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[*] posted on 4-10-2015 at 13:20


exam ? Exam ? EXAM ?!?!

Eeeeek !

Where to begin? The square hippopotamus is equal to the sum of the bowels of the quantum well.

Oh deary me.




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[*] posted on 4-10-2015 at 13:51


Quote: Originally posted by aga  
exam ? Exam ? EXAM ?!?!

Eeeeek !

Where to begin? The square hippopotamus is equal to the sum of the bowels of the quantum well.

Oh deary me.


Of course you could waste the course money (sorry, no refunds!) and throw your life away but you don't want to do that, right?




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[*] posted on 4-10-2015 at 14:50


Quote: Originally posted by aga  
So the atoms really do have an 'effective' charge (as seen from the angle at which bonding takes place) that is not actually 1, although overall, mathematically, taking the system as a whole, it is 1.


As long as it helps you understand better, I suppose you could look at it that way. Just keep in mind that, yes, it is the same -1 charge overall. The only thing changing between fluoride and iodide is the area the charge is distributed over, which does effectively make a difference. It's kind of like taking a pencil and placing it in your palm eraser-end down and setting a heavy textbook on top of it. It doesn't hurt or break your skin, right? Okay, now take the pencil and flip it over, putting the sharp end down into your palm. Now place the same heavy textbook on top of it. This time it'll hurt and likely puncture the skin. In both cases the force exerted onto your palm was the same since the weight of the textbook never changed; however, what did change was the area over which that force was distributed. A similar thing is going on with fluoride and iodide, only it's charge and not force. The smaller the area, the stronger and more dense the charge will be at any given point within the area.

Now try to apply this to organic chemistry, for example to see why carboxylate ions are much weaker bases than alkoxide ions despite them both having the same overall charge of negative one. I've made the following for you:

for aga.bmp - 215kB

While I'm sure you're well aware, the top one is acetic acid and the bottom one is ethanol. Now let's remove a proton to give an acetate ion and an ethoxide ion, respectively. I don't know if you've learned about resonance yet, so the dashed lines on the acetate ion may not make much sense, but let's just say that the lines indicate that the charge is distributed evenly across the two oxygen atoms. Unlike the ethoxide ion (bottom) whose -1 charge is more or less centered completely on the one oxygen, in acetate each oxygen really only has a -1/2 charge. So while those charges may add up to an overall charge of -1, you could say that, effectively, acetate's -1 charge is not nearly as strong as ethoxide's -1 charge, even though overall they have the same numerical value.

Make sense?

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[*] posted on 4-10-2015 at 15:57


For aga's benefit (and because we haven't done much with resonance structures), in the top right structure of the carboxylate ion, one orbital (2 electrons) 'stretches' from one oxygen atom to the other, that 'stretched orbital' is represented by the dotted line.

This has two consequences:

1. Both C-O bonds are identical.

2. The unit charge of -1 is distributed over that O-C-O sequence, so that each O atom has a charge that is really only about -1/3. That makes them less attractive to electrophiles like the proton H<sup>+</sup> and thus weaker bases.

A corollary is that simple alcohols are extremely weak acids, while carboxylic acids are only weak acids, much stronger than alcohols.

[Edited on 5-10-2015 by blogfast25]




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[*] posted on 5-10-2015 at 09:06


So ..... would it be right to imagine that the electron on the O-C-O structure is whizzing about in a New Orbital, common to all three atoms, or is leaping from one atom's discrete orbital to another ?



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[*] posted on 5-10-2015 at 14:42


Quote: Originally posted by aga  
So ..... would it be right to imagine that the electron on the O-C-O structure is whizzing about in a New Orbital, common to all three atoms, or is leaping from one atom's discrete orbital to another ?


Personally I prefer to think of the dotted line as a molecular orbital (TWO electrons) stretching from one O to the other O.

