« first day (1633 days earlier)      last day (2759 days later) » 
01:00 - 09:0009:00 - 20:00

1:21 AM
@pH13-YetanotherPhilipp credits?
Wait, you drew them?
 
 
1 hour later…
user228700
2:41 AM
Hi everyone :-)
 
user228700
I've got a quick question; is Nitrogen capable of forming four bonds?
 
@Kaumudi hi
@Kaumudi yes, e.g. in ammonium ion (NH4+)
 
user228700
@DHMO Yes, alright. I was trying to draw the structure of the azide ion. Thanks :-) [And it seems like u're earlier than usual!]
 
(from Wikipedia)
@Kaumudi I just discovered this:
Azide ion contains 5*3+1 = 16 valence electrons
But it contains 3 atoms which needs 3 octets = 24 valence electrons
(24-16)/2 = 4 bonds need to be formed
 
user228700
@DHMO Hang on, what is that rule u used just now, to calculate the number of bonds to be formed?
 
2:46 AM
@Kaumudi I just made up this rule just now
Might be helpful
 
user228700
@DHMO Huh? What is it?
 
@Kaumudi I don't have a name for it lol
Just see how I calculated the number 4
I don't know how to describe the rule, so just give me another molecule
 
user228700
:-P Okay. #of bonds to be formed=[#of valence electrons needed for all atoms to complete octet- #of valence electrons available]/2
 
@Kaumudi yes
 
user228700
@DHMO Cool. I'll check if this holds up most of the time.
 
2:49 AM
and of course simple rules like this come with exceptions
this only holds for covalent compounds
 
user228700
@DHMO Yes, of course.
 
@Kaumudi and a dative bond counts as one bonds
using this rule, the number of bonds needed for the nitrate ion (NO3^-) is 4
 
user228700
Also, while trying to draw the structure of the azide ion, I discovered that the net charge on a molecule needn't be localized on a single atom; in the azide ion, the central Nitrogen actually has a positive charge, while the two terminal Nitrogen atoms have one negative charge each.
 
which even holds up with the bond order considering molecular orbitals
 
user228700
@DHMO Now I'm starting to wonder if u made it up just now :-P
 
2:51 AM
@Kaumudi yes, I made it up just now
I wouldn't be surprised if such a rule is already described on the internet
but at least I have never come across it
@Kaumudi how is that not localized?
 
user228700
@DHMO It's not localized on a single atom.
 
@Kaumudi it is indeed localized
on three atoms, alright
 
user228700
Yes, on three atoms. I didn't know that this also is a possibility. Anyway, back to my textbook...
 
I thought you mean they are delocalized...
 
Hello
 
2:57 AM
@MelanieShebel hi
 
user228700
@DHMO Ah, OK...
 
user228700
@MelanieShebel Hi :-)
 
user228700
@DHMO: For an element like mercury, whose atomic number is 80, how do we go about writing the electronic configuration in shells?(2,8,18,32 etc.)
 
@Kaumudi 2,8,18,32,18,2
 
user228700
We have 20 atoms to fill even after 32 and the total number of atoms that can be accommodated in the next shell is 50.
 
user228700
3:02 AM
@DHMO Yes, how did u write the last two?
 
@Kaumudi have you learnt orbitals?
 
user228700
@DHMO Yep.
 
@Kaumudi then use the orbitals
 
user228700
@DHMO Well, I do know how to write the electronic configuration in terms of orbitals. I was wondering how u wrote that 18, 2 at the end.
 
@Kaumudi write it in orbitals first
 
user228700
3:04 AM
@DHMO I did...now what?
 
@Kaumudi what is it?
 
user228700
Ugh, MathJax.
 
user228700
Hang on...
 
very beautiful $\LaTeX$
:32931561 now, could you sort the list according to the first number, then the type of orbital?
 
user228700
One sec...
 
user228700
3:11 AM
$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{10}$
 
user228700
There, that's better.
 
@Kaumudi now, could you sort the list according to the first number, then the type of orbital?
And do you know what the first number means?
 
user228700
Uhh, well, even w/o sorting, we see that 6 being the last orbital has only 2 electrons, so its valency would be 2...
 
correct
 
user228700
Now what..?
 
