der Waals equation

Van der Waals  had stated that at high pressure, pressure exerted by ideal gas on the walls of container is more than real gas.
However in this graph, one can clearly see that, the at high pressure volume of real gas tends to be more than volume of ideal gas. Hence real gas should be exerting more pressure on the walls of the container, as it occupies more volume. So why does this not hold true?

It's because in ideal gases there are no intermolecular forces whereas in a real gas the force with which it collides is reduced due to intermolecular forces.Now for the volume occupying part ideal gas particles are point sizes but in a real gas the molecules do occupy some volume.That's why some corrections were made to volume and pressure in the ideal gas equation.
Note that the above were assumptions of the Kinetic theory of gases.

But see the graph above,  it says that volume occupied by the Real gas is more than the volume occupied by ideal gas . So thus , the pressure exerted by real gas must be more than that of ideal gas , simply because it occupies more volume.

Can you tell me how more volume means more pressure?
Because they are inversely proportional its if there's very high pressure wouldn't the molecules clump together(i know repulsion) and the volume decreases  Aka Boyle's law?

In a same container( keeping the volume constant) , a real gas will tend to occupy more volume than ideal gas.  So since we are restricting it's volume,  the real gas should exert more pressure on the walls of the container.  It's that simple

i don't quite understand first you say volume is constant , then real gas occupies more volume ? Is volume property of the container or molecule
what's the reasoning behind "real gas "should" exert more pressure"?

See the graph properly and the line I drew to mark the pressure.  At a constant pressure,  the volume occupied by real gas is more than that of ideal gas. Now if we try to confine the both gases in a container, the gas that tends to acquire larger volume will tend to exert more pressure. ( Try to draw a line from the volume axis to the graph ,it will cut the real gas and ideal gas graphs at different points.  It means that at a constant and confined volume real gas exerts more pressure on wall than ideal gas)

I think the original conclusion is wrong-> correct one is, At high pressure means (when)they both exert same pressure , volume occupied by real gas is more than ideal gas . See pressure here is the pressure exerted by the gas not by the piston(if it has one to control the conditions) of the container in which they are present.

Let us continue this discussion in chat.


text:
Why are d orbitals used in the hybridisation of PCl5 and similar compounds?
Phosphorus uses its $\ce{3s}$, three $\ce{3p}$ and one $\ce{3d}$ orbitals for its hybridization in $\ce{PCl5}$ and similar compounds.
However, shouldn't it use its $\ce{4s}$ rather than $\ce{3d}$ orbitals as the former is of lower energy?
I initially thought that this was not the case because hybridization cannot take place between orbitals of different principal quantum numbers, but then learnt that this indeed was possible, for example in coordination compounds with $\ce{dsp^2}$ hybridization such as $\ce{[Pt(NH3)2Cl2]}$.

The use of d-orbitals in hybridisation schemes has often be disproved, they do not play a significant role. The bonding is better described by multi-centre bonds. Unfortunately it is very often still taught, even though it is wrong.

Regarding hybridisation in coordination compounds: https://chemistry.stackexchange.com/q/76726/16683

Regarding main group compounds, see e.g. J. Am. Chem. Soc. 1990, 112 (4), 1434–1445

https://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5

I linked that one as a duplicate because it shows the currently accepted theory.

https://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit


text:
Does the name 4-nitrodiphenylmethane follow the IUPAC rules?
Does the name 4-nitrodiphenylmethane follow the IUPAC rules? It doesn't make sense to me. How can you assign a locant of 4 when your stem name is just methane?

I feel like the stem is "diphenylmethane" not just methane

The preferred IUPAC name for the compound that is given in the question is 1-benzyl-4-nitrobenzene; however, that does not necessarily mean that other names are not in accordance with IUPAC nomenclature.
The name diphenylmethane actually is in accordance with current IUPAC recommendations (see Subsection P-44.1.2.2 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)), although the preferred name is 1,1'-methylenedibenzene (see also this answer).
The name diphenylmethane treats the phenyl groups alike as substituents of methane.
If now one of the simple phenyl substituents in diphenylmethane is replaced by a compound 4-nitrophenyl substituent, the analogous name of the resulting compound is (4-nitrophenyl)(phenyl)methane and not ‘4-nitrodiphenylmethane’.

Ooh, thanks! What if the other phenyl group had been substituted with NO2 para to the methylene substituent? Do I name it bis(4-nitrophenyl)methane?


text:
Food which does not produce urea
My professor of bioengineering said that all foods produce urea. Do foods exist which does not produce urea?
Thank you very much.

Sugar, fat, any other food that does not contain nitrogen will do.

Food doesn't produce urea, your body produces urea from the nitrogen content of the food you eat (mostly comes from proteins). So you can eat e.g honey, which contains minimal, if any, amount of nitrogen.

Of course, as your body breaks down dead cells (e.g. red blood cells, life expectancy 3-4 months), urea will also be produced from non-food sources. And you'll need to replace these cells, so you can't eliminate proteins from your diet to eliminate urea production. That said, most diets are very rich in proteins.

@MSalters Agree. While "very rich" is maybe overstatement (there are many types of food with low protein content), it is true that the whole point of eating food is to eat nutrients like protein.

In addition to Greg's answer, surprisingly, a high sodium diet causes the body to produce more urea in order to conserve water.  The point being, that there is no direct correlation between food consumed and the production of urea.  The body uses urea in it's water/salt balance mechanism.
https://www.nih.gov/news-events/nih-research-matters/how-body-regulates-salt-levels

"A high salt diet increased glucocorticoid levels, causing muscle and
  liver to burn more energy to produce urea, which was then used in the
  kidney for water conservation."

Good point and interesting addition.


text:
Why does lead(II) acetate does not ionize appreciably?
As this page in this book says, lead(II) acetate is soluble, but does not ionize appreciably. It seems contradictory to me, since, to call a substance soluble means it has high $K_\mathrm{sp}$ (when compared with other substances with the same elemental ratios in their formula units). But how can the book say it doesn't ionize appreciably? Why call it soluble then?
EDIT: I forgot that it is wrong to say that "to call a substance soluble means it has high $K_\mathrm{sp}$", so I revised it accordingly.
Here's the picture of the actual paragraph for those who can't access the link:

I recognize that this question actually reflects my inability to distinguish between dissociation and dissolution, now that I have cleared it from me, I rephrased the question to reflect my actual question. I did not erase the previous version to be able to show the reference of MEL Science's and alphonse's answers.
ACTUAL QUESTION: Why does lead(II) acetate not dissociate appreciably in water, considering that the ionic forms of acetate and lead(II) ion can be hydrated well due to the presence of partial charges in the water molecule?

It's not true that a soluble substance is one with a high $K_{\mathrm{sp}}$. $K_{\mathrm{sp}}$ describes the solubility of substances that dissociate in the solvent (see link). Not all substances dissociate on solvation. Sucrose in water is one such example. So I'd say it probably solves as a neutral molecule, but I don't know why.

Related: https://chemistry.stackexchange.com/questions/48502/why-do-some-insoluble-salts-undergo-slow-chemical-reaction-instead-of-dissociati

How can a book say?  A book can say that giraffes are purple, six-legged and can fly. The better question is: what is the book saying? I don't know. A cursory glance at the Wikipedia articles (you apparently aren't aware of the need to clearly communicate the identity of the chemical you wish to discuss. I have no idea whether "lead acetate" refers to the Pb(II) or Pb(IV) compound, or something else) shows that both acetates are soluble in polar organic solvents. This indicates that both are capable of dissolution without dissociation. Confusing dissociation (which generally occurs in solvents with high dielectric strength, such as water) with dissolution (which can occur in almost any liquid) is a newbie mistake. Whether the book you refer to (your link doesn't work) is actually claiming this or something else, I don't know. You don't provide enough information and I am not familiar enough with the solution chemistry of lead to make any judgement. Lead(II) acetate is used to precipitate anions (in aqueous solution) so I'd guess that the book is simply wrong - IF - it meant to imply that there was negligible dissociation of the dissolved salt.

