Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

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.

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_z2 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 ψ² (at a given location) as proportional to the probability of finding the electron at that location. In areas of high ψ² you're more likely to find the electron.

[Edited on 2-9-2015 by blogfast25]

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(r₁ and V(r₂ 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(r₁₂ which represents the potential energy caused by mutual electrostatic repulsion between both electrons 1 and 2. (Note: I believe the term V(r₁₂ 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<sub>ℓ</sub> and m<sub>s</sub>.

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_l = 0, then as long as those two electrons have different quantum spin numbers m_s then that’s okay? Yes!

Note that n = 1, l = 0, m_l = 0 is the Ground State of Hydrogen, aka 1s, and that m_s 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 1s¹, the ground state of Helium by 1s². 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 1s¹ 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 1s²? 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, 1s² is fine but 1s³ 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]

Quote: Originally posted by blogfast25

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

Quote: Originally posted by blogfast25

The He SE can be solved only numerically with so-called iteration algorithms

Quote: Originally posted by blogfast25

The Pauli Exclusion Principle (PEP)

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) ?

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!

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.

Quote: Originally posted by aga

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

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.

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_l, 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 1s² is allowed but 1s³ 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:

https://en.wikipedia.org/wiki/Aufbau_principle#Madelung_ener...

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 1s²2s¹ and for Be we get 1s²2s² (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).

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^st Period: 1s²
Electrons from 2^nd Period: 2s² and 2p⁶
Electrons from 3^rd Period: 3s² and 3p⁶


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


Fourth period filling starts with 4s² but then, following the Diagonal rule continues by filling 3d, with the 6 remaining electrons, so 3d⁶. 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]5s²4d¹⁰5p⁴



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 (4s² 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]

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 ?

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 .

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

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?

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?

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...

Quote: Originally posted by blogfast25

I've been trying to 'transition' my understanding of QM in R<sup>3</sup> to the Hilbert space

Quote: Originally posted by blogfast25

What do you mean by 'my notes'?

Quote: Originally posted by annaandherdad

Quote: Originally posted by blogfast25   What do you mean by 'my notes'? I wrote them, for a class I used to teach.

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_ss bond in Molecular Dihydrogen

Imagine two hydrogen atoms in the ground state (1s¹ 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 ψ² 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_ss (σ for short) has formed and a H₂ is born.

But in the bottom part for ψ² 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^*.

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]

Quote: Originally posted by aga

Wooohoo ! Applications ! Is sigma ss the sum of the two s orbitals ?

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_x, p_y and p_z:

• 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_x AOs:



π bonding and anti-bonding MOs formed from two p_y (or p_z, 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_x atomic orbital.
• 2 half filled p_x atomic orbitals.
• 2 half filled p_y or p_z atomic orbitals. </sub>

These bonding MOs are respectively named:

• sigma (σ) s-s bond. σ_2s for short.
• sigma (σ) s-p bond. σ_sp for short.
• sigma (σ) p-p bond. σ_2p for short.
• Pi (π) p-p bond. π for short.

Here are some ‘artist’s impressions of:

σ_2p:

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:

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!

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 !

Double bonds and triple bonds

Bonding MOs can be combined, between the same bonded nuclei. The double bond in CH₂ = CH₂ (ethene) is made of one σ s-p and one π p-p bond.

The triple bond in ethyn (acetylene, C₂H₂ 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 (C₂H₆ as two bonded methyl groups: H₃C-CH₃. 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: H₂C=CH₂ then the π bond prevents rotation of the methylene (CH₂ groups. Similarly, the CH groups in ethyn (C₂H₂, acetylene) cannot rotate because of the two π bonds.


Conjugated double bonds and Aromaticity

Lets look at the molecule 1,3 butadiene, CH₂=CH-CH=CH₂ 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_2p 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 CH₃-CH₂-CH=CH-CH=CH-CH₂-CH₃ (3,5 octadiene), this merging occurs. Perhaps the best known example is benzene, C₆H₆, 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]

Molecular shapes and molecular orbital repulsion

Let’s look the bond structure and shape of the water molecule. Oxygen’s ground state is 2s² 2p⁴. According to Hund, the four 2p electrons are distributed as 2 paired in p_x and the two remaining unpaired ones in a p_y and a p_z orbital. These latter ones form σ(sp) with the 1s¹ 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_n -type molecules (X = an exoatom, like H or a halogen):





[Edited on 21-9-2015 by blogfast25]

Quote: Originally posted by aga

Yous should be Paid well to be a lecturer somewhere.

