The electronic configuration should not be in this form:2,8,8........ but in the orbital form.
2006-06-12 02:19:31 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3dn (where n=1 to 10)
There's the first Transition series for you. It can also be shown as:
[Ar],4s2, 3dn
2006-06-12 02:55:45 · answer #1 · answered by unclefrunk 7 · 1⤊ 0⤋
You can get better information with fewer words on a search. I gave yahoo main page search "electronic configuration of transition metals" and this popped out as the second link.
There aren't that many : your question was the third link.
The trick is fill them as you go, but remembering that a half full or full d orbital gives more stability than a full s orbital, so there is no 4s2 3d4 or 4s2 3d9, they turn out to be 4s1 3d5 and 4s1 3d9, thats the only trick (and why its such a favourite exam question)
2006-06-13 07:29:07 · answer #2 · answered by The_Otter 3 · 1⤊ 0⤋
Don't forget that if the metal has a charge or is bonded to anything that the 3d level becomes lower in energy (will get filled first) before the 4s level.
2006-06-12 07:36:33 · answer #3 · answered by Anonymous · 1⤊ 0⤋
Here's the first row. Add one to the numbers for each subsequent period.
http://wwwchem.uwimona.edu.jm:1104/courses/TMintro.html
2006-06-12 02:53:29 · answer #4 · answered by TheHza 4 · 1⤊ 0⤋
use the formula 2nsquare.otherwise use orbital method s,p,d,f orbital.
2006-06-12 02:28:23 · answer #5 · answered by Prakash 2 · 1⤊ 0⤋