Ahh yes. Maybe it's my problem of distinguishing between dissociation and dissolution. This means that lead(II) acetate actually dissolves in water, but it doesn't dissociate in it.  So now the real question is why lead(II) acetate does not dissociate in water?

A source in my native language says that lgK of PbAc2 is 4.08. It seems that lead acts as a central atom with two acetates as two-dentant ligands and two molecules of water acting as ligands too.
That means that lead acetate has its lattice ruined during dissolving, but it still forms a coordination compound.
 I am not really good in coordination chemistry of lead, but I just found out that lead wants to have octahedral geometry around it, so I would assume the structure of that would be
All bonds from lead ion are not really covalent rather than coordination bonds. Mind the charges!
P. S. aq=water
P. P. S. People in the comments are saying lead is too large for acetate to be bidentate,thus I would assume acetate just forms coordination compound with lead just like a simple monodentate ligand.

Picture is wrong Pb 2+ is too big to be chelated by acetate.

Sure, I don't pretend the structure must be 100% true, that's just my assumption. Perhaps acetate acts as a monodentate or bridge ligand depending on the concentration.


text:
What is the difference between ignition temperature and flash point?
I am trying to clarify this for Class 8 Science. Their textbook says "It is the lowest temperature at which a substance catches fire". However, it does not clarify whether it is on its own or due to an external source. I read through various websites and there are both kinds of definitions put up! Some also mentioned a flash point. So wanted to know what the proper definition of ignition temperature is and how it differs from flash point. Thanks!

From Wikipedia:
Autoignition Temperature: the lowest temperature at which a substance spontaneously ignites without a source of ignition (such as a flame or spark).
Flash Point: the lowest temperature at which the vapours of a (volatile) material will ignite, with an ignition source.
Ignition temperature means the same thing as autoignition temperature according to Merriam Webster.

The question talks about "ignition temperature", not "autoignition temperature". To be a reasonable answer to the question, you would have to demonstrate that "ignition temperature" is almost certainly a mis-reading of "autoignition temperature." (i.e. that "ignition temperature" isn't a thing distinct from "autoignition temperature".)

@R.M. The resolution might be the frequent use of overly casual language. Many people might causally say "ignition temperature" when they mean autoignition temperature.

I always heard the two used interchangeably, and figured the second one was shortened for ease of speech.


text:
Why is isopropylbromide formed as the major product but not n-propylbromide from 1-propene?
When 1-propane reacts with HBr, the major product formed is isopropyl bromide and minor product is n-propyl bromide. I know that this is said to happen because the reaction mechanism favours the formation of a more stable product which is isopropyl bromide in this case which is again due to the formation of a stable reactive intermediate-secondary carbocation in this case. (For n propyl bromide,it is a primary carbocation that is formed as a reactive intermediate less stable and more reactive than isopropyl carbocation). This is based on markonikoff’s rule. I know this much but I have a question. My question is when a secondary carbocation is stable than primary, it means that secondary is less reactive than primary. So the major product should be n propyl bromide because it is formed from a primary carbocation which is more reactive than isopropyl carbocation. But that's not the case. Why? Kindly someone clarify my doubt. Suggestions for some good books in this topic are much appreciated.

The primary cation may be the more reactive but the fastest reaction it can undergo is the internal proton transfer to give the secondary cation.

Thank you so much. I am just a beginner in chemistry learning the subject out of interest. So kindly elaborate.....

The proton shift is an internal reaction so it can happen very fast, it does not require the molecule to bump into a second molecule as the formation of the primary bromide would.

If I helped you, please consider accepting my answer by clicking on the checkmark. Of course, you may also wait in case more answers are written to select the best one.

Well yes, a primary carbocation is more reactive than a secondary carbocation. But both are still so extremely reactive that they have lifetimes as short as is physically possible. Any reactions with both are basically, for all intents and purposes, diffusion limited.
Furthermore, the fastest reaction a primary carbocation can perform is a 1,2-hydride shift to form the corresponding secondary carbocation. Since this is not a diffusion process but basically intramolecular tunneling, it is even faster. This is why it is said that primary carbocations do not even form in the first place.


text:
Why is hexafluoridocobaltate(III) the only common high-spin cobalt(III) complex?
Almost all $\ce{Co^3+} (\mathrm{d^6})$ complexes are low spin, including $\ce{[Co(H2O)6]^3+, [Co(ox)3]^3–, [Co(NH3)6]^3+}$ and $\ce{[Co(CN)6]^3–}$. The only common high-spin cobalt(III) complex is $\ce{[CoF6]^3–}$. Why is the hexafluoridocobaltate(III) complex in a high-spin state?

This is a question that the crystal field theory cannot answer because of its inherent shortcomings.
The crystal field theory assumes point-type charges approaching the central metal from the coordinate axes leading to the destabilisation of those $\mathrm d$ orbitals that do not feature a nodal plane in the coordinate axes: $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$. The others have zero contribution in all $x,y$ and $z$ direction and are thus not affected by point-sized ligands approaching from there. As simple as this model is, it only gives a crude approximation of why field splitting occurs and which orbitals are likely to end up higher in energy. For everything that involves quantitative discussions, the full molecular orbital approach is needed.
Below you will find the fullest molecular orbital scheme of an $\ce{[ML6]}$ complex including σ and π interactions. A simplified version without the π interactions can be found here; you should attempt to understand it before adding the π interactions to the picture. In this question, however, we need to understand the π effects and thus I cannot leave them out.

Scheme 1: Molecular orbital scheme of an $\ce{[ML6]}$ complex featuring both σ and π interactions.
The system we are looking at — $\ce{[CoF6]^3-}$ contains the π-basic fluoride ions (meaning that their $\mathrm p$ orbitals that can interact with the metal in a π symmetric manner, are fully populated). The interaction of the $\mathrm{t_{2g}}$ fluorine group orbital with the metal’s $\mathrm{t_{2g}}$ orbitals will create a bonding-antibonding interaction. With fluoride most likely having lower-lying orbitals, this interaction should raise the energy levels of cobalt’s $\mathrm{t_{2g}}$ orbitals which thereby turn into $\mathrm{t_{2g}{}^*}$ orbitals.
In this case, fluorido is the only strongly π-basic ligand you are considering in your comparison. And chloride does not form a hexachlorido complex like fluoride would but a tetrahedral tetrachloridocobaltate(III) for steric reasons. Thus, we don’t know whether the fact that fluoride is π-basic enough to raise the $\mathrm{t_{2g}}$ levels high enough so spin pairing does not occur is because of its π basicity alone or whether further elements come into play. Oxalato, ammin and aqua ligands have only weak π-basic effects while cyanido, if it displays π effects, is a π acid.

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


text:
How to calculate energy gap when determining electron configuration?
The weird electron configuration of Niobium was well explained on this post:
How can one explain Niobium’s weird electronic configuration?
It was pointed out that electrons fill orbitals with higher exchange energy and lower repulsion energy, only if the sum of the positive exchange energy and negative repulsion energy overcomes the energy gap between orbitals.
The only problem is that I do not know how to quantify the energy gap.
How can I calculate the energy gap between orbitals?

Long story short, you can't.

@IvanNeretin So would the value have to be taken out of a data booklet?

Basically, yes.

You could calculate the energy of your initial electron configuration, then do the same for your final state and work out the difference. The Los Alamos Flexible Atomic Code (https://github.com/flexible-atomic-code/fac) can do that for you. There should be no state where you fill shells from the ground up that has a lower energy than what we now call the ground state of nobium.


text:
Oxymercuration of alkynes using alcohol
Alkynes can be transformed into ketones or aldehydes, via the corresponding enol, by addition of water in the presence of $\ce{Hg^2+}$.
Similarly, one would expect alkynes to be transformed into enol ethers by using an alcohol instead of water. However I've looked through most organic chemistry textbooks, and they do not seem to make any reference to this. A search on Google also doesn't turn up much. I expected the reaction between alkynes and alcohol would be more commonly mentioned.
Is the reaction between alkyne and alcohol using $\ce{Hg^2+}$ only theoretical that doesn't actually in real life, since textbooks do not introduce this reaction?

