## Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

Pages:  1    3    5  ..  9 blogfast25 - 1-9-2015 at 13:50

 Quote: Originally posted by aga 1. So the Orbital is a plot of the probability of finding an electron at, say n=1 given a number of variables, such as What ? When ? 2. If so, then the isosurface is an imagined nice smooth surface, imagining a convenient boundary so it can be all nice and shiny for the graphical representation, whereas the reality is a fuzzy blob occupying approximately that shape. 3.A bit like a line drawing of a cat, omitting the furry bits.

2. and 3. are correct.

1. is imprecise.

ψ<sup>2</sup> is not "probability of finding an electron at", it's the probability density. To find the probability of the electron you need to calculate it, using ψ<sup>2</sup> and a small volume element around the location you're interested in. Obviously, areas of high ψ<sup>2</sup> tend to be areas where the probability of finding the electron is high. The 'lobes' of the orbitals are such areas. The only variable here is r and the angles theta and phi, the position with respect to the nucleus. Remember, we're using spherical coordinates here!

[Edited on 1-9-2015 by blogfast25]

aga - 1-9-2015 at 14:57

Proability Density is a bit hard for my brain.

Also the Toroidal formations of some orbitals cause brain-pain.

blogfast25 - 1-9-2015 at 15:17

 Quote: Originally posted by aga Proability Density is a bit hard for my brain. Also the Toroidal formations of some orbitals cause brain-pain. Please explain.

The toroidal formations cause everyone brain-pain! Look at the equation of the angular wave function Y for the 3d<sub>z2</sub> orbital (that's one of those with a torus):

http://winter.group.shef.ac.uk/orbitron/AOs/3d/equations.htm...

Don't lose your mind over it!

And I'll let you off on the probability thingy: just look at ψ<sup>2</sup> (at a given location) as proportional to the probability of finding the electron at that location. In areas of high ψ<sup>2</sup> you're more likely to find the electron.

[Edited on 2-9-2015 by blogfast25]

blogfast25 - 2-9-2015 at 07:48

Multi-electronic elements: Helium and beyond

So far we’ve only dealt with the simplest of all atoms, hydrogen, which contains only one electron. Helium, the next one up in the Periodic Table, contains two electrons.

We’ve seen that obtaining the eigenstates (wave functions and energy levels) can be obtained by solving the Schrӧdinger equation:

HΨ=EΨ

Where H is the Hamiltonian operator of the atom (the “total energy” operator, if you will).

An expression for the Helium Hamiltonian can be found here (Eq.9.2 to 9.5):

http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mecha...

Where V(r<sub>1</sub> and V(r<sub>2</sub> represent the potential energies caused by the electrostatic attractions between the nucleus and electron 1 and 2.

But note that there’s now an additional term V(r<sub>12</sub> which represents the potential energy caused by mutual electrostatic repulsion between both electrons 1 and 2. (Note: I believe the term V(r<sub>12</sub> should carry a positive sign, and not a negative on as on that page, because it is a repulsion. But the sign matters little for our purposes here).

For atoms with three electrons more potential energy terms need to be added, see Eq.9.7 for the general case with i electrons.

The main problem is that even for the case of helium the resulting Schrӧdinger equation is no longer analytically solvable. No finite steps algorithm exists to find general expressions for Ψ and E (the eigenstates).

The He SE can be solved only numerically with so-called iteration algorithms (to great precision) but numerical methods do not yield algebraic expressions for Ψ and E like the ones we obtained for the Pi1DB, the Quantum Harmonic Oscillator and the Hydrogen atom.

In order to solve the problem of electron occupation in multi-electron atoms physicists have resorted to developing a number of rules, often heuristically, sometimes empirically, sometimes formally, that help determine the electron configuration of atoms from He to beyond.

As an end to this installment I’ll introduce one of the most important of these rules:

The Pauli Exclusion Principle (PEP):

 Quote: It is impossible for two electrons of a poly-electron atom to have the same values of the four quantum numbers (n, ℓ, m and ms.

Let’s read this kind of in reverse: it indicates that if 2 electrons share three quantum numbers, e.g. n = 1, l = 0, m<sub>l</sub> = 0, then as long as those two electrons have different quantum spin numbers m<sub>s</sub> then that’s okay? Yes!

Note that n = 1, l = 0, m<sub>l</sub> = 0 is the Ground State of Hydrogen, aka 1s, and that m<sub>s</sub> can only take on two values: + ½ or – ½.

So the PEP is telling us that the 1s orbital can accommodate 2 electrons, as long as they have different quantum spin numbers.

The ground state of Hydrogen is represented by 1s<sup>1</sup>, the ground state of Helium by 1s<sup>2</sup>. Because electron spin is often represented by an upward or downward arrow, these ground states are often depicted as:

The box symbolizes the 1s orbital, the arrows the electrons and their quantum spins. Note that for 1s<sup>1</sup> the arrow could also be pointing down, without violating anything.

But there are more ramifications for electron configurations flowing from the PEP. For instance, could we cram another electron into 1s<sup>2</sup>? The answer is no: the third electron would inevitably share its quantum spin number with either of the two others, thereby violating the PEP.

In short, 1s<sup>2</sup> is fine but 1s<sup>3</sup> is forbidden, verboten, haram, kaput. We’ll later see that this principle of electrons in atomic orbitals of forming doublets is general and which other rules are needed to explain the electron configurations of multi-electron atoms and even explain the Periodic Table itself!

[Edited on 3-9-2015 by blogfast25]

aga - 2-9-2015 at 11:28

 Quote: Originally posted by blogfast25 http://winter.group.shef.ac.uk/orbitron/AOs/3d/equations.htm... Don't lose your mind over it!

Too late - i looked.

OK, i've come to terms with the fact that i cannot follow the maths, however your clear explanations of what they mean are easy to follow.

 Quote: Originally posted by blogfast25 The He SE can be solved only numerically with so-called iteration algorithms

Does that mean that the algebra doesn't work out properly, so the equations have to be tried with real values over and over again to see what comes out ?

 Quote: Originally posted by blogfast25 The Pauli Exclusion Principle (PEP)

Can i safely assume that basically says that two particles cannot occupy the same space at the same time (they would not fit) ?

blogfast25 - 2-9-2015 at 12:17

An example of an iteration method for finding the roots of a function f(x), if calculus fails:

https://en.wikipedia.org/wiki/Secant_method#The_method

By repeating the process the iteration values x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, ...,x<sub>i</sub>, converge to a stable, 'true' value of x (the sought root). Of course the SE is a differential equation and requires a specific iteration approach but the philosophy is the same.

 Quote: Can i safely assume that basically says that two particles cannot occupy the same space at the same time (they would not fit) ?

Specifically, no 2 electrons can occupy the same state (here: orbital) if they have the same values for their four quantum numbers.

But take an orbital with n = 2, l = 1 and m<sub>l</sub> = 0, which is a p-suborbital. It can contain 2 electrons, provided one is spin up and one is spin down. That way both electrons occupy the same state, except for their quantum spin state.

[Edited on 2-9-2015 by blogfast25]

aga - 2-9-2015 at 12:58

So, given that their properties are identical in that n = 2, l = 1 and ml = 0, they must be different in some other way, hence spin Up and spin Down, introducing another property in which way they can be different.

Starting to sound kinda familar.

Darkstar - 2-9-2015 at 16:40

 Quote: Originally posted by aga Can i safely assume that basically says that two particles cannot occupy the same space at the same time (they would not fit) ?

They'll "fit" alright. With enough pressure, you can fit as many of them little bastards in the same space as you want. Under pressure, the first particles to go will be the protons and electrons. These will fuse with one another to create a bunch of neutrons, which are far more resistant to collapse. Now, in the event that the pressure is strong enough to overcome even the degeneracy pressure of those neutrons, they, too, will be forced to occupy the same space. And after those neutrons go, there will be nothing left to prevent or resist further collapse.

This is how most black holes form, by the way.

blogfast25 - 2-9-2015 at 17:41

Final instalment of Part I by the end of this week. Phew!

