The Periodic Table - 2016-10-23 (page 1 of 2)

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1:31 AM

@MAFIA36790 Oops, sorry, I missed this message. Thanks :) Well, not all my content is good. Sometimes I am on mobile, or sometimes I just want to fire off a quick answer before heading off somewhere. But I personally really like teaching (in real life as well) and writing lousy answers here makes me feel like I am not being helpful to anybody else be it OP or future visitors. So I generally try to keep my content at a somewhat high standard.

As for "Can someone write the HW comment for me?", that was just me asking if someone could leave a comment informing the guy of the HW policy. I'm trying to make sure that there's some guidance for new users if stuff gets closed (especially if it's moderator unilateral closure - which I've reserved for homework dumps - but sadly had to do quite a few times already...).

user228700

2:12 AM

Hey everyone :-)

@Kaumudi hi

user228700

My textbook has provided a tree diagram for the classification of organic compounds. In this diagram, it has used the term "aliphatic" synonymously with the term "Acyclic/Open chain". Was wondering if this is correct...

@Kaumudi that is not in alignment with the common definition

user228700

That's what I thought too, after doing some googling...

2

A: What are the differences and similarities between an aliphatic and an alicyclic compound?

Aromatic compounds (wiki) (about.com) (study.com) (chemguide) (khan) Aromatic compounds are compounds that contain one or more aromatic rings. Each aromatic ring, in its unsubstituted form, has the chemical formula $\ce{C_nH_n}$ (e.g. benzene), or with one or more of the $\ce{CH}$ groups replac...

@Kaumudi self-promotion at its best

user228700

2:16 AM

:-P Alright, I'll check it out. Thanks :-)

@ringo Are you around? Could you do me a favour?

Head over to this question, and tell me what you see on the newest answer - does it write "This answer was marked as spam or offensive and is therefore not shown - you can see the revision history for details"?

Or do you see my embarrassingly bad attempt at constructing a rubbish answer? :D

Sorry for pinging you, I had to find a 10k user who wasn't a mod, most of them don't come to chat, so that pretty much left you, Jan, and Wildcat, and you were the most recently active.

Nighty nights - I'm off to bed. No hurry about my question, I am just a little bit curious.

Hmph. In fact - never mind. I found the answer. Sorry and thanks anyway.

1 hour later…

3:38 AM

@orthocresol sorry I missed you

Yes it does say "This answer was marked as spam or rude or abusive and is therefore not shown - you can see the revision history for details."

user228700

I've a quick question; everything non-aromatic is aliphatic, correct?

@Kaumudi yes; read my answer

they must be hydrocarbons though

user228700

Yes. I had read it before but I went to do further reading...

user228700

@DHMO Lol, of course :-P

user228700

So, where as aromatic compounds have a specific definition/defining properties, aliphatic compounds are just "everything not aromatic", yeah? (Not that these don't have properties of their own...emphasis on the phrase defining properties)

user228700

3:47 AM

..?

@Kaumudi "hydrocarbons not aromatic"

user228700

Yes, obviously. Am I correct, tho?

yes

user228700

OK, thanks ^.^

user228700

Also, when defining heterocyclic compounds, it is given that "...consists of atoms of at least two different elements as members of its ring"

user228700

3:56 AM

Members = Within the ring, yeah?

@Kaumudi yes...

user image

like this

4:13 AM

hey @M.A.R. , which edit are you talking about , Thanks for letting me know will keep that in mind.

user228700

@DHMO (Y)

user228700

4:31 AM

@DHMO: If asked to define the term conjugation, what would ur answer be?

4:44 AM

@Kaumudi maybe filled orbital donates electron to empty p orbital?

user228700

Ah, I think u mean backbonding. No, I was checking to see how significant the distinction b/w the terms conjugation and conjugated system is...there is no proper definition to be found of the former, except on a Wikipedia tangent.

user228700

Every definition of conjugation essentially just describes what a conjugated system is, but there is a different definition on Wikipedia...

what..

user228700

en.m.wikipedia.org/wiki/Conjugated_system

no idea dont ask me

user228700

4:48 AM

OK.

user228700

10:26 AM

Oh, man, where is everybody?

10:37 AM

I'm here

:-)

user228700

Ah, hi :-)

user228700

Hey, do u know any geometry+are u not busy?

user228700

Oh, never mind. I'll do some Chemistry.

user228700

10:52 AM

Since I've moved on to Chemistry for the time being, (of course) I have a small question...

user228700

My textbook says, with regard to MO theory: "The molecular orbitals, like the atomic orbitals, are filled in accordance with Aufbau principle, Pauli's exclusion principle and Hund's rule of max. multiplicity. But the filling of these molecular orbitals is always experimentally decided; there is no rule like $(n+l)"

Oh damn sorry I didn't see your message >.>

user228700

That's OK...

user228700

Do u know what my textbook is trying to say..?