Try searching for "alkyne alkoxymercuration", reject Google's suggestion of "alkyne oxymercuration" (if it pops up)


text:
Flammable lighter than air gases (safeish)
I'm an artist currently using propane and butane in my show where I put the gas into a soap bubble and light it to produce a nice flame.
But I'd like to start working with a lighter than air flammable gas, that's relatively easy to get hold of, and as safe as it possibly can be.
The propane/butane I use comes in small canisters and the biggest risk, as I see it, is forgetting to close the tap.
Any ideas?

Ever considered hydrogen? Not sure how easily available that'll be for you... I just electrolyze water to get mine O:) Besides, if you were to electrolyze water to get some hydrogen, might as well collect the oxygen produced in the same container as well... the result is Oxyhydrogen, and it's pretty flammable. [ https://en.wikipedia.org/wiki/Oxyhydrogen ]

flammable && sefeish = False;

@paracetamol hydrogen would be a lot of "fun" for another audience (paramedics I guess). Mixed with oxygen it rather detonates than burns, so most likely there will be a boom and very little fire.

@andselisk I was under the impression hydrogen deflagrates (and doesn't usually explode). Besides, you have a point there! :O

@andselisk Also, I think I've seen the "trick" the OP does with flammable gases and soap bubbles before. Considering the size of the bubble (could've fit in my palm...) and the amount of (oxy)hydrogen in it, I don't think it'll be much of a threat to the OP's existence. The stored hydrogen is another matter though... :D

@paracetamol Yep, but one needs to get H2 : O2 ratio right. Minor deviation can cause troubles.

@andselisk True, and yes... oxyhydrogen detonates, I  clean forgot that >_<

I would propose to use hydrogen, as it is pretty easy to obtain right there where you need it and it's the lightest gas ever.
You can even watch a  video where the whole setup for the bubble experiment is described  (at the end).
Besides hydrogen, methane looks good to me, and it is also in our everyday life, being the natural gas. But the setup for using household methane is imagined more complicated to me than the one with hydrogen.
Just be very careful with any flammable gas, don't make it another Hindenburg :)

Methane gas is lighter than air and you can do practically everything with it that you can do with propane/butane. I can’t comment on how easy it is to acquire because that depends on your jurisdiction.
In the Christmas and carneval lectures at my university, some fun with large methane-filled balloons that were then ignited was always had, but you can also ignite soap bubbles.

You can't liquefy methane with relatively moderate pressure and store it in canisters, that's where it differs from propane/butane.


text:
Why one gramm of O and one gramm of O2 icludes same amount of oxygen atoms?
So there is a problem I can't understand why one gram of O and one gram of O2 includes same amount of oxygen atoms.

Which of the following contains most oxygen atoms?a) 1,0 g O-atomeja, b) 1,0 g O2(g), c) 1,0 g O3(g). So there is clear question. And the right answer is that all of them includes same amount. And i don't understand how i can count this, because if i count in my way answer is wrong,

@paracetamol I guess I see your point, but this question somehow reminds me of "1kg of feathers vs 1kg of lead" problem:)

@Daria They all have the same number of atoms. Number of particles, will be different though ;)

@DariaKostiniuk No no, that's not what we meant. If it helps, one can show it analytically, e.g. presenting number of oxygen atoms as following: $$\begin{align}N(\ce{O}) &= n(\ce{O}) \times N_A = \ &= \frac{x \times m(\ce{O_x}) \times N_A}{M(\ce{O_x})} = \ &= \frac{x \times m(\ce{O_x}) \times N_A}{x \times M(\ce{O})} = \ &= \mathrm{const}\end{align}$$

https://chemistry.stackexchange.com/questions/28615/do-i-have-to-consider-the-molecular-mass-of-the-oxygen-atom-or-the-diatomic-oxyg https://chemistry.stackexchange.com/questions/23808/why-is-o2-enough-to-form-a-mole-of-oxygen

Possible duplicate of Would 2 C₂ be 2 moles of diatomic carbon, or 4 moles of carbon?

Suppose you are able to put $1.0$ gram Oxygen atom in a flask. Now you order all atoms of this sample to join to another one, so as to make couples. After that you weigh the content of you flask. it will have the same weight.

I think this question needs an eye-opener in non-chemical terms.
Counter-question: What contains more chestnuts?
a) 1 kg of "Cn" (chestnuts)

b) 1 kg of "Cn2"

b) 1 kg of "Cn3"

c) 1 kg of "Cn4"


text:
Does MOPAC allows calculation of Standard Enthalpy of reaction in this way?
I am using MOPAC2016 to calculate $\Delta H^\circ_\mathrm{r}$ (and then $\Delta G^\circ_\mathrm{r}$) for a reaction in aqueous phase, at $\pu{298 K}$. I will explain how, and then the question. Finally I give an example. Suppose this simple reaction:
\begin{align}
\ce{A(aq)<=>B(aq)}
\end{align}
Then, $\Delta H^\circ_\mathrm{r}$ can be obtained as:
$$\Delta H^\circ_\mathrm{r} = \Delta H_\mathrm{f}^\circ(\ce{B}) - \Delta H_\mathrm{f}^\circ(\ce{A})$$ where each of those values is what we get on a thermodynamic calculation (also in an optimization) with the name of H.O.F. (heat of formation, HOF).
Formally the question should be: Is that correct to calculate $\Delta H^\circ_\mathrm{r}$ (and then $\Delta G^\circ_\mathrm{r}$) by using the value of HOF for each reactant?
Example
Here is an example of MOPAC output for a thermodynamic calculation, suppose it is the one for molecule $\ce{B}$:
                    CALCULATED THERMODYNAMIC PROPERTIES
                                          *
   TEMP. (K)   PARTITION FUNCTION   H.O.F.    ENTHALPY  HEAT CAPACITY ENTROPY
                                   KCAL/MOL   CAL/MOLE    CAL/K/MOL  CAL/K/MOL

 298.00  VIB.     0.2155D+06                  8205.6617    53.3353    51.9400
         ROT.     0.3572D+07                   888.2813     2.9808    32.9651
         INT.     0.7697D+12                  9093.9430    56.3161    84.9051
         TRA.     0.4009D+28                  1480.4688     4.9680    42.5424
         TOT.                     -670.232   10574.4119    61.2841   127.4474

   *: NOTE: HEATS OF FORMATION ARE RELATIVE TO THE
            ELEMENTS IN THEIR STANDARD STATE AT 298K
              (=  Standard Enthalpy of Formation).

            Hvib:  Zero-point energy is not included.
                   frequencies of less than zero cm-1 are not included.
            Hrot = (3/2)RT
            Htra = (3/2)RT + pV = (5/2)RT

            Heat capacity is Cp, not Cv

          For more information, see: HTTP://OpenMOPAC.net/Manual/thermochemistry.html

With that data table $\Delta G^\circ_\mathrm{f}$ of the molecule $\ce{B}$ should be:
$$\Delta G^\circ_\mathrm{f}= \mathtt{H.O.F.} - TS^\circ= -670.232\times1000 - 298\times(127.4474- S_{atoms})$$
$S_{atoms}$ will be canceled when calculating $\Delta G^\circ_r$.
On this mopac site, $\Delta H_\mathrm{f}^\circ$ appears to be obtained in a similar way, for the $\ce{NH4+}$ molecule.
Am I correct?

I have no experience with MOPAC, and I don't completely understand the output, but your method looks correct to me.

great, @Martin-マーチン, and thanks for the comment. For some reasons I don't understand most of people working on it don't use this approach..

Semi-empiric calculations are much too inaccurate for many systems and many people use other methods instead. DFT and other ab initio methods allow to calculate an absolute value for the electronic energy and thermodynamic corrections and using those to calculate the relative energies. (The linked manual section is very worth reading, there is more information on how thermochemistry is treated in MOPAC.)

