[Rutherford]

Why is water diamagnetic? It has 10e^-:
1s2 1*s2 2s2 2*s2 2p2 there are unpaired electrons in 2p or maybe not?

[Dan]

The orbitals you are listing are atomic orbitals - these apply to atoms. For molecules, we use molecular orbitals.

See: Molecular Orbital Theory, check out some inorganic chemistry textbook chapters on it.

The fact that you are asking in high school chem makes me think you are delving far outside the curriculum - good for you, a sharp question. Basically, molecular orbitals (MOs) are constructed by combining different atomic orbitals of the atoms that form the molecule. The MO diagram for water looks like this:



As you can see, all the electrons are paired.

[Rutherford]

But I wrote molecular orbitals, I added a ''*'' symbol for the anti-bonding orbitals.

[Jasim]

To answer very simply, diamagnetic (as opposed to paramagnetic) means that all of the electrons are in pairs. Paramagnetic means you have a lone electron somewhere that isn't paired up. How this affects magentism is a little more complicated, but essentially the electron spins cancel each other out when they are paired up.

Also, to clarify what you have posted and what Dan responded with... Atomic orbitals are designated as 's', 'p', 'd', 'f'...and so on, while molecular orbitals are designated as 'sigma' and 'pi'.

Water only has single bonds, so we only have 'sigma' molecular orbitals, denoted by the Greek letter 'sigma' or σ.

In molecules we have hybrid orbitals, they are mixtures of atomic orbitals. Also, you said that water has 10 electrons. You are correct, in total there are 10 electrons, but only 8 of them are valence electrons. The molecular orbitals are going to be mixtures of all valence electrons. From the two hydrogens you have two 1s electrons and from oxygen you have two 2s and four 2p electrons, for a total of eight valence electrons available for filling the hybrid molecular orbitals.

Also keep in mind that anti-bonding orbitals don't form bonds, if you have enough anti-bonding orbitals filled in a molecule, that molecule won't exist. In water we only have bonding molecular orbitals that are filled. In the molecular orbital diagram, everything below the center line is bonding, everything above is anti-bonding. The height of the orbital indicates the relative energy of that orbital. Electrons flow 'down-hill'. Water is stable because it's molecular orbitals that are filled are all lower in energy than if the electrons simple stayed in their hydrogen and oxygen atomic orbitals - this is clearly seen in the molecular orbital diagram above.

Notice on the molecular orbital diagram that two of the filled orbitals are from the oxygen p shell, which has no electrons. This means that all four of those electrons are coming from the oxygen, these are the lone pairs on the oxygen.


Going from the bottom of the molecular orbital (MO) diagram to the top, the orbitals are as follows...

The bottom orbital is composed of a hydrogen 1s and an oxygen 2sp³ atomic orbitals (note where the dashed lines lead to, this MO has 's' and 'p' character from the oxygen). This MO is indicative of a sigma bond between the oxygen and one of the hydrogens and is filled by one electron from each atom.

The second one up is composed only of 2p from oxygen and 1s from hydrogen (the upper 1s, where hydrogen has no electrons). This MO is indicative of a lone pair on oxygen.

Continuing up we see that the next MO is composed like the first MO, it has oxygen 2sp³ and hydrogen 1s (the lower 1s where hydrogen has electrons!). This MO is indicative of the second oxygen and hydrogen sigma bond.

The last filled MO consists of oxygen 2sp³ and the upper 1s hydrogen and again is indicative of a lone pair on oxygen where hydrogen has contributed no electrons.

I hope this explains it a bit better. A very good question, but I agree that this is far beyond high school chemistry. To really explain the nitty-gritty behind molecular orbitals requires some complicated mathematics.

[Rutherford]

Thank you, and can you tell me what MO exist other than σ and pi?

[Jasim]

Read over my post above again, I edited it a bit.

Sigma bonds are single bonds between atoms, pi bonds are double and triple bonds between atoms. Those are it. Other MOs are anti-bonding, that is they are repulsive molecular orbitals. When anti-bonding orbitals have electrons in them, they are putting pressure on that molecule to 'fly apart' in a sense. If you look at the MO diagrams of the noble gases you will see that ALL of the MOs are filled, including all of the anti-bonding orbitals, this indicates that noble gases can't form molecules.


