Chemistry 101

[Karlos]

Well, getting back to QM, there is no motion (in the common sense of the word) in the orbital. It isn't even an orbit (which implies a classical style momentum) :-)

Spin is the name of a quantum property for which the best analogy is something spinning about an axis - but it is just an analogy, not the actual nature of the beast, just as an orbit does not really describe the nature of the QM orbital.
Just think of it as another quantum state. Spin is unusual in that it requires fractional values of n to satisfy the 2n+1 rule; for electrons, the values are +1/2 and -1/2. Your other basic quantum values are all integer.

As for the orbital shape, that's a bit more complex.

What you basically do to derive these shapes is to first derive the wavefunction for the orbital in question. The solutions for all the existing orbitals shapes (which are properties of the basic quantum numbers for orbital angular momentum etc.).

Now, you may (or may not) know that the probability of finding the electon at a specific point in 3D space is proportional to the square of the wavefunction at that point.

Once you know this, you can derive the "shape" of the obitals by defining a boundary surface that represents eg a 95% probability volume for the electron. That is to say there is a 95% chance the electron is somewhere inside that volume.

Naturally you cannot use a 100% probability volume as it would effectively fill the entire universe ;-)

[Karlos]

Off the top of my head, no chance :-D

An s orbital, would me more managable...

The wavefunction for the orbital where quantum n=1 (corresponding to a 1s orbital) looks like

Phi = (1/(pi.a0^3))^(1/2) . e^(-r/a0)

where a0 is the Bohr radius and r the radial distance (This is for the single charged nucleus only - for higher nuclei you need factor in the effect of increasing nuclear charge Z, and but the symmetry of the function doesn't change).

What you should observe is that this function is totally independent of the angle, so it is spehrically symmetrical.

The probability density is proportional to Phi^2, which for a 1s orbital gives us an e^-(2r/a0) curve.

You can also evaluate the probability of finding the electron in a shell of thickness dr at radius r. The volume of this shell is 4.pi.r^2.dr

Using this we can get

P.dr = Phi^2 . 4.pi.r^2.dr

Which tells gives us (for a 1s orbital)

P proportional to r^2.e^-(2r/a0)

If you plot this (P vs r), you find it has maximum probability at r=a0 - ie the classical Bohr radius corresponds to the most probable radius in quantum terms.

Now, for a p orbital it is a bit more complex. The wavefunction, Phi, is no longer simple and spherically symmetrical. The reason is that the p orbital has overall non-zero quantum angular momentum. If we choose our z-axis such that the p orbital for ml=0 has zero angular momentum about this axiz (but naturally will have it about x or y since it is non zero overall), we get a wavefunction with the following dependency

Phi = f(r).cos(theta)

...where theta is the 'polar' angle (that is, the angle subtended with the z-axis). This is our "pz" orbital wavefunction. The behaviour of this wavefunction gives us a node for the xy plane with "lobes" either side. The lobe on the +z side the xy plane is positive, on the other it is negative and is zero anywhere in the xy plane.

The probability is again determined by Phi^2, so we find we have a a cos^2(theta) in our probability function. The node becomes a region of zero probability and you end up with your familiar 2pz orbital shape.

[Karlos]

bjjones37 wrote:

Hey, I could have asked for an f orbital.:-D

Or a g, h, etc...

You can derive orbitals for any value of n. Also, there 2n-1 values for l:

n=1, l=1 (s orbital)
n=2, l=3 (3 equivalent p orbitals)
n=3, l=5 (5 equivalent d orbitals)
n=4, l=7 (7 equivalent f orbitals)

etc..

-edit-

BTW, I corrected an earlier remark I made about the 1s orbital. I said that for 2s you had to factor in the charge.

This is wrong - you can, of course, deterimine any orbital you want for any value of Z - I meant to say you need to factor in Z for higher nuclear charge - I chose the hydrogen case simply because Z=1 simplifies the wavefunction.

I mention it here because you quoted the 'incorrect' part. As it goes, I mention the effects on a 1s orbital for Z=2 later on in this post...

-/edit-

Could you clarify a0 (Bohr radius) and r (radial distance), using the nucleus as a frame of reference?

Yes. a0 is the classical radius, defined as distance from the centre of mass (nucleus) for the electron in a normal hydrogen atom - something like 5.291x10^-11 m.

r is defined as the distance from the nucleus to the point of interest.

Since r is the only parameter involved, you can tell the symmetry is spherical.

For other orbitals you get theta/phi angles (think polar coordinates relative to a z-axis) in the wavefunction. Ultimately it is these that affect the overall shape and symmetry of the orbitals.

Would not the effect of the nuclear charge Z be reducing with each increase of n, or is this assuming a larger number of nucleons in a stable atom?

It's a bit more complex than that. There is an expression in the full wavefunction (the hydrogen case has Z=1 and so the complexity falls away, which is why I chose it - even then it is complex enough!) that is proportional to Z^3/2.