@Martin-マーチン I mean that they use mopac but calculate gibbs free energy in a different way. I didn't specify how because it is long..; this methods aren't too accurate but are parametrized to work good, so, they are quite accurate for most purposes (and also are fast)..

$\Delta G_f^\circ$, like, $\Delta H_f^\circ$, is relative to elements in their standard state, so the entropy is missing a term, but that will also cancel when you compute $\Delta G_r^\circ$.  So, for 298K, the equation for $\Delta G_f^\circ$ will give the right $\Delta G_r^\circ$.  For other temps you also need $(H_T - 10574.4119)$ where $H_T$ is the corresponding total $H$ for temperature $T$.

@JanJensen right..so gibbs free energy of reaction can be calculated directly with $\Delta G^\circ_f$ of the species in the reaction and don't need to be relative to gas phase species?(I thought you were calculating by this approach)

The free energy change of a reaction occurring in solution can be computed using a thermodynamic cycle:
$$\Delta G^\circ_{r,soln} = \Delta G^\circ_{r,gas} + \Delta\Delta G^\circ_{r,solv} $$
$\Delta G^\circ_{r,soln}$ can also be written as
$$\Delta G^\circ_{r,soln} = G^\circ_{soln}(B)- G^\circ_{soln}(A)$$
where
$$G^\circ_{soln}(X) = G^\circ_{gas}(X) + \Delta G^\circ_{solv} $$
and
$$G^\circ_{gas}(X) = E_{gas}(X) + G^\circ_{gas,RRHO}(X)$$
The solvation energy is usually computed using a continuum as the difference in electronic energy with and without the continuum.
$\Delta G^\circ_{solv}(X) = E^\prime_{soln}(X)-E^\prime_{gas}(X)$
The parameters in the continuum model are usually optimized for a particular level of theory, which may not be appropriate for the computation of $\Delta G^\circ_{r,gas}$, so $E_{gas}(X)$ is not necessarily the same as $E^\prime_{gas}(X)$
However, if $E_{gas}(X) = E^\prime_{gas}(X)$ then
$$G^\circ_{soln}(X) = E_{soln}(X) + G^\circ_{gas,RRHO}(X) $$
The parameters in the continuum model are usually determined using the gas phase geometry $(X)$.  But if $E_{gas}(X) = E^\prime_{gas}(X)$ then many people often choose to use the geometry optimized in solution
$G^\circ_{soln}(X_{soln}) = E_{soln}(X_{soln}) + G^\circ_{soln,RRHO}(X_{soln})$.
For semiempirical calculations, the electronic energy corresponds to the enthalpy of formation at 298K, so if one simply makes the substitution
$$E(X) \rightarrow \Delta H^\circ_{f,298}(X)$$
then (at 298K)
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) -TS^\circ_{gas,RRHO}(X)$$ $G^\circ_{soln}(X)$ doesn't correspond to a measurable free energy because $S^\circ_{gas,RRHO} \ne S^\circ_{f}$ but it can still be used to compute $\Delta G^\circ_{r,soln}$ since the missing terms cancel.
For temperatures other than 298K
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) + (H^\circ_{gas,RRHO}(X,T)-H^\circ_{gas,RRHO}(X,T=298K)) -TS^\circ_{gas,RRHO}(X)$$

I don't understand why the gas-solution cycle is needed, although I understand the procedure you wrote. But if mopac give  us "$G_f$
"for each reactant in solution, why don't we direcly sum those values?(with corresponding index for reactant and products) –

the cycle is needed because the continuum model has parameters that need to be fitted to measured values.  You can't measure $G^\circ_{soln}(X)$, but you can measure $\Delta G^\circ_{solv}(X)$. So you need the cycle to relate $\Delta G^\circ_{solv}(X)$. to $G^\circ_{soln}(X)$


text:
Why do n AOs only form n MOs?
The textbooks I have read introduce LCAO by considering the H2 molecule. In this example, there are only two possible combinations of the two 1s orbitals (in phase and out of phase).
When there are more than two valence orbitals, however, a greater amount of combinations arise. A common description of the pi-MOs in butadiene is shown below:
There are two (and I think only these two, assuming all combining p-orbitals are identical) possible combinations of the p-orbitals which (apparently) do not contribute to the bonding in butadiene. These are shown at the bottom of the illustration above.
Why is it that n AOs only form n MOs, even when more than n LCAO are possible? You would think that all possible combinations would contribute a possible electron energy state to the molecule.

I believe that is more about the number of nodes than the look of MOs. Moreover, symmetry plays not the last part in such distribution.

Yes, symmetry is very important (all MOs must transform according to the system’s irreps). Also, this is a conservation-type thing. But the MO guys will answer soon enough.

While symmetry is important, the question will be perfectly valid in a system without any symmetry at all.

related https://chemistry.stackexchange.com/questions/23578/why-do-two-atomic-orbitals-form-two-molecular-orbitals

If you have $n$ functions (e.g. AOs) you can make a maximum of $n$ new linearly independent functions (e.g. MOs).  If you try to make $n+1$ MOs, then any one of them can be expressed as a linear combination of the other $n$ MOs.
The most usual way to enforce linear independence is to enforce orthogonality, i.e. all your MOs have to have zero net overlap with each other. All the "correct" MOs you drew are orthogonal to each other, but the "incorrect" MOs are not orthogonal to any of the "correct" MOs.

Upvoted! MOs are required to be orthogonal, orthogonality implies linear independence, and one can not build more than $n$ linear independent functions (MOs) by making linear combinations of $n$ basis functions (AOs). The only problem is that I have no idea how to explain all this at the introductory MOT level. :|

You're correct in saying that p-orbitals are identical, and it follows that the following configurations are the same:

You must only consider their relative alignments (i.e. their symmetry).
To consider your question, approach it using the Hückel method. The wavefunction of a molecular orbital is given as a linear combination of the atomic orbitals, which mathematically looks like:
\begin{equation} \tag{1}
\Psi = c_1\phi_1 + c_2\phi_2 + c_3\phi_3 + c_4\phi_4
\end{equation}
where $c_i$ is the coefficient of $\phi_i$ in the LCAO-MO picture. Using the Hückel method we build the secular determinant for the molecular orbitals (see here if you're unfamiliar with this):
\begin{equation} \tag{2}
\begin{vmatrix}
x & 1 & 0 & 0 \\
1 & x & 1 & 0 \\
0 & 1 & x & 1 \\
0 & 0 & 1 & x
\end{vmatrix}
= 0
\end{equation}
where $x$ is defined as
\begin{equation} \tag{3}
x = \frac{\alpha - E}{\beta}
\end{equation}
The expansion of the secular determinant is trivial and gives the following roots:
\begin{equation} \tag{4}
x = \pm 1.62 \hspace{0.5cm} \text{or} \hspace{0.5cm} x = \pm 0.62
\end{equation}
Since the 4x4 secular determinant is made from 4 atomic orbitals, it is mathematically restricted to have 4 solutions. Using Equation 3 one may rearrange these solutions to find the orbital energies in terms of $\alpha$ and $\beta$. These roots are actually the eigenvalues of the secular matrix $ \ \textbf{H} - E\textbf{S} \ $ used in the Hückel method. The corresponding eigenfunctions are the coefficients of the atomic orbitals in Equation 1. The molecular orbitals for butadiene are found to be:
\begin{align} \tag{5}
  \Psi_a &= \phantom{-}0.372\phi_1 + 0.602\phi_2 + 0.602\phi_3 + 0.372\phi_4 &
  E &= \alpha + 1.618 \beta \\
  \Psi_b &= -0.602\phi_1 - 0.372\phi_2 + 0.372\phi_3 + 0.602\phi_4 &
  E &= \alpha + 0.618 \beta \\
  \Psi_c &= -0.602\phi_1 + 0.372\phi_2 + 0.372\phi_3 - 0.602\phi_4 &
  E &= \alpha - 0.618 \beta \\
  \Psi_d &= \phantom{-}0.372\phi_1 - 0.602\phi_2 + 0.602\phi_3 - 0.372\phi_4 &
  E &= \alpha - 1.618 \beta
\end{align}
If you look closely at the signs of each coefficient, you can see that they correspond to the phases in the allowed molecular orbital configurations. The origin of this behaviour follows from the discrete nature of quantum mechanics. If we call the coefficients amplitudes of a sine wave fitted to the length of the molecule this is more clear:

