MO Theory & OH: Bond Order & Nonbonding Electrons

[jolly_math]

Homework Statement

Assume that the MOs result from the overlap of a pz orbital from oxygen and the 1s orbital of hydrogen (the O-H bond lies along the z-axis). Only the 2p orbitals of oxygen interact significantly with the 1s orbital of hydrogen. The MO diagram is shown below.
a) Estimate the bond order for OH.
b) Predict whether the bond order of OH+ is greater than, less than, or the same as that of OH.

Relevant Equations

MO diagram
bond order = (# of bonding electrons – # of antibonding electrons) / 2

a) The solution says that there are 2 bonding electrons and that the 2px and 2py electrons have no effect on the bond order. I don't understand why this is case.

b) Why is it that, to form OH+, specifically a nonbonding electron is removed from OH, not a bonding electron?

Thank you.

 

[TeethWhitener]

a) You’ve been given that the bonding MO is formed from the interaction between the H 1s orbital and the O 2pz. This means that the px and py orbitals on O don’t participate in bonding. These are usually called “non-bonding orbitals,” in contrast to bonding and antibonding orbitals.

b) Hint: the y-axis of the MO diagram is energy.

 

[jolly_math]

b) Hint: the y-axis of the MO diagram is energy.

Is it that the 2s orbital (nonbonding orbital) has the lowest energy, so an electron from there is easiest to remove?

 

[jolly_math]

Also, if all valence electrons were used to create the molecular orbital (e.g. for
NO+), does that mean that all valence electrons are used in bonding (the 2s, and two 2p orbitals)?

 

[TeethWhitener]

Electrons are removed from the highest energy orbitals first. This is because it takes energy to remove an electron, so it would take more energy to remove a lower energy electron than a higher energy one. This is exactly analogous to the fact that it takes more energy to lift something from the ground to the top of a roof than it goes to lift something from one inch below the roof to the top. In the former case, the potential energy is initially lower than in the latter case, but the change in potential energy from the initial state to the final state is much larger.

 

[TeethWhitener]

Also, if all valence electrons were used to create the molecular orbital (e.g. for
NO+), does that mean that all valence electrons are used in bonding (the 2s, and two 2p orbitals)?

Not necessarily. The number of molecular orbitals must match the number of atomic orbitals (this is a technicality resulting from the fact that the AO to MO transformation is a change of basis). Sometimes the resulting molecular orbitals are non-bonding. This happens, for instance, in carbonyl groups and is a major defining feature of their visible light spectroscopy.

For NO+ specifically, you have a system which is isoelectronic to N2, so all the valence electrons will be in bonding orbitals.

 

[jolly_math]

Electrons are removed from the highest energy orbitals first. This is because it takes energy to remove an electron, so it would take more energy to remove a lower energy electron than a higher energy one. This is exactly analogous to the fact that it takes more energy to lift something from the ground to the top of a roof than it goes to lift something from one inch below the roof to the top. In the former case, the potential energy is initially lower than in the latter case, but the change in potential energy from the initial state to the final state is much larger.

That makes sense now, so an electron from the 2px or 2py orbitals would be removed? Why wouldn't an electron from pz be removed, would nonbonding electrons be easier to remove? Thank you.

 

[TeethWhitener]

That makes sense now, so an electron from the 2px or 2py orbitals would be removed? Why wouldn't an electron from pz be removed, would nonbonding electrons be easier to remove? Thank you.

The vertical direction on the MO diagram is energy. Which is higher in energy, the px and py non bonding orbitals, or the bonding sigma orbital formed from the pz and the H1s?