Answer
405.3k+ views
Hint: 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 $pH > 7$) 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
$Be,Mg,Ca,Sr,Ba,Ra$
-In which $Ra$ is the only radioactive element.
-Now these element have general electronic configuration as $n{s^2}n{p^0}$
-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 $s$ 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 $I{E_1}$
$M \to {M^ + } + {e^ - }$
-For removing the second electron $I{E_2}$ will be provided leading to the formation of ${M^{2 + }}$
${M^ + } \to {M^{2 + }} + {e^ - }$
The net reaction is as follow
$M \to {M^{2 + }} + 2{e^ - }$
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.
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 $pH > 7$) 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
$Be,Mg,Ca,Sr,Ba,Ra$
-In which $Ra$ is the only radioactive element.
-Now these element have general electronic configuration as $n{s^2}n{p^0}$
-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 $s$ 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 $I{E_1}$
$M \to {M^ + } + {e^ - }$
-For removing the second electron $I{E_2}$ will be provided leading to the formation of ${M^{2 + }}$
${M^ + } \to {M^{2 + }} + {e^ - }$
The net reaction is as follow
$M \to {M^{2 + }} + 2{e^ - }$
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)