H3 Chemistry 9813 · Study focus: H3 Chemistry: Construct and interpret diatomic MO diagrams, including HOMO and LUMO
H3 Chemistry: Construct and interpret diatomic MO diagrams, including HOMO and LUMO
Start from the governing chemical model, test it against evidence, then transfer the reasoning to an unfamiliar case.
Your success criteria
- Construct and interpret diatomic MO diagrams, including HOMO and LUMO
- Use named chemical evidence.
- Transfer the governing reason to an unfamiliar case.
Orientation and prerequisite retrieval
A diatomic MO diagram is filled from lowest energy using the molecule's valence-electron count.
The HOMO is the highest-energy occupied molecular orbital.
Definitions and exact language
The LUMO is the lowest-energy unoccupied molecular orbital.
H2 has σ1s as its occupied HOMO and σ*1s as its LUMO.
Bond order is half the number of bonding electrons minus antibonding electrons.
Unpaired electrons in occupied MOs predict paramagnetism.
Text alternative: Text alternative: O2 has two unpaired electrons in degenerate π* orbitals in the qualitative diagram. A complete interpretation states configuration, frontier orbitals and any bond-order or magnetic consequence asked.
Detailed molecular explanation
Bond order is half the number of bonding electrons minus antibonding electrons.
Unpaired electrons in occupied MOs predict paramagnetism.
Worked leaf-specific case
O2 has two unpaired electrons in degenerate π* orbitals in the qualitative diagram.
Removing an electron from the HOMO and adding one to the LUMO are different operations.
- A candidate writes, “Electrons are placed randomly on a diatomic MO ladder.” Which correction should replace it?
Open the feedback checkpoint after attempting
- A diatomic MO diagram is filled from lowest energy using the molecule's valence-electron count.
Guided evidence check
Degenerate π orbitals must be filled according to Hund's rule.
Core and valence levels must not be double-counted when using a valence-only diagram.
- Which labelled diagram, spectrum or calculation would most directly disprove “In H2 the occupied σ1s orbital is the LUMO.”?
Open the feedback checkpoint after attempting
- H2 has σ1s as its occupied HOMO and σ*1s as its LUMO.
Independent transfer
A highest drawn empty line is not necessarily the LUMO if a lower empty line exists.
A complete interpretation states configuration, frontier orbitals and any bond-order or magnetic consequence asked.
- Design a specific chemical check that would expose the error in “O2 has all electrons paired in its π* orbitals.”
Open the feedback checkpoint after attempting
- O2 has two unpaired electrons in degenerate π* orbitals in the qualitative diagram.
Misconception repair
Misconception: Any empty orbital is a LUMO regardless of energy.
Repair: A highest drawn empty line is not necessarily the LUMO if a lower empty line exists.
- Write leaf-specific feedback for the misconception “Any empty orbital is a LUMO regardless of energy.”
Open the feedback checkpoint after attempting
- A highest drawn empty line is not necessarily the LUMO if a lower empty line exists.
Practice, unseen assessment, re-test and next step
Complete the diagnostic questions, review the feedback, then attempt the final check.
After delayed re-test, continue to Construct and interpret pi-MO diagrams for benzene and polyenes at /learning/h3-molecular-orbitals-diagrams-conjugated-lesson.html.
- Review the feedback for each diagnostic question, then attempt the final check.
Open the feedback checkpoint after attempting
- Before moving on, state this boundary: A complete interpretation states configuration, frontier orbitals and any bond-order or magnetic consequence asked.