3:14 AM
Do you know what the first number means?
 
user228700
Yes, of course; it stands for the principal quantum number of the electron; it gives the shell number.
 
so how many electrons are in the fifth shell?
 
user228700
Ah, 18.
 
user228700
Okay, I see what u did.
 
user228700
Thank you! :-D
 
3:17 AM
you're welcome
 
user228700
4:08 AM
@DHMO: U there?
 
@Jan what is the issue with the current answer to your bountied question?
 
@Kaumudi yes, though I'm answering 2 main questions, so expect slower reply, but ask nevertheless.
 
user228700
@DHMO OK, no problem. Consider $NOCl$. The steric number of the central atom, which is Nitrogen, comes out to be 3, meaning that it has $sp^2$ hybridization.
 
@Kaumudi yes
@Kaumudi Finished answering the first question:
0
A: If sigma bond is stronger than pi, then why bond length for sigma is longer and less energy?

DHMOSome points to clarify: Sigma bonds is indeed stronger than pi bonds. Therefore, sigma bonds have a lower energy than pi bonds (although sigma bonds do indeed have a higher dissociation energy than pi bonds). Double bonds (usually) contain $1$ sigma bond and $1$ pi bond. Triple bonds (usually) ...

 
user228700
However, I don't understand how this works out when considering the excitations and everything.
 
4:11 AM
@Kaumudi could you elaborate on your doubt?
 
user228700
@DHMO Sure. Give me one second...
 
user228700
 
user228700
Correct?
 
@Kaumudi correct
 
user228700
OK, lemme write the excitations thing...
 
user228700
4:17 AM
 
user228700
Correct..?
 
wuh?
 
user228700
@ringo Hybridization..?
 
@Kaumudi do you know what hybridization means?
 
Are you confusing the concept of hybridization with electron transitions?
 
user228700
4:20 AM
@ringo No, this is just how I've been taught about it; to have $sp^{anything}$, we need the s-orbital to be half empty.
 
@Kaumudi How many sp2 orbitals are there?
@Kaumudi and no, that's not how that works
 
user228700
And it's not half-empty, so we promote one of the two electrons to one of the p-orbitals. And then, the s-orbitak can hybridize with any number of p-orbitals to form equivalent number of hybrid orbitals.
 
user228700
@DHMO Sigh. 3.
 
@Kaumudi nope.
Pretend there's no electrons
how many orbitals are there now? Which orbitals?
3 sp2 orbitals, and what?
now draw them, and don't fill the electrons yet
by "draw" i really mean just draw the boxes
 
user228700
Just the two rectangles? OK..?
 
4:24 AM
@Kaumudi Just draw them
By the way, I finished my second question:
0
A: Does double or triple bond have higher priority?

DHMOAccording to R-$3.1.1$ of IUPAC, Commission on Nomenclature of Organic Chemistry. A Guide to IUPAC Nomenclature of Organic Compounds (Recommendations 1993), 1993, Blackwell Scientific publications, Copyright 1993 IUPAC: The presence of both double and triple bonds is similarly denoted by e...

 
user228700
Okay, done. Now what..?
 
user228700
@DHMO (Y)
 
@Kaumudi now send here
 
user228700
 
@Kaumudi no, not s x 1 and p x 3
I asked you to draw the boxes for sp2 x 3 and p x 1
 
user228700
4:29 AM
 
nice, now you can fill electrons according to electron-filling rules
 
user228700
Can u show me how..?
 
Sigh I was just looking around on Math.SE and I saw 1 question with 7 answers
We really don't get that on here
I wonder if it's because there are more ways to approach a math problem than a chemistry problem
 
two rules: (electrons go to)
1. lower energy orbitals before higher energy orbitals
2. empty sub-orbitals before half-filled sub-orbitals
@ringo link?
 
15
Q: Proving that the sum of any two sides of the triangle is greater than the third side

AmritanshuI know it's easy to prove with the help of linear inequalities, but this time I want to prove it with the help of trigonometry. Is it possible? If yes, then how?

