Answer
$ \psi_a(x) \rm \;antibonding \;orbital$.
$ \psi_b(x) \rm \;bonding \;orbital$.
Work Step by Step
Recalling that:
$\rm Bonding \;orbitals$: In these orbitals, the electron density is concentrated between the atoms. This leads to a net attractive interaction between the atoms, which stabilizes the molecule.
$\rm Antibonding \;orbitals$: In these orbitals, the electron density has nodes between the atoms, which introduces repulsive interactions and destabilizes the molecule.
Now let's analyze the wavefunctions:
$\bullet$ $\underline{\text{ Wave function $ \psi_a(x) $}}$: This wave function has nodes (points where the wave function crosses zero) between the atoms. The electron density is low or zero between the atoms, which means there is a lack of electron density holding the atoms together, thus destabilizing the molecule.
Therefore, it is an $\rm antibonding \;orbital$.
$\bullet$ $\underline{\text{ Wave function $ \psi_b(x) $}}$: The wave function is smooth and continuous across the atoms, with no nodes (zero crossings) between the atoms. The electron density (related to the square of the wave function) is higher between the atoms, which indicates a bonding orbital.
Therefore, it is a $\rm bonding \;orbital$.