What's the fullomh ?

user228700

11:05 AM

Oh, crap! Typo. That was supposed to to be "the filling".

user228700

@orthocresol: Thanks! :-)

Uh I'm not sure what they mean, I've always filled my MO according to theory. I guess @orthocresol knows more than I do though :P

user228700

@Hippalectryon What dyou mean "according to theory"?

@Kaumudi Using only the Aufbau principle, Pauli's exclusion principle and Hund's rule of max. multiplicity. I've never found an example when those rules gave me a result different from what we get experimentally

user228700

OK...

user228700

11:08 AM

Thanks ^.^

@Kaumudi What was your geometry problem ? I might be able to help

user228700

Hang on, can u tell me how the $\sigma$ bond is one in which the electron distribution is symmetrical about the internuclear axis?

user228700

@Hippalectryon Are u a member of MSE? I've asked my question at the chat. It's right at the front, don't worry...

@Kaumudi alright I'll look at it later

user228700

OK :-) What about $\sigma$ bonds? Can u pls tell me..?

11:12 AM

@Kaumudi I don't get your question; that's part of the definition of sigma bonds

user228700

Yes, how is this definition valid? I'm asking u how it is that electron distribution is symmetrical about the internuclear axis in $\sigma$ bonds...

user228700

user image

A bond is a sigma bond when the electron distribution is symmetrical about the internuclear axis

user228700

Yes, that represents a $\sigma$ bond, yes?

That's how we define signa bonds. How can that definition be "invalid" ?

yeah

user228700

11:15 AM

How is the electron distribution symmetrical about the internuclear axis?

Don't you see on the picture above that it's symmetrical with respect to the horizontal axis ?

user228700

But it mostly looks like it's symmetrical about the vertical axis. Hang on, how are $\pi$ bonds defined?

From wiki, a good definition : pi bonds (π bonds) are covalent chemical bonds where two lobes of one involved atomic orbital overlap two lobes of the other involved atomic orbital. Each of these atomic orbitals is zero at a shared nodal plane, passing through the two bonded nuclei. The same plane is also a nodal plane for the molecular orbital of the pi bond.

The orbital above doesn't have any nodal plane

user228700

U mean that the $s$-orbital doesn't have a nodal plane..?

The image of a s-orbital you've posted above doesn't have a nodal plane, so it can't be a pi-orbital

user228700

11:21 AM

Riight, but aah, doesn't it look like the $\sigma$ bond is symmetrical about both axes?

@Kaumudi It is, but it's only symetrical by rotation around one axis

user228700

I mean, it would make a lot more sense (to me, at least) if it were the $\pi$ bonds that were defined as being symmetrical about the internuclear axis :/

user228700

@Hippalectryon Huh. Can u pls elaborate a bit..?

The 2D image is kind of misleading, I'm trying to find a 3D one

user image

Alright, here you can see that the orbital is both symmetrical with respect to the plane of your screen, the horizontal plane going through the center and the vertical plane going through the center

@Kaumudi when we say symmetrical, it means rotational symmetry; and the internuclear axis is horizontal

11:28 AM

But in terms of rotation, it's only symetrical around the internuclear axis

user228700

Ah, right, yes...

@Kaumudi It means, experimentally, you may get 1.92 electrons in one orbital and 0.08 in another, while according to theory it should be 2 and 0.

user228700

Huh? O.o

user228700

@Hippalectryon: OK, I get it,thank you :-)

The n+l is a reference to diagonal filling, which states that the order of filling is decided by n+l, e.g. 3d before 4s

user228700

11:30 AM

Yes, right...

@Kaumudi what is not clear?

user228700

What u said about 1.92 and 0.8 and all. What did u mean?!

are you familiar with the 2 MOs of H2?

user228700

Yep.

user228700

Wait, what do u mean?

11:32 AM

We call them $\sigma$ orbital and $\sigma*$ orbital

user228700

Yeah...

in theory, orbital 1 should have 2 electrons and orbital 2 should have 0.

user228700

Yeah..?

but experimentally you might find that orbital 1 has 1.92 electrons and orbital 2 has 0.08 electrons

user228700

Riight.

user228700

11:34 AM

Okay, thanks :-)

4

A: What is the hybridisation in dimolybdenum?

