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04:00 - 14:0014:00 - 00:00

2:00 PM
@DHMO: I'm stuck now. Still thinking. Do u have a theory?
@Kaumudi same as yours, but still a guess
I was finding online while you were typing your hypothesis
and I just found something:
> In general, more polar bonds are stronger, because the bonding electrons are held more tightly at one end of the bond. This means to redistribute those electrons evenly throughout the bond so it can then be broken homolytically would require more energy than to just break a bond which already has evenly distributed electrons (i.e., a nonpolar bond).
> The O-H bond in acetic acid is more polar than the alpha C-H bonds, meaning it's harder to break homolytically, and thus has a higher BDE. However, it's much easier to break the O-H bond heterolytically than it is to do the same to the C-H bonds. See the deal?
Lol, I was looking up the same thing :-P
@hBy2Py I wouldn't have much insight to offer, too, but considering that for most of the functionals the amount of votes is around 70% only, it seems like it might be already enough if you just put in the functionals for whatever functional you use most. And as I said earlier, if one of the most hated functionals ends up on #3, what's the whole thing worth in the first place...
Hm, emphasis on the word homolytically.
@Kaumudi BDE must be homolytical
2:04 PM
Yep. Hm...
by the way, source
Well, with this definition in mind, I think it's safe to say that the atoms now attract each other some more and hence, the bond length decreases.
Gimme one second.
i suppose so
@hBy2Py SEDE can be quite terrifying and you really need to work yourself into the whole thing. It's like learning a language. You could start here:data.stackexchange.com/help The disadvantage is, it's not real time. Searching is, and the advanced keywords are given here. It's easy to use, too.
2:11 PM
Welcome to The Periodic Table @MelanieShebel! Here are our chat guidelines and it's recommended that you read them. If you want to turn Mathjax on, follow the instructions in this answer. Happy chatting!
Yes, so I guess that makes sense. Shall we move on..?
@Kaumudi surely
We're done with bond lengths. Let's talk bond angles.
@Kaumudi surely
2:13 PM
Firstly, it depends on the hybridization of the central atom, of course.
Let's not discuss this point since the dependence is obvious (not so much when u think about it along the lines of QM, but never mind that, for now)
@DHMO Copy, paste, copy paste? :-P
@Kaumudi oh, sh!t, I didn't even notice it
OK, I believe u :-P The next point is the number of lone pairs on the central atom.
2:18 PM
Wait, I thought I had this one figured out...huh.
@DHMO: Anything? I'm still thinking.
@Kaumudi What is the question?
You didn't even bring out a statement.
Oh, but I did:
5 mins ago, by Kaumudi
OK, I believe u :-P The next point is the number of lone pairs on the central atom.
"the number of lone pairs on the central atom" is not a statement
2:22 PM
Well, I didn't tell how my textbook thinks it'll affect the bond angle.
