What orbitals form sigma bonds?
1 Answer
POSSIBLE ORBITAL COMBINATIONS TO GENERATE SIGMA MOLECULAR ORBITALS
For simplicity, if we examine only the
Let's also suppose that we are ignoring
Then, orbitals that can capably overlap with each other to form
#s + p_z -> sigma# (bonding)#s - p_z -> sigma^"*"# (antibonding)#s + d_(z^2) -> sigma# (bonding)#s - d_(z^2) -> sigma^"*"# (antibonding)#s + d_(x^2-y^2) -> sigma# (bonding)#s - d_(x^2-y^2) -> sigma^"*"# (antibonding)#p_z + d_(z^2) -> sigma# (bonding)#p_z - d_(z^2) -> sigma^"*"# (antibonding)#p_x + d_(x^2 - y^2) -> sigma# (bonding)#p_x - d_(x^2 - y^2) -> sigma^"*"# (antibonding)#p_y + d_(x^2 - y^2) -> sigma# (bonding)#p_y - d_(x^2 - y^2) -> sigma^"*"# (antibonding)
where the
We would also suppose that the antibonding molecular orbitals are unoccupied so that the bond is a standard single bond.
HOW TO DEPICT/IMAGINE THESE ORBITAL OVERLAPS
When you sketch these orbital overlaps:
- All
#s# orbitals are spheres. The only way these can change sign is if the whole thing changes sign. - The
#p_z# orbitals can be approximated as dumbbells, regardless of their#n# , without losing the essence of the#sigma# MOs generated (head-on overlap). One lobe is the opposite sign to the other. - The
#d_(z^2)# look almost like#p_z# orbitals, except there is a donut in the middle. You can also approximate these as dumbbells, regardless of their#n# , without losing the essence of the#sigma# MOs generated (head-on overlap). Both lobes are the same sign. - The
#d_(x^2-y^2)# can be approximated as four-leaf clovers, essentially, on the#xy# -plane, with the lobes aligned along the#x# and#y# axes. The opposite lobes along each axis are the same sign. Therefore, they overlap with the#p_x# and#p_y# , which also lie long those axes.
Since they are all aligned along the same axis (
(Since
Of course, there exist