The electronic configuration of $\ce{Mo2}$ is given in Bursten, B.E.; Cotton, F.A.; Hall, M.B. Dimolybdenum: nature of the sextuple bond. J. Am. Chem. Soc. 1980, 102 (20), 6348–6349: $$\ldots\mathrm{ 9\sigma_g^{1.88} 5\pi_u^{3.78} 2\delta_g^{3.42} 10\sigma_g^{1.92} 9\sigma_u^{0.08} 2\delta_u^...

look at an example here

@Kaumudi the first electronic configuration is experimental

user228700

@DHMO Oh, wow. Didn't know about this.

user228700

Thanks, again!

@Kaumudi No problem, it's just a self promotion XD

user228700

:-P Are u familiar w/ grade and ungerade orbitals, by any chance?

11:48 AM

yes

user228700

How does it work for $\sigma$ and $\sigma$* MOs?

$\sigma_g$ is bonding, $\sigma_u$ is antibonding

user228700

Yeah, but how does it work?

user image

user228700

I had a look at that diagram before...

user228700

11:52 AM

Those things represent the MOs?!

the p-p sigma MO

eg in F2

user228700

p-p sigma?

yes

3

Q: How derive g and u symmetry labels for orbitals?

When asked whether a molecule has an inversion center, we "invert" the coordinates of all atoms; i.e. move each atom from its position through the center of symmetry and to a new position equidistant to the center of symmetry compared to its initial position, and see if the final configuration is...

molecular-orbital-theory symmetry

5

Q: How to determine gerade & ungerade symmetry of a MO orbital?

J.D.Lee writes in his book Concise Inorganic Chemistry: [...] An alternative method for determining the symmetry of the molecular orbital is to rotate the orbital about the line joining the two nuclei and then rotate the orbital about the line perpendicular to this. If the sign of the lobes r...

molecular-orbital-theory

actually, pz-pz

user228700

11:54 AM

I checked MAF's question...it is kind of unrelated.

user228700

I'll check out the first link now...

yeah, it is not the same

forget about that double rotation thing - it is strictly not correct

inversion turns a 3D point (x,y,z) into the point (-x,-y,-z), if that helps in visualising. the origin (0,0,0) is the centre of inversion of the molecule, i.e. right smack in the middle of the molecule.

user228700

12:49 PM

Hm, yes, that does help, thanks :-)

user228700

This:

user228700

user image

@Kaumudi what is wrong with this?

user228700

How is it that two positive lobes of $s$-orbitals combine to give on negative and one positive molecular orbital?

simple

you subtract them

user228700

12:54 PM

(This is what I'd asked before when I asked u "how it works")

user228700

And..?

so the second becomes negative

user228700

That's not making sense to me...

do you know what orbitals are?

user228700

Omg, really?

12:55 PM

@Kaumudi ??

user228700

Dude, it's not like u're talking to me for the first time!

then what are orbitals?

user228700

Just regions around the nucleus where there is a reasonable probability of finding the electron.

no, that is how you draw the orbitals

but not the orbitals themselves

In quantum mechanics, an atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus. The term, atomic orbital, may also refer to the physical region or space where the electron can be calculated to be present, as defined by the particular mathematical form of the orbital. Each orbital in an atom is characterized by a unique set of values of the three quantum numbers n, ℓ, and m, which...

> In quantum mechanics, an atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom.

> This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus.

alright, the term "atomic orbital" also refers to the region

but I'm not talking about the region here

user228700

OK.

user228700

12:59 PM

U're talking about the solutions to Schrödinger's equation, yeah?

so the orbitals combine in phase to form bonding orbitals and out of phase to form anti-bonding orbitals

yes

and the combination is a quantum phenomenon

user228700

Yeah, OK, never mind then. I had a hunch that asking this question wouldn't help me very much right now.

exactly

user228700

Thanks anyway.

I don't understand it very much either

user228700

1:01 PM

Yeah :/ Hopefully in university...

sure

what's the policy here, can I just ask my question or should I ask if I can ask about something?

@towc just ask

cool, thanks

I was revising for my highschool chemistry course, and I think I really don't understand salt hydrolysis

I googled a bit, don't get me wrong

@towc just ask

1:03 PM

in my notes, I have written that the salts formed by a strong acid and strong base give a salt with a pH of 7

that is the ideal case

and it says that it's because the ions move freely and don't interact with the water molecules or other ions

but how does that make sense? Doesn't the salt always interact with water?

that is the simplified picture

and isn't the salt the combination of 2 ions?

how can I understand it better?