[Edited on 5-10-2015 by blogfast25]




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[*] posted on 5-10-2015 at 16:28


Quote: Originally posted by aga  
So ..... would it be right to imagine that the electron on the O-C-O structure is whizzing about in a New Orbital, common to all three atoms, or is leaping from one atom's discrete orbital to another ?


I look at it as the electrons in the pi bond becoming delocalized. In other words, the double bond that used to be localized in acetic acid between carbon and oxygen has now become delocalized and stretched across all three atoms. You'll learn later that this is a result of resonance stabilization, as you could represent the acetate ion with a negative charge on either oxygen simply by alternating which oxygen has the double bond between it and carbon. In reality, however, the bond isn't actually alternating or "resonating," which is why the more correct way to draw an acetate ion is as a hybrid of the two structures.

[Edited on 10-6-2015 by Darkstar]
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[*] posted on 5-10-2015 at 16:43


Quote: Originally posted by Darkstar  
You'll learn later that this is a result of resonance stabilization, as you could represent the acetate ion with a negative charge on either oxygen simply by alternating which oxygen has the double bond between it and carbon. In reality, however, the bond isn't actually alternating or "resonating," which is why the more correct way to draw an acetate ion is as a hybrid of the two structures.



For the reason you outlined in your last paragraph, I wasn't planning on writing on "resonance" (I didn't in the segment on conjugated bonds and benzene either), so in that sense there will be no 'later'. If you wish to spend a few words on it then feel free (of course) and I'll include it in the final table of contents (as well as any other additional material).

I see Feynman still refers to resonance on benzene.




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[*] posted on 5-10-2015 at 18:15


I didn't necessarily mean later in this course, just later in his journey of learning organic chemistry. A brief segment on resonance does sound like a good idea, though. I may add something about it soon.
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[*] posted on 5-10-2015 at 23:50


Thanks for the clarifications.

The notion of delocalisation seems straightforward - the electron(s) are not centred around 1 nucleus, they're centred around 2 or more.

Imagining it in simple terms of the orbital volume, it seems logical that those delocalisd electrons are less likely to become involved with an atom external to that system, as they are not as concentrated in the 'bonding area' as they would be if they were concentrated into a smaller orbital.

Would 'resonance' be the notion of how the delocalised orbital Looks, and which atoms are involved ?




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[*] posted on 6-10-2015 at 04:09


Quote: Originally posted by aga  
Would 'resonance' be the notion of how the delocalised orbital Looks, and which atoms are involved ?


I made this to show you what's going on when acetic acid is deprotonated. The red arrows represent the flow of electrons:

carboxylate resonance.bmp - 641kB

As you can see, you could draw the acetate ion with a negative charge on either oxygen simply by moving the double bond. These two structures are called resonance structures, as they give the appearance that the molecule is "resonating" back and forth between the two. In reality, the molecule isn't actually going back and forth, and at no time is the double bond ever truly localized between any two atoms. Instead, the molecule always exists as a single structure, basically a combination of all the possible resonance structures. This combination is called a hybrid structure, and is the more correct way to depict the molecule.
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[*] posted on 6-10-2015 at 08:42


Again for the benefit of aga, I've 'zoomed in' on the left hand top part of the diagram, showing also the three lone pair (non-bonding) electron pairs (orbitals) on the O-nucleus of the OH<sup>-</sup> anion:

Alkaline attack.png - 12kB

One of these lone electron pairs latches on to the H atom of the acid and that H atom then let's go of the MO that bonded it to the acid. The result is a water molecule and a carboxylate ion.




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[*] posted on 6-10-2015 at 10:26


So 'resonance' is basically seeing that the resulting structure can flip between two or more states, kind of 'oscillating' back and forth (although it isn't).

That's kind of what i meant about the bonding orbital shape, although i didn't explain what i meant very well.

I suppose the $1m question is why does the H prefer to be with the OH- so much that it releases it's bonding with the C-O in order to do so.

Are the orbitals involved in the OH different to those in the C-O-H (it's an O to H bond after all in both cases).




