
What happens when aluminum fills its valence shell?
A) It will belong to group \[17\]
B) It will have 4 electrons in its outermost shell
C) It will attain noble gas configuration
D) None of these
Answer
462.9k+ views
Hint: We have to remember that the valence shell represents the outermost shell of an element. The electrons present in this valence shell can be termed as valence electrons. Noble gas has a stable electronic configuration i.e. completely filled orbitals and it belongs to the group \[18\].
Complete step by step answer:
We are going to calculate electronic configuration for \[Al\] first.
We have to know that the electronic configuration of Aluminum is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}\]
Symbol: \[Al\]
Atomic number: \[13\]
In order to write the electronic configuration of \[Al\], we need to know the number of electrons in an element, Since Atomic number= No. of electrons
Thus, the total number of electrons \[ = 13\], so to write the configuration we need to fill the orbitals around the\[Al\]atom.
Since, \[s\] orbital can hold up to $2$ electrons so we will start filling the electrons from \[1s\] orbital to \[2s\] orbital. Now comes p orbital, $6$ electrons can be filled in p orbital thus \[2p\] orbital will be filled next. Then we will move to the next orbital i.e. \[3s\] followed by \[3p\] on the basis of its electron capacity and the no. of electrons left for filling.
So,
Electronic configuration: \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}\]
Now what happens if the outermost shell of Aluminum is filled completely, i.e.
Electronic configuration will turned out to be- \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}\], which is the configuration of noble gases that means \[Al\] will attain its stability as noble gas which belongs to group 18 in periodic table.
Option A) is incorrect as when \[Al\] fills up its electron according to group \[17\] i.e. Halogen will have this configuration\[1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}\].
Option B) is incorrect as on filling up its outermost shell completely it will have \[8\] electrons in it.
Option C) is correct as, the stable noble gas configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}\] so when outermost shell for \[Al\] is filled completely it will attain noble gas configuration.
Option D) is incorrect since Option C) turns out to be right.
Hence, Option C is the correct answer.
Note: As we know that the noble gas configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}\] in which all the orbitals are completely filled and the elements of group $18$ does not react with others due to having completely filled outermost orbital. Group \[17\] belongs to Halogen family having one electron less in its outermost shell leading to an electronic configuration \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}\].
Complete step by step answer:
We are going to calculate electronic configuration for \[Al\] first.
We have to know that the electronic configuration of Aluminum is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}\]
Symbol: \[Al\]
Atomic number: \[13\]
In order to write the electronic configuration of \[Al\], we need to know the number of electrons in an element, Since Atomic number= No. of electrons
Thus, the total number of electrons \[ = 13\], so to write the configuration we need to fill the orbitals around the\[Al\]atom.
Since, \[s\] orbital can hold up to $2$ electrons so we will start filling the electrons from \[1s\] orbital to \[2s\] orbital. Now comes p orbital, $6$ electrons can be filled in p orbital thus \[2p\] orbital will be filled next. Then we will move to the next orbital i.e. \[3s\] followed by \[3p\] on the basis of its electron capacity and the no. of electrons left for filling.
So,
Electronic configuration: \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}\]
Now what happens if the outermost shell of Aluminum is filled completely, i.e.
Electronic configuration will turned out to be- \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}\], which is the configuration of noble gases that means \[Al\] will attain its stability as noble gas which belongs to group 18 in periodic table.
Option A) is incorrect as when \[Al\] fills up its electron according to group \[17\] i.e. Halogen will have this configuration\[1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}\].
Option B) is incorrect as on filling up its outermost shell completely it will have \[8\] electrons in it.
Option C) is correct as, the stable noble gas configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}\] so when outermost shell for \[Al\] is filled completely it will attain noble gas configuration.
Option D) is incorrect since Option C) turns out to be right.
Hence, Option C is the correct answer.
Note: As we know that the noble gas configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}\] in which all the orbitals are completely filled and the elements of group $18$ does not react with others due to having completely filled outermost orbital. Group \[17\] belongs to Halogen family having one electron less in its outermost shell leading to an electronic configuration \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}\].
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