Molecular Orbitals Molecular Orbital Theory A molecular orbital explicitly describes the spatial distribution of a single electron around two or more nuclei. This is an electrondelocalized approach to bonding.

s orbitals We form (approximate) molecular orbitals by adding and subtracting atomic orbitals. We add and subtract the 1s orbitals in H2 :

This is a (mostly) general recipe: MO(1)

=

AO(atom A) + AO(atom B)

MO(2)

=

AO(atom A) − AO(atom B)

For hydrogen, we name these molecular orbitals σ orbitals. We have a bonding orbital. σ1s ≈ 1s(A) + 1s(B)

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And an antibonding orbital. ∗ σ1s ≈ 1s(A) − 1s(B)

There is a node in electron density between the atoms, and more negative charge away from the two nuclei than between them.

Notation: σ describes any molecular orbital that is cylindrically symmetric around the internuclear axis; that is, if you look down the bond, it does not change on rotation. * indicates that an orbital is antibonding (there is no star if the orbital is bonding).

Energy Level Diagrams ∗ σ1s is lower in energy than the 1s atomic orbitals, and σ1s is higher in energy. Draw this out using an energy level diagram:

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These are our new molecular orbitals, and we fill them up (lowest energy first) according to the Pauli principle: two electrons per orbital. For H2 , this puts both electrons (one spin up, one spin down) in the σ1s bonding orbital. We define the bond order using bond order =

# bonding electrons − # antibonding electrons 2

That is, each pair of electrons in a bonding orbital makes a bond, but each pair in an antibonding orbital cancels a bond. For H2 , we have the simple 2−0 =1 bond order = 2 for a single bond. He2 has bond order 0 [(2 − 2)/2 = 0], and we can make H+ 2, − H2 , and He+ 2 , all of which have bond order 1/2.

We can write out ground-state electron configurations using molecular orbitals, like so H+ 2 H2 He+ 2 He2

(σ1s )1 (σ1s )2 ∗ 1 (σ1s )2 (σ1s ) ∗ 2 (σ1s )2 (σ1s )

1 2

1 1 2

0

Once we hit the second row of the periodic table, we open up the 2s and 2p orbitals for bonding, and the 1s electrons are now core electrons. Core electrons stick close to the nucleus and do not participate in binding (and, even if they did, the bond order would be 0, because all bonding and antibonding orbitals would be filled.)

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There are always the same number of molecular orbitals as atomic orbitals, and always the same number of electrons to put in them.

The 2s-2s bonds also give σ and σ ∗ MOs, as do 3s, 4s, etc. Alkali diatomics (Li2 , Na2 , etc.) are equivalent to H2 ; alkali earth diatomics are equivalent to He2 and do not bond. For anything to the right of the alkali earths, the p orbitals get involved (and things get more interesting).

MOs from s and p orbitals p orbitals have three ml sublevels, and so 3 orientations.

By convention, we choose the z axis to lie along a chemical bond. If we draw two 2pz orbitals side by side and bring them together we get a node.

This is an antibonding orbital, again higher in energy: ∗ σ2p = 2pz (A) + 2pz (B) z

If we flip the sign of one of the orbitals, however, we get a bonding orbital: σ2pz = 2pz (A) − 2pz (B)

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Note that these orbitals are still symmetric around the bond, and therefore σ orbitals. This is not so with the other 2p orbitals, which are perpendicular to the bond. Combinations of these give us π and π ∗ orbitals:

We write π2px

=

2px (A) + 2px (B)

∗ π2p x

=

2px (A) − 2px (B)

With similar orbitals for 2py , 3p, 4p, etc. Here the π tells us the orbital is not cylindrically symmetric around the bond, and in fact has one planar node where the plane includes the bond. (δ orbitals would have 2!)

The energy-level diagram is much fancier:

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We can also use p orbitals on one atom and s orbitals on another:

It is also somewhat more variable, in that the σ2s and σ2pz orbitals are less shielded and thus more strongly affected by increasing core charge as we move from Li2 to F2 . They are quite far apart in Li2 , giving a diagram like the one above; for N2 they are quite close:

and O2 is the “tipping point” where the σ2pz orbital becomes lower in energy than the π2px and π2py orbitals:

We can fill in orbitals for C2 :

Note that C2 has a double bond, and therefore carbon does not have an octet.

and O2

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Note that we can still use Hund’s rule, and it tells us that O2 has two unpaired electrons (and is thus paramagnetic). N2 does not:

Note that the bond orders of O2 and N2 are 2 and 3, respectively.

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