EDIT:
I should say that those are it as far as I know and the only ones you will need to be concerned with unless you go into graduate school to study chemistry. I'm not that familiar with how the 'd' and 'f' orbitals produce molecular orbitals.

[Jasim]

I noticed in my post that I said "Notice on the molecular orbital diagram that two of the filled orbitals are from the oxygen p shell, which has no electrons..." What I meant was that no electrons are coming from hydrogen. Obviously, there are electrons there, but they are all coming from oxygen.

[Rutherford]

Read over my post above again, I edited it a bit.

Sigma bonds are single bonds between atoms, pi bonds are double and triple bonds between atoms. Those are it. Other MOs are anti-bonding, that is they are repulsive molecular orbitals. When anti-bonding orbitals have electrons in them, they are putting pressure on that molecule to 'fly apart' in a sense. If you look at the MO diagrams of the noble gases you will see that ALL of the MOs are filled, including all of the anti-bonding orbitals, this indicates that noble gases can't form molecules.


EDIT:
I should say that those are it as far as I know and the only ones you will need to be concerned with unless you go into graduate school to study chemistry. I'm not that familiar with how the 'd' and 'f' orbitals produce molecular orbitals.

Now I am confused. Why the pi bonds have to be double or triple when they form between 2py or 2pz AO?

[Jasim]

Between any two atoms that are singly, covalently bonded you will have a sigma bond. This is where the orbitals overlap in a linear fashion. Pi bonds only exist when the atoms are covalently bonded double or triple. Pi bonds do not form in a linear fashion.
Try this: http://www.youtube.com/watch?v=ree49ge4VA4


The reason is because with two atoms you only have 3 planes were the orbitals can overlap (x, y, z). With the first bond between them, they use up the plane that goes through the center of both atoms - that is a sigma bond. For consecutive bonds the planes orthogonal to the adjacent atom must be utilized for bonding. It's a matter of three-dimensional geometry and spatial constraints.

[Rutherford]

I found that molecule B2 is paramagnetic while having the same number of e^- as water. 3e^- from the 1st atom and 3e^- from the 2nd atom will be paired( sp hybridisation if I am correct) so why is it paramagnetic? Finding the difference will be a big help for me.

[Dan]

For diatomics the symmetry is such that you get degenerate orbitals (multiple orbitals at the same energy level), each of which are singly occupied before electron pairing (just as in the order of filling of atomic orbitals).

MO diagram for diboron, note the two degenerate pi orbitals are each singly occupied, rather than just one of them holding a pair of electrons:

[Rutherford]

This diagram I understood completely, but the diagram for water is different. Can you post a diagram for water like the one for diboron?

[fledarmus]

Hmmm, isn't that what Dan posted in his first response?

[Dan]

Water is more complicated because there are three atoms, it changes the symmetry. The MO diagram for water (posted 26-01-2012) looks different for this reason, the atomic orbitals (AOs) for oxygen are combined with the AOs for both hydrogens simultaneously (this is shown pictorially to the right of the MO diagram).

You can look at it by considering the two combinations of the H AOs, which are ++ or +-, called A₁ and B₁ which are not equivalent in energy. Note that these are the same combinations that make up the dihydrogen molecule, the interactions are just lower in magnitude because the H atoms are not as close together. These two combinations are then in turn combined with O. A₁ and B₁ interact differently with 2p_x, 2p_y and 2p_z, so the MOs have different energy levels. You can look at it as a combination of O and "H₂".

In the case of B₂ (and other diatomics) the interaction of p orbitals in the y,z plane are equivalent (if we arbitrarily say the B-B bond lies along the x-axis, as has been applied to the diboron MO diagram I posted). That is, the 2p_y-2p_y combination is equivalent in energy to the 2p_z-2p_z. This is because the spatial orientation of the 2p_y AO of one B atom relative to the 2p_y AO of the other B atom (side-to-side) is the same as the spatial orientation of the 2p_z AO of one B atom relative to the 2p_z AO of the other B atom (side-to-side) - this gives energetically equivalent MOs of pi symmetry. The 2p_x-2p_x is end-to-end, and gives MOs of sigma symmetry.

[Rutherford]

Thanks Dan & Jasim for helping me with this. Nothing more I need here.