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Posted (edited)

Can someone explain to me why the d-orbitals of a transition metal when they are in a complex are filled before the s-orbital? thanks.

 

 

* i also have another question*

 

How come the maximum oxidation state of Ni does not exceed 4 since it has many more valence electrons than 4.

Edited by hellomister
added question
Posted

All i know is that its more stable, but does anyone know exactly why?

 

and when you write the electron configuration of a transition metal in a complex do you write the configuration in the order that its filled or do you still write the s in front of the d?

Posted

If there was only a single electron in the atom, the subshells would degenerate (all at the same energy as each other within each shell). However, as you get more electrons, the subshells start to split in energy.

 

The s-subshells get closest to the nucleus, so they prevent the p- and d- orbitals from feeling the full impact of the attraction of the nucleus. This is called screening, or shielding. What this means is that the s-subshells have lower energy than the p-orbitals. The same is true for p- and d-subshells.

 

To start with, for the 2s and 2p subshells, the difference is small, but it does mean the 2s subshell is filled first.

 

When we get to the 3s, again we see it's lower than the 3p. The surprise is that the next subshell is the 4s, and not the 3d subshell. The 4s subshell is reduced in energy so much by its proximity to the nucleus that it actually ends up lower than the 3d subshell. but not by very much, which is why we get some weird electron configurations and other phenomena in transition metals.

 

You average textbook contains a lot of useful diagrams and probably a better explanation as well.

  • 2 weeks later...

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