High spin or low spin

Science · Chemistry · ages 19-21

Name ______________________ Date ____________

Which pair sits at the strong field end of the series?
- Fluoride and water
- Cyanide and carbon monoxide
- Iodide and bromide, the largest halides
How do you decide between high spin and low spin?
- Compare the splitting energy with the pairing energy
- Compare the number of d electrons with the number of ligands
- Compare the metal charge with the total charge on the ligands
Ligand field strength follows directly from ligand charge.
Circle one: True False
The same d5 ion is forced low spin by cyanide. Type its number of unpaired electrons.
Answer: ______________
An octahedral d5 ion is high spin. Type its number of unpaired electrons.
Answer: ______________
A complex has a spin-only moment near 3.87 Bohr magnetons. How many unpaired electrons does it hold?
- 4
- 2
- 3
A tetrahedral cobalt complex is probably low spin.
Circle one: True False
An octahedral iron(III) complex, which is d5, has one unpaired electron. What do you conclude?
- High spin with a weak field ligand
- Low spin with a strong field ligand
- Tetrahedral with a weak field ligand

Answer key

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

Cyanide and carbon monoxide · Cyanide and carbon monoxide sit at the strong field end of the taught series. They give large splittings and often favour low spin.
Compare the splitting energy with the pairing energy · The cheaper option, spreading out or pairing up, wins.
False · Neutral carbon monoxide outranks fluoride despite having no charge.
1 · Low spin d5 pairs four electrons and leaves one single.
5 · High spin d5 spreads one electron per level with none paired.
3 · With mu = √(n(n+2)), three unpaired electrons give √15, about 3.87 Bohr magnetons.
False · The tetrahedral gap is tiny, so pairing almost never pays.
Low spin with a strong field ligand · One unpaired d5 means low spin, which needs a strong field ligand.