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What I was thinking is that in 3d subshell (l=2) we have two electrons with $$m_l=-2$$ (spin up and down) and if we move to 3p we will fill the last vacant position - that is $$m_l=1$$ with spin down position.

and that doesn't obey the selection rule of $$ \triangle m_l=0,-1,1$$

but I saw in my notes that the transition $$3p^53d^1 \to3p^63d^0$$ is allowed

so I don't understand something fundamental in there...

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While the selection rule is important, I think you also need to see it in terms of energy and electronic stability of the subshell configuration. The $3p^63d^0$ configuration is more favorable from energy point of view compared to $3p^5d^1$. –  JKL Mar 8 '13 at 0:00
    
This is because a half filled sub shell or completely filled sub shell is more stable that's why $(3p^6)(3d^0)$ since p orbital is completely filled and also according to Bohr's/Bury Rule the sub shell with lower value of$n+l$ will be filled first. –  Akash Mar 8 '13 at 4:28
    
so If I got this right - 3d^2 is more stable then 3d^1 ? and if thats the case, why is the second transition allowed? how I see it : if an electron is alone in 3d (i.e 3d^1) then m_l=-2 and it moves to fill the last vacant position of 3p - which has m_l=1 ? –  YNWA Mar 8 '13 at 7:54

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