Darkstar, can you point to the page that contains that Table of Content you so kindly made? I'll update it myself here, before going to Part II. Thanks!

aga: please note that Darkstar is correct but to observe the things he describe requires electron pressures many, many orders of magnitude higher than what can be supplied by the electrostatic attraction between the nucleus and the atom's electrons.

[Edited on 3-9-2015 by blogfast25]

Darkstar - 2-9-2015 at 19:28

 Quote: Originally posted by blogfast25 Darkstar, can you point to the page that contains that Table of Content you so kindly made? I'll update it myself here, before going to Part II. Thanks!

Here is the post. Be sure to read the part in red before you add to it, though. I initially copied the link to each post directly out of my address bar only to end up having to redo them because they have a reference to the page the post is on. Since my posts per page is set to 50, the page I see a post on may not always be the same page someone else sees it on.

Anyway, just make sure the links have "&goto=search&" between the topic ID and post ID and not "&page=some number." You can get the correct link by simply right clicking on the "posted on..." link above the target post and selecting "copy link location" (or whatever it is for your browser). So for example, the link to my table of contents post should look like this:

and not this:

aga - 3-9-2015 at 10:42

 Quote: Originally posted by Darkstar They'll "fit" alright. With enough pressure, you can fit as many of them little bastards in the same space as you want.

That doesn't make much intuitive sense.

It would suggest that all the 3-dimensional matter in a black hole was compressed into the same place at the same time, which clearly is mind boggling.

Would that not be a case where the Pauli Exclusion Principle is violated ?

blogfast25 - 3-9-2015 at 11:17

 Quote: Originally posted by aga Would that not be a case where the Pauli Exclusion Principle is violated ?

Not really. The PEP actually applies to a broader class of subatomic particles called fermions (to which electrons belong). But at the pressure Darkstar is referring to electrons are squeezed into the protons, forming neutrons. Neutrons aren't fermions though, so they don't have to obey the PEP...

[Edited on 3-9-2015 by blogfast25]

aga - 3-9-2015 at 11:22

Electrons squashed into protons, now neutrons squashed into other neutrons ...

Is the core of a black hole actually a true singularity ?

Can this even be modelled by anything we have available, including QM ?

blogfast25 - 4-9-2015 at 09:32

 Quote: Originally posted by aga Is the core of a black hole actually a true singularity ? Can this even be modelled by anything we have available, including QM ?

I'll only answer questions that I feel comfortable with answering and this is not one of these.

But in about an hour (or so) the final instalment of Part I will be published.

blogfast25 - 4-9-2015 at 11:58

The Aufbau Principle:

As we’ve seen from the Pauli Exclusion Principle (PEP), a pair of electrons can occupy the same orbital or suborbital (same n, l and m<sub>l</sub>, as long as the quantum spin numbers of both electrons are different ([+1/2,-1/2] or [-1/2,+1/2]). So the ground state 1s<sup>2</sup> is allowed but 1s<sup>3</sup> violates the PEP.

So in the case of the third element Li, where does that third electron go? And what about the fourth for Be etc etc?

The so-called ‘Aufbau’ or filling of the atomic orbitals as more electrons are added to non-hydrogenic atoms is governed by a second rule, known as the Madelung Energy Ordering Rule, often referred to as the diagonal rule:

Following this rule (follow the right arrows for the filling order), the third Li electron ends up occupying 2s, so Li’s electron configuration becomes 1s<sup>2</sup>2s<sup>1</sup> and for Be we get 1s<sup>2</sup>2s<sup>2</sup> (where in accordance with the PEP that fourth Be electron has different spin from that previous third one).

The electron configurations of Li and Be (ground states) thus can be represented as:

In our arsenal for successful ‘Aufbau’ we need one more element: Hund’s rule(s).

 Quote: Hund's rule states that: 1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. 2. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

The principle is well explained and illustrated here (but I'm not sure the link doesn't render properly here):

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurations/Hund's_Rules

With the Pauli Exclusion Principle, the Diagonal Rule and Hund’s Rules we have the tools needed to predict the electron configurations of the elements of the Periodic Table.

Example: Fe (Z = 26)

Electrons from 1<sup>st</sup> Period: 1s<sup>2</sup>
Electrons from 2<sup>nd</sup> Period: 2s<sup>2</sup> and 2p<sup>6</sup>
Electrons from 3<sup>rd</sup> Period: 3s<sup>2</sup> and 3p<sup>6</sup>

That’s 2 + 8 + 8 = 18 electrons so far.

Fourth period filling starts with 4s<sup>2</sup> but then, following the Diagonal rule continues by filling 3d, with the 6 remaining electrons, so 3d<sup>6</sup>. The distribution of these 6 electrons over the five 4d-suborbitals is given by Hund’s rules:

Obviously for high Z this can be a bit of a kerfuffle, so we can apply a shortcut. Take e.g. Te (Z = 52). Find the nearest Group 18 (Noble gases) element with a Z < 52, that is Kr (Z = 36) which is a Period 4 element. The electron configuration of Te is then that of Kr, with the added electrons to get to 52 electrons, so:

[Te] = [Kr]5s<sup>2</sup>4d<sup>10</sup>5p<sup>4</sup>

Importance of Outer electrons in Chemical Bonding

Getting near the end well of Part I this is probably a good place to remind readers that chemical bonds (the subject of Part II) only involve the outermost electrons (the so-called valence electrons) because the inner electrons are far too tightly bound to the nucleus to get involved in molecular orbitals (or ionic bonds).

A brief explanation of the Periodic Table of the Elements

When Dimitri Mendeleev in 1869 presented his version of the Periodic Table this ‘Eureka!’ moment was also accompanied by much wonderment as to what really caused this arrangement to be correct. It was only with the advent of QM that the explanation was found to be the way atomic orbitals fill up.

Using the modern representation of the Periodic Table, like here:

http://sciencenotes.org/printable-periodic-table/

... we can distinguish different areas, aka ‘blocks’ which are the direct consequence of atomic orbital filling (‘Aufbau’) in accordance with the laws and rules of QM.

Groups 1 and 2 (formerly IA and IIA) as known as the s-block because the outer (valence) electrons of the atoms of the elements in that group are assigned to s-orbitals.

Groups 13 through to 18 (formerly IIIA to VIIIA) are called the p-block because the outer valence orbital have been filled up with (three) p-orbitals.

Groups 3 through to 12 are called d-block elements. From Period 4 something interesting happens: when calcium’s outer orbital 4s is full (4s<sup>2</sup> the diagonal rule tells us that the orbital filling now continues at 3d (and not 4p as might otherwise be expected). There are five d-suborbitals, each accommodating 2 electrons, so the d-block is 10 groups wide. The same effect repeats at periods 5,6 and 7 where orbital filling is respectively via 4d, 5d and 6d.

But in periods 6 and 7 something similar happens. For elements 57 to 71 (Lanthanides) orbital filling proceeds via 4f, for elements 89 to 103 (Actinides) orbital filling proceeds via 5f. There are seven f-suborbitals, each accommodating 2 electrons, so the f-block is 14 groups wide (together the Lanthanides and Actinides are the f-block).

[Edited on 4-9-2015 by blogfast25]

aga - 4-9-2015 at 13:56

 Quote: [url=http://che...

Missed the closing square bracket on the starting url bit is all, and put an = in there instead.

How on earth did Mendeleev work out the PT ?

Presumably not QM ?

[Edited on 4-9-2015 by aga]

blogfast25 - 4-9-2015 at 14:13

Quote: Originally posted by aga
 Quote: [url=http://che...

Missed the closing square bracket on the starting url bit is all, and put an = in there instead.

How on earth did Mendeleev work out the PT ?

Presumably not QM ?

Nope. Still doesn't render.

People had been working for a PT type arrangement for a long time with mixed results, as had Mendeleev. It kind of 'came to him', in a moment of inspiration.

blogfast25 - 4-9-2015 at 14:16

Part I – Basic Wave Mechanics

aga - 4-9-2015 at 14:17

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_C...

