The electronic configuration of Scandium is: $\ce{[Ar] 4s^2 3d^1}$

What is the electronic configuration of Scandium ion?

Also about the electronic configuration of Cobalt, which is $\ce{[Ar] 4s^2 3d^7}$. So What is the electronic configuration if Cobalt ion?

Is it $\ce{[Ar] 4s^2 3d^6}$ or $\ce{[Ar] 4s^1 3d^7}$ ? And why?

Thanks for your help.


closed as off-topic by ringo, Klaus-Dieter Warzecha, bon, Curt F., Jannis Andreska Mar 5 '16 at 13:49

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  • $\begingroup$ Strictly speaking, in general you can't determine the ground state electronic configuration of an atom through qualitative arguments, and for ions it gets even harder. We're lucky that the aufbau principle works as well as it does for how simple it is, but it makes many people overconfident in its strength. $\endgroup$ – Nicolau Saker Neto Mar 5 '16 at 6:49
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    $\begingroup$ Please use latex, and please shorten the configuration to nearest noble gas for brevity. $\endgroup$ – Max Payne Mar 5 '16 at 8:00

The electronic configuration of the cobalt is $\ce{[Ar] 3d^1 4s^2}$ the electrons with the higher energy will be snatched and the electronic configuration of Argon is very stable then with scandium you will have easely $\ce{Sc^{3+}}$.

For the Cobalt it is a bit more difficult, like you write the configuration you cannot find a correct answer. For all element first you do first the configuration using the Klechkovsky rule and after you put all orbitals by principal quantum number growing.

So the cobalt configuration is $\ce{[Ar] 3d^7 4s^2}$ then if you snatch two electrons from the $\ce{4s}$ orbital you have a stable configuration for the $\ce{Co(II)}$ ion. You cannot snatch them from the $\ce{3d}$ orbital (even if $\ce{[Ar] 3d^5 4s^2}$ looks stable because the $\ce{3d}$ orbital is half full then the spin is maximal) because its energy is lesser than the energy of the $\ce{4s}$ orbital.

You can find the Cobalt at different oxydation states from $\ce{+I}$ to $\ce{+IV}$ but it depends on what you have in your solution or in your gas if you have a gas.

NB : Remember that the configuration of the elements are given if gas phase, then for example the more stable configuration of the copper iron is for $\ce{Cu^+}$ and not for $\ce{Cu^2+}$, $\ce{Cu^2+}$ is stable in water, so the answer may depends on the problem you have.

Explaination for the copper :

Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^{2+}}$ ion has a greater charge density than the $\ce{Cu^+}$ ion and so forms much stronger bonds releasing more energy.

The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu^+}$ ion loses an electron to become $\ce{Cu^{2+}}$ which can then release this hydration energy.

I hope it can help you !


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