Question

Figure 6.11 shows the atomic s and p orbitals in a chain
of atoms and how these are combined to form the bonding and antibonding
states. For the s band, the bonding state is formed with the atomic wave
functions on all sites combined in phase, corresponding to a wave vector
k = 0, and the antibonding state is associated with a sign change of the wave
function on every other site, corresponding to a wave vector at the Brillouin
zone boundary. Now consider the ? bands in graphene, which are formed
from the pz orbitals, as shown in Figure 6.15b. (a) How should these orbitals
be combined to form bonding and antibonding ? states? (b) If we associate the bonding/antibonding states with the energy extrema of the ? band, it
appears that both are found at the Brillouin zone center (Γ). Is this consistent
with your result from (a)?

Es Ys First Brillouin zone Figure 6.11 Bands for a one-dimensional for k 0 and k =π/a. The black dots sym- solid calculated in the tight-binding approx bolize the position of the nuclei and the two imation. The right-hand side also shows the different shades of gray symbolize the sign

Answer 1

a) side by side overlap of 2 p - orbitals gives rise to a pi bounding molecular orbital and pi* anti bonding molecular orbital (in the molecular orbitals of diatomic molecular each, atom also has 2 set of p orbitals oriented side by orbitals combine to create 2 pi orbitals \r\n the pi_p y and pi_p z bidals are identical and have the same energy: they are dagenorable orbitals energy they are dagenorable orbitals the pi* p y and pi^* p x anti bonding orbitals are also degenerate and identical except for their orientation