# Atomic orbitals

1. Mar 3, 2009

### yyttr2

Can someone please direct me to a site that explains atomic orbitales well.
Thanks.

2. Mar 6, 2009

### yyttr2

come on :( someone has to know

3. Mar 6, 2009

### alxm

What? This is a basic concept. It's in every general/introductory chemistry book. There are dozens of sites that'll give an introduction. Just google for it.

Or to put it another way, try asking something more specific.

Last edited: Mar 6, 2009
4. Mar 7, 2009

### yyttr2

I have yet to find a site that accurately explains these.

5. Mar 9, 2009

### alxm

Okay. Well I'll explain them.
The solution to the Schrödinger equation for the Hydrogen atom is essentially (leaving out some factors):
$$\psi_{nlm}=e^{-\frac{4}{na_0}}L^{2l+1}_{n-l-1}Y^m_l(\theta,\phi)$$

Where L and Y are mathematical sets of functions, the Legendre and Laguerre polynomials.
The pictured orbitals are these solutions plotted in space. You can view them as the electron density clouds. (which is technically the square of the wavefunction, but that looks essentially the same, geometrically) The solution depends on three quantum numbers, n, l and m.
n is the principle quantum number, an integer, 0,1,2.. etc. Chemically it corresponds to a shell (n=0,1,2 is K,L,M shells respectively). l is the azimuthal quantum number, it has an integer value less than n. (if n=0 then l=0). It corresponds to a sub-shell (l=0 is an 's' orbital, l=1 is a 'p' orbital, l=2 is a 'd' orbital etc)
m is the magnetic quantum number, an integer from -l to +l.
In addition to these you have spin which is either up or down, so each orbital can contain two electrons.

So you have:
K shell, n=0:
Then l=0, m=0 -> a single 's' subshell which contains two electrons
L shell, n=1:
If l=0 m=0 -> an 's' subshell with two electrons
If l=1 m=-1,0,+1 -> three 'p' subshells with six electrons --> a total 2+6+8 electrons in the L shell
M shell, n=2
If l=0 m=0 -> an 's' subshell with two electrons
If l=1 m=-1,0,+1 -> three 'p' subshells with six electrons l
If l=2 m=-2,-1,0,+1,+2 -> four 'd' subshells with 8 electrons - a total of 2+6+8+10 electrons in the M shell

And so you get the structure of the whole periodic table out of this.

Now, hydrogen only has one electron. And if you have more than one electron, they interact with each other through electrical repulsion. So for any other atom the real orbitals don't look exactly like this. But it's enough to give an idea. This, in turn, can be used to visualize and understand chemical bonding through how they combine to form 'molecular orbitals'. Which is a good enough model to explain most chemistry qualitatively. (The Lewis structures one learns first, OTOH, don't go very far)

Just to give an example, with a metal atom that has octahedral coordination (ligands on the lines of the x, y and z axises) the Dxy, Dxz and Dyz will be lower in energy (due to the ligands) than the other two d orbitals. With a metal atom that has tetrahedral coordination (no bonds on axis-lines) it's the opposite. And that difference in energy often corresponds to the visual range of the spectrum, which is why metal complexes are so often colored.

6. Mar 11, 2009

### LongLiveYorke

The above poster went into some of the math.

In modern atomic theory, the electron isn't in any one place at a time. Rather, we can't ever say for sure where the electron actually IS. The best we can do is determine where it is very likely to be, or where it spends most of its time. Thus, we can only determine the probability that an electron is in a particular place at a particular time.

The pictures that you've shown are contours of equal probability for different states of the electron. In other words, the electron can exist in different states of different shapes and energies in the atom (let's for now focus on hydrogen because it's so simple). The shapes are two dimensional surfaces in three dimensions. The interpretation is that the electron is equally likely to be found on any part of that surface than any other. They are meant to give you a sense of what the probability function of the electron looks like (it would be impossible to draw it out completely because there are three dimensions and we assign a probability to every point, so we would need four dimensions to express what we want to).