 
4:33 AM
I'm sure we had a problem with quite many answers here
but I have no idea where it is now
 
It was a pretty cool question, with a lot of cool answers
Yeah a few, like why is gold gold and why can you smell HCN but...
I don't know
 
@ringo mathematics and science are too different and two different subjects
 
Just checked Area 51, our answer ratio is .9 lower than theirs
 
@ringo because we have more things that we don't know than mathematics
We can't even solve the SE for helium
@Kaumudi are you here?
 
@Kaumudi if I can be of any assistance... when more than one bond is formed to one atom, these bonds always lie in p orbitals
 
user228700
4:38 AM
This is where I'm coming from:
 
user228700
 
Excited... ?
Oh well, hybridization isn't even real
 
user228700
And this is how it's taught here so I'm positive that it's not incorrect.
 
user228700
@ringo Yes, excited.
 
I wasn't taught that they were excited, I was taught that the orbitals energies changed when they were mixed with other orbitals
 
4:41 AM
@Kaumudi complete bullshit
 
I guess that may be referring to the same concept
When 1 s and 2 p orbitals mix, they for 3 hybrid orbitals of degenerate energy, called sp2 orbitals
 
user228700
@DHMO How are u so sure?! This is how it's taught here, so I'm sure that it's not incorrect!
 
@Kaumudi complete bullshit
 
user228700
>.<
 
Here... just look at this
 
4:47 AM
@ringo I don't think you want to reveal the answer...
 
ok
Just thought we could use it as a teaching tool
Maybe another molecule is in order then
 
@Kaumudi do you understand?
Can you do it or do you want the answer?
 
user228700
@DHMO No, I don't. I'm finding it very difficult to believe that that is "complete bullshit".
 
@Kaumudi $\Huge\mathrm{very\ difficult}$
 
lol
 
4:50 AM
@Kaumudi just follow my (our) method, lol
now could you fill in the 5 electrons in the 3 sp2 orbitals and the p orbital?
 
user228700
@DHMO No! I need to understand why my method is wrong! My textbook doesn't generally spouse bullshit.
 
Okay, how I learned it was:
 
@Kaumudi Because, you would get 3 electrons in the 3 sp2 orbitals and 2 electrons in the p orbital according to your method
and then I have no idea how to form a double bond form that
 
@JohnRennie welcome!
sorry,
 
user228700
@JohnRennie: Help help! :-O
 
4:53 AM
lol
 
I'm just looking back to see where this all starts ...
 
user228700
@DHMO This is how:
 
It's a mess...
 
@Kaumudi what's the problem anyway?
 
This?
15 mins ago, by Kaumudi
user image
 
user228700
4:54 AM
Yes.
 
see I wouldn't say complete bullshit, just maybe a little misworded
 
@Kaumudi how?
 
Ok, so what's the problem?
 
@ringo but, you see, it wouldn't be dramatic enough
 
lol
 
4:55 AM
@JohnRennie the problem is that I said paragraph 2 is complete bullshit and s/he disagrees.
 
Well in the presence of an external constraint on the geometry the spherical harmonics are no longer eigenfunctions of the Schrodinger equation.
 
user228700
Right, like u explained yesterday...
 
@ringo but, if I follow my method, I would end up with 5 electrons in the sp2 orbitals and no electron in the p orbital, so no pi bond also
 
instead we get eigenfunctions that reflect the triangular symmetry
 
LOL
 
4:57 AM
And those eigenfunctions are the $sp^2$ orbitals.
 
user228700
@JohnRennie Yup.
 
@JohnRennie Hybridization has long been disproven by q.m.
It's just a useful teaching tool
 
@ringo what?
there is often s-p mixing in q.m. ...
 
But we can write the $sp^2$ orbitals as a sum of the spherical harmonics:
$$ \psi_{sp^2} \approx a\psi_{2s} + b\psi_{2p_x} + c\psi_{2p_y}+ ...$$
For some constants $a,b,c,etc$
 
In actuality everything mixes. There is no $sp^2$ orbital in q.m.
Every atom is bonded to every other
 
5:00 AM
When we say the $sp^2$ are hybrid orbitals is just means they can be written as a weighted sum of the spherical harmonics.
 