That one gets close enough.

Little_Ghost_again - 4-9-2015 at 14:51

I am following but most of it is above my level! I am enjoying it alot though. I just thought I would make you aware I was lurking but trying not to ask questions that made me sound stupid .
blogfast25 - 4-9-2015 at 15:18

Thanks, L_G_A, I appreciate it!
Little_Ghost_again - 5-9-2015 at 01:56

I wont disturb this thread but I might have a question in a couple of days.
I have a old book on valence theory, so while following some of this thread I have also read some of the book. In places the book seems to be??? (wrong isnt the right word but close), I am wondering if its more a case that a pretty old book on this is now too out of date to be much use?
I cant formulate a decent question at the moment as I dont understand some of the maths (I am working on it), but as soon as I can identify what I think the discrepancy is I will post the question.
I know its a science forum but a maths section would be cool! some of us dont have great maths but want to learn

Darkstar - 5-9-2015 at 04:15

 Quote: Originally posted by Little_Ghost_again I have a old book on valence theory, so while following some of this thread I have also read some of the book. In places the book seems to be??? (wrong isnt the right word but close), I am wondering if its more a case that a pretty old book on this is now too out of date to be much use?

What level is the book? I'm guessing it's more of a beginner-level book? It could just be that the explanations given here are just slightly more advanced than what is written in the book, making the book's explanations seem almost wrong but not quite. This is a common issue in just about all of the natural sciences, with physics being the most notorious of the bunch. Basically, many times the explanation you're given when first introduced to a concept is seen as not really being correct from a more advanced level of view. To make matters worse, sometimes even the more advanced level's view is also considered incorrect once you get even more advanced. It's kind of a pain in the ass, honestly.

For example, as I mentioned in an earlier post, you're initially taught that particles also behave like waves, and exhibit this sort of "wave-particle duality" where they behave like both waves and particles at the same time. That's all fine and dandy when you're just starting out, as it helps you to stop thinking about particles in the classical way that you're used to; however, once you start getting more advanced, you learn that the truth of the matter is that, while a quantum object does sometimes behave like a wave and other times like a particle, the vast majority of the time it behaves like neither. And when that particle does "behave like a wave," it isn't literally a real, physical wave that propagates through three-dimensional space, but rather one that exists in an abstract mathematical space called Hilbert space. The problem, however, is that the latter explanation is not exactly beginner-friendly.

A similar example in chemistry would be how acids are initially taught. At first, you're taught that things like HCl dissolve in water to give a bunch of free protons in solution. It's only later that you learn it actually protonates the water molecules and produces a bunch of hydronium ions. And even then, once you get slightly more advanced, you find out that there's still more to it. (Grotthuss mechanism)

To put it simply: you have to crawl before you walk!

When I was a student I learned a lot from an old book by Pauling and Wilson, "Introduction to Quantum Mechanics with Applications to Chemistry" or something like that. It's available as a Dover reprint pretty cheap. I thought it was really well written and easy to understand.

Darkstar---I wouldn't say that the waves are in Hilbert space, but you are right that they are not in real (3-dimensional) space. Rather, the waves are in configuration space. For example, the wave function for the electrons in helium move in a six-dimensional space (configuration space) with coordinates (x1,y1,z1,x2,y2,z2), the xyz coordinates of the two electrons.

Hilbert space is an abstract space in which a single point represents an entire wave function.

Also, it would be equally valid to say the wave lives in momentum space. See section 2 of my notes,
http://bohr.physics.berkeley.edu/classes/221/1112/notes/hilb...

blogfast25 - 5-9-2015 at 08:16

 Quote: Originally posted by Little_Ghost_again I have a old book on valence theory, so while following some of this thread I have also read some of the book. In places the book seems to be??? (wrong isnt the right word but close), I am wondering if its more a case that a pretty old book on this is now too out of date to be much use?

Text books (as opposed to peer reviewed papers) often place too much emphasis on valence theory to explain which compounds can exist and which not.

For a compound to be 'stable' (in itself a relative and subjective term) it's Change of Free Energy of Formation needs to be negative. Although considerations of AO and MO definitely play a part in this, other factors like entropy changes and lattice energies do to.

That's why simple 'rules' like the Octet Rule often work but have many exceptions. Nature doesn't always comply to our sometimes reductionist theories.

blogfast25 - 5-9-2015 at 08:23

 Quote: Originally posted by annaandherdad Also, it would be equally valid to say the wave lives in momentum space. See section 2 of my notes, http://bohr.physics.berkeley.edu/classes/221/1112/notes/hilb...

What do you mean by 'my notes'?

I've been trying to 'transition' my understanding of QM in R<sup>3</sup> to the Hilbert space and am finding it hard. Will try and read these notes. Ta.

aga - 5-9-2015 at 12:20

 Quote: Originally posted by blogfast25 I've been trying to 'transition' my understanding of QM in R3 to the Hilbert space

What's R<sup>3</sup> ?

Edit:

Apologies. Looked it up and found it means Euclidean Space.

Hilbert Space looks pretty much like the bunny, almost.

[Edited on 5-9-2015 by aga]

 Quote: Originally posted by blogfast25 What do you mean by 'my notes'?

I wrote them, for a class I used to teach.

blogfast25 - 7-9-2015 at 07:12

 Quote: Originally posted by blogfast25 What do you mean by 'my notes'?

I wrote them, for a class I used to teach.

Wow, so we have a real QMechanic in our midst, perhaps even two with Darkstar. I'm impressed.

blogfast25 - 12-9-2015 at 11:50

Part II: Applications of Wave Mechanics in Chemical Bond Theory

This section will possibly be even more ‘math-lite’ than Part I and will treat chemical bonds mainly from a VSEPR (Valence Shell Electron Pair Repulsion) point of view.

The Mother of all Covalent Bonds: the σ<sub>ss</sub> bond in Molecular Dihydrogen

Imagine two hydrogen atoms in the ground state (1s<sup>1</sup> approaching each other as in the diagram below:

Remember that both wave functions + ψ and – ψ satisfy the Schrödinger equation. In the top part of the diagram the wave functions have the same sign (either +,+ or -,-) and are said to be ‘in phase’. In the bottom part the wave functions have the opposite sign (either +,- or -,+) and are said to be ‘out of phase’.

Like actual waves, the wave functions can now show reinforcing interference when they are in phase (top) or negative interference when they out of phase (bottom).

In the top situation a bonding molecular orbital has formed. See how in the top part the value for ψ<sup>2</sup> between the nuclei (represented as dots) is positive. This means there is considerable probability of finding the electron(s) in that region and this is where they reduce the electrostatic repulsion between the (both positively charged) nuclei. A σ<sub>ss</sub> (σ for short) has formed and a H<sub>2</sub> is born.

But in the bottom part for ψ<sup>2</sup> between the nuclei is zero and with no electron presence in that region there is nothing to prevent the electrostatic repulsion between the nuclei from ripping the ensemble apart. This orbital is called an anti-bonding molecular orbital, noted as σ<sup>*</sup>.

Graphical representations of bonding an anti-bonding molecular can be found at the following links:

Bonding:

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

Anti-bonding:

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

Molecular orbitals must also obey Pauli’s Exclusion Principle, so one electron is the σ orbital must be spin up and the other spin down. Also, the molecular orbital cannot accommodate a third electron.

In box representation we can write:

[Edited on 12-9-2015 by blogfast25]

aga - 12-9-2015 at 12:34

Wooohoo ! Applications !

Is sigma ss the sum of the two s orbitals ?

blogfast25 - 12-9-2015 at 14:30

 Quote: Originally posted by aga Wooohoo ! Applications ! Is sigma ss the sum of the two s orbitals ?

