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I have came across the following statement : "the oxidation state $+II$ is stable for the elements of the alkalin earth metals" , the same is true for the elements of the first group , i've looked for a possible explanation and heard that it has a relation with the charge densities of the elements , if that's true what is that relation ? otherwise , what could explain the oxidation states for each of the groups in the periodic table ?

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    $\begingroup$ Can you clarify "the same is true for elements of the first group"? If you mean "the oxidation state +2 is stable for the elements of the alkali metals" then the statement is not correct. $\endgroup$ – bobthechemist Sep 27 '13 at 20:04
  • $\begingroup$ @bobthechemist no , i mean that the oxidation states +1 and + 2 are stable for the alkali and the alkali earth metals respectively . $\endgroup$ – user22323 Sep 28 '13 at 11:49
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The 'preferred' oxidation states of s-block metals can be explained by looking at their electron configurations:

Mathematica graphics

Note that for the alkali metals, each element has one more electron than a noble-gas configuration. This electron is very weakly bound to the nucleus as evidenced by the ver low ionization energy:

Mathematica graphics

It is much harder to remove the 2nd electron, since it is contained within a completely filled shell. (Please ignore the 2nd IE for Fr, as it doesn't exist in my data source.)

Therefore, the alkali metals have a tendency to lose just one electron, forming the +1 oxidation state.

The same thought process can be applied to the alkaline earth metals. Here are the first three ionization eneriges for each:

Mathematica graphics

Hopefully, it is clear that in the case of the alkaline earth metals, it is the 3rd electron that is in the completely filled shell, and that electron is extremely difficult to remove. Thus, the metals in this group tend to stay in the +2 oxidation state.

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