I'm not sure what you mean by explain. I assume you are talking about the 3d series (there are several transition series; i.e., 3d, 4d, 5d, 4f, etc)The easy way to remember the electron configuration for the 3d series is this.
The number of electrons in the 3d orbitals will the same as the second number of the atomic number EXCEPT for Cr and Cu. Those have 1 more than you expect in the 3d and 1 less in the 4s. This is for the neutral atom. For ions it varies since the charge of the ion varies.
Can you explain the electron configuration for the transition metal ions?
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Well Dr Bob can you explain the rule for where you take from or add in transition metal ions?
We take from the 4s level first and go to the 3d to form the ions from the neutral atom. For example,
Cu is 1s2 2s2 2p6 3s2 3p6 3d10 4s1.
Remove the 4s electron to form Cu+ and remove the 4s1 and 1 of the 3d to make Cu^2+.
Cr is 1s2 2s2 2p6 3s2 3p6 3d5 4s1
Remove 4s1 to form Cr+; remove 4s1 and one 3d to form Cr^2+ and remove 4s1 and 2 3d to form Cr^3+. I hope this helps. The easy way to remember the electronic configuration of the IONS is
the number of electrons in the 3d shell = the second number of the atomic number (with no exceptions) for the 2+ ion.
So for Cu^2+ I would write 1s2 2s2 2p6 3s2 3p6 3d9
Cu is 1s2 2s2 2p6 3s2 3p6 3d10 4s1.
Remove the 4s electron to form Cu+ and remove the 4s1 and 1 of the 3d to make Cu^2+.
Cr is 1s2 2s2 2p6 3s2 3p6 3d5 4s1
Remove 4s1 to form Cr+; remove 4s1 and one 3d to form Cr^2+ and remove 4s1 and 2 3d to form Cr^3+. I hope this helps. The easy way to remember the electronic configuration of the IONS is
the number of electrons in the 3d shell = the second number of the atomic number (with no exceptions) for the 2+ ion.
So for Cu^2+ I would write 1s2 2s2 2p6 3s2 3p6 3d9
1 answer