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[*] posted on 6-10-2015 at 11:18


Quote: Originally posted by aga  


I suppose the $1m question is why does the H prefer to be with the OH- so much that it releases it's bonding with the C-O in order to do so.



That's really a matter of molecular orbital energy levels, kind of the subject of the next instalment, coming up in about 15'.




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[*] posted on 6-10-2015 at 11:25


Bond Enthalpies:

So far I’ve avoided using molecular orbital diagrams, writing mostly from VSEPR point of view. But let’s go back to the beginning of Part II, in particular the interaction between two ground state hydrogen 1s<sup>1</sup> orbitals to form a bonding σ MO. In Molecular Orbital Theory this is often schematised as follows in an MO diagram:

MO diagram for H2.png - 15kB

(if you’re wondering what the u and g symbols stand for, u stands for ‘ungerate’ which is German for ‘odd’ and in the parlance of this course means ‘out of phase’. g stands for ‘gerate’, which is German for ‘even’ and in the parlance of this course means ‘in phase’.)

One of the beauties of the MO diagram is that it shows the relative energy levels of the orbitals involved. We can see that energy of the filled bonding MO, noted as σ<sub>g</sub>(1s), is lower than the energies of the atomic orbitals ψ<sub>1</sub> and ψ<sub>2</sub> of the individual H atoms. The anti-bonding MO σ<sup>*</sup><sub>u</sub>(1s) remains of course empty.

In practice this means that the reaction:

H. + .H ===> H - H (or 2 H === >H<sub>2</sub>;)

... is exothermic, i.e. ΔE < 0. ΔE is the bond enthalpy of the formation of the σ<sub>g</sub>(1s) MO bonding orbital.

Bond enthalpies for the main types of covalent bonds between various atoms have been experimentally determined and a sample of values is shown below:

Bond ethalpies.png - 40kB

Note that in the case of formation of bonds, we have to assign a negative sign to these values, in the case breaking bonds a positive sign has to be assigned.

Furthermore, while the bond enthalpies between monovalent atoms are absolutes, for bonds between multivalent atoms the real bond enthalpy may vary a bit from context to context: a C-C bond in ethane is not exactly the same as in butane (e.g.). In that case the bond enthalpies have to be considered approximate, average values.

Estimating reaction Enthalpies from bond Enthalpies:

These values can be used to estimate the reaction enthalpies of simple reactions. Here, mainly for illustrative purposes, the estimated Enthalpy of formation of ethane:

2 C + 3 H<sub>2</sub> === > C<sub>2</sub>H<sub>6</sub>, ΔH<sub>f</sub><sup>298K</sup>

To find this value write this equation as a sum of ‘sub reactions’ using Hess’ Law:

1) 2 C === > C – C, yields - 368 kJ

2) 3 X [H<sub>2</sub> === > 6 H.], costs 3 X + 435 = + 1305 kJ

3) C – C + 6 H. === > H<sub>3</sub>C – CH<sub>3</sub>, yields 6 X – 414 = - 2484 kJ

Adding up: ΔH<sub>f</sub><sup>298K</sup> = - 368 + 1305 – 2484 = -1547 kJ per mol C<sub>2</sub>H<sub>6</sub>. NIST webbook lists the value of ΔH<sub>f</sub><sup>298K</sup> as – 1560 kJ/mol, about 1 % difference.

Note that where phase changes happen, e.g. had ethane been a liquid at 298 K, a correction would have to be applied for the phase change enthalpy but that was not the case here.