@ringo In actuality you might see something like $s^{1.5}p^{2.3}$ orbitals, which still count.
 
user228700
What is happening?!
 
In reality atomic and molecular orbitals don't exist at all because electron correlation causes them to mix with each other.
It's just a question of how purist you want to be.
 
^
Let's just give up on chemistry
 
At pre-university level I don't think it does any harm to use the approximation that hybrid orbitals exist.
 
5:02 AM
@Kaumudi bullshit is happening
 
user228700
@DHMO That...doesn't even make sense.
 
@Kaumudi what doesn't?
 
@DHMO I don't think it's useful to use the term bullshit. It's just a level of approximation, and there are many levels of approximation.
 
@JohnRennie but to her it's just a bunch of bullshit
 
user228700
I am showing u how my method works, HANG ON!
 
5:03 AM
Simple approximations like hybrid orbitals have their uses.
 
@Kaumudi you hybridized the wrong p orbitals, lol
 
user228700
@DHMO To whom?
 
@Kaumudi to you who only cares about the sp2 hybridization, lol
let's go back to the sp2
 
user228700
@DHMO I am a her, not a him, fyi.
 
@Kaumudi are you trying to write down an expression for the $sp^2$ orbitals?
 
5:05 AM
@Kaumudi done
 
user228700
Wait! I will post a picture of what I'm trying to do!
 
@JohnRennie no, an orbital diagram
 
@JohnRennie For heaven's sake, she's just trying to figure out how NOCl works, in terms of hybridization
 
Wild trip this has been
Not everyday you get a visit from the god of physics.se either
Very nice to have you
 
@ringo every day
 
5:07 AM
No, I don't visit the Chem chat that often though I'm in the h bar most days.
 
😒
 
I started out as a physical chemist - solid state photochemistry ...
 
No kidding
 
@JohnRennie what are the other parts of your name?
 
5:09 AM
Highly Strung
 
LOL
I'm taking p chem next semester. I'm really excited, because I've been told it has a lot of math, which I like
I've only ever really known organic chemistry, so it will be fun to try something new
 
It was the math that pulled me into physical chemistry. In my final undergrad year I specialised in quantum chemisty.
 
I'm also taking pchem next semester. I'm looking forward to pchem II more.
 
user228700
 
I can recommend a career as a physical chemist as you get to do awesome experiments e.g. neutron diffraction and Rutherford backscattering.
 
5:12 AM
@Kaumudi I think the safer approach is to hybridize before excitation
 
@Kaumudi I would draw the $sp^3$ together since they're degenerate in energy
 
user228700
@DHMO I don't even know how to do this!
 
@ringo that ^
With tetrahedral symmetry you get four orbitals of the same energy
 
@ringo she's doing the excitation before the hybridization
 
So you end up with one electron in each orbital
 
user228700
5:14 AM
@JohnRennie: Did u read that paragraph from my textbook..?
 
@MelanieShebel loved your post about computational chemistry, btw
I fought them closing it:)
 
@Kaumudi Hang on, let me scroll up and have a look ...
 
I'm trying to clean it up, but I can't figure a way to make it less...
I KNOW it can be done, but I don't know how.
Thank you @ringo
 
I'm actually in a somewhat similar situation. My computational chemistry degree requires a project as well
 
5:16 AM
@Kaumudi Ah, OK, I think that first paragraph is being a bit misleading. The symmetry dictates the orbital structure. It isn't the case that you first excite an electron then as a result of this the $sp^2$ orbitals form.
 
 
^^^^
 
user228700
@JohnRennie OK..?
 
I'm REALLY trying to get an organic chemistry project done before I finish undergrad, but I'm really sick so wet chem is a no-go. I may never be able to do wet chem which scares me because one of the comments on my post said, "honestly though, chemistry is all about getting wet." ... "The future is interdisciplinary. You've got to have both."
 
user228700
@DHMO How did u end up with $sp^2$ hybridized orbitals that have two electrons each in two of the orbitals?
 
5:17 AM
I hope that's not the case.
 
John Rennie is a pchemist. Maybe he can verify
 
@Kaumudi refer to my electron filling rules
 
(by trade)
 
You have 2 $s$ electrons and 2 $p$ electrons, but as you bring in the three atoms and perturb the symmetry these mix to give you the four $sp^3$ orbitals.
 
user228700
@JohnRennie So no excitations involved?
 
