Why is the electron configuration of Ni [Ar] 4s2 3d8 instead of [Ar] 4s0 3d10, while the configuration of Cu is [Ar] 4s1 3d10? The configuration of Pd is also [Kr] 5s0 4d10. I'm confused as to why this pattern doesn't hold up for Nickel. In addition, for all 3d transition elements, the s electrons are always at a higher energy correct? So when a question refers to the valence electron of a transition metal it means the electron in the 4s, 5s, etc.?
Electron Configuration/quantum numbers question....
- Thread starter katniss09
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It's been so long since I've looked at any of this, so take this with a grain of salt. There's a few exceptions for periodic trends for transition metals that bring half-filled d shell stability and filled-d shell stability. Both of these are favorable ground states for these atoms to be in. Here are the exceptions worth noting:
Transition metals in the 4th column (Cr, Mo, W) exhibit half-filled d shell stability. Take Chromium (Cr) for example. Ignoring these exceptions for a moment, if we were to fill the orbitals from lowest to highest energy, we would get this electronic configuration: [Ar]4s2 3d4 -- But what actually happens in these odd-ball exceptions is that one of the electrons in the 4s orbital is promoted to the higher energy d orbital which increases the stability of the atom. (This 'promotion' is not the same as the promotion of an excited electron which occurs when a photon of the right energy strikes the atom). In this case, the correct configuration is [Ar]4s1 3d5. The same can be said for transition metals in the 9th colum (Cu, Ag, Au) -- these metals exhibit filled d shell stability. Let's use copper (Cu) this time. Rather than an electronic configuration of [Ar]4s2 3d9 we instead get: [Ar]4s1 3d10 for the same reason (increased stability). Once again, these are exceptions to the general rule.
Notice, Nickel is in the 8th row and isn't one of the exceptions to this rule. So it follows the general trend of filling orbitals from lowest to highest energy: [Ar] 4s2 3d8.
Palladium is also in the same row as Nickel. This also follows the general trend and so the correct electron configuration for Pd is: [Kr]5s2 4d8.
You have to be careful though. If the question mentions that one or more electrons were excited to the next available energy level -- because ground state and excited state configurations are entirely different. It's possible the electron configuration you provided for Pd was an excited state configuration, in which case: [Kr]5s0 4d10 is a valid answer choice.
Hope this helps.
Transition metals in the 4th column (Cr, Mo, W) exhibit half-filled d shell stability. Take Chromium (Cr) for example. Ignoring these exceptions for a moment, if we were to fill the orbitals from lowest to highest energy, we would get this electronic configuration: [Ar]4s2 3d4 -- But what actually happens in these odd-ball exceptions is that one of the electrons in the 4s orbital is promoted to the higher energy d orbital which increases the stability of the atom. (This 'promotion' is not the same as the promotion of an excited electron which occurs when a photon of the right energy strikes the atom). In this case, the correct configuration is [Ar]4s1 3d5. The same can be said for transition metals in the 9th colum (Cu, Ag, Au) -- these metals exhibit filled d shell stability. Let's use copper (Cu) this time. Rather than an electronic configuration of [Ar]4s2 3d9 we instead get: [Ar]4s1 3d10 for the same reason (increased stability). Once again, these are exceptions to the general rule.
Notice, Nickel is in the 8th row and isn't one of the exceptions to this rule. So it follows the general trend of filling orbitals from lowest to highest energy: [Ar] 4s2 3d8.
Palladium is also in the same row as Nickel. This also follows the general trend and so the correct electron configuration for Pd is: [Kr]5s2 4d8.
You have to be careful though. If the question mentions that one or more electrons were excited to the next available energy level -- because ground state and excited state configurations are entirely different. It's possible the electron configuration you provided for Pd was an excited state configuration, in which case: [Kr]5s0 4d10 is a valid answer choice.
Hope this helps.
