Welcome, Guest. Please login or register.

Author Topic: Chemistry 101  (Read 5205 times)

Description:

0 Members and 1 Guest are viewing this topic.

Offline bjjones37Topic starter

  • Hero Member
  • *****
  • Join Date: Oct 2004
  • Posts: 524
    • Show only replies by bjjones37
Chemistry 101
« on: January 04, 2005, 11:34:04 PM »
Okay, now you guys got me back into it, you have to put up with me pestering you with questions.  I am going back through my college textbook so that is the source for everything I am asking about.

The 'spin' of an electron within an orbital.  How can it have a direction if it motion is random within the orbital?  Or is it's motion random?  It says that the electrons must spin in opposite directions, but then says that this is mainly an analogy.  Is there a better description of it yet?  

Also the book describes and illustrates the shape of the electronic orbitals, but does not explain how they were derived. You know - s, p, d, and f.  How do we know their shapes?
Any obstacle can be an opportunity, try a different perspective.
 

Offline Speelgoedmannetje

  • Hero Member
  • *****
  • Join Date: Oct 2002
  • Posts: 9656
    • Show only replies by Speelgoedmannetje
Re: Chemistry 101
« Reply #1 on: January 04, 2005, 11:56:21 PM »
digging deep in my (partially destroyed) memory,
electrons always swing around the core of an atom in pairs
so, if there's only one electron (for instance, H) electrons of other atoms are being attracted, and then it kinda works like gear wheels.
so I guess the direction of the electrons around the core is determined by it's surrounding variables.

but correct me if necessary, KennyR :-)
And the canary said: \'chirp\'
 

Offline Karlos

  • Sockologist
  • Global Moderator
  • Hero Member
  • *****
  • Join Date: Nov 2002
  • Posts: 16879
  • Country: gb
  • Thanked: 5 times
    • Show only replies by Karlos
Re: Chemistry 101
« Reply #2 on: January 05, 2005, 12:03:49 AM »
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 ;-)
int p; // A
 

Offline bjjones37Topic starter

  • Hero Member
  • *****
  • Join Date: Oct 2004
  • Posts: 524
    • Show only replies by bjjones37
Re: Chemistry 101
« Reply #3 on: January 05, 2005, 12:48:48 AM »
Quote

Karlos wrote:

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.



Any chance you could demonstrate the math used to derive, say, a p orbital? I don't really know what the wave function would look like. I assume these are standard equations derived for each of the orbitals to define their statistical boundaries.
Any obstacle can be an opportunity, try a different perspective.
 

Offline Karlos

  • Sockologist
  • Global Moderator
  • Hero Member
  • *****
  • Join Date: Nov 2002
  • Posts: 16879
  • Country: gb
  • Thanked: 5 times
    • Show only replies by Karlos
Re: Chemistry 101
« Reply #4 on: January 05, 2005, 01:45:17 AM »
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.
int p; // A
 

Offline bjjones37Topic starter

  • Hero Member
  • *****
  • Join Date: Oct 2004
  • Posts: 524
    • Show only replies by bjjones37
Re: Chemistry 101
« Reply #5 on: January 05, 2005, 02:43:26 AM »
Quote

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

An s orbital, would me more managable...



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

Quote


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 1s case only - for 2s etc you have to factor in the effect of increasing nuclear charge Z, and but the symmetry of the function doesn't change).



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

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?

Any obstacle can be an opportunity, try a different perspective.
 

Offline Karlos

  • Sockologist
  • Global Moderator
  • Hero Member
  • *****
  • Join Date: Nov 2002
  • Posts: 16879
  • Country: gb
  • Thanked: 5 times
    • Show only replies by Karlos
Re: Chemistry 101
« Reply #6 on: January 05, 2005, 10:45:58 AM »
Quote

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-

Quote


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.

Quote

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.
int p; // A
 

Offline Cymric

  • Hero Member
  • *****
  • Join Date: Nov 2002
  • Posts: 1031
    • Show only replies by Cymric
Re: Chemistry 101
« Reply #7 on: January 05, 2005, 01:20:55 PM »
Quote
Karlos wrote:
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.

Slight addition: it is only possible to solve the Shroedinger equation analytically in one-electron systems. Add a second electron, and you have to use a computer, or resort to various approximations which, despite their inaccuracy, yield surprisingly good physical insights into the quantum world of electrons  :-). (You can for example argue on purely theoretical grounds that HeH+ exists, but that He_2 does not. I've always found that a fascinating result.)
Some people say that cats are sneaky, evil and cruel. True, and they have many other fine qualities as well.
 

Offline Karlos

  • Sockologist
  • Global Moderator
  • Hero Member
  • *****
  • Join Date: Nov 2002
  • Posts: 16879
  • Country: gb
  • Thanked: 5 times
    • Show only replies by Karlos
Re: Chemistry 101
« Reply #8 on: January 05, 2005, 01:54:51 PM »
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 :-)
int p; // A
 

Offline bloodline

  • Master Sock Abuser
  • Hero Member
  • *****
  • Join Date: Mar 2002
  • Posts: 12113
    • Show only replies by bloodline
    • http://www.troubled-mind.com
Re: Chemistry 101
« Reply #9 on: January 05, 2005, 03:40:31 PM »
I used to love MO theory, it was one of the few things I was any good at :-D

Offline Karlos

  • Sockologist
  • Global Moderator
  • Hero Member
  • *****
  • Join Date: Nov 2002
  • Posts: 16879
  • Country: gb
  • Thanked: 5 times
    • Show only replies by Karlos
Re: Chemistry 101
« Reply #10 on: January 05, 2005, 03:50:04 PM »
Quote