5:18 AM
And the electrons occupy these new orbitals
@Kaumudi No
 
Do you have to do labwork as a physical chemist @JohnRennie
 
user228700
@JohnRennie Which electrons?!
 
45 mins ago, by DHMO
two rules: (electrons go to)
1. lower energy orbitals before higher energy orbitals
2. empty sub-orbitals before half-filled sub-orbitals
 
I made this. Probably dumb, but I think it's hilarious: twitter.com/MelanieShebel/status/783140219520425984
 
In all honesty I'm not quite sure what they do aside from math
 
5:19 AM
@Kaumudi all electrons
 
Lemme just upload the pic.
 
BAHAHA
 
user228700
@DHMO Dude, I know these rules! I don't understand which electrons!
 
@Kaumudi all electrons?
 
5:20 AM
@Kaumudi we generally assume the innermost electrons are little perturbed, so in carbon we ignore changes to the $1s$ orbitals (though they do indeed change) and only consider the $2s$ and $2p$.
You're probably thinking this is all a bit vague, but that's because it's an approximation.
 
let's just use MO, lol
 
The vagueness creeps in because we aren't using a fully rigorous approach.
 
user228700
@JohnRennie Right. Okay, this is not how we've been taught! We aren't taught about the steric number in school; we do this excitation business to figure out what type of hybridization is involved.
 
@Kaumudi and that's OK. All of education involves discovering that what you were taught before wasn't strictly true.
 
@Kaumudi you should determine the hybridization from the geometry instead
 
5:22 AM
But the approximations are used because they're a good way to learn. You can't start out learning Hartree-Foch-CI theory.
 
user228700
@JohnRennie Sigh. I guess. God, I am so mad at my high school!
 
@JohnRennie Are you familiar with MOT?
 
user228700
@DHMO Write the Lewis structure, figure out geometry from VSEPR, then figure out hybridization accordingly?
 
@Kaumudi Just think of it as a foundation
 
@Kaumudi yes. then you would have stacked a false theory (hybridization) on a false theory (VSEPR) on a false theory (Lewis structure)
 
5:24 AM
@DHMO MOT?
 
Molecular Orbital Theory
 
Ah, yes.
 
user228700
@DHMO Lol.
 
@JohnRennie Are you familiar with MOT?
 
@DHMO Yes
 
user228700
5:25 AM
OK, thanks, all three of u!
 
@JohnRennie how do I stack any orbitals together at all to create anything with D3h symmetry?
 
I can't remember what D3h is - that's the symmetry of an equilateral triangle isn't it?
 
@JohnRennie yes
or anything with tetrahedral symmetry?
or do I need to rely on degerate orbitals?
 
I'm not sure what you're asking. That sounds more like LCAO.
In MO we just calculate eigenfunctions for the whole molecule.
 
@JohnRennie yes, I'm asking about LCAO
@JohnRennie could you teach me?
 
5:29 AM
So you're asking how to construct $sp^2$ type orbitals from the spherical harmonics?
 
@JohnRennie not sure what you mean by sp2-type orbitals
 
Orbitals with a triangular symmetry
As in: $$ \psi_{sp^2} \approx a\psi_{2s} + b\psi_{2p_x} + c\psi_{2p_y}+ ...$$
 
@JohnRennie so we are still relying on degenerate orbitals?
 
Man, this chat is hoppin!
 
Well LCAO involves contructing approximate bonding orbitals by combining atomic orbitals. And the atomic orbitals are approximately the hydrogenic orbitals.
 
5:34 AM
You know what they say about teaching hybridization
 
@JohnRennie yes
 
So the question is how to construct orbitals with a triangular symmetry starting from the hydrogenic orbitals that don't have any triangular symettry.
And the answer is ...
... I can't remember :-)
To be fair I learned this stuff nearly forty years ago.
I'm sure that nice Mr Google could find out though.
 
Oh wow, one of my old questions on Chem.SE is THE MOST viewed question on chem.se. It was on finding hybridization. My very second question on the site. That's an awesome feel!
 
