Special Electron Configuration Q

[Meas]

---

Members don't see this ad.

I dont really understand why the electron configuration for tin will be [Kr] 4d10 5s2 5p2. Isn't half-filled & full orbitals have lower energy (more stable)? If so, shouldn't the answer be 4d10 5s1 5p3 so that we have one 5s half-filled & one 5p half-filled?

The half-filled orbital idea makes sense for Cr: [Ar] 4s1 3d5 but does not really make sense for tin.

Thanks in advance!

 

[dentista123]

Half filled or full D-orbitals is how I understand it. The exceptions occur in the d's. But P-orbitals you have to look out for 8+ octets.

 

[cyrano]

Not exactly sure why it doesn't exhibit that property completely, but it does a little bit. Check out this abstract

http://www.iop.org/EJ/abstract/0022-3719/20/9/025

You can see that it tends towards having an s electron spend a bit of time in the p orbital but not enough for it to actually be defined as being located there. Weird huh?

5s1.785 p2.22

 

[Sublimation]

For the first row of the transition metals the 4s and 3d are at the same energy. This rule only applies to the first row of the transition metals.

 

[ttran9229]

I dont really understand why the electron configuration for tin will be [Kr] 4d10 5s2 5p2. Isn't half-filled & full orbitals have lower energy (more stable)? If so, shouldn't the answer be 4d10 5s1 5p3 so that we have one 5s half-filled & one 5p half-filled?

The half-filled orbital idea makes sense for Cr: [Ar] 4s1 3d5 but does not really make sense for tin.

Thanks in advance!

This rule applies to both d and f sublevels although there are some elements that do not follow this general rule. I will not name them all, but an example would be Ni, Pd, and Pt. Ni has electronic configuration of 4s2 3d8 while Pd and Pt are 4d10 and 6s1 5d9, respectively. All are in the same column of group 8B.

 

[BTtheman777]

For the first row of the transition metals the 4s and 3d are at the same energy. This rule only applies to the first row of the transition metals.

Isnt the s sublevel at a lower energy than the d sublevel. In other words for the s sublevel, its energy is (n+ l) = (4+0) = 4. While the d sublevel is (3+2)=5. Thus the s sublevel is lower and it fills before the d sublevel even though it is one principal quantum number higher in the 4th period?