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 ;-)
int p; // A
 

Offline bjjones37Topic starter

  • Hero Member
  • *****
  • Join Date: Oct 2004
  • Posts: 524
    • Show only replies by bjjones37
Re: Chemistry 101
« Reply #11 on: January 06, 2005, 07:26:02 PM »
I know this may be one of those "just the way it is" things, but I would like a little insight on something. I was taught that covalent bonds are formed because elements like to have their valence shells filled - 1s, 2s, 2p and so forth.  At least this was the rationale I was given.  I don't remember any particular rational given for ionic bonding.  Some elements are more likely to form ionic bonds, others covalent bonds, and some do both. Calcium, which has it's valence shells filled up thru 4s is extremely reactive, while Gold which has an available 6s electron is rather nonreactive.  (This observation seems to be rather consistent for the IIA and IB elemental groups.)  Could you comment on why this is so?
Any obstacle can be an opportunity, try a different perspective.
 

Offline bloodline

  • Master Sock Abuser
  • Hero Member
  • *****
  • Join Date: Mar 2002
  • Posts: 12113
    • Show only replies by bloodline
    • http://www.troubled-mind.com
Re: Chemistry 101
« Reply #12 on: January 06, 2005, 08:50:25 PM »
VSEPR theoy is just a "rule of thumb", it breaks down after atomic number 20 or there abouts.

The funny thing is that I could never work with VSEPR but had no problems with MO, for most people it's the other way around!

Offline bjjones37Topic starter

  • Hero Member
  • *****
  • Join Date: Oct 2004
  • Posts: 524
    • Show only replies by bjjones37
Re: Chemistry 101
« Reply #13 on: January 06, 2005, 08:56:36 PM »
Quote

bloodline wrote:
VSEPR theoy is just a "rule of thumb", it breaks down after atomic number 20 or there abouts.

The funny thing is that I could never work with VSEPR but had no problems with MO, for most people it's the other way around!


Could you clarify VSEPR and MO please? Not sure what the letters stand for.
Any obstacle can be an opportunity, try a different perspective.
 

Offline Cymric

  • Hero Member
  • *****
  • Join Date: Nov 2002
  • Posts: 1031
    • Show only replies by Cymric
Re: Chemistry 101
« Reply #14 on: January 06, 2005, 10:18:52 PM »
This is actually quite a good question which caused me to scratch my head a lot. There are several issues at work here, and it is not always clear which one dominates the overall outcome.

First, gold is not so inert as you might think. If you take gold powder of flakes, and drop them into a cyanide solution through which you bubble air (hardly extreme conditions), the gold dissolves to form [Au(CN)2]- ions. Gold is oxidised by plain air at elevated temperatures (100 degrees C) but the oxide Au2O3 is not particularly stable and decomposes back to solid Au and O2 at slightly higher temperatures. You can prepare Au2Cl6 by heating gold in the presence of chlorine gas at about 200 degrees C. From this compound you can go on to organometallic compounds which are quite useful. So while a little stubborn, you can get it to react.

Now as to the mystery of Aus lone 6s1 valence electron. You would expect: higher quantum number, more distance, less attraction, easier gotten rid of. But that's not how it works. You have to work out a couple things here. One is that orbitals become bigger as their principal quantum number increases. It simply puts them further away from the nucleus, but not in a linear fashion. Second, electrons repel each other, so they don't want to be in each other's neighbourhood. Third, electrons tend to 'shield' one another from the attractive force of the positive nucleus. Electrons which are in higher orbitals therefore experience 'less' positive charge. Where electrons end up is of course where all these effects are balancing each other. If you do the math, you will end up with the rather surprising result that despite its high principal quantum number, the 6s1 electron of Au is more tightly bound to the nucleus than Ca's two 4s electrons! In other words, the increase in Z is not offset sufficiently by higher shielding and higher quantum number to make it experience less attraction.

And just once you think you understand it, in steps another effect, namely that of orbital overlap. In order for atoms to bond properly, their orbitals have to overlap sufficiently for a bond to appear. If you take Au+ (so without its 6s1 electron), it can only offer 5d electrons to anything whishing to cuddle up. (Well, perhaps a little 5p too, it depends.) Atomic oxygen can only offer 2p orbitals, and the overlap between a 2p and 5d orbital is not good. That is why in Au2O3, there are [O2]- entities rather than O2- ones. [O2]- is simply bigger and can thus provide better overlap. Chlorine can offer 3p electrons, still not perfect, but better than 2p ones. And so forth.

You can go a step further and look at the particular shape of the 5d-orbitals available for bonding, as some will not overlap, and others will. If you take out electrons from the 5d orbital, some of the remaining ones will slightly shift position, becoming more elongated or contracted to accomodate for the 'gap' (and thus change in repulsion). This complicates overlap calculations (and thus which compounds will form) considerably.

In effect, what you learned that orbitals must be filled for ions to be stable is only an approximation to the real thing. For example, you can let the noble gas Xe react with PtF6, or even pure F2 under relatively mild conditions, even though the simple approach predicts that the filled valence shell of Xe is immune to reactions. It is not: apparently the system can exist in a lower energy state by overlapping and mixing some of their orbitals, and is thus susceptible to chemical bonding.

Finally, the terms 'ionic' and 'covalent' bonds are just descriptions of the extremes of orbital overlap. Sometimes the shape of the resulting bonding orbital is centered predominantly on one atom. That is an 'ionic' bond, because effectively, one atom becomes negatively charged, and one positively. Sometimes it is centered right between two atoms. That is a covalent one: neither atom has a particular charge.

Now I have to cross my fingers and hope a real chemist doesn't frown too much at what I wrote... :-P
Some people say that cats are sneaky, evil and cruel. True, and they have many other fine qualities as well.