Ever consider that you are dealing with atoms of different proton number?

Greetings fellow Chemists,

Can anyone provide a logical explanation behind the order of Molecular orbitals ( in terms of energy ) cos I think they sometimes differ. In the text Im using ( Zumdahl ) , they sometimes use Pi (2p ) before Taw(2p) , and sometimes they invert them. Why? And why do Pi MO have a space of 4 electrons not 2? The entire concept seems very illogical to me , there's no math at all! IS this how Chemists operate? by predicting? Shouldn't we prove our models mathematically ?

Are you serious? MO Theory is all straight up quantum mechanics!

Read "Chemical Structure and Bonding" Dekock and Gray for an introduction.

Because to handle the quantum mechanics of MO theory you would need to know first year calculus, multi-variable calculus and differential equations. If G-Chem students came in with that amount of knowledge then it might be appropriate. The degree of s-p mixing is the answer to your other question.

Well, to answer that question you would need to know the energies of the orbitals you are working with, which requires some pretty hefty calculations.  My inorganic professor once provided us with the energy values and had us draw "to scale"(energy wise) MO diagrams and we could see that the s and p orbitals are farther apart as you move from left to right in a period.  When the energies of the s and p orbitals are closer together in energy, you get the s and p mixing and your pi orbitals are lower in enrgy than your sigma orbitals.  When they are farther apart in energy, you don't get the mixing so the sigma orbitals are lower in energy than the pi orbitals, as "logic" dictates.  

So it is all about energy.  Simply, you just need to know that for certain molecules, namely nitrogen, carbon, and boron, you need to have the pi orbitals lower in energy than the sigma orbitals.  For oxygen and fluorine you don't.  In general chemistry, these are probably the only elements you will work with anyhow.  Also, it really depends on your professor/teacher, I have seem some professors not really care about the s-p mixing, they just want you to draw a basic diagram and get a bond order.  

Hope this helps

No problem, happy to help

I am curious as to the locality of the two 1s carbon electrons.  What "space" do they occupy?  What "shape" is their cloud?  And how do the four sp³ orbitals affect their behaviour?

1s atomic orbitals are spherical and centred around the nucleus. They are closer from it than the 2s orbitals. See http://en.wikipedia.org/wiki/Atomic_orbitals for pictures of atomic orbitals.

In CH₄, only the 2s and 2p electrons from the carbon are involved in covalent bonds. The 1s electrons remain unchanged in a spherical shape around the nucleus. They are screened by the 2s and 2p. Only valence electrons are forming bonds while core electrons remain unchanged.

Explained in another way: atomic orbitals interact to form molecular orbitals when they are not too far energetically from each other. However, 1s orbitals from carbon are much lower in energy than 1s from hydrogen (because carbon has 6 protons while hydrogen has only one  1s electron are more attracted by the 6 protons, they are lowered in energy). Therefore, 1s electrons from hydrogen interact with valence electrons from carbon, leaving the 1s  carbon electrons unchanged.

The reason I ask this is because all the atomic/molecular models that I have seen depict the carbon, in a molecule of methane, as having four sp³ hybridised orbitals without accounting for the inner two 1s electrons.

There is no need to draw them because core electrons remain unchanged. Everybody knows they are there, so drawing them would bring no extra information. And as well, core electrons are much lower in energy and would go out of the energy scale in orbital diagrams.

Yes, same principle.

In an atomic or molecular orbital diagram, you have an energy scale going up from negative energies towards 0. The 0 energy is for a free electron which is not attracted by any proton. When you add protons nearby, there is an electrical attraction with the electron  the electron is more stable in this position, its energy decreases (It represents the amount of energy you have to provide in order to extract the electron from the protons influence and restore it as a free electron). Naturally, the more protons you have in the nucleus, the stronger electrons are attracted, the lower their energies become.

Iodine atoms have 53 neutrons and 53 electrons. The electrons 1s occupy a spherical shape around the nucleus and they "feel" the direct attracting effect of the 53 neutrons  they are very down in the "basement" of the orbital diagram energy scale. And so, you would need a hell load of energy to extract them.
But, for the 7 valence electrons (5s²5p⁵), the nucleus electrical charge is partially hidden by the 46 electrons that are closer to the nucleus  The attraction is lower and the electron energy is higher.


Orbitals f are occupied by some of the heaviest elements such as lanthanides and actinides, for example uranium.
No known elements have orbitals g that are occupied in the ground state. The website you mentioned says "However these orbitals may be populated in some excited states". Perhaps, but I have never heard of any reports about it. In fact, one could play a little mathematical game and calculate the shape of orbitals h, i, j,. But it wouldn't have any chemical meaning. If you take a heavy element like uranium and start to excite one of the valence electron to go in higher energy orbitals, you may be able to reach the orbital g. But as you reach higher energy orbital, you go further away from the nucleus. Then, comes a point where the electrical attraction energy from the nucleus is not enough to keep the excited electron from breaking free.

Do these orbitals all occur simultaneously?  If this is so, atomic space is very crowded, is it not?

Yes, all occupied orbitals occur at the same time.
No, it isn't so crowded. You should remember that matter is mostly made of vaccum. Electron and proton diameters are very small compared to atomic radii. If you consider for example the density of neutron stars http://en.wikipedia.org/wiki/Neutron_star, then you realise how "crowded" it can get...
From Wikipedia, "This density is approximately equivalent to the mass of the entire human population condensed into the size of a sugar cube". Hong Kong is not so densely populated after all !