@DHMO So..? That's the next factor upon which the bond angle depends.
that is plain VSEPR
so where is your doubt?
I'm trying to frame it in a way that makes sense.
2:23 PM
just use VSEPR, lol
:-P I'm making notes. I can't write "Just use VSEPR, lol".
@Kaumudi "for the case of number of lone pairs, just use VSEPR, lol"
The increased density of electrons around the central atom causes the repulsions b/w these electrons and the bond pairs. So, the bond angle decreases?
Hmm... I wonder if there is a fast back of the envelope way to use MO for getting the same results as VSEPR
@Kaumudi yes, which is essentially VSEPR
@Zhe no
2:26 PM
@Kaumudi NOOO
VSEPR is a failed theory
Bond angle decreases
oops, missed that
No, I'm just talking about bond angles
Yeah, typo :/
2:27 PM
I don't get it, tho.
Can you quickly indicate that moving to $C_{3}$ point group from $T_{h}$ will reduce bond angles...
Probably not...
How come the bond angle decreases?
$C_{3}$ and $T_{h}$. Dunno these notations.
Anyone good at stoichiometry ?
@AaronAbraham state your question
2:32 PM
..? What about my question?! :-o
Regarding chemical equivalence; I'm told that an equivalent of an element is the amount of that element that combines with a mole of Hydrogen atoms. Right so far? @DHMO
@Kaumudi well, consider AX4 vs AX3, right
@Kaunudi Because the lone pair pushes the other bonds closer
@AaronAbraham Depends on the reaction
@Zhe Why does it push the other bonds closer?
equivalent is relative to another reactant
2:34 PM
@Zhe what kind of 'dependence' exactly?
In VSEPR, you would've said that the lone pair is more energetic and takes up more space
@Zhe OK..?
@AaronAbraham depends on your reaction. What is it?
@Kaumudi If it takes up more space, the bonds take up less space so they get pushed together
@Zhe Hm, OK...
That's why it's called Valence Shell Electron Pair Repulsion
It's just another heuristic that works a lot of the time
2:35 PM
@DHMO: Next point?
@Kaumudi ok
Now we're talking about the size of the central atom.
As the size increases, the electrons now have more room to occupy and hence, the bond angle increases. Makes sense?
@Zhe Alright, if it does depend on the reaction, I guess that clears stuff up a bit. The book I'm reading cites H2O ( @DHMO ) and then proceeds to say that the equivalent weight of oxygen is 8g ; but then H2O2 popped into my mind and that's what caused the muddle . From H2O2, I reckon oxygen's eq.wgt, should be 16. So I'm right?
2:38 PM
@Kaumudi decrease?
@orthocresol Hiya!
@AaronAbraham embarrassed
@DHMO Increase...
@Kaumudi for example?
That's not how we use equivalent
Equivalents has a reserved meaning for when you run a reaction with an excess reagent, and the number of equivalents is the mole ratio between one reagent and another
Generally you set the number of equivalents of the limiting reagent to 1
2:41 PM
@DHMO Don't have one :/ Why do u think it decreases? .
Oh you're Mods now! I only just found out. Congrats @Loong and @orthocresol !! [ @Loong The Bundeswehr would be proud of you :'/ ]
Q: Why does bond angle decrease in the order H₂O, H₂S, H₂Se?