@MelanieShebel link?
 
27
Q: How do I figure out the hybridization of a particular atom in a molecule?

Melanie ShebelI'm learning how to apply the VSEPR theory to Lewis structures and in my homework I'm being asked to provide the hybridization of the central atom in each Lewis structure I've drawn. I've drawn out the Lewis structure for all the required compounds and figured out the arrangements of the electro...

So long ago, before I even changed my major to chem. (From communication hahahahaha!)
 
5:39 AM
@ringo physical chemistry is a huge field. I did a mixture of wet chemistry and working with high tech kit. You can choose whatever subfield you prefer to do your PhD.
After my PhD I moved on to colloid science and that was lots of bench work.
 
@ringo I don't understand why this one isn't being voted to close: chemistry.stackexchange.com/questions/6720/…
 
But some of my contemporaries went into surface spectroscopy and that was all ultrahigh vacuum stuff.
 
So I must be doing something with my wording on my orgo/comp question that's getting close votes.
 
user228700
@DHMO: U still here?
 
5:48 AM
@Kaumudi yes
 
user228700
OK, I'm doing it the way u had asked me to.
 
user228700
$CO_3^{-2}$
 
user228700
Here:
 
user228700
 
user228700
And this is wrong, because I need to fill that p orbital too?
 
5:52 AM
yes
 
user228700
(Y)
 
@MelanieShebel different times, different users; and Chemistry was a small beta site then
 
True that
 
For example, we have just found this comment:
Hi Josh, and welcome to Chemistry Stack Exchange! On homework questions, we usually prefer to give a hint rather than a direct solution… — F'x Oct 25 '12 at 7:27
Today, the community thinks that we don't want "hints" as answers.
 
@pH13-YetanotherPhilipp wow, nice
 
5:58 AM
@DHMO no, nothing special ... believe me
 
@pH13-YetanotherPhilipp how did you draw them?
 
I see
 
but actually, I don’t know what you want to show with them
 
@pH13-YetanotherPhilipp how the orbitals combine together
 
6:00 AM
@DHMO but then the resulting MOs are more interesting, than the pure AOs
 
@pH13-YetanotherPhilipp could you draw them?
 
I am repeating another calculation j4f
 
user228700
@DHMO: How would u do it for $SO_3$?!
 
user228700
Here:
 
user228700
 
6:11 AM
@Kaumudi You know, sp3d2
 
user228700
But:
 
user228700
 
I need to start thinking about grad school. :S
 
user228700
Those are all examples of $sp^2$! :>.<
 
8
A: Why is the bond order in the SO₃ molecule 1.33 and not 2?

Martin - マーチンThe bonding situation in $\ce{SO3}$ is a tough nut to understand. In a historical context this molecule belongs to the species of hypervalent molecules, that disobey the octet rule. The concept of hypervalence is still very much debated. Recently there was a question raised by ron, seeking for mo...

Let's talk about quantum mechanics
 
user228700
6:17 AM
@DHMO Are u talking to me in particular?
 
@Kaumudi yes
 
user228700
I don't have any time to talk about QM :/
 
@Kaumudi did you read the answer?
 
user228700
I did, but I didn't understand very much.
 
user228700
Never mind. I'ma do something else for awhile. Thanks.
 
6:30 AM
You're welcome
 
7:06 AM
@Rubisco now I have to think of Romanesco ... thank you
 
7:38 AM
@pH13-YetanotherPhilipp You need to stop thinking of things.
 
@Rubisco good point
 
 
1 hour later…
8:52 AM
@9-BBN Pas vraiment mais il y a de la bio pour ceux qui veulent. Pour l'instant je suis en stage mili pour 6 mois.
 
@Hippalectryon what is stage mili?
 
@pH13-YetanotherPhilipp stage militaire. An "internship" in a military place
 
are you forced to or do you want to?
 
@pH13-YetanotherPhilipp Kind of forced to :-) the school I went in has a mandatory internship for th 6 first months, and there are not enough places in civil internships for all those who want one
 
01:00 - 09:0009:00 - 20:00

« first day (1633 days earlier)      last day (2759 days later) »