There is no solution to our secular determinant which allows for orbital symmetry in the way which you queried. This is explained by the picture formed by the sine waves, along with the particle in a box model.
References:

P. Atkins & R. Friedman, Molecular Quantum Mechanics, Oxford University Press, Oxford, 5th edn., 2011.
https://chem.libretexts.org/


text:
Other liquids for collecting gas by displacement method for SO2
I am looking for a low cost method of quantifying gases evolved from a reaction. The gas mixture is expected to have $\ce{H2}$, $\ce{SO2}$ and $\ce{CO2}$ but likely others too. I am considering using a syringe but am somewhat apprehensive about syringe friction and back pressure distorting volume. Commonly used water displacement method obviously would not work due to soluble and reactive gases in the mix. Is there an inert replacement for water for this method to work with $\ce{SO2}$?
I thought about allowing the $\ce{SO2}$ to react in the water and then calculating the amount through pH change. However, $\ce{CO2}$ is also soluble (and others too perhaps)
Without access to chromatographic or other elemental analysis equipment, is there a primitive way of just measuring the volume of this gas mixture?

I guess you could you easily use many typical solvents instead of water

Thanks @Mithoron. Is it reasonable to assume that the likes of toluene or acetone will not react with carbon dioxide uncatalysed and at atmospheric conditions? Same for SO2?

Non-volatile, non-polar solvent would be best, I guess.


text:
How to pronounce G°?
To denote standard state like enthalpy of formation in standard state, or change in Gibbs free energy in standard state, we use a degree-like symbol above the $G$ as in $\Delta G^0$. How should I refer to this superscript in spoken language?

That's not a degree at all, but a zero.

Then how should i pronounce it ? "g zero" ?

What about G nought (or G null or G sub zero) and G prime?

I heard people pronounce like that "G nought" and "G prime" . But I am not sure these are the right ones.

There is no authority-on-high for English language usage. I suspect that "G nought" and "G prime" were understood (or are perfectly understandable) by one and all, so are quite OK to use. Nobody will issue you a language ticket...

G-zero and G-prime.
G-zero has also in my presence been referred to as "G-ought", "G-nought", "standard-G" or "G at standard state". I am particular towards "G-zero" myself. (Mostly because of a witty limerick I heard during my studies. Zero rhymes with hero.)


text:
How to differentiate magnetite from metallic iron?
Both metallic iron and magnetite, $\ce{Fe3O4}$, are magnetic. Now, iron is grey in color, whereas hematite is black, so that they can be distinguished visually.
Now, I have a powdery mixture of iron/magnetite and lighter grains so that the mixture appears grey, making visual differentiating a bit tricky. I assumed it was iron, but someone else claimed it could also be magnetite, and now I'm not sure.
Is there a really easy way to be sure what I am dealing with? I don't want to do an XRD just for this.

Hematite is red.

oh my goodness, I meant magnetite! changing it now

Do the XRD. Unless you have to pay for it or something.

These are the methods that come to mind:

LIBS
XRF
Streak color. When scratching on unglazed porcelain, magnetite leaves charcoal-black traces, whereas iron leaves shiny steel-gray streaks. Having a powdery mixture this test can be tricky, unless there are reasonably big shards left. Alternatively, one can try to assemble a bulky piece by gluing the powder together with silicate adhesive or cyanoacrylate (as paracetamol suggested in the comments, "rubbing some of the mysterious powder between two shards of porcelain" can also do).
Compare density of your sample with $\rho(\ce{Fe}) = \pu{7.874 g cm-3}$, $\rho(\ce{Fe3O4}) = \pu{4.9 .. 5.2 g cm-3}$ and, assuming this mechanical mixture consists of iron and magnetite exclusively, calculate the percentage of each phase.

Magnetite in its mineral form often contains $\ce{(Fe,Mg)O}$, $\ce{TiO2}$ and isomorphous impurities of $\ce{Mn}$, rarely $\ce{Cr}$, $\ce{V}$; therefore physical methods of analysis (1, 2) are more preferable. Due to the same reason I'm not suggesting treatment with $\ce{HCl}$ and other methods based on chemical transformations as primary for this analysis.

"Alternatively, one can try to assemble a bulky piece by gluing the powder together with silicate adhesive or cyanoacrylate." That's a very novel method :D

@paracetamol Well, when means are limited, one must improvise. Colorless adhesive that doesn't change composition of phases shouldn't influence the outcome, and if a single phase is dominating, then it will become clear quite fast. But again, it depends on the grain size -- maybe this gluing thing isn't necessary.

How about rubbing some of the mysterious powder between two shards of porcelain? A little messier, maybe...


text:
Solutions of Group 1 and Group 2 metals in Ammonia
When we add Group-1 and Group-2 metals to liquid ammonia, they dissolve to form metal cations and solvated electrons.
$$\ce{M + NH3(liq) -> M+ + e-}$$
Now, when the G-1 solutions evaporate, we get the metal back, but in the case of G-2 solutions (except Beryllium), they give a metal complex, $\ce{[M(NH3)6]}$.

"Evaporation of the ammonia from solutions of Group 1 metals yields the metal, but with Group 2 metals evaporation of ammonia gives hexammoniates $\ce{[M(NH3)6]}$ of the metals. These slowly decompose to give amides."
Page No. 335, Chapter 11-'Group 2 - the alkaline earth elements', Concise Inorganic Chemistry

and,

"The alkaline earth metals except beryllium form similar solutions [to alkali metals] readily, but upon evaporation, a solid 'ammoniate', $\ce{[M(NH3)x]}$ is formed."
Page No. 249, Chapter 9-'Chemistry in Aqueous and Nonaqueous Solvents', Inorganic Chemistry: Principles of Structure and Reactivity

What is the reason for this contrasting behaviour, and do all Group 2 metals(except beryllium) exhibit such a property?
Source:
1. Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998
2. Huheey, James E., Keiter, Ellen A., Keiter, Richard L.; Inorganic Chemistry: Principles of Structure and Reactivity, 4th Edition; Pearson Education Inc.; 1993

related https://chemistry.stackexchange.com/questions/41669/what-is-the-reason-for-the-blue-color-of-an-ammoniated-electron https://chemistry.stackexchange.com/questions/80431/how-to-think-of-solvated-electrons

As per P. Bahadur's Objective Chemistry 1st Year Program for IIT-JEE and All Other Engineering Entrance Examinations, under the chapter "s-Block Elements - The Alkaline Earth Metals", under Solubility in Liquid Ammonia, it reads:

Like Alkali Metals, alkaline earth metals also dissolve in liquid ammonia and form coloured solutions. When such a solution is evaporated, hexa-ammoniate, $\ce{M(NH3)6}$ is formed, which slowly decomposes to amides.
  $$\ce{M(NH3)6 -> M(NH2)2 + 4NH3 +H2}$$

(I noticed that it is almost exactly the same as JD Lee's Concise Inorganic Chemistry text.)
So far, it doesn't specify whether this metal $\ce{M}$ is only Beryllium or is the representation of the other alkaline earth metals as well... Since this is the case, I can safely assume that this applies to all the alkaline earth metals, and not just beryllium.
Another point mentioned in the book by Bahadur, under the chapter "s-Block Elements - The Alkali Metals", under Complex Ion Formation is that the Group 1 metals, i.e., the alkali earth metals have a larger size, low nuclear charge and hence don't tend to form complexes too easily. On the other hand, the Alkaline Earth metals are smaller, have higher nuclear charge, and hence, have a greater tendency to accept electrons, forming complexes. In Bahadur's Book, under "s-Block Elements - The Alkaline Earth Metals", under Formation of Complexes, he writes that:

However, $\ce{Be^{2+}}$ on account of smaller size forms many complexes such as $\ce{BeF3-}$, $\ce{BeF4^{2-}}$ and $\ce{Be(H2O)4^{2+}}$. Of the others, only $\ce{Mg}$ and $\ce{Ca}$ show much tendency to form complexes in solution and these are usually with oxygen donor ligands.