In a sense. In textbooks you'll read that 'MOs are linear combinations of AOs'. This 'LCAO' approach is an approximation that works remarkably well.

blogfast25 - 14-9-2015 at 11:20

Bonding and anti-bonding MOs between p-orbitals and between s and p-orbitals

First, a short note on the x, y, z orientation of (the three) p atomic suborbitals. There are three p suborbitals, denoted as p<sub>x</sub>, p<sub>y</sub> and p<sub>z</sub>:

• when the lobes of the p suborbital lie on the same axis that connects the nuclei of the atoms being bonded, we call that orientation x.
• when the lobes of the p suborbital are perpendicular to the axis that connects the nuclei of the atoms being bonded, we call that orientation y or z.

Whether or not atomic orbitals can combine into molecular orbital depends (mainly) on two factors:

1. Symmetry considerations which are best illustrated below:

In the cases (a) and (b), lack of symmetry prevents a molecular orbital from being formed. Only in the case (c) is there sufficient symmetry for bonding to occur.

2. Energy requirements:

For an MO to be formed both contributing AOs need to be of energies not too different (case (c)). MOs resulting from interaction between s an p orbitals are for that reason comparatively rare (an example is HF, hydrogen fluoride).

σ bonding and anti-bonding MOs formed from two p<sub>x</sub> AOs:

π bonding and anti-bonding MOs formed from two p<sub>y</sub> (or p<sub>z</sub>, that makes no difference here) AOs:

In both figures same-coloured lobes are in phase, different coloured ones are out of phase.

Summarising (as well as over-simplifying) so far we have:

A bonding Molecular Orbital (MO) contains 2 electrons (one spin 'up', one spin 'down'). One is 'donated' by the first atom in the bond, the second by the second atom in the bond. Bonding MOs only form between electrons that are part of the valence electrons, i.e. the outermost electrons. Inner electrons are too tightly bound to the nucleus to be able to take part in bonding MOs.

Bonding MOs arise between:

• 2 half filled s atomic orbitals.
• a half filled s atomic orbital and a half filled p<sub>x</sub> atomic orbital.
• 2 half filled p<sub>x</sub> atomic orbitals.
• 2 half filled p<sub>y</sub> or p<sub>z</sub> atomic orbitals. </sub>

These bonding MOs are respectively named:

• sigma (σ) s-s bond. σ<sub>2s</sub> for short.
• sigma (σ) s-p bond. σ<sub>sp</sub> for short.
• sigma (σ) p-p bond. σ<sub>2p</sub> for short.
• Pi (π) p-p bond. π for short.

Here are some ‘artist’s impressions of:

σ<sub>2p</sub>:

http://winter.group.shef.ac.uk/orbitron/MOs/N2/2pz2pz-sigma/...

π:

http://winter.group.shef.ac.uk/orbitron/MOs/N2/2px2px-pi/ind...

[Edited on 14-9-2015 by blogfast25]

[Edited on 15-9-2015 by blogfast25]

aga - 14-9-2015 at 12:31

Cor blogers. You'r into mindf**k territory now.

Assuming that AO means atomic orbitals and MO means Molecular Orbitals.

Do Kinetics comes into play seeing as the Symmetry rule means that no bond can form by kinky approaching atoms, and only head-on approaches form bonds (vast generalisation). ?

I mean, is the Bond type also influenced by the angle of approach, or is that a function of the structure of the reacting species ?

The naming and the ordering of the bondings seems remarkably clear.

Remarkable in that it seems simple.

blogfast25 - 14-9-2015 at 16:11

aga:

For now try and forget about kinetics and 'approach angles': right now we're only using QM/WM to predict which molecular orbitals (and thus also which overall molecular shapes) are possible, not how they might come about in chemical reactions. That's a related yet also different topic.

Crawling, standing up, walking... maybe running much later!

aga - 14-9-2015 at 22:35

Oh. So it's more about predicting what can react with what, and if/where they could form bonds.

Is this anything to do with the meta / ortho / para stuff in OC (which i believe describes where on a ring bonds are formed) ?

Slitherin' before crawling !

[Edited on 15-9-2015 by aga]

blogfast25 - 15-9-2015 at 06:07

 Quote: Originally posted by aga 1. Oh. So it's more about predicting what can react with what, and if/where they could form bonds. 2. Is this anything to do with the meta / ortho / para stuff in OC (which i believe describes where on a ring bonds are formed) ? 3. Slitherin' before crawling !

1. At this point the only thing we are predicting is what kind of molecular orbital structures are 'allowed' (by QM/WM rules/laws). Of course that's related to chemical reactions but so far I've only really mentioned one chemical: dihydrogen! Chemical reactions involve the breaking and forming of chemical bonds.

Think of what I'm doing now as playing with ball and stick models.

2. Aromatic ring structures coming up very soon now.

3. You're very much part of the human zoo!

[Edited on 15-9-2015 by blogfast25]

aga - 15-9-2015 at 08:47

I await the next instalment with bated gill O2 absorbtion.

[Edited on 15-9-2015 by aga]

blogfast25 - 16-9-2015 at 08:25

Double bonds and triple bonds

Bonding MOs can be combined, between the same bonded nuclei. The double bond in CH<sub>2</sub> = CH<sub>2</sub> (ethene) is made of one σ s-p and one π p-p bond.

The triple bond in ethyn (acetylene, C<sub>2</sub>H<sub>2</sub> is made of one σ s-p and two π p-p bonds.

π bonds in fact never occur on their own. That would never be a stable spatial arrangement, as there are no electrons to shield the nuclei from each other: it would fly apart.

The double bond (e.g. C = C bond) can thus be represented as below, where the sigma p-p electron cloud (MO) provides the shielding of the nuclei, to prevent the thing falling apart from electrostatic repulsion between the positively charged nuclei:

Double and triple bonds also impart torsional rigidity to bonds. Imagine ethane (C<sub>2</sub>H<sub>6</sub> as two bonded methyl groups: H<sub>3</sub>C-CH<sub>3</sub>. The methyl groups can rotate with respect to each other around the inter-nuclear axis. But if we add a double bond (then we get ethene: H<sub>2</sub>C=CH<sub>2</sub> then the π bond prevents rotation of the methylene (CH<sub>2</sub> groups. Similarly, the CH groups in ethyn (C<sub>2</sub>H<sub>2</sub>, acetylene) cannot rotate because of the two π bonds.

Conjugated double bonds and Aromaticity

Lets look at the molecule 1,3 butadiene, CH<sub>2</sub>=CH-CH=CH<sub>2</sub> and number the atoms 1 to 4 from left to right.
The bonds between 1 and 2 and between 3 and 4 are double bonds: a σ<sub>2p</sub> and a π bond. Studies show unequivocally that the two π bonds do not exist separately but that instead the electron clouds merge:

The hyphens represent the unaffected sigma bonds. Both lobes of the merged π bonds together contain a total 4 electrons (2 spin up, 2 spin down). The bond lengths of 1-2, 2-3 and 3-4 are identical.

Wherever double bonds are conjugated, for example CH<sub>3</sub>-CH<sub>2</sub>-CH=CH-CH=CH-CH<sub>2</sub>-CH<sub>3</sub> (3,5 octadiene), this merging occurs. Perhaps the best known example is benzene, C<sub>6</sub>H<sub>6</sub>, where three conjugated bonds merge into a single π torus electron cloud:

Hydrogen atoms, each bonded with one sigma bond to the C atoms, are not shown here. Both annular electron clouds contain a total of 6 electrons (three spin up, three spin down). All bond lengths are identical.

However, both in the case of conjugated dienes and benzene , this electron cloud overlapping must not be seen as a linear combination or the respective π MOs. Such a linear combination would strongly violate Pauli’s rule of only two spin-opposing electrons per MO.

Below an image of the iso-probability density surfaces for the π electrons (as based on a real calculation with ORCA software). The smoothness and continuity of the orbital surfaces creates the illusion of orbital merger but in reality the three π orbitals continue to exist, each with only two paired electrons.

[Edited on 16-9-2015 by blogfast25]

aga - 19-9-2015 at 11:11

So, if i got this right, the nucelii in a molecule are sandwiched together by electrons whizzing about in the various orbitals that surround them.

No orbital can contain more than 2 electrons : 1 in spin Up the other in spin Down state.