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]2s² 2p¹. Hybridisation causes a new type of atomic orbital (AO) to form, termed sp². There are three sp² orbitals, each contain 1 unpaired electron and all three orbitals are at the same energy level.

Spatially, they are arranged like this:



The three sp² orbitals much resemble the 'half lobes' of a p_x 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_x atomic orbitals, like p_x orbitals do.

sp² hybrid orbitals and the bonding MOs they form explain why B forms planar, triangular molecules like BH₃ and BCl₃.

sp² hybridisation also explains the shape of molecules like CH₂=CH₂.

The Ground State for carbon is 2s² 2p². As sp² hybridisation causes one of the 2s to jump to the 2p orbital, and 3 of the 4 electrons then form half-occupied sp² orbitals but the p_z orbital remains unaffected (and half-occupied). The 3 sp² orbitals lie in the plane and are represented here by thick black lines, the remaining p_z orbital is in green and is vertical to the plane:



When the atoms approach, the sp² orbitals form a σ MO and the pz orbitals a π MO. The remaining 4 sp² orbitals bond with 1s¹ hydrogen AOs.

sp³ 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 CH₄, CX₄ (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).

NH₃ is another example. Nitrogen's valence electrons 2s² 2p³ also undergo an sp3 hybridisation, albeit with a small difference. One of the sp³ orbitals contains a full pair of electrons (paired up and down), the three others each one unpaired electron (three half-filled sp³ orbitals).

The latter three orbitals combine with 3 hydrogen 1s1 atomic orbitals to form NH₃ 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 (NH₄⁺ exist: with a proton (H+, no electrons) the lone electron pair of NH3 can form a bonding sigma orbital. NH₄⁺ is also a tetrahydron. All four N-H sigma bonds in NH>sub>4</sub>⁺ 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:

H₂O(aq) + H₂O(aq) < === > H₃O⁺(aq) + OH^-(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]

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.

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?)

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!

Higher Hybridisations

Let’s look at the structure and 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.

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 Al³⁺ 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(H₂O)₆]³⁺ 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(H₂O)₆]³⁺ 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_z2 and d_x2-y2 suborbitals that have strong electron densities along the x, y and z axis.

The t group contains the d_xy, d_yz and d_xz 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_photon = 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 H₂O, NH₃, Cl^- and many others.


[Edited on 26-9-2015 by blogfast25]

Quote: Originally posted by blogfast25

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.

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_z orbital... Or has that now half-filled 2p_z¹ 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 Al³⁺ sp³d² 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.

Electronegativity:

Electronegativity, χ^P, 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]

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]

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 CF₄:



The difference in χP between C and F is about 1.4 and the C-F bonds are significantly polarised. But despite this, CF₄ 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]

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.

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...

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.

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?

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 ?

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 ?

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.

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.

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^- 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.

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.

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 1s¹ 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_g(1s), is lower than the energies of the atomic orbitals ψ₁ and ψ₂ of the individual H atoms. The anti-bonding MO σ^*_u(1s) remains of course empty.

In practice this means that the reaction:

H. + .H ===> H - H (or 2 H === >H₂

... is exothermic, i.e. ΔE < 0. ΔE is the bond enthalpy of the formation of the σ_g(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 H₂ === > C₂H₆, ΔH_f^298K

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 [H₂ === > 6 H.], costs 3 X + 435 = + 1305 kJ

3) C – C + 6 H. === > H₃C – CH₃, yields 6 X – 414 = - 2484 kJ

Adding up: ΔH_f^298K = - 368 + 1305 – 2484 = -1547 kJ per mol C₂H₆. NIST webbook lists the value of ΔH_f^298K 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_nH_2n+2(g) + Cl₂(g) === > C_nH_2n+1Cl(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 Cl₂ 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]

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 H₂ 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_σ - 2 E_ψ

Entropy? Huh? Who mentioned that?

[Edited on 7-10-2015 by blogfast25]

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]

Quote: Originally posted by aga

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

Quote: Originally posted by aga

So odd means No. Antibond means No Bond Forms. Out-of-phase means No bond. OK. Got it.

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_z 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_x,y orbitals.

2) Bonding π molecular orbitals are formed from the interaction of ungerate (u) p_z orbitals.

To paraphrase:

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]