$\ce{NH3}$ is a ligand, and hence acts as a powerful complexing agent. This is why the $\ce{NH3}$ coordinates with these metals, forming hexa-ammoniates.
Source:

Bahadur, P.; Objective Chemistry 1st Year Programme for IIT-JEE & All Other Engineering Entrance Examinations; GRB Books; 2016
Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998

To add to this Eu and Yb, which form reasonably stable 2+ ions, form blue solutions in ammonia which according to https://doi.org/10.1016/0022-5088(69)90184-2 decompose to the amide and imide. So while I can't quickly find a reference to the hexa-ammoniates it is consistent with the above as Eu and Yb are smaller than Ba, and Eu2+ and Yb2+ are smaller than Ba2+. Other lanthanides dissolve in NH3 to form blu solutions, but after a full 5 minutes searching I can't find decomposition products for Sm in particular.

@abhigyan According to your answer, the group 1 metals won't form complexes due to their large size and lower charge density. But, the second book(Principles of…) explicitly states that beryllium won't form  the complex with ammonia, which seems a bit contradictory to me.

@Ayushmaan I agree with your point, but sadly, I don't have any way of providing you an answer for this. I've looked through just about any textbook I could find, but most of them are copied off either from Concise Inorganic Chemistry or off Inorganic Chemistry: Principles of Structure and Reactivity. P. Bahadur's book is a little different, but as per that book, beryllium forms various complexes. Therefore, I'm truly unaware... And I'm really sorry

@Ayushmaan I found the answer to your question regarding whether beryllium forms complexes or not. It turns out, beryllium is able to form complexes of various types, and has a strong tendency to do so. However, it is unable to form the "hexa" ammoniate because it's maximum coordination number is 4, and can't be extended to 6. What I wrote about the alkali metals (group 1) still stands the same.

I can't answer your two questions. As to your first question, which if I'm understanding it, is "Why are Group I and Group II metals different?" - I don't understand it. Why are red rubber balls and blue rubber balls different? Why are giraffes and crocodiles different? We actually can get into the details of what their differences are, but that's so obvious as to suggest that that is not what you have attempted to articulate. As a broad rule, Science isn't very good at answering "Why" questions, it is much better at answering "How" questions. I suspect you know why I and II elements are different. While I am not a big fan of the Periodic Table, we can see that I elements have an s electron, and II elements have two (full sub-shell) s electrons (in their valence shells). Would you predict, knowing just that, that they'd behave the same? I sure wouldn't.
As to your second question, (which I also can't really answer), well either Huheey, et al are wrong or not. I couldn't find much on-line about it, especially for Ra and Fr. What little I did find was decades out of date and itself admitted that the chemistry wasn't (then) well understood. (Cotton & Wilkinson, Adv. Inorg. Chem. 1972(!!)) From that is claimed that A. The II ammines are only "moderately stable". B. Dilute solutions of I & II metals in NH3 are blue, more concentrated solutions (e.g. 3M) are yellow/gold. This and other evidence suggests that in more concentrated solutions metal clusters rather than lone M+ ions exist. C. M+n + ne- is in equilibrium with  M(NH2)n + (n/2)H2 and D. the M° coordination compounds are non-stoichiometric and only approach M(NH3)6.
I re-emphasize that this info is 45 years old and I know that a lot of progress has been made since then, especially in the area of M° coordination cmpds.
As an aside, I was tempted to "answer" your question by pointing out that everything has an opportunity cost. That is, the "reason" why the group I metals didn't form M° hexammines would be either because either the metal and (liquid? gaseous?) ammonia has a more negative free energy OR the amide is somehow kinetically retarded, contradicting the claimed equilibrium OR because the amide has a more negative free energy - compared to the M° complex's crystal structure. Opportunity cost. But since that is obvious (I hope!) I didn't think it would be helpful. (It's a tautology)

G-2 metals are smaller in size, so more charge density and thus have more (deltaO) value, so they form more stable complexes compared to G-1.
G-1 metals complexes are uncommon because they are so elctropositive and can rarely act as lewis acids. Although there are certain cage like complexes formed by G-1 metals. (Refer JD Lee Chapter-Alkaline metals).
 Berillium is very small so it cannot accomodate 6 ammonia molecules around it. Also on a CFT point of view larger metals have larger (deltaO) value so they are more stable.
(You can refer to Huheey Keter , chapter- Coordination Compounds on what are the factors that effect (deltaO) values.)


text:
What are SN1ʹ and SN2ʹ reactions?
I went through the problems 119 and 120 from chapter 4 Reaction mechanism of Singh's Conceptual problems in organic chemistry [1, p. 190] targeting nucleophilic substitution reactions for 3-bromo-3-methylcyclohex-1-ene:

The answer of the problem 119 is $\mathrm{S_N1'}$, and that of 120 is $\mathrm{S_N2'}$. I actually can't figure out what $\mathrm{S_N1'}$ ($\mathrm{S_N1}$ prime) and $\mathrm{S_N2'}$ ($\mathrm{S_N2}$ prime) are.
Also, how does the answer change by changing the nucleophile?
Reference

Singh, D. K. Conceptual Problems in Organic Chemistry: For Engineering and Medical Entrance Examinations, 3rd ed.; Pearson India, 2013. ISBN 978-93-325-8207-1.

The ' indicates that the actual substitution site is not where the leaving group is attached. In the cases presented, the nucleophile would add at the upper end of the double bond.

The primed version of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ can only occur if there is a double bond in the vicinity of the leaving group as in your example. I have drawn both possibilities for the $\mathrm{S_N2}$ case in the scheme below.

Scheme 1: Comparison of the reaction products in an $\mathrm{S_N2}$ (top) and $\mathrm{S_N2'}$ (bottom) pathway.
The $\mathrm{S_N2}$ reaction is as you expect it to be. The $\mathrm{S_N2'}$ reaction uses the double bond as an electron relay system. Instead of the nucleophile (here: $\ce{OMe-}$) directly interacting with the $\unicode{x3c3}^*(\ce{C-Br})$ orbital, the π system interacts with the σ* orbital. The nucleophile then interacts with the π system’s LUMO (corresponding to the middle orbital of the allyl π system) to perform the attack. Since the nucleophilic attack and the leaving group are on different carbon atoms, relayed by the π system, this is not a direct $\mathrm{S_N2}$ but a derivative of it ($\mathrm{S_N2'}$).
The same logic can be applied to $\mathrm{S_N1'}$: here, the intermediate carbocation is not a localised one but an allyl cation.
The primed versions here are observed because not only the bimolecular attack on the highly substituted tertiary carbon ($\mathrm{S_N2}$) is unlikely — this carbon is tertiary which usually already implies no $\mathrm{S_N2}$ — but also the capturing of the carbocation under $\mathrm{S_N1}$ conditions is more likely to happen at the sterically much less hindered carbon atom.
As to why methanol predominantly attacks according to an $\mathrm{S_N1}$-type mechanism while methanolate predominantly follows an $\mathrm{S_N2}$-type mechanism I will refer you back to your textbook; the reasons are stated pretty often.


text:
Grams of bicarbonate vs realeased CO2?
So, it's been a good two years since I've done chemistry and I have my first university lab in a few days and I'm sort of freaking out because I can't even remember how to do stoichiometry properly. This a question off of my pre-lab review and I'm having troubles remembering what to do. I believe this is a limiting reagent question.

In Experiment 1, you will place and AlkaSeltzer tablet in each of the dilute $\ce{HCl}$ solutions you prepared. The reaction between $\ce{HCl}$ and $\ce{NaHCO3}$ should take place quickly and release gaseous $\ce{CO2}$. You will then use the mass lost for each flask to determine the amount of $\ce{NaHCO3}$ that reacted.
A student performs this experiment and determines that $\pu{1.60 g}$ of $\ce{CO2}$ were lost. How many grams of $\ce{NaHCO3}$ would have reacted to produce this amount of $\ce{CO2}$?

