Molecular Orbital Theory for Homonuclear Diatomics · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
En anglais

Two atoms sharing whole orbitals

Science · Chemistry · ages 18-19
Name ______________________   Date ____________

  1. How is bond order calculated from bonding and antibonding electron counts?

    • (bonding + antibonding) / 2
    • (bonding - antibonding) / 2
    • (bonding x antibonding) / 2

  2. Two identical atomic orbitals with suitable overlap form bonding and antibonding combinations. Where is the bonding level?

    • Below the parent levels in energy
    • At the parent levels in energy
    • Above the parent levels in energy

  3. Equal numbers of bonding and antibonding electrons give bond order zero in the elementary MO model.

    Circle one:   True   False

  4. Build C2's ground-state valence scheme from 8 electrons. Counts follow sigma(2s), sigma*(2s), pi(2p) pair, sigma(2p), pi*(2p) pair, sigma*(2p). Which occupation, bond order and unpaired-electron count result?

    • 2,2,2,2,0,0; bond order 2; unpaired 2
    • 2,0,4,2,0,0; bond order 4; unpaired 0
    • 2,2,4,0,0,0; bond order 2; unpaired 0

  5. A diatomic has 8 bonding and 2 antibonding electrons. What is its bond order?

    Answer: ______________

  6. For comparable bonds between the same elements, what usually follows from a higher bond order?

    • A longer bond with a lower dissociation energy
    • A shorter bond with exactly unchanged dissociation energy
    • A shorter bond with a higher dissociation energy

  7. A student fills F2's 14 valence electrons but leaves bonding sigma(2p) empty and fills sigma*(2p). Which corrected ground-state occupation and conclusions follow? Counts follow sigma(2s), sigma*(2s), sigma(2p), pi(2p) pair, pi*(2p) pair, sigma*(2p).

    • 2,2,2,4,4,0; bond order 1; unpaired 0
    • 2,2,2,4,2,2; bond order 1; unpaired 2
    • 2,2,2,4,4,0; bond order 2; unpaired 2

  8. Ground-state oxygen is attracted to a magnetic field, although its usual Lewis structure pairs all electrons. How does its MO configuration explain this?

    • Both pi* electrons pair in one orbital
    • Each pi* orbital holds one unpaired electron
    • The empty sigma* orbital supplies two unpaired electrons
LightMySky · lightmysky.comW1-mt_wRBDDmRYka-s1

Answer key

For grown-ups. Fold this page away before handing over the rest.

Two atoms sharing whole orbitals W1-mt_wRBDDmRYka-s1

  1. (bonding - antibonding) / 2 · Bond order is (bonding electrons - antibonding electrons) / 2. Magnetism requires a separate count of unpaired electrons.
  2. Below the parent levels in energy · The bonding combination increases electron density between the nuclei and lies below the identical parent levels.
  3. True · Equal counts cancel in the subtraction before division by two. This arithmetic does not require a universal rule about the sizes of orbital energy shifts.
  4. 2,2,4,0,0,0; bond order 2; unpaired 0 · C2 uses bonding pi below sigma(2p). Its complete set counts are 2,2,4,0,0,0. Six bonding and two antibonding electrons give order 2. Both pi orbitals are filled, leaving no unpaired electrons.
  5. 3 · Subtract first, then halve: (8 - 2) / 2 = 3.
  6. A shorter bond with a higher dissociation energy · For comparable bonds between the same elements, higher bond order usually means a shorter bond and more energy needed to separate the atoms.
  7. 2,2,2,4,4,0; bond order 1; unpaired 0 · F2 uses bonding sigma(2p) below pi(2p). Filling fourteen electrons gives 2,2,2,4,4,0. Eight bonding and six antibonding electrons give order 1. All electrons are paired.
  8. Each pi* orbital holds one unpaired electron · One electron occupies each equal-energy pi* orbital. These two unpaired electrons explain oxygen's paramagnetism.
Worksheet · LightMySky