Yomal AmarathungeI know that bond angle decreases in the order $\ce{H2O}$, $\ce{H2S}$ and $\ce{H2Se}$. I want to know the reason. I think this is because lone pair repulsion but how?

@Kaumudi my thought was that larger central atom -> more diffuse side atoms -> less repulsion
@DHMO What do u mean by more diffuse side atoms?
@Kaumudi diffuse = distant from the central atom
@AaronAbraham Hi, thanks.
2:46 PM
You mean more diffusED side atoms?
@Kaumudi ........
That's just me being a pedantic idiot, sorry.
@Kaumudi no
characterized by great length or discursiveness in speech or writing; wordy.
widely spread or scattered; dispersed.
Botany. widely or loosely spreading.
Optics. (of reflected light) scattered, as from a rough surface (opposed to specular ).
the 7th sense
@orthocresol Welcome! Least I ought to do for my favourite fellow Common-Wealth seal ;)
OK, sorry. Also, thanks :-P
2:51 PM
Riight, OK, I see ur point. Alright, I think that makes sense.
The next point is regarding the electronegativity of the central atom.
@Kaumudi what....
Uh, how the angle depends on the electronegativity of the central atom. What's wrong?
@Kaumudi how?
Well, when the electronegativity of the central atom is high, the electrons (constituting the bond) spend more time near the nucleus. So, the repulsion b/w bond pairs increases. Hence, the bond angle increases.
2:57 PM
@DHMO Carbon quadruple bond example?
@Zhe read the comments
Oh, crap, I must go to have my dinner. @DHMO: Will u still be here in about 30 mins?
@Kaumudi depends
@Zhe in chemistry there is only one rule: there is no rule.
2:59 PM
Okay. I'll risk it. In case u're not here, just wanted to thank you for spending time to discuss with me :-)
you're welcome
@DHMO strongly disagree
@Zhe i'm just kidding
Actually, I strongly agree
Because all of the "rules" are just to generalize
We can't handle the complexity
in that case, i'm not just kidding. by all means i am serious
3:00 PM
QM isn't a rule
what is it?
anyway meeting time
always, nice chatting with you
It a firm theoretical grounding!
@Zhe I think you should make your first comment as an answer
3:15 PM
@M.A.R. o/
I again missed Zhe.
Haha the one above is full of Trump speech jokes in the comments.
3:47 PM
@DHMO: Have u gone?
@Kaumudi yes
Ah, so u're busy now?
no, i was just kidding
OK :-P So, yeah, electronegativity of the central atom.
for example?
3:49 PM
Do u agree with my line of reasoning?
OK, look:
I thought we discussed polarity before
3:52 PM
@DHMO We'd discussed with regard to bond length, not angle.
Why don't u agree with my line of reasoning?
@Kaumudi unfair, all are less electronegative than H
@Kaumudi well, decreased bond length -> less diffuse -> more repulsion -> larger angle
Uuugh. I'm unable to think of other examples.
I also have no idea
But I'm pretty sure...okay, Googling.
3:57 PM
BTW, I still don't think that "less diffuse" is the correct way to use that word.
@Kaumudi correction + source?
I think it should be less diffusion. They diffuse to a lesser extent would be correct.
diffuse is an adjective
using diffuse as a verb or diffusion as a noun would not be the same meaning
No, diffuse is a verb.
do i have to cite the same link to you again?
the verb "diffuse" and the adjective "diffuse" have different meanings
4:09 PM
Not so different...
quite different
"diffuse" as a verb is a movement
"diffuse" as an adjective is a location
Hm, OK...
Little confusing it was.
4:12 PM
OK, I'm reading this:
Q: What affects bond angles?

DissenterI have three factors in mind: 1) vdW repulsions 2) Bond length 3) Electron-electron repulsions. I know that vdW repulsions between ligand atoms push the atoms apart. And bond length obviously influences bond angle; if the bonds were very long, then the bond angle can shrink without bringing...

Did u read answered David BasedMathematician Coven's answer?
@Kaumudi yes
no idea why
Why don't u agree with my line of reasoning?
increased electronegativity of the central atom might lead to less polarity
4:19 PM
How so?
if the side atom is more electronegative than the central atom
increased electronegativity of the central atom would be catching up
Oh, dude, I've been saying increased(/decreased) electronegativity of the central atom, this whole time!
let's say central atom is -2.98
while side atom is -3.50
so you increase the negativity of the central atom
it becomes -3.33 vs -3.50
and the polarity decreased
I'm talking about those cases when the central atom is more electronegative than the terminal atom. My reasoning only works for those cases...
you didn't mention it
4:23 PM
Thought it might be obvious, sorry.
NP. I really dunno if we can comment on the bond angle when the electronegativity is higher but not higher than the terminal atoms so u've made a good point...
And lastly, the size/electronegativity of the terminal atom:
are you sure that it is not because of the size?
4:37 PM
What is not?
never mind
Tell me, what did u mean..?
I meant, in your tiny experiment, you changed two factors...
Which experiment?
4:41 PM
comparing PF3 and PCl3 and PBr3 and PI3
Two factors?
the electronegativity difference as well as the size of the terminal atoms
Well, yeah...
We can't really do it any other way.
4:49 PM
Hm, alright. This took too long, I'd argue, but I guess it was worth it. I hope u benefited too, at least a little...
Thank you :-)
thank you too
OK, goodnight :-)
5:05 PM
I'm enjoying the closehammering.
I do not like assumptions when there is no evidence provided. You need to first prove that Seems this problem is much more common ... much more correct answers being downvoted with comments pointing out problems that not exist. I understand your frustration but you haven't linked to any credible sources to back up your answer. Saying this as an untrained eye, but I do not see why the two answers don't align well. You speak of perfectly flat surfaces, and as that is hard to come across, hBy2Py neglected the fact that it might happen. That doesn't make his answer wrong. 'Wrong' is too strong. — M.A.R. 5 mins ago
@orthocresol And I'm MAR. Nice to meetcha
5:20 PM
Q: Organic chemistry Empirical formula Homework