Firstly, I wrote out part of the equation but I'm also beginning to think it's incorrect too:
$$\ce{NaHCO3(aq) + HCl(aq) <=> CO2(g) + NaCl + H2O(l)}$$
The initial amount for $\ce{NaHCO3}$, I know, is unknown and the final amount for $\ce{CO2}$ is $\pu{1.60 g}$. Other than that though I'm at a loss for what to do next. Please help if you can, thank you!
Edit (see answer): To calculate the amount of moles in $\ce{CO2(g)}$ I used the formula $m/M = n$, which gave me 0.36355 moles for carbon dioxide. Now I use 0.36355 moles for $\ce{NaHCO3(aq)}$ since it's a 1:1 ratio.

For starters, carbon dioxide is $\ce{CO2}$ not $\ce{CO3}$. The products of the reaction are sodium chloride (which you have), carbon dioxide (not quite right in your equation), and water (we don't make hydrogen). I would start there, and try to fix your balanced equation.

x grams of CO2 are so-many moles of CO2, the same amount of moles of carbonate are on the other side of the equation, ...

So, now you apply the stoichiometry from your balanced reaction exactly like Karl says: $\mathrm{mol}\ \ce{NaHCO3} \propto \mathrm{mol}\ \ce{CO2}$.

Thank you for all the help, I've figured out the answer:
$$\ce{NaHCO3(aq) + HCl(aq) -> NaCl(aq) + CO2(g) + H2O(l)}$$
Since we're given $\pu{1.60 g}$ of $\ce{CO2(g)}$ we can use the mass to calculate the number of moles with the equation:
$$n = \frac{m}{M} = \frac{\pu{1.60 g}}{\pu{44.01 g mol-1}} = \pu{0.036355 mol}$$
Since it's a 1 : 1 ratio between $\ce{NaHCO3(aq)}$ and $\ce{CO2(g)}$, we can directly solve for the mass of $\ce{NaHCO3(aq)}$ using the same formula:
$$n = \frac{m}{M} $$
$$m = n \times M = \pu{0.036355 mol} \times \pu{83.42 g mol-1} = \pu{3.0509116 g}$$
Round to two decimal places: $\pu{3.05 g}$

Great work! Keep in mind you can accept your own answer. As an advice, it's generally better to round up to significant digits at once.


text:
Does the dipole moment increase or decrease by increasing the bond length?
It is established the dipole moment is a result of multiplication of the magnitude of charges (Q) and the distance between them (r). What I understand is that when an electron and a proton get closer (bond length decreases), polarity decreases hence the dipole decreases. When they move further away from each other(bond length increases), the polarity increases and hence the dipole moment increases. However, in case of halides (H-X), let's take for example HF and HI, HF has a bigger dipole moment than HI, it is stated that the bond length of HF is smaller than HI. Why do I find this contradictory?
Can you help me with What I am missing here?

First part of your first statement?

If you relate and compare two properties  (here bond lenght and dipole moment) quantitatively, you have to look if other relevant properties (here the effective charge!) are changing. Which they are.

related https://chemistry.stackexchange.com/questions/6780/why-do-c-f-bond-has-less-dipole-moment-than-c-cl-bond

Strictly speaking, you can neither decrease nor increase the bond length a slightest bit.$\dagger$ All molecules of a given compound are equal. It is not like you may grab one with tiny pincers of some sort, stretch it a little and see what happens.
You may, however, consider a molecule of a different compound; indeed, that's what you do in your question. But that's not just a different bond length; it is different atoms with different properties, and hence different everything.
Most importantly for the question at hand, iodine has much lower electronegativity than fluorine, so the effective charges on $\rm H$ and $\rm I$ in $\ce{HI}$ will be much lower than those on $\rm H$ and $\rm F$ in $\ce{HF}$. It just turns out that the increase in bond length can't compensate for this effect, hence the lower dipole moment.

$\dagger$ For those who might want to point out possible vibrational excitation or some other subtle (or not-so-subtle) effect: don't. Let's keep it simple.

Point taken for no vibrations, but what about going from e.g. $\ce{CH3-NH2}$ to $\ce{CH2=NH}$?

@Jan Just like I said, these are different molecules with different everything.

If the same point charges are separated by a larger distance, that's a higher dipole moment.
In the real world, you have to regard the distance and the actual charge difference and distribution.
In the case of HF and HI, fluorine has the highest electronegativity, which means the bond in HF is more polarised than in HI. So, in spite of the lower bond lenght, HF can have, and indeed has, a higher dipole moment.

but why distance isnt taking into consideration ?

@adosar please explain what you mean by "consideration". I don't get it.

Why hf has greater dipole moment if hcl bond length is bigger ? How can we know that q1d1>q2d2 only from the electronegativity difference?

@adosar I explain exactly that above. It's a fact, we know the distance and the dipole moment from different measurements, so we can calculate the charge difference! And we know about the high electronegativity of fluorine, so the result makes sense. That's nature, I can't change it.


text:
Platinum Orbitals in Cisplatin Forming Bonds with LCAOs of Chlorine
My inorganic chemistry professor placed two questions on the exam review that he did not cover in class; I haven't been able to figure out how to answer them and could really use some help in learning how to think about this problem.
The two questions are:

Which orbitals of $\ce{Pt}$ can form a $\sigma$ bond with the LCAO $\left[\Psi((\ce{Cl}')_{\mathrm p_z})+\Psi((\ce{Cl}'')_{\mathrm p_z})\right]$ of the two $\ce{Cl}$ atoms in cisplatin?

Which $\mathrm d$ orbital of $\ce{Pt}$ can form a $\pi$ bond with the LCAO $\left[\Psi((\mathrm{Cl}')_{\mathrm p_z})-\Psi((\mathrm{Cl}'')_{\mathrm p_z})\right]$ of the $\mathrm p_{z}$ orbitals of the two $\ce{Cl}$ atoms in cisplatin?

How do I arrive at the corresponding oritals?

You should make it a little clearer what part of the question is causing you trouble. Is it the LCAO, is it determining pi or sigma bonds, what cis-platin is etc?

Hi Tyberius! I know what cis-platin is, I'm just unsure how to determine which orbitals are in-phase with the chlorine LCAO.

got you. Best of luck in getting an answer. As a new user I would recommend taking the [tour] and visiting the [help]. One thing useful thing to know is usually comments you send to a person won't notify them unless you put the "at" symbol in front of their username. This will notify that user, assuming they have already commented on the post.

I figured it out!
We have to start by looking at d orbital shapes:

To form bonds, the d orbitals of platinum must have the same symmetry (i.e. shape) as the molecular orbital that's approximated by the LCAO.
We start with the bonding orbital, which looks like this: (sorry about the pencil-and-paper phone picture).

We can see we need a d orbital with exactly one lobe in the zx plane (x is pointing to the right in the above). The only d orbital that fits the requirement is the $\textrm{dz}^{2}$.
Moving on  to the anti-bonding orbital, which I can't post because I don't have at least 10 reputation, we see that it has four lobes with two nodal planes. It looks like a $\textrm{d}_{xz}$ orbital, in fact.
We can see we need a d orbital in the zx plane that has two lobes. Clearly, this is the $\textrm{d}_{xz}$ orbital.
Easier than I thought once I got the concepts down!

