How come the trend doesn't stand for Barium to Radium

  • Thread starter Thread starter Ed Aboud
  • Start date Start date
Click For Summary
SUMMARY

The trend in ionization energies typically decreases down a group in the periodic table; however, an exception occurs between Barium (Ba) and Radium (Ra). Barium has an ionization energy of 502 kJ/mol, while Radium has a higher ionization energy of 510 kJ/mol. This anomaly is attributed to the presence of f orbitals, which are less effective at shielding nuclear charge compared to d, p, and s orbitals. As a result, the increased nuclear charge in Radium leads to a greater ionization energy.

PREREQUISITES
  • Understanding of periodic trends in ionization energy
  • Knowledge of atomic structure, including orbitals
  • Familiarity with the Lanthanide series and its impact on ionization energy
  • Basic concepts of shielding effect in atomic physics
NEXT STEPS
  • Research the role of f orbitals in atomic structure and their shielding effects
  • Study the ionization energies of elements in the Lanthanide series
  • Explore the periodic trends of transition metals, particularly the anomalies
  • Learn about the relationship between nuclear charge and ionization energy
USEFUL FOR

Chemistry students, educators, and professionals interested in atomic theory and periodic trends, particularly those focusing on ionization energy anomalies.

Ed Aboud
Messages
200
Reaction score
0
The trend in ionization energies going down a group is that it decreases between each element. How come the trend doesn't stand for Barium to Radium. Barium has an ionization energy of 502 kilojoules per mole but Radium has an ionization energy of 510 kilojoules per mole. Why is this?
Thanks for any help.

edit:
Just noticed it occurs for La to Ac, Mo to W, Tc to Re, Ru to Os, Rh to Ir, Pd to Pt, Ag to Au, Cd to Hg, In to Tl, Sn to Pb as well.
 
Last edited:
Chemistry news on Phys.org
I'd guess it has to do either with the appearance of all those extra elements in the Lanthanide series, or with the appearance of F orbitals, in between the pairs of elements you mention.

But more specifically, why this would cause a higher I.E. I don't know.
 
Redbelly98 said:
I'd guess it has to do either with the appearance of all those extra elements in the Lanthanide series, or with the appearance of F orbitals, in between the pairs of elements you mention.

But more specifically, why this would cause a higher I.E. I don't know.

It is because f orbitals are less effective at shielding than d orbitals which are less effective than p and ultimately s orbitals. The nuclear charge increases but the shielding effect is lessened resulting in a greater ionization energy.
 
  • Like
Likes   Reactions: Astronuc
Ah makes sense now. Thanks for the help.
 

Similar threads

What is Plasmonic Photocatalysis and How Can It Help Remove Hydrogen Sulfide?
  • · Replies 1 ·
Replies
1
Views
2K