This is a better picture:

user image

Source:

8

A: Does water ionically bond to chloride ion?

The most common mechanism for when $\ce{NaCl}$ dissolves is as folows: As water is a polar molecule, as per the UC Davis ChemWiki page All About Water, shown below: The caption reads: the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in ...

1:04 PM

the water strips away the ions from the salt, I got that

but that means that the ions are interacting with the water molecules

yes

what your notes meant was probably that after the hydration, the ions don't interact with other water molecules

hmm

ok, so the idea is that they go away silently without unbalancing anything, and that's why they give a pH of 7?

as i said, the pH of 7 is ideal

while in weak acid + strong base salts there is some unbalancing, and that's why the pH changes?

@towc I don't think you have the right picture

1:06 PM

yeah sure, I'm talking about a basic level of understanding

@DHMO I really don't

Alright

for example, let's look at sodium acetate CH3COO-Na+

if you can read LaTeX, $\ce{CH3COO- Na+}$

ce?

6

A: MathJax in chat (ChatJax offshoot)

Based on the same principle, this little bookmarklet loads cancel in addition to mhchem: javascript:(function(){if(window.MathJax===undefined){var%20script%20=%20document.createElement(%22script%22);script.type%20=%20%22text/javascript%22;script.src%20=%20%22http://cdn.mathjax.org/mathjax/latest...

ce is chemistry package

click the link and put the script as bookmark

alright, setting it up, brb

alright, got it

what about it?

@DHMO

so, why is it slightly alkaline?

1:10 PM

it's from a weak acid and a strong base?

$\ce{CH3COOH\+NaOH}$

yes, but that is not the explanation

that is just a way to memorize it

which is what I'm trying to understand

Why is it slightly alkaline?

yes

so what does it mean that acetic acid is weak?

it doesn't completely dissociate in water?

oooh I think I see

@towc what do you see?

1:12 PM

but the NaOH does, so there will be leftover Na+ ions that will react with water and make the solution more alkaline?

just a guess

what makes things alkaline is not Na+

So, acetic acid being weak means the following reaction happens: $$\ce{CH3COOH <<=> CH3COO- + H+}$$

yes

adding $\ce{OH-}$ to both sides, we get: $$\ce{CH3COOH + OH- <<=> CH3COO- + H2O}$$

sure

Now, sodium acetate, just like other salts, dissociate (almost) completely in water: $$\ce{CH3COO- Na+ -> CH3COO- + Na+}$$

1:16 PM

ok

now, we know that $\ce{CH3COO-}$ interacts with $\ce{H2O}$ according to the above equation

$$\ce{CH3COO- Na+ + H2O -> CH3COO- + Na+ + H2O <=>> CH3COOH + Na+ + OH-}$$

Therefore, a large amount of $\ce{OH-}$ is produced, making the solution alkaline

why is a large amount of OH- produced?

because the equilibrium favours the product side

that is what it means for acetic acid to be weak

and there's more OH- than Na+?

it means that there is much more acetic acid molecule than acetate ion in water

@towc the concentration of Na+ does not matter a bit

what makes a solution alkaline is OH-

1:20 PM

yes, but Na+ and OH- can combine into NaOH...

but there's so little Na+ compared to OH- that the Na+ doesn't matter?

that only happens very rarely in water, because NaOH is a strong alkaline

@towc no, Na+ never matters

strong alkalines can't form in water?

what makes an alkaline strong is the fact that it dissociates almost completely in water

$\ce{NaOH -> Na+ + OH-}$

oh, I see

This has to be the strangest glitch ever:

1:22 PM

the backwards reaction is just very very unlikely

!!img/methylammonium chloride

user image

@towc touché

ok, so the idea is that the CH3COOH can form back into itself, while the NaOH can't, therefore there's some left over OH- that makes the solution alkaline?

@towc yes

1:24 PM

oh, wow, thanks!

I just filed a report lol

so, in strong acid and strong base, neither can turn back, therefore the leftover H+ and OH- just combine themselves back into water?

or rather, it's highly unlikely for them to turn back

@towc you can view it that way

oh, do you have a simpler explanation?

@towc you can't have both simple and correct...

1:26 PM

correct enough for highschool :P

I don't think you need an explanation for highschool

well played

by the way, in highschool you are taught that not all H+ and OH- combine to water

doesn't ring any bell

are you taught pH?

1:27 PM

it's what we're doing now

are you taught the meaning of pH=7?

yeah

what does it mean?

we're doing indicators, buffer solutions and titrations

[OH-] = [H+]

what does the 7 mean?

