SME
General Chemistry I · Problem type
Bond order and magnetism from an MO diagram
Use this for “Draw the MO diagram of O₂ from its 12 valence electrons. Find the bond order and the unpaired electrons: is O₂ paramagnetic?”
Diagram
O₂: 12 valence electrons, filled from σ2s up; from O₂ on, σ2p lies below π2p.
Bonding 8, antibonding 4 (lit red): bond order 2.
2 unpaired electrons (lit): paramagnetic.
This is an example. Type your own values to start a new problem.
e
Valence electrons
example
b
Bonding electrons
calculated
a
Antibonding electrons
calculated
BO
Bond order
calculated
u
Unpaired electrons
calculated
Formulas
b
=
bonding
electrons
of
e
8 = bonding electrons of 12
a
=
antibonding
electrons
of
e
4 = antibonding electrons of 12
BO
=
b
−
a
2
2
=
8 − 4
2
u
=
unpaired
MO
electrons
of
e
2 = unpaired MO electrons of 12
Assumptions
•
Only valence electrons go in (2s and 2p): add each atom’s, plus one for each negative charge and less one for each positive charge. O₂⁺ has 11.
•
Through N₂ (10 valence electrons) s–p mixing puts π2p below σ2p; from O₂ on, σ2p is lower. The MOs fill from the lowest.
•
Bond order is half of bonding less antibonding electrons. Zero means no bond: Be₂ and Ne₂ are not stable molecules.
•
Any unpaired electron makes the molecule paramagnetic (pulled into a magnet); none makes it diamagnetic.
Step-by-step
WE KNOW
e = 12
FIND
bonding electrons (b), antibonding electrons (a), bond order (BO), unpaired electrons (u)
Step 1 · Find bonding electrons (b)
Use b = bonding electrons of e
Fill σ2s, σ∗2s, then the 2p MOs from the lowest. Add the electrons in σ2s, π2p and σ2p.
b
=
bonding
electrons
of
12
σ2s 2 + σ2p 2 + π2p 4
b
=
8
Step 2 · Find antibonding electrons (a)
Use a = antibonding electrons of e
Add the electrons that land in σ∗2s, π∗2p and σ∗2p.
a
=
antibonding
electrons
of
12
σ∗2s 2 + π∗2p 2
a
=
4
Step 3 · Find bond order (BO)
Use BO = (b − a) ÷ 2
Bonding less antibonding electrons, halved: each pair of bonding electrons is half a bond.
BO
=
8 − 4
2
BO
=
4
2
BO = 2
Step 4 · Find unpaired electrons (u)
Use u = unpaired MO electrons of e
Count the single arrows: the two orbitals of a π pair each take one electron before either takes a second.
u
=
unpaired
MO
electrons
of
12
u
=
2
Check
Put the numbers back in. Both sides should match.
8 = bonding electrons of 12 ✓
4 = antibonding electrons of 12 ✓
2
=
8 − 4
2
✓
2 = unpaired MO electrons of 12 ✓
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