In case of alkaline earth metals, the oxidation state more than two is not observed because

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They have only two electrons in the outer most shell

The s-orbital can accommodate only two electrons

The removal of third electron involves breaking up of noble gas configuration and the energy needed for this purpose is extremely high

None of the above

answer is C.

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Detailed Solution

_{12}Mgleft[ {1{s^2};2{s^2};2{p^2};3{s^2};} right];xrightarrow{{;; - {e^ - };;}}{;_{12}}M{g^ + }left[ {left[ {Ne} right];3{S^1}} right];xrightarrow{{;; - {e^ - };;}}{;_2}M{g^{ + 2}}left[ {mathop {left[ {Ne} right]}limits_{Stable;inert;gas;E.C} ;3{S^0}} right]

The alkaline earth metals in their valency shell contain only "2" electron hence they lose two electrons easiliy and shows +II oxdiation state by forming stable M⁺² ion.But they can't show oxidation state more than two
Reason
M⁺² ion attain stable noble gas electronic configuration and hence the removal of 3^rd electron is highly difficult , and the energy required for this purpose is extremely high .

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