1:29 PM

[OH-][H+] = 10e-14 by experiment, therefore [H+] = [OH-] = 10e-7 in water, which is what neutrality means

yes

that is correct enough for highschool

so you can see that there is still H+ and OH- in neutral solutions

mostly bonded toghether

yes

but 1e-7 (not 10e-7) of them is not

oh shit, right

and wait, why aren't they? as [H+]=[OH-], assuming they also are in the same quantity, they can all easily pair up and form H20...

you see, chemistry is complicated

1:33 PM

any way, I don't need to know that for highschool. Although I hate to say it, but I'll keep that knowledge away from me

the ultra-simple explanation is that they just can't meet each other

I hope you don't mind if I ask some more questions

oh, that would make sense

sure

but as I have said, that is ultra-ultra-ultra-simple

so, looking at titration graphs, in class we "identified" end points and half equivalence points. The end points are an approximation for the equivalence points (n of mol of acid/base = n of mol of base/acid), right? And the half-equivalence points are where [acid/base] = [salt], as well as being where half of the volume of titrant compared to the equivalence point has been added. Is this correct?

we're supposed to understand what the pH at the equivalence point is based on the strength of the titrand and titrant. My easy way to remember it is getting an approximate pH of both, average it, and then find out whether the result is >, <, = to 7

but ofc we're meant to know more

maybe this can help you

chemistry.tutorvista.com/analytical-chemistry/…

1:39 PM

I went through that during the googling

but there are some things that I don't know if they are equivalent to what we are supposed to know

for example, it doesn't talk about salts at the half-equivalence point

and according to the guide, at the half-equivalence point the amount of acid/base is twice the amount of base/acid

and I don't know if that means the same thing as amount of salt is same as amount of base/acid

sorry, I am not too familiar with half-equivalence point

I'm guessing that if A is 2*B, then 1A and 1B combine to give a salt, so we end up with 1S + 1B, therefore [S] = [B]?

oh, right, thanks anyway

@towc i think you are right

are you familiar with buffer solutions at all then?

still in the context of pH and titrations

@towc a bit

1:45 PM

hmm, I don't think I have any questions about those actually

well, you've been of great help :)

you are welcome

!!greet/@towc

Welcome to The Periodic Table @towc! 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!

cheers

I love how the guidelines are a question rather than a post ;)

@towc there is no post

everything is in Q&A style

well, questions are still not semantically correct for guidelines

1:48 PM

then don't view it as questions

afterall, it isn't the main site

it is meant to be internal

I mainly frequent stackoverflow btw. We're all programmers, so we make our sites if we need anything, and we care about semantics

either way, was just a funny note

but it is still Q&A even in stackoverflow

yes, but for guidelines and similar we make our stuff

javascriptroom.github.io/rules

@towc i see

well, thanks again, I'll probably be back at some point :) I'll be in here if you feel lonely

1:52 PM

@towc looks like I need to log in there

maybe you would like to find me here when you have questions about chemistry or physics or mathematics

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...

Would anyone care to take a swing at this

2

Q: Explaining differences in rate of propagation based on monomer structure

The values given below have been taken from my lecture hand out, and they weren't cited so I don't know where my instructor got them from The $k_p$ (rate constant for propagation in radical polymerisation ) step for the following monomers is given below: Styrene: 106 $\mathrm{mol^{-1} s^{-1}}$...

organic-chemistry kinetics polymers

Still no answers, a few good comments though..

1 hour later…

3:03 PM

@Vedant Generally, edits of yours I reviewed. Unfortunately there isn't a system to show why your edits were rejected and why, unless you dig them up yourself in your profile. I noticed that you edit some things and leave out other important stuff that needs editing.

So I probably rejected one or two of those.

3:26 PM

@getafix I think porphyrin is on the right track.

@M.A.R. Thanks will keep that in mind, though I dont think i commited such a trivial mistake

4:19 PM

@Vedant Nobody makes mistakes on purpose

chemistry.stackexchange.com/review/suggested-edits/44349

That one, for instance

I rejected so it'd send a message to you, but it failed.

Two very important things you missed

1. The post is an on-hold or soon-to-be-on-hold-at-that-time HW dump.

You shouldn't edit crap like that.

It's like polishing turds.

2. You didn't edit parts that direly need editing

No mathjax, no markdown, no paragraphing.

4:42 PM

Hello fellas!

Hey

for quite a long time i haven't been here.

Yeah, welcome back

they automatically changed my avatar?

Gravatar did.

I like your new avatar better.

It has a more solid color to it.

4:46 PM

I see

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