(a) the 2s orbital on each;
Recall that an s-orbital (l=0) is spherical in shape; therefore, orbital overlap between s-orbitals on two atoms is nondirectional. We draw this case as follows:
Notice that the maximum overlap of the 2-2s orbitals occurs in the area directly between the nuclei.
(b) the 2pz orbitals on each (assume that the atoms are on the z-axis)
Recall that there are three orbitals in the 2p subshell (we have three ml values: -1, 0, 1.) The p-orbitals (l=1) are directional in nature; one orbital lies along the x-axis, one along the y-axis, and one along the z-axis. If we are considering orbital overlap of 2-2p orbitals along the z-axis (i.e., the 2pz orbitals), then we imagine the atoms approaching along the z-axis as follows:
again, note that the maximum orbital overlap occurs between the nuclei; as we will see, when the region of maximum orbital overlap (and electron density) occurs in the area between the nuclei, we have the formation of a sigma bond. In the graphic above, the x and y-axes are shown at 90o to the z-axis at each nucleus; recall that there are also px and py orbitals lying along these axes. Sketch these orbitals in on the graphic above. If the 2-py orbitals (or the 2-px orbitals)overlap above and below the nuclei, then we have the formation of a pi bond.
(c) The 2s orbital on one and the 2pz orbital on the other.
We assume that the maximum orbital overlap will occur if the atom with the 2s orbital approaches the atom with the 2pz orbital along the z-axis:
Again, this arrangement results in the formation of a sigma bond. in closing, a question: of the three cases presented here, which bond would be the longest? Why?