
Why is the iron KLMN configuration $ 2,8,14,2 $ and not $ 2,8,8,8? $
Answer
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Hint: AS we know that electronic configuration is the arrangement of electrons in different orbits and its orbitals. While writing it, we follow a special procedure where the energy number or shell number is written first and then we write subshell and the number of electrons it contains is written in subscript. We can say that the electronic configuration is written in KLMN format, where the K,L,M,N represent the energy shells. So here we will first understand the ground state configuration and then the reason why is it not possible.
Complete answer:
We know that Iron is a part of the d-block in the periodic table. It is a transition metal.
The atomic number of Iron is $ 26 $ .
So the ground state configuration is written as
$ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6} $ .
But we know that Iron is not going to use the above configuration because Iron cannot skip $ 3d $ orbitals and use $ 4p $ orbitals, because we already know that $ 4p $ orbitals are higher in energy than $ 3d $ orbitals. So the $ 3d $ orbitals are fully occupied.
We should know that the K,L,M,N shell is further divided into subshells named as $ s,p,d,f $ subshells. These subshells consist of a specific number of orbitals.
We know that each $ s $ subshell can have a maximum number of two electrons, $ p $ subshell can have $ 6 $ electrons and so on.
So here in case of iron, it has completely filled $ d $ subshell due to which it has configuration $ 2,8,14,2 $ not $ 2,8,8,8 $ .
Let us understand this in simple term:
We know that only a certain number of electrons can revolve in certain orbits . For example in K shells almost $ 2 $ electrons can be there. Through some experiments it is found that in the valence shell almost $ 8 $ electrons can be there.
So if Iron $ \left( {Fe} \right) $ electronic configuration would be like $ 2,8,16 $ , then we can see that the valence shell would not be a stable one.
So the electronic configuration goes like
$ 2,8,14,2 $ .
Here we can see that the atoms is table since it does not exceed the number
$ 8 $ .
Hence this is the reason for the Iron configuration to be like $ 2,8,14,2 $ .
Note:
We should know that iron having $ + 3 $ oxidation is called a ferric ion and the one having $ + 2 $ oxidation rate is known as a ferrous ion. We should note that removal of ions will always take place from the last filled orbitals regardless of whether it is half filled, partially filled and completely filled.
Complete answer:
We know that Iron is a part of the d-block in the periodic table. It is a transition metal.
The atomic number of Iron is $ 26 $ .
So the ground state configuration is written as
$ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6} $ .
But we know that Iron is not going to use the above configuration because Iron cannot skip $ 3d $ orbitals and use $ 4p $ orbitals, because we already know that $ 4p $ orbitals are higher in energy than $ 3d $ orbitals. So the $ 3d $ orbitals are fully occupied.
We should know that the K,L,M,N shell is further divided into subshells named as $ s,p,d,f $ subshells. These subshells consist of a specific number of orbitals.
We know that each $ s $ subshell can have a maximum number of two electrons, $ p $ subshell can have $ 6 $ electrons and so on.
So here in case of iron, it has completely filled $ d $ subshell due to which it has