Where those systems are stable defines which and how the molecular nucleii can be contained inside that system.

Not sure why the py and pz bond is called a Pi bond, seeing as it's just a Sigma bond with different orbitals, with the naming being seemingly arbitrary ...

The dz<sup>2</sup> orbital appears to be the first torroidal-containing orbital, which would kind of cement the whole orientation thing, fixing where x y and z have to be.

Ok. Hands up.

Still not clear why the py/pz or py/py or pz/pz bond is called a Pi bond instead of a Sigma pyz/pzz/pyy bond.

[Edited on 19-9-2015 by aga]

aga - 19-9-2015 at 12:43

Aha !

Is it the case that a Sigma bond forms first, then there simply is no geometric room/space left to afford anything other than a Pi bond ?

I.e. the 'mating' px-es are occupied, leaving only the py or pz available for further bonding along that axis, and in that direction.

[Edited on 19-9-2015 by aga]

blogfast25 - 20-9-2015 at 11:38

aga:

The difference between σ and π MOs is non-trivial and not just a matter of fancy names. It should become even clearer after the next instalment but for now I’ll say this.

In all cases 1) to 5) below:

• Nuclei approach via the x-axis (x-axis is inter-nuclear axis)
• Atomic Orbitals contain 1 spin up electron (first atom) and 1 spin down (second atom).
• Atomic Orbitals are in phase.

1) Two 1s orbitals approach and interact to form a bonding σ(ss) MO.
2) One 1s orbital and 1 p<sub>x</sub> orbital approach and interact to form a bonding σ(sp) MO.
3) Two p<sub>x</sub> orbitals approach and interact to form a bonding σ(pp) MO.
4) Two p<sub>y</sub> orbitals approach and interact to form a bonding π(pp) MO.
5) Two p<sub>z</sub> orbitals approach and interact to form a bonding π(pp) MO.

(note that p<sub>y</sub> or p<sub>z</sub> orbitals cannot interact with s orbitals, due to poor symmetry (lack of overlap), see also next instalment)

In short, σ MOs are formed when the interacting atomic orbitals have electron density symmetry around the inter-nuclear axis. p<sub>y</sub> and p<sub>z</sub> have electron densities that lie above and below the inter-nuclear axis, so they can’t form σ MOs, only π MOs.

[Edited on 20-9-2015 by blogfast25]

aga - 20-9-2015 at 11:47

OK. Thanks for the explanation.

Seems pretty clear (best check) :-

s-s, s-px, px-px bonds are Sigma bonds.

py-py, py-pz, pz-pz bonds are Pi bonds.

blogfast25 - 20-9-2015 at 11:57

Amen to that!
aga - 20-9-2015 at 12:52

Sorry if this sounds silly : it seems clear that a Pi bond could never be a simple shape like a sigma bond orbital could be.

The maths are beyond me, however the principles (and pictures) seem very clear.

Given that the Py and Pz orbitals are oriented at angles to the Px axis, and are dislocated/unconnected in any case, the toroidal orbital above-and-below seems like a fair guess at how they'd look.

Personally i imagine it would be a ring, but not uniform in radius or width.

As there is now a second attractant (the second nucleus) surely the opposite side of the orbital would be shortened, creating more of an eliptical orbital rather than a spherical one.

Hmm. Even with an s-s bond the distance from the 'edge' of the orbital (if there is such a thing) to the nucleus should be reduced when compared to an unbonded atom.

blogfast25 - 21-9-2015 at 11:19

aga:

Try not get too hung up on the shapes: those shiny, hard looking surfaces are only surfaces of iso ψ<sup>2</sup> values. Orbitals are 'fuzzy' objects, they have no clear boundaries.

We use these shapes to demonstrate and illustrate how AOs interact with each other to form MOs.

[Edited on 21-9-2015 by blogfast25]

blogfast25 - 21-9-2015 at 11:27

Molecular shapes and molecular orbital repulsion

Let’s look the bond structure and shape of the water molecule. Oxygen’s ground state is 2s<sup>2</sup> 2p<sup>4</sup>. According to Hund, the four 2p electrons are distributed as 2 paired in p<sub>x</sub> and the two remaining unpaired ones in a p<sub>y</sub> and a p<sub>z</sub> orbital. These latter ones form σ(sp) with the 1s<sup>1</sup> hydrogen orbitals, which would strongly suggest a bond angle (between the H-O bonds) of 90 degrees, as in the Lewis structure below (note that Lewis notation isn’t really intended to show bond real angles):

In reality, the structure is closer to that of a tetrahedron, with the two hydrogens in two corners and the two lone paired AOs in the remaining two corners. That would give a bond angle (between the HO bonds) of 109.5 degrees.

Experiment finds these bond angles to actually be slightly smaller: 104.5 degrees.

The reason that the bond is neither 90 nor 109.5 degrees is due to electrostatic repulsion between the electron pairs (AOs or MOs): as electrons are negatively charged and each AO/MO contains 2 of them, electrostatic repulsion between them is inevitable.

The repulsion between electron pairs (orbitals) decreases in the following order:

Lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair.

The larger repulsion between the two lone pair electrons, for instance in the case of water, causes the HO bond angle to be 104.5 degrees.

Molecular shapes often reflect this: molecular shape tries to minimise inter-MO repulsion because that decreases potential energy of the structure.

Below are the expected molecular shapes for EX<sub>n</sub> -type molecules (X = an exoatom, like H or a halogen):

[Edited on 21-9-2015 by blogfast25]

aga - 21-9-2015 at 11:41

Cor bloggers. Your're laying out so flat even i can follow it !

Yous should be Paid well to be a lecturer somewhere.

blogfast25 - 21-9-2015 at 12:04

 Quote: Originally posted by aga Yous should be Paid well to be a lecturer somewhere.

Nah, I'm rubbish at crowd control.

aga - 21-9-2015 at 12:55

No need.

I feel sure you'd be able to make 2-Chlorobenzalmalononitrile for that.

If you're an IOC purist, H2S would do.

blogfast25 - 22-9-2015 at 07:32

Hybridisation

The valence electrons of the elements Be through to C often undergo so-called hybridisation.

Hybridisation helps explain bond structure and molecular shape.

Take the example of Boron, [He]2s<sup>2</sup> 2p<sup>1</sup>. Hybridisation causes a new type of atomic orbital (AO) to form, termed sp<sup>2</sup>. There are three sp<sup>2</sup> orbitals, each contain 1 unpaired electron and all three orbitals are at the same energy level.

Spatially, they are arranged like this:

The three sp<sup>2</sup> orbitals much resemble the 'half lobes' of a p<sub>x</sub> orbital and are represented here by thick lines. The orbitals lie in a single plane and at 120 degree angles of each other.

They form sigma bonds with s or p<sub>x</sub> atomic orbitals, like p<sub>x</sub> orbitals do.

sp<sup>2</sup> hybrid orbitals and the bonding MOs they form explain why B forms planar, triangular molecules like BH<sub>3</sub> and BCl<sub>3</sub>.

sp<sup>2</sup> hybridisation also explains the shape of molecules like CH<sub>2</sub>=CH<sub>2</sub>.

The Ground State for carbon is 2s<sup>2</sup> 2p<sup>2</sup>. As sp<sup>2</sup> hybridisation causes one of the 2s to jump to the 2p orbital, and 3 of the 4 electrons then form half-occupied sp<sup>2</sup> orbitals but the p<sub>z</sub> orbital remains unaffected (and half-occupied). The 3 sp<sup>2</sup> orbitals lie in the plane and are represented here by thick black lines, the remaining p<sub>z</sub> orbital is in green and is vertical to the plane:

When the atoms approach, the sp<sup>2</sup> orbitals form a σ MO and the pz orbitals a π MO. The remaining 4 sp<sup>2</sup> orbitals bond with 1s<sup>1</sup> hydrogen AOs.

sp<sup>3</sup> hybridisation: 4 unpaired electrons each in one in four iso-energetic orbitals. Spatially they arrange as a tetrahedron with the nucleus at the centre and the lobes pointing to the corners of the tetrahedron. Explains the tetrahedral structure of CH<sub>4</sub>, CX<sub>4</sub> (carbon halides) and the shape of alkanes in general.

