
Compare the first ionization energy of Aluminium $(Z = 13)$ with that of Magnesium $(Z = 12)$. Justify your answer.
Answer
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Hint:Aluminium is present in the group $13$ and Magnesium is present in the group $2$ which is also known as alkaline earth metals. The size of Magnesium is more than Aluminium.
Complete answer:
Magnesium Is the 2nd member of group$2$having the following electronic configuration ($Z = 12$)$1{s^2}2{s^2}2{p^6}3{s^2}$.
It has two electrons in its outermost shell.
Ionization energy is defined as the energy required to pull out an electron from an isolated gaseous atom.
For Magnesium the reaction can be written as,
$Mg \to M{g^ + } + {e^ - }$
When an electron is pulled out from the outermost shell of Magnesium which is $3s$ orbital, the electron left will be one.
So the ionization energy required here will be high here as the fully filled $3s$ orbital is reduced to half-filled configuration and so the stability also decreases.
Ionization energy is inversely proportional to the size of elements. But here moving from Magnesium to Aluminium there is an introduction of a new outer shell which is $3p$ orbital. So the comparison will not be according to the size of element.
Considering Aluminium the 2nd member of group$13$, which is also a p-block element, the electronic configuration will be as follows-
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}$.
As we know that s orbital is closest to the nucleus, so the nuclear attraction is very strong for s orbital as compared to p orbital, it will be easier and more convenient to pull out electron from the outermost shell of Aluminium which is not even half filled. After the ionization Aluminium will have configuration like ground state Magnesium which is more stable.
For Aluminium the reaction can be written as,
$Al \to A{l^ + } + {e^ - }$
Thus the ionization energy required to pull out electrons from Aluminium will be very low as compared to Magnesium.
Hence the first ionization energy ofAluminium $(Z = 13)$ is less than that of Magnesium $(Z = 12)$.
Note:
Hydroxides of Magnesium is very basic as Magnesium is a metal and hydroxide ofAluminium is amphoteric in nature due to its semi metalloid properties.
Magnesium exists in only $ + 2$ oxidation state. With this oxidation state Magnesium acquires noble gas like confirmation which is highly stable. For Aluminium the most stable oxidation state by acquiring noble gas like configuration is$ + 3$.
Complete answer:
Magnesium Is the 2nd member of group$2$having the following electronic configuration ($Z = 12$)$1{s^2}2{s^2}2{p^6}3{s^2}$.
It has two electrons in its outermost shell.
Ionization energy is defined as the energy required to pull out an electron from an isolated gaseous atom.
For Magnesium the reaction can be written as,
$Mg \to M{g^ + } + {e^ - }$
When an electron is pulled out from the outermost shell of Magnesium which is $3s$ orbital, the electron left will be one.
So the ionization energy required here will be high here as the fully filled $3s$ orbital is reduced to half-filled configuration and so the stability also decreases.
Ionization energy is inversely proportional to the size of elements. But here moving from Magnesium to Aluminium there is an introduction of a new outer shell which is $3p$ orbital. So the comparison will not be according to the size of element.
Considering Aluminium the 2nd member of group$13$, which is also a p-block element, the electronic configuration will be as follows-
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}$.
As we know that s orbital is closest to the nucleus, so the nuclear attraction is very strong for s orbital as compared to p orbital, it will be easier and more convenient to pull out electron from the outermost shell of Aluminium which is not even half filled. After the ionization Aluminium will have configuration like ground state Magnesium which is more stable.
For Aluminium the reaction can be written as,
$Al \to A{l^ + } + {e^ - }$
Thus the ionization energy required to pull out electrons from Aluminium will be very low as compared to Magnesium.
Hence the first ionization energy ofAluminium $(Z = 13)$ is less than that of Magnesium $(Z = 12)$.
Note:
Hydroxides of Magnesium is very basic as Magnesium is a metal and hydroxide ofAluminium is amphoteric in nature due to its semi metalloid properties.
Magnesium exists in only $ + 2$ oxidation state. With this oxidation state Magnesium acquires noble gas like confirmation which is highly stable. For Aluminium the most stable oxidation state by acquiring noble gas like configuration is$ + 3$.
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