First Ionization energy exceptions

  • Thread starter henry3369
  • Start date
  • #1
194
0

Main Question or Discussion Point

I'm having a hard time understanding the first ionization energy exception between the s and p block. My book tries to explain it but I feel like it is a poor explanation. It states that the ionization energy for Boron is actually lower than Beryllium because of the fact that the p block is higher in energy. Is there another explanation someone can provide because this doesn't really explain why. If this was the case then the entire p block should have a lower ionization energy. I understand why ionization energy increases as you approach the right side of the periodic table due to an increase in Zeff, but I'm not sure what is creating this exception between the s and p block. Also I'm not sure why a higher energy would result in less energy required to remove an electron. Wouldn't something with more energy be more difficult to remove?
 

Answers and Replies

  • #2
Bystander
Science Advisor
Homework Helper
Gold Member
5,173
1,173
If you compare energy levels of 2S and 2P orbitals in atoms with more than one electron, you will find that they have "split," are no longer equal in energy as is implied by the principle quantum number, the P being at a higher level than the S. As you move across a row filling orbitals, the S orbitals at the lower energy are filled first (releasing more energy), then the P orbitals. The first P orbital to be filled, boron, is at enough higher energy that the energy required for the first ionization is less than the energy required to remove an electron from the 2S orbital of Be.
 

Related Threads on First Ionization energy exceptions

Replies
8
Views
3K
Replies
13
Views
45K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
13
Views
15K
Top