Hybridisation can be interpreted as the need to minimise the electrostatic repulsion between MOs. In each case (sp, sp2 and sp3) the angles between the resulting MOs is such that electrostatic repulsion between them is minimal (but not zero).

NH<sub>3</sub> is another example. Nitrogen's valence electrons 2s<sup>2</sup> 2p<sup>3</sup> also undergo an sp3 hybridisation, albeit with a small difference. One of the sp<sup>3</sup> orbitals contains a full pair of electrons (paired up and down), the three others each one unpaired electron (three half-filled sp<sup>3</sup> orbitals).

The latter three orbitals combine with 3 hydrogen 1s1 atomic orbitals to form NH<sub>3</sub> by means of sigma bonds.

The result is a tetrahydron with three N-H sigma bonds pointing 'downwards' and one lone electron pair pointing 'upwards'.

The lone electron pair that is left after hydrogen bnding explains why ammonium ions (NH<sub>4</sub><sup>+</sup> exist: with a proton (H+, no electrons) the lone electron pair of NH3 can form a bonding sigma orbital. NH<sub>4</sub><sup>+</sup> is also a tetrahydron. All four N-H sigma bonds in NH>sub>4</sub><sup>+</sup> are identical in all respects.

Water and the oxonium ion

In the preceding section I told a little white lie. The structure and shape of the H2O molecule can also explained by means of sp3 hybridisation of the 2s2 2p4 valence electrons of oxygen. Two of the four tetrahydrally oriented sp3 orbitals are fully filled (2 opposing spin electrons), two contain a lone electron. The latter two form sigma bonds with the 1s1 electrons of two hydrogen atoms. And as explained above, the high electrostatic repulsion between the two unbonded sp3 orbitals then forces the H-O bond angle to 104.5 degrees.

If one of the full sp3 orbitals forms a sigma bond with a proton (H+), this is the oxonium (aka hydronium) ion H3O+.

Special mini-interlude for aga:

Because I know he’s been dying to see some chemical reaction explained by QM/WM, here we have our first simple example: the auto-dissociation of water:

H<sub>2</sub>O(aq) + H<sub>2</sub>O(aq) < === > H<sub>3</sub>O<sup>+</sup>(aq) + OH<sup>-</sup>(aq)

This (much left leaning) equilibrium is made possible by one of these full but unbonded sp3 orbitals ‘donating’ its electrons to a proton and forming a covalent σ, identical to the two other H-O bonds in water.

[Edited on 23-9-2015 by blogfast25]

aga - 22-9-2015 at 13:19

The hybridised thing troubles my addled mind, but not so much as the Equilibrium.

What powers the constant to-and-fro of elemental/hydronium ion formations ?

If we assume that electrons are always in motion, then that is a clue i suppose.

Why does hybridisation happen ?

Would an opposing pair of electrons simply prefer their own room ?

(i would: i hate sharing my room with an opposing electron).

blogfast25 - 22-9-2015 at 14:03

Oooopppssieie: massive typo alert (already fixed), that reaction should have been (OF COURSE):

2 H<sub>2</sub>O(aq) < === > H<sub>3</sub>O<sup>+</sup>(aq) + OH<sup>-</sup>(aq)

The equilibrium constant for this reaction is:

K<sub>w</sub> = [H<sub>3</sub>O<sup>+</sup>] x [OH<sup>-</sup>] = 10<sup>-14</sup>

Hybridisation happens because it provides outcomes of minimum inter-MO electrostatic repulsion, therefore minimum potential energy (stored in the molecule as electrostatic potential).

[Edited on 22-9-2015 by blogfast25]

aga - 22-9-2015 at 14:09

One did not wish to trouble one with such simplicities, seeing as the subject matter is of far greater import.
blogfast25 - 22-9-2015 at 14:16

 Quote: Originally posted by aga One did not wish to trouble one with such simplicities, seeing as the subject matter is of far greater import.

You say that but this could have been the biggest SM bloopers/outtakes since Fry and Laurie!

Blooper free: without hybridisation, the shapes for BH3, CH4, H2O and NH3 would predict 90 degrees angles between the E-H bonds.

[Edited on 22-9-2015 by blogfast25]

aga - 24-9-2015 at 14:14

Why does Hybridisation happen ?

Is it simply that electrons prefer to NOT share an orbital ?

Edit:

Sorry.

[Edited on 24-9-2015 by aga]

Metacelsus - 25-9-2015 at 04:40

I've been reading more about quantum mechanics and wavefunctions. Could someone explain to me what a Hamiltonian is, and how to use one? (I know it's something related to the total energy of the system, but how does it relate to the Schrödinger equation?)
blogfast25 - 25-9-2015 at 06:29

Let's look at the Schrödinger equation in one dimension:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/scheq.htm...

(try and keep this window open)

It's a second order differential equation and now look at the different terms, from left to right:

a) First term:

-(ћ<sup>2</sup>/2m) δ<sup>2</sup>ψ(x)/δx<sup>2</sup>

(With ћ = h/(2π) (“h bar”))

It tells you to take the second derivative of the system's wave function ψ(x) and multiply the result with -(ћ<sup>2</sup>/2m).

b) Second term:

U(x)ψ(x) with U(x) the potential energy function.

So this term means 'multiply the wave function with the potential energy function of the system'.

We often combine the first and second term using operator algebra:

[-(ћ<sup>2</sup>/2m)δ<sup>2</sup>/δx<sup>2</sup>+U(x)]ψ(x) = Hψ(x), where H is the Hamiltonian Operator, which acts on the wave function ψ(x).

c) Third term, right to the equal sign:

Eψ(x) where E is the total energy of the system.

This term means 'multiply the total energy of the system with the wave function'.

Put it all together and the SE means:

'The Hamiltonian Operator H acting on the wave function ψ(x) yields the product of the total energy of the system E and the wave function ψ(x)'.

In mathematical formulation you get one of the most famous equations in physics:

Hψ=Eψ

[Edited on 25-9-2015 by blogfast25]

 Quote: Originally posted by Cheddite Cheese I've been reading more about quantum mechanics and wavefunctions. Could someone explain to me what a Hamiltonian is, and how to use one? (I know it's something related to the total energy of the system, but how does it relate to the Schrödinger equation?)

In quantum mechanics observables are associated with operators that act on wave functions. So anything you observe, the position of an electron, its energy, the angular momentum of a molecule, the spin of a nucleus, the momentum of an atom, the electric or magnetic field at a point of space---there is an operator corresponding to each of these.

In classical mechanics an observable is just a function of the dynamical state of the system. For example, the kinetic energy of a particle is (1/2) mv^2. The "dynamical state of the system" means its position and velocity, and the kinetic energy is a function of the velocity. The total energy is the kinetic energy plus the potential energy, V(x), which depends on the position x. So the total energy is a function of the position and velocity of the particle, that is, of its dynamical state. Similarly, the z-component of angular momentum is m(xv_y - yv_x), another function of the dynamical state. The linear momentum is mv.

In quantum mechanics the operator corresponding to the position of the particle is multiplication by x, that is, the x operator converts the wave function psi(x) into x psi(x), a new wave function. The operator corresponding to momentum (in one dimension) is -i hbar d/dx, a differential operator. When momentum acts on psi(x), it produces -i hbar d psi(x)/dx.

These rules for associating these two observables (position and momentum) with operators are not supposed to be obvious. They were worked out over a period of twenty years or so during the development of quantum mechanics.

But once these rules for position and momentum are known, then the operators corresponding to other observables become known. For this purpose it is convenient to express the classical observable as a function of position and momentum, rather than position and velocity. The conversion is easy, because p=mv. For example, the classical energy, expressed as a function of position and momentum, is

E = p^2/2m + V(x)

To get the corresponding operator in quantum mechanics, we just replace the classical x and p by their operators, x -> multiplication by x, p -> -i hbar d/dx. So, the energy operator, what is usually called the Hamiltonian and written H instead of E, becomes

H= - (1/2m) hbar^2 d^2/dx^2 + V(x)

It is a second order differential operator.