You need to take another look at the orbitals you are using and the coordinate system. $\mathrm d_{z^2}$ is nonbonding with respect to both $\Psi$ of the question. You also need to define which coordinate axes the chlorine atoms are on before you can ascertain that $\mathrm d_{xz}$ is bonding with respect to $\Psi_2$

It is important to first take a look at your coordinate system. Cisplatin is a square planar complex. In these types of complexes the principal axis ($z$ axis) is defined as the one out of the ligand plane. Without loss of generality I will assume that one chlorine is in positive $x$ direction and the other in positive $y$ direction. Since we are dealing with a complex whose shape derives from the octahedron, ligands will approach along the coordinate axes, not between them.
Next you should consider which type of symmetry your chlorine group orbitals have. As I already mentioned, the $z$ axis is perpendicular to both $\ce{Pt-Cl}$ bonds. Therefore, the $\mathrm p_z$ orbitals of chlorine will have a nodal plane that completely contains the bond axes.
This immediately answers our question 1: there is no orbital on platinum that can form a σ bond with the chlorine group orbital because σ bonds cannot have a nodal plane along the bond axis.
In an extension to the question, I want to consider which platinum orbitals can form bonds of any kind with this $\Psi_1$.
In the positive part of the coordinate system, we can only have a single nodal plane in the $xy$ plane. This points us towards the $\mathrm d_{xz}$ and $\mathrm d_{yz}$ orbitals — but each of these would only be π symmetric towards one chlorine atom and δ-symmetric towards the other which means effectively nonbonding. We can, however, form a linear combination of the two $\Phi_1 = \phi(\mathrm d_{xz}) + \phi(\mathrm d_{yz})$ which would point between the $x$ and $y$ coordinate axes and be able to build up a π-symmetric bond.
Of course, the $\mathrm p_z$ orbital of platinum is another trivial (and much more likely) answer. These two are shown in figure 1.

Figure 1: potential bonding orbitals towards the first chorine group orbital $\Psi_1$. Top: linear combination of two platinum $\mathrm d$ orbitals. Bottom: platinum $\mathrm p$ orbital.
Now let’s move on to $\Psi_2$. This not only has the (π-symmetry invoking) nodal plane containing the bond axes that the previous group orbital had; it also has a second nodal plane through the plane of symmetry, bisecting the platinum atom. This may initially seem like an ideal case for a $\mathrm d$ orbital but alas, the $\mathrm d$ orbitals are pointing in the wrong direction. We need to form the opposite linear combination of above, $\Phi_2 = \phi(\mathrm d_{xz}) - \phi(\mathrm d_{yz})$ which would have the correct orientation as shown in figure 2 below.
Using only uncombined platinum orbitals results in $\Psi_2$ being overall nonbonding again.

Figure 2: linear combined platinum orbital capable of interacting with the second chlorine group orbital.

There are no such things as non-bonding orbitals as there is no intermediate between having a nodal plane perpendicular to the bonding axis or not.

@Martin-マーチン I took nonbonding to mean symmetry-forbidden interaction in this answer.


text:
How do the toxicities of arsenate and arsenite compare?
So I was browsing the detoxification metabolistic mechanisms for E. coli, and came across an arsenic detoxification mechanism that converted arsenate to arsenite using a glutaredoxin. So my question is, how does the toxicity of arsenate compare to arsenite? And what is the use of this pathway in E. coli?

that's weird because I found a source that says arsenite is more toxic than arsenate. (ncbi.nlm.nih.gov/pmc/articles/PMC3183630/ see introduction). That might be for humans though.

@S.Chevalier I think that it can go both ways, but your point is interesting

I see now.
According to the same website with the reaction mechanism,

Although the arsenite formed is more toxic than arsenate , it can be extruded from some bacteria by EC 3.6.3.16 , arsenite-transporting ATPase; in other organisms, arsenite can be methylated by EC 2.1.1.137 , arsenite methyltransferase, in a pathway that produces non-toxic organoarsenical compounds. cf EC 1.20.4.4, arsenate reductase (thioredoxin) .


text:
How does 1-methylcyclohexene reacting with KMnO4 produce 1-methyl-1,2-cyclohexanediol?
I am interested in knowing how diols are produced when you introduce a reagent like potassium permanganate to an alkene. What is the mechanism for these reactions? The hydroxyl substituents happens to be at the 1,2 position each time.

There's a syn-addition of $\ce{-OH}$.

That's a particular case of syn-dihydroxylation, also known as Wagner reaction. "Syn-dihydroxylation" means that a double bond is cleaved into a single bond and two OH groups are attached to the carbon atoms that were involved into a double bond from the same side. Anti-addition is not possible as permanganate anion can't be attached to the double bond from different sides of the bond due to steric reasons.
There is a Khan Academy video  that describes the mechanism of this particular addition, but that video is all about different cases of syn-addition.
In brief, the mechanism can be also drawn as following:


text:
Is chlorine the central atom in NOCl?
We've been told that the in the structure of $\ce{NOCl}$ the central atom is $\ce{Cl}$ (rather than $\ce{N}$) due to some exception.
Looking up on internet I could not find anything to support this. I would like to know if it really exists in such a state due to some exception. If yes, what is that exception.

Related: https://chemistry.stackexchange.com/questions/53568/what-is-the-structural-formula-of-alkali-hypohalite-mox-or-mxo

@Tyberius I don’t quite see how the structure of an inorganic molecule has to do with that of hypochlorite …

Wikipedia gives a structure with a central nitrogen atom:

The nitrogen atom (blue, using the standard color scheme) can form all three covalent bonds without building up an excess positive or negative charge, and so is favored in the central position through which all three bonds pass.

The only situation where I can imagine $\ce{N-Cl-O}$ bond is N-perchloryls ($\ce{R2NClO3}$); from [1]:

The perchlorylamines were obtained as carbon tetrachloride solutions, with no impurities detectable by nmr. Thus piperidine, diethylamine, and dipentylamine gave N-perchlorylpiperidine, N-perchloryldiethylamine, and N-perchloryldipentylamine, respectively. [...]
$$\ce{2R2NH + Cl2O7 ->[CCl4] R2\color{red}{NClO3} + R2NH2+ClO4-}$$

There are only few coordination compounds with amidoperchlorate group, e.g. $\ce{KH\color{red}{NClO3}}$ and catena(aqua-bis($\mu_2$-(N-cyclohexylamidoperchlorate)-silver(I))) $\ce{AgC6H11\color{red}{NClO3} * 0.5 H2O}$ [2]; crystal structure of the latter is shown below (hydrogen atoms are omitted, numbers show interatomic distances in Å):
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#C0C0C0}{\Large\bullet}~\ce{Ag}$.

References

Beard, C. D.; Baum, K. J. Am. Chem. Soc. 1974, 96 (10), 3237–3239 DOI: 10.1021/ja00817a034.
Pritzkow, H. Zeitschrift für Naturforschung B 1976, 31 (9), 1170–1174 DOI: 10.1515/znb-1976-0903. (In German, free access)


text:
Find NMR equivalent atoms
I'm trying to find a program/algorithm/API that can find the equivalent atoms in a molecule when given a molecule (knowing the complete geometry with x, y and z coordinates, and bonds between the atoms). Equivalent in the sense that they show up as one peak in the NMR spectrum.
Example: I supply ethanol and the program needs to return that the three H's on one C are equivalent and that the two H's on the other C are also equivalent. So that my program knows that it should only have 3 (sets of) peaks in my NMR spectrum and not 6.

Without 3D structure, this will be a heuristic at best. Also, equivalency is a much stronger condition that show up as a single peak.There is also chemical equivalence and magnetic equivalence. What are your actually asking for?.

As Zhe said. The two hydrogens on the oxy-carbon are actually enantiotopic, not homotopic, so in a chiral environment they will give two peaks.

@Zhe I do have the complete geometry of the molecule with all the x,y and z coordinates of the nuclei plus the bonds that are present (and if they are single, double or triple).
I am personally more interested in the magnetic equivalence since this creates a different spectrum than just chemical equivalence, but if an algorithm for magnetic equivalence is too expensive then I may settle for an algorithm that just calculates the chemical equivalence (if it is cheaper).

Some drawing programs come with a heuristic prediction feature for NMR. Try ChemDraw and /or ChemSketch.

For chemical equivalence, you just need to check the environment of the atom, i.e., the spatial arrangement of all other atoms, and compare that to the environment of the other atom. In principle, chirality matters, though in general, NMR is achiral. The big caveat is watching out for conformational changes that are fast on the time scale of NMR. That would lead to an averaging that could merge some peaks.