For a 1s orbital, the effect of increasing Z is to pull in the radius of "highest probability" relative to hydrogen. Consequently, the effect is that the electron becomes more strongly bound, harder to ionize etc.

If you apply the full wavefunction to the Z=2 case for a 1 electron system (ie the He+ ion), you find it predicts accurately the properties (spectra, ionization potential etc) of the system.

[Karlos]

That's why I stuck to one electron examples.

Additional electrons increase the complexity of the wavefunction functions geometrically - the maths is at breaking point just trying to explain the 2 electron helium like system.

Most of the existing models work by treating each electron in turn as a single electron system, then applying the average effects of all the other electrons.

As it goes, the approximations work extremely well for single atoms.

You can then go on to approximate the mixing of wavefunctions that in turn gives rise to hybridisation, such as sp, sp2 and sp3 orbitals.

The same sort of approximations are used to derive the wave equations for multiple atom systems, from which we derive our quantum molecular orbitals, from which in turn we derive our theoretical basis for the chemical bond.

Likewise, these give extremely good predictions.

Mixing 2 orbitals (symmetry allowing) from 2 atoms gives you 2 new molecular orbitals - one of which is lower in energy (bonding) and one of which is higher (antibonding). You then populate these with the total number of electrons involved.

The bonding order is simply the sum of electrons occupying bonding orbitals minus the sum of those forced to enter the higher energy antibonding orbitals.

You normally assign a bond order of 1 to a pair of bonding electrons. Therefore the overall bond order is

((sum of electrons in bonding orbitals) - (sum of electrons in antibonding orbitals))/2

Just considering the mixing of a pair 1s orbitals in H/He, you can get

(H2)+ : 1 bonding : 0 antibonding : order = (1-0)/2 = 0.5
H2 : 2 bonding : 0 antibonding : order = (2-0)/2 = 1
HeH+ : 2 bonding : 0 antibonding : order = (2-0)/2 = 1
HeH : 2 bonding : 1 antibonding : order = (2-1)/2 = 0.5
He2 : 2 bonding : 2 antibonding : order = 0

Note that He2 would have a bond order of 0, which is why it does not exist. All of the others exist, albeit transiently for systems like HeH+/HeH (since they readily break down into H2 / He).

The prediction of fractional bond orders might seem bizzare, but they are the only models that can accurately describe the bonding in systems like boranes - you can't explain the bonding in B2H6 using any other system.

Quantum molecular orbitals and their application in chemical reaction (so called frontier orbital theory) also predict the nature of chemical reactions very well. Things like the reactivity of aromatics, heterocycles and many stereochemical phenomena come straight out of the maths.

It's great :-)

[Karlos]

bloodline wrote:
I used to love MO theory, it was one of the few things I was any good at :-D

Thankfully for us chemists, it is a lot easier to use than it is to derive from first principals ;-)

[Karlos]

To simplify some of Cymrics points, "sheilding" and "penetration" are two principal factors regarding electon-nucleus attraction at work when you scan across a period in the table. There are several others too, but I'll stick with these for now.

Sheilding is an effect electons have on each other as a result of their charge. Each electron effectively "screens" each other electron from some of the nuclear charge, reducing the effective strength with which the electrons are bound. This effect is not linear with respect to the electron count, and furthermore depends on the orbital of the electron you are looking at.

Penetration is a property of some wavefunctions for higher values of n that have regions of increased density closer the nucleus. If you took a slice through a 3s orbital, you'd find it had regions of increased density like concentric circles. Where you get such a region close to the nucleus, the effect is that that electron becomes more tightly bound.

The nuclear charge across a period increases linearly. This causes all of the electrons to bind more tightly, hence overall the ionization energy increases. However, thanks to the 2 effects above, you find this is not a simple function of Z. It depends on which orbital an electron is being removed, what the overall electronic configuration is etc. But you do see a pattern. The pattern overall shows that full and half full shells are favoured, which when analysed from a QM approach can be quantitatively shown to be more stable. Consequently oxidation states that lead to full (or half full which is common for d-orbital configurations in transition metals) shells tend to crop up.

When it comes to actual bonds, as cymric says there is a lot more to consider. From MO, a good 'covalent' chemical bond depends on

1) Good overlap between the respective atom's orbitals. This in turn requires the correct symmetry, size and wavefunction phase. This will produce a set of molecular orbitals that have a decent energy between their bonding and antibonding levels.

2) There should be fewer electrons than it takes to fill all the molecular orbitals (ie you want a ratio of bonding to antibonding greater than 1)

The classic covalent and ionic bonds are just different ends of the spectrum for MO. You get covalent character where the electron density is evenly spread between the atoms. You can gradually move to cases where the electron density begins to shift more and more to one of them, until you reach a point where you have a strong seperation of charge. This is the ionic end of the spectrum.

In reality, there are plenty of systems at both ends and in between. Best to "unlearn" the classic bond definitions as a rule and regard it as a handy approximation.

Go Quantum. Embrace Uncertianty.