In classical mechanics there is a sense in which the energy is the "generator" of time evolution, that is, the small correction to the dynamical state of a system after a small time dt can be expressed in terms of the energy. This idea carries over to quantum mechanics. If psi(x,t) is the wave function at time t, then the change in psi when t goes from t to t+dt is

psi(x,t+dt) - psi(x,t) = - dt (i/hbar) H psi(x,t)

where H is the Hamiltonian operator. Dividing this by dt, you get a differential equation in time,
i hbar partial psi(x,t)/partial t = H psi(x,t).

This is the Schro"dinger equation.

This is explained more carefully in my notes,

http://bohr.physics.berkeley.edu/classes/221/1112/notes/tevo...

These notes are for a second (graduate) course in QM, not an introductory course, so they are more advanced than you would find in an introductory book. However, I was at some pains to explain the subject carefully, so you may find them useful.

blogfast25 - 25-9-2015 at 06:53

A reminder for any 'Johnnie-come-latelys':

Table of content of Part I

Metacelsus - 25-9-2015 at 11:08

So, basically the Hamiltonian is an operator on the wavefunction, and its eigenstates represent allowed energy levels. I think I get it now.

Also, I just learned that whether or not operators commute is representative of whether or not observable quantities have an uncertainty relation. Quite interesting!

blogfast25 - 25-9-2015 at 11:21

 Quote: Originally posted by Cheddite Cheese So, basically the Hamiltonian is an operator on the wavefunction, and its eigenstates represent allowed energy levels. I think I get it now. Also, I just learned that whether or not operators commute is representative of whether or not observable quantities have an uncertainty relation. Quite interesting!

Yes, H is the Total Energy operator.

Other operators.

Oh, and it appears AAHD and me wrote our responses at almost exactly the same time. Her's is for the time dependent SE, mine's for the time independent SE.

For time independent eigenfunctions ψ(x) of the TISE with total energy E, you can find the time dependent equation as:

ψ(x,t) = ψ(x)e<sup>-iEt/ћ</sup> and at t = 0, ψ(x,0) = ψ(x). But it's only valid for eigenstates ψ(x), E.

[Edited on 25-9-2015 by blogfast25]

blogfast25 - 26-9-2015 at 06:00

Higher Hybridisations

Let’s look at the structure and bonding of the well known aluminium hexaqua cation [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>.

The 'naked ' Al<sup>3+</sup> ion has lost its valence electrons 3s2 3p1, so its level 3 orbitals are now empty.

The empty 3s orbital, three empty 3p suborbitals and two empty 3d suborbitals hybridise into six sp3d2 orbitals, all empty.

Each water molecule now donates one of its lone electron pairs (filled orbitals) to these empty hybrid atomic orbitals, resulting in the following structure.

Four of the bonds (represented by thin lines, water molecules not shown) lie in a plane, one points perpendicularly 'upwards', one perpendicularly 'downwards' (both represented by thick lines, water molecules not shown).

An ‘artist’s impression’ of a full sp3 water MO approaching an empty z-oriented sp3d2:

This kind of bonding is also referred to as dative and electron pair donors like water, ammonia and many others as Lewis bases. Conversely Al<sup>3+</sup> is a Lewis acid (electron pair receptor). The principle of dative bonding explains the structure of many hydrated cations.

A partial explanation of the colours of transition cation complexes

As an example we'll use the strongly blue coloured hydrated titanous(III) [Ti(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup> cation. The blue colour indicates that the green component of white light has been absorbed.

Titanium's electron configuration is [Ar]3d2 4s2 and the naked Ti3+ ion has lost one 3d and two 4s electrons, leaving it with [Ar]3d1.

The next orbitals are 4p, 5s and 4d orbitals which hybridise to six sp3d2 and these empty hybrid atomic orbitals accept one of the lone electron pairs of each H2O molecule, resulting in an octagonal structure similar to that of the [Al(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup> ion.

The 5 suborbitals of the unfilled 3d orbital, like the px, py and pz suborbitals of any p orbital, are in principle degenerated because they all have the same energy level.

The five d suborbitals are categorised into two subgroups: e and t.

A reminder of the shapes and orientations of the 3d suborbitals, here:

http://winter.group.shef.ac.uk/orbitron/AOs/3d/index.html

In the first figure orbital lobes are represented by very thick lines. The e group contains the d<sub>z2</sub> and d<sub>x2-y2</sub> suborbitals that have strong electron densities along the x, y and z axis.

The t group contains the d<sub>xy</sub>, d<sub>yz</sub> and d<sub>xz</sub> suborbitals that have strong electron densities in the xy, yz and xz planes but not along the x, y and z axis.

The presence of filled sp3d2 hybrids in the hydrated cation causes an energy difference to occur between the e and t group 3d suborbitals because of the difference in electron repulsion experienced between the sp3d2 orbitals and the e and t group suborbitals. The e suborbitals experience greater electron repulsion by the sp3d2 filled hybrids than the t suborbitals.

This small energy difference causes the lone 3d1 electron to be able to absorb energy (E<sub>photon</sub> = hf) from visible light photons (here, in the green part of the VIS spectrum) and 'toggle' between the higher and lower energy states of the e and t group suborbitals.

This general principle also applies to many other transition metal complexes with ligands like H<sub>2</sub>O, NH<sub>3</sub>, Cl<sup>-</sup> and many others.

[Edited on 26-9-2015 by blogfast25]

aga - 30-9-2015 at 13:39

 Quote: Originally posted by blogfast25 bonding of the well known aluminium hexaqua cation [Al(H2O)6]3+. The 'naked ' Al3+ ion has lost its valence electrons 3s2 3p1, so its level 3 orbitals are now empty. The empty 3s orbital, three empty 3p suborbitals and two empty 3d suborbitals hybridise into six sp3d2 orbitals, all empty.

How can an electron orbital be empty ?

With no electron, surely there is zero probability of finding one there.

blogfast25 - 30-9-2015 at 15:34

 Quote: Originally posted by aga How can an electron orbital be empty ? With no electron, surely there is zero probability of finding one there.

Ah, a question after my own heart: on existence/non-existence, mathematical objects, the 'Real' and the meaning of life!

Consider the transition from the ground state of carbon to its first excitation:

2s2 2p2 is carbon's ground state and 2s1 2p3 the first excitation (that explains C's tetravalence).

In the ground state, one of the three 2p orbitals is empty. So does this empty orbital really exist? Well, when the transition occurs, a 2s2 electron 'jumps' into the empty 2p<sub>z</sub> orbital... Or has that now half-filled 2p<sub>z</sub><sup>1</sup> been 'poofed' into existence, just for the occasion?

IMHO it matters not one iota: orbitals are mathematical objects that exist as descriptions of a Quantum state, whether they contain electrons or not. So perhaps the empty Al<sup>3+</sup> sp<sup>3</sup>d<sup>2</sup> hybrid orbitals exist or maybe they jump into existence when the dative lone electron pairs from water approaches them? I see no useful distinction here.

 Quote: With no electron, surely there is zero probability of finding one there.

That's a weak argument: even in electron filled orbitals there are poinst where electron probability: the areas between the lobes, known as nodes:

http://winter.group.shef.ac.uk/orbitron/AOs/2p/index.html

[Edited on 30-9-2015 by blogfast25]

blogfast25 - 3-10-2015 at 08:21

Electronegativity:

Electronegativity, χ<sup>P</sup>, was defined by Pauling as ‘the power of an atom in a molecule to attract electrons to itself’.

I won’t go into the quantitative definition of electronegativity because although Pauling’s version can be measured, it’s not an ‘objective’ measurement like ‘velocity’ or ‘mass’.

