Why does niobium have a #d^4 s^1# electron configuration but vanadium has a #d^3 s^2# electron configuration?
2 Answers
It only looks strange. Energetically, it is filling orbitals as normally as any other element.
Explanation:
It follows the normal rules for energy minimization for stability. It is not "weird", except possible from the perspective of numbered shell filling, which is common for all the "transition" elements. The 4d orbital is actually at a higher energy state than the 5s. Hund's Rule dictates the unpaired filling.
See also:
https://www.webelements.com/niobium/
for more properties.
It's not obvious. There are two opposing trends:
ns-(n-1)d ENERGY GAP
- The
#3d# orbital of#"V"# is#"2.79 eV"# lower in energy than the#4s# (see Appendix B.9). - The
#4d# orbital of#"Nb"# is only#"1.64 eV"# lower in energy than the#5s# (see Appendix B.9).
For visual comparison, I had graphed data from the above reference:
As a result, one might expect that purely based on energy, at some partway filling point, the fourth valence electron would favor going into the
ORBITAL SIZE
The
This would favor putting the fourth valence electron in the
Due to the first trend favoring putting the fourth valence electron in to the
CONCLUSION
Apparently, what I'm seeing is that there is some cancellation between:
-
having bigger orbitals, reducing electron repulsions and making it more favorable to continue on filling the
#d# orbitals with the fourth valence electron. -
having
#(n-1)d# orbitals closer in energy to the#ns# than#"V"# does, making it more favorable for the fourth electron to be placed into the#ns# instead of the#d# .
And as it turns out, a