AlexxskieHey need help with this: Find the empirical formula of the compound which contains 0.5 g of hydrogen and 6 g carbon and 8 g of oxygen. Thanks.

This guy managed to beat the closehammer.
Not bad.
BTW I'm gonna post the AMIRITE policy discussion in thirty minutes or what.
That is, if I don't get distracted.
That's great. Is there any chance you could post it earlier? I gotta go off in around 45min.
Then don't chat so I focus on the post.
@AaronAbraham Just saw your post. What's the question regarding equivs again?
5:46 PM
@ortho ortho, I posted.
Q: Policy on AMIRITE questions

M.A.R.Intro Consider two likely scenarios: Q: How many grams of $\ce{O2}$ are there in one mole of oxygen gas? I know that O is 16, so it should be 16 grams, AMIRITE? Please halp ugrant kthxbye Q: How many grams of $\ce{O2}$ are there in one mole of oxygen gas? I know that O is 16...

What should be our policy on AMIRITE questions? Please weigh in.
If needed I'll convert my comment to an answer properly another time.
@orthocresol The only problem here is that @Mart is going to remind me every 15 minutes beginning tomorrow for the next month that meta is slow and I should give it time.
6:02 PM
I just hope more people pitch in, we know that lots of people do head round to meta, we just need them to say something
Even if it's just "I agree with XYZ because <short reason>."
Well, we don't have an election anymore, so meta is going to be dead quiet.
Like, I don't think anybody was opposed to your HW experiment. It's just that there was no sign of agreement.
But Mart, Brian, Jan, Todd and Klaus are the only people who're gonna chip in anyway.
@orthocresol Well, people started not voting to close the next hour I posted that thing.
When people disagree with you on meta and they're not telling you, they keep doing what they're doing until you realize they disagree.
Nothing of the sort happened.
You could cheat, and simply create several answers with the possible solutions, and see which gets upvoted most. That lets people voice opinions without having to bother justifying themselves.
Yeah, but I just felt too lazy all of a sudden.
6:06 PM
I know what you mean.
Well, I've gotta go off, all the best
I haven't been writing a meta question post thingy for a Loong time.
@orthocresol Cya
BTW @Ortho could you feature that post so we'll get results quicker? We have a heavy backlog of things to do.
6:38 PM
@M.A.R. Were you trying to get a hold of me before?
Hi All
Out of methanoic and ethanoic acid, which would give a brown precipitate when potassium permanganate solution was added?
@NoahP what do you think?
I believe that methanoic acid will turn brown and ethanoic will not
@Zhe but I'm not 100% sure and wanted to confirm
OK. So why do you think that?
Because the methanoic acid can be considered to contain an aldehyde group
Which I'm unhappy about - but apparently, that's the case
And thus, it oxidises
6:51 PM
sounds good to me
So it would be the methanoic?
@Zhe Were I?
7:06 PM
"I again missed Zhe."
Yeah, that's because I did miss you
Throws something at Zhe
I again missed Zhe.
7:31 PM
3 hours later…
10:04 PM
@M.A.R. tsk tsk
Seems like there's no coffee.
I've had 5 double espressos today and I'm ready to go
What's on?
time to give my unqualified opinion on AMIRITE
Excellent :)
10:28 PM
I'm out
11:18 PM
@M.A.R. Should we make it into a poll?
04:00 - 14:0014:00 - 00:00

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