Below are the Pauling electronegativity values of some important elements, the higher the number the higher the power of an atom in a molecule to attract electrons to itself:

Although ‘electropositivity’ is an antonym of electronegativity, no value can be assigned to ‘electropositivity’. It’s therefore better to state:

‘Element E1 is less electronegative than element E2’ than to state ‘Element E1 is more electropositive than element E2’.

Polar diatomic molecules:

Consider the structure of a simple diatomic molecule like HBr (hydrogen bromide):

In it the Br nucleus is surrounded by three lone (non-bonding) electon pairs (orbitals) and one σ bonding molecular orbital (shown in red) which it shares with the H nucleus.

However, the Pauling electronegativity values of Br and H are 3.0 and 2.2 respectively, so Br is more electronegative than H and attracts these σ electrons more than does H. Quantum mechanically this means that the σ electron density (probability) will be higher on the left hand side of the (green) line of symmetry than on the right hand side.

This causes the Br to be partially negatively charged, indicated by δ-, and the H atom to be partially positively charged, indicated by δ+. By δ here should be understood 0 < δ < 1 (with respect to the unitary charge).

HBr is a permanent electrical dipole.

[Edited on 3-10-2015 by blogfast25]

aga - 3-10-2015 at 11:48

 Quote: That's a weak argument

Ooops. After another Can i see that it is a pointless (!) argument.

Given what's been explained so far, a 0 probability is just as valid as a 1 probability as it isn't a Static system in any way, shape or form.

The 0 probabilities just define a rough limit to the extent of an orbital.

Electronegativity :

This implies that the effective positive charge on a Proton is greater than the negative charge on an Electron, by some small amount.

(maybe the distance from the protons in the nucleus modulates the relative charge effect of electrons in different orbitals)

Does it follow that within a column of the periodic table that the more Protons an atom has then the more electronegative it is ?

blogfast25 - 3-10-2015 at 12:22

 Quote: Electronegativity : This implies that the effective positive charge on a Proton is greater than the negative charge on an Electron, by some small amount. (maybe the distance from the protons in the nucleus modulates the relative charge effect of electrons in different orbitals) Does it follow that within a column of the periodic table that the more Protons an atom has then the more electronegative it is ?

The charge of a proton and an electron are exactly the same (opposite in sign, of course).

The two electrons in the σ bonding orbital (represented by the red thickish line) spend on average more time left of the line of symmetry (the Br side of the molecule) than right of the line of symmetry (the H side of the molecule). That makes the left side partially negatively charged and the right side partially positively charged.

 Quote: Does it follow that within a column of the periodic table that the more Protons an atom has then the more electronegative it is ?

Look at the table and the values for χ of the halogens. Does that bear out your assertion? No.

It's more complicated than that but you're actually fairly close. Electronegative elements tend to be those whose outer atomic orbitals (valence electrons) tend to be close to the nucleus. These and bonded electrons then tend to feel the pull of the nucleus more strongly (electrostatic attraction being inversely proportional to electron/nucleus distance, squared).

Another important factor is shielding: the inner electrons (non-valence) tend to shield the outer ones from the nuclear electrostatic attraction. That explains why the alkali metals are low on the electronegativity scale.

χ is really hard to predict on purely theoretical grounds, so the tabled values are ALL empirical.

[Edited on 3-10-2015 by blogfast25]

aga - 3-10-2015 at 12:57

Huh ? Pauling Measured them ?!?!

There's a lot to think about here.

blogfast25 - 3-10-2015 at 13:15

 Quote: Originally posted by aga Huh ? Pauling Measured them ?!?! There's a lot to think about here.

Well, I'm sure he had a lot of help doing that! But he's one of the fathers of the concept.

[Edited on 3-10-2015 by blogfast25]

blogfast25 - 4-10-2015 at 06:06

Polar and non-polar molecules:

We saw with the example of H-Br a molecule that has a net electrical dipole, due to bond polarisation as a result of differing values of χP of H and Br. Such a molecule aligns itself in an electrical field with the negative side facing the anode and the positive side facing the cathode.

Since as most elements have different values for χP, bond polarisation is common. Now look at the following case of CF<sub>4</sub>:

The difference in χP between C and F is about 1.4 and the C-F bonds are significantly polarised. But despite this, CF<sub>4</sub> is not a polar (dipole) molecule and it doesn’t orient itself permanently in an electrical field. The four polarised bonds can be thought of as charge vectors all pointing in the same direction with the same magnitude and cancelling each other out.

Below are some examples of polar and non-polar molecules, above the line are the polar ones, below the non polar ones:

[Edited on 4-10-2015 by blogfast25]

Darkstar - 4-10-2015 at 10:10

Quote: Originally posted by blogfast25
 Quote: Does it follow that within a column of the periodic table that the more Protons an atom has then the more electronegative it is ?

Look at the table and the values for χ of the halogens. Does that bear out your assertion? No.

It's more complicated than that but you're actually fairly close. Electronegative elements tend to be those whose outer atomic orbitals (valence electrons) tend to be close to the nucleus. These and bonded electrons then tend to feel the pull of the nucleus more strongly (electrostatic attraction being inversely proportional to electron/nucleus distance, squared).

Another important factor is shielding: the inner electrons (non-valence) tend to shield the outer ones from the nuclear electrostatic attraction. That explains why the alkali metals are low on the electronegativity scale.

While we're on the topic of electronegativity and its relationship to atomic size and number of protons, I suppose it might be a good time to also point out the effect that atomic size has on charge strength/charge density of ions. For example, just like how fluorine is more electronegative than iodine because it is smaller in size with fewer electrons between its outer electrons and inner non-bonding electrons, the negative charge on a fluoride ion is also stronger than the negative charge on an iodide ion.

So while fluoride and iodide may both have the same overall charge of negative one, that charge is distributed over a much smaller area in fluoride than it is in the much larger iodide. Because of this, despite both ions having the same formal charge of negative one, the negative charge on fluoride is much, much, much stronger than it is on iodide. This same concept also applies to positive charges as well, for example the positive charge on a tiny lithium ion being much stronger than the positive charge on a giant cesium ion.

This explains why the bond strength of LiF is the strongest of the alkali-halide salts, and the bond strength of CsI the weakest. Also, this concept will later help you understand why the larger halides like bromide and iodide make better nucleophiles than the smaller ones like fluoride and chloride, as well as why carbon-fluorine bonds are so strong and carbon-iodine bonds are so weak.

blogfast25 - 4-10-2015 at 11:03

@Darkstar:

High central electrical field strength of small, multivalent cations like Al<sup>3+</sup> and Fe<sup>3+</sup> (list not exhaustive) also explains their ability to expel protons from their water mantle (for the solvated ions). i.e. undergo hydrolysis.

[Edited on 4-10-2015 by blogfast25]

aga - 4-10-2015 at 11:27

 Quote: Originally posted by Darkstar Another important factor is shielding: ... ... fluoride and iodide may both have the same overall charge of negative one, that charge is distributed over a much smaller area in fluoride than it is in the much larger iodide...

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.

blogfast25 - 4-10-2015 at 13:12

aga:

I'll let Darkstar answer that one.

Good news, though: 2 more instalments till X-mas, I mean end of course. Then THE EXAM! So start cramming now!

aga - 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.

blogfast25 - 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?

Darkstar - 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:

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?

blogfast25 - 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]

aga - 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 ?
blogfast25 - 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]

Darkstar - 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]

blogfast25 - 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.

Darkstar - 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.
aga - 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 ?

Darkstar - 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:

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.

blogfast25 - 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:

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.

aga - 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).

blogfast25 - 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'.

blogfast25 - 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:

(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:

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]

aga - 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]

blogfast25 - 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:

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?

[Edited on 7-10-2015 by blogfast25]

blogfast25 - 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]

aga - 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 ?

blogfast25 - 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:

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]

aga - 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]

blogfast25 - 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]

aga - 7-10-2015 at 14:38

So odd means No.

Antibond means No Bond Forms.

Out-of-phase means No bond.

OK. Got it.

blogfast25 - 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.

blogfast25 - 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).

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!

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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