Problem 9.40

Draw sketches illustrating the overlap between the following orbitals on two atoms:

(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 2p_z 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 m_l 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 2p_z 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 90^o to the z-axis at each nucleus; recall that there are also p_x and p_y orbitals lying along these axes. Sketch these orbitals in on the graphic above. If the 2-p_y orbitals (or the 2-p_x orbitals)overlap above and below the nuclei, then we have the formation of a pi bond.

(c) The 2s orbital on one and the 2p_z 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 2p_z 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?