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Question: Alkaline earth metal form ions of the formula: A) \({M^ + }\) B) \({M^ - }\) C) \({M^{2 + }}\)...

Alkaline earth metal form ions of the formula:
A) M+{M^ + }
B) M−{M^ - }
C) M2+{M^{2 + }}
D) M2−{M^{2 - }}

Explanation

Solution

Alkaline earth metals are the elements present in the second group (column) of the periodic table. As we know that all elements of a group will have the same number of electrons in the outermost shell, by examining the general electronic configuration of these elements, we will be able to identify how many electrons need to be added/removed in order for them to attain stability and form ions.

Complete step by step solution:
-Alkaline earth metals are placed in group two of the periodic table and are named such because they form alkaline solution (solution having pH>7pH > 7) when dissolve in water
-While the word Earth refers to the oxides of these metals which are heat resistant
-As the word earth was used previously by scientist for compounds which does not undergo any change when heated using apparatus that were available during those days
-The members of group two are as follow
Be,Mg,Ca,Sr,Ba,RaBe,Mg,Ca,Sr,Ba,Ra
-In which RaRa is the only radioactive element.
-Now these element have general electronic configuration as ns2np0n{s^2}n{p^0}
-Which indicates they have two electrons in their outermost shell hence these elements can achieve inert gas configuration by either losing two electrons or by gaining six electrons (which is impossible)
-So, they will achieve the inert pair configuration by removing two electrons present in their outermost ss orbital as follows
-The removal of both electrons will be done in two steps
-In the first step an amount of energy which will be necessary in order to remove the outermost electron will be provided this energy is known as Ionization Energy or IE1I{E_1}
M→M++e−M \to {M^ + } + {e^ - }
-For removing the second electron IE2I{E_2} will be provided leading to the formation of M2+{M^{2 + }}
M+→M2++e−{M^ + } \to {M^{2 + }} + {e^ - }
The net reaction is as follow
M→M2++2e−M \to {M^{2 + }} + 2{e^ - }
Hence the option ‘C’ is the correct solution.

Note:
-Alkaline earth metals are less electropositive (electropositivity is the tendency to give away/push electrons away from themselves) than the alkali metals hence require a higher Ionization energy than the former.
-Also, due to their high value of Ionization energy they are less reactive than alkali metals, which are placed in group one of the periodic table.