Chapter 6, Problem 1P

Determine the number of nodes along the internuclearaxis for each of the $\sigma$ molecular orbitals for ${\text{H}}_{\text{2}}^{\text{+}}$ shown in Figure 6.5.

Interpretation Introduction

Interpretation: The number of nodes along the internuclear axis for each of the $\sigma$ molecular orbitals for ${\text{H}}_{\text{2}}{}^{+}$ is to be determined.

Concept introduction: Born-Oppenheimer approximation state that nuclei are heavier than electrons that considered fixed in space whereas the electrons move constantly around them. The Born−Oppenheimer approximation solves the electronic Schrodinger equation for ${\text{H}}_{\text{2}}{}^{+}$ for fixed internuclear separation $R_{\text{AB}}$ . The result is one-electron $\text{MO}$ (molecular orbitals), that is analogous to one-electron hydrogen $\text{AO}$ (atomic orbital). The quantum picture for $\text{MO}$ (molecular orbitals) of ${\text{H}}_{\text{2}}{}^{+}$ illustrates that first eight $\text{MO}$ for ${\text{H}}_{\text{2}}{}^{+}$ that show their shapes and characterize them by their energies as well as symmetry The $\text{MO}$ (molecular orbitals) are characterized by angular momentum along the internuclear axis and analogy to hydrogen atom These are called $\sigma$ for orbital with cylindrical symmetry and $\pi$ for molecular orbitals with one nodal plane that have internuclear axis.

Answer to Problem 1P

The number of nodes in $1{\sigma }_g$ is 0, in $1{\sigma }_u{}^{*}$ is 1, in $2{\sigma }_g$ is 2, in $2{\sigma }_u{}^{*}$ is 3, in $3{\sigma }_g$ is 2 and in $3{\sigma }_u{}^{*}$ is 3.

Explanation of Solution

Node is point in molecular orbitals where probability to find the electron density is zero. Nodal plane is the imaginary plane where probability to find electrons is zero.

  

In the figure, node is the oval region. It is formed when plane white surface cut the spherical orbital into two equal halves having opposite symmetry. Also, there are six $\sigma$ orbitals, $1{\sigma }_g$ , $1{\sigma }_u{}^{*}$ , $2{\sigma }_g$ , $2{\sigma }_u{}^{*}$ , $3{\sigma }_g$ and $3{\sigma }_u{}^{*}$ .The energy level of $\sigma$ orbital is that $1{\sigma }_g$ has the lowest energy and $3{\sigma }_u{}^{*}$ has highest energy.

In $1{\sigma }_g$ orbital, there is no oval region on the spherical orbital. Hence number of nodes is 0.

In $1{\sigma }_u{}^{*}$ orbital, there is one cut on the spherical orbital. Hence number of nodes is 1.

In $2{\sigma }_g$ orbital, there are two oval regions on the spherical orbital. Hence number of nodes is 2.

In $2{\sigma }_u{}^{*}$ orbital, there are three oval regions formed by three cuts on the spherical orbital. Hence number of nodes is 3.

In $3{\sigma }_g$ orbital, there are two oval regions formed by two cuts on the spherical orbital. Hence number of nodes is 2.

In $3{\sigma }_u{}^{*}$ orbital, there are three oval regions formed by three cuts on the spherical orbital. Hence number of nodes is 3.

Conclusion

Therefore, number of nodes in $1{\sigma }_g$ is 0, in $1{\sigma }_u{}^{*}$ is 1, in $2{\sigma }_g$ is 2, in $2{\sigma }_u{}^{*}$ is 3, in $3{\sigma }_g$ is 2 and in $3{\sigma }_u{}^{*}$ is 3.