Further simplified this method can also be used for the estimation of reaction Enthalpies of simple substitutions. Take the substitution of a single hydrogen atom in an alkane molecule by a chlorine atom:

C<sub>n</sub>H<sub>2n+2</sub>(g) + Cl<sub>2</sub>(g) === > C<sub>n</sub>H<sub>2n+1</sub>Cl(g) + HCl(g)

To effectuate this we need to:

1) break 1 mol of C-H bonds, which costs + 414 kJ
2) break 1 mol of Cl<sub>2</sub> bonds, which costs + 243 kJ
3) form 1 mol of C-Cl bonds, which yields – 331 kJ
4) form 1 mol of H-Cl bonds, which yields – 431 kJ

Which gives a total of -105 kJ/mol of estimated reaction enthalpy at 298 K. In reality that value will depend a bit on the alkane, the substitution position and possible phase changes.

And with this we’ve arrived at something really quite fundamentally important: chemical reaction Enthalpies are the consequence of the breaking and forming of molecular bonds, when molecular orbitals rearrange into energetically lower (more favourable) configurations.

[Edited on 7-10-2015 by blogfast25]




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[*] posted on 6-10-2015 at 14:38


Jees bloggers ! That's a lot of bits all at once !

Firstly the odd/even diagram thing : Bonding / Anti-bonding is unclear to me, unless it just means that a Bond is formed when the resulting system is at a Lower energy state than it would be otherwise (edit: for a given Energy level).

Is this the Entropy thing where stuff always goes from a High to Low energy state if it can ?





[Edited on 6-10-2015 by aga]




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[*] posted on 6-10-2015 at 16:31


Quote: Originally posted by aga  
Jees bloggers ! That's a lot of bits all at once !

Firstly the odd/even diagram thing : Bonding / Anti-bonding is unclear to me, unless it just means that a Bond is formed when the resulting system is at a Lower energy state than it would be otherwise (edit: for a given Energy level).

Is this the Entropy thing where stuff always goes from a High to Low energy state if it can ?


Bonding / anti-bonding reminder:

http://www.sciencemadness.org/talk/viewthread.php?tid=62973&...

Yes, the system H<sub>2</sub> is of lower energy than the system of H. + .H (H. is an unbonded hydrogen atom), that's what the diagram shows. The difference ΔE is the bond Enthalpy, literally the energy released when the bond forms. We can even write ΔE more explicitly as:

ΔE = E<sub>σ</sub> - 2 E<sub>ψ</sub>

Entropy? Huh? Who mentioned that? :D

[Edited on 7-10-2015 by blogfast25]




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[*] posted on 7-10-2015 at 04:08


Another simple example demonstrating the use of bond enthalpies.

Estimate the Enthalpy of formation of water, ΔH<sub>f</sub><sup>298K</sup>.

H<sub>2</sub>(g) + 1/2 O<sub>2</sub>(g) === > H<sub>2</sub>O(l)

We're only interested in Enthalpy, so thermodynamics, and not in kinetics or how precisely we carry out this reaction. But we know for certain that it involves:

1) Breaking 1 mol of H<sub>2</sub> bonds: H<sub>2</sub>(g) === > 2 H(g). This costs + 435 kJ.

2) Breaking 1/2 mol of O<sub>2</sub> bonds: O<sub>2</sub>(g) === > 2 O(g). This costs 1/2 X + 498 = + 249 kJ.

3) Forming bonds 2 mol of HO: 2 H(g) + O(g) === > H<sub>2</sub>O(g), This yields 2 X - 465 = - 930 kJ.

- 930 + 249 + 435 = - 246 kJ/mol. The NIST webbook is - 286 kJ/mol but that is for liquid water. So to our value we have to add the Latent Heat of Condensation of -41 kJ/mol, so we get - 287 kJ/mol. Not too shabby!

[Edited on 7-10-2015 by blogfast25]




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[*] posted on 7-10-2015 at 12:11


Wow !

So the data works out almost perfectly.

Would this be a bad time to ask what Energy actually is ?




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[*] posted on 7-10-2015 at 12:53


Quote: Originally posted by aga  


So the data works out almost perfectly.

Would this be a bad time to ask what Energy actually is ?


It's not hard to find some examples where it doesn't work so sterlingly. But for reasonable estimates 'playing Lego' with bond Enthalpies works well.

Energy: for all intents and purposes practical:

Something has energy if it has the capacity to do useful work.

Useful work in the physics sense of the word:

Work.png - 2kB

The force F displaces a object by a distance Δx in the same direction as F. The useful work done on the object is:

W = FΔx, assuming F was constant over the interval Δx.

If F is not constant then:

W = ʃF(x)dx, integrated from x = 0 to x = Δx.

Example: in internal combustion engines chemical energy of the fuel and air is converted to combustion Enthalpy, which is in turn used to drive the piston upward and do useful work on it.

We now know that 'Molecular Orbital Energy' would be a better term for chemical energy!

[Edited on 7-10-2015 by blogfast25]




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[*] posted on 7-10-2015 at 13:12


Valliant effort, and gratitudes for it.

The actuality of Energy Itself might lay in a distant corner of unexplored agaspace.

Who knows ? Back to mainstream and safer ground ...

Is the out-of-phase antibonding orbital result of the wavefunction predicting that a bond will NOT form, or is it called 'antibonding' to denote that a bond Might form, wheras the bonding orbital (in-phase) denotes that a bond Will form ?

Just that the word 'antibonding' suggests that a bond will not form.


[Edited on 7-10-2015 by aga]




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[*] posted on 7-10-2015 at 13:29


Quote: Originally posted by aga  

Just that the word 'antibonding' suggests that a bond will not form.



It means that no bond is formed. In an anti-bonding molecular orbital the bulk of the electron density doesn't lie between the nuclei and provides no shielding against the electrostatic repulsive forces between the nuclei: that's a non-bonding arrangement.

Example: σ<sup>*</sup><sub>u</sub>(1s) representation, scroll down and see how one orbital is red (+) and the other blue (-), they are ungerate:

http://winter.group.shef.ac.uk/orbitron/MOs/H2/1s1s-sigma-st...


[Edited on 7-10-2015 by blogfast25]




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[*] posted on 7-10-2015 at 14:38


So odd means No.

Antibond means No Bond Forms.

Out-of-phase means No bond.

OK. Got it.




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[*] posted on 7-10-2015 at 15:06


Quote: Originally posted by aga  
So odd means No.

Antibond means No Bond Forms.

Out-of-phase means No bond.

OK. Got it.


Yes, correct.

Ahem. Cough. There's a small complication. For π bonds it's the opposite: ungerate orbitals form bonding MOs, gerate orbitals form anti-bonding orbitals.

I'll now dedicate a mini-excursion on that gerate/ungerate business. Need to make a diagram. Bear with me.




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[*] posted on 7-10-2015 at 15:44


Gerate and ungerate orbitals:

Gerate/ungerate relates to the concept of wave function parity we discussed in Part I.

Look at the diagram below. In it orbitals or lobes of orbitals are coloured red or blue, corresponding to the sign of the wave function, + or - (or the other way around).

gerate ungerate.png - 10kB

Whether the orbitals are gerate or ungerate is determined as follows.

Draw an arrow through any point of symmetry:

If the colour (sign) doesn't change when crossing the symmetry point (S) the lobes are gerate.

If the colour (sign) does change when crossing the symmetry point (S) the lobes are ungerate.

In the diagram the top s orbitals are ungerate and form anti-bonding σ molecular orbitals.

The bottom p<sub>z</sub> are also ungerate but form bonding π molecular orbitals.

What you really need to take home from this:

1) Bonding σ molecular orbitals are formed from the interaction of gerate (g) s or p<sub>x,y</sub> orbitals.

2) Bonding π molecular orbitals are formed from the interaction of ungerate (u) p<sub>z</sub> orbitals.

To paraphrase:

Quote:
So odd means No.


Odd means No when it's σ, except when it's π because then it's Yes! :D:cool:

Und das ist ze geschichte der gerate und ungerate, Herr Aga! Ein bier, bitte... ;)



[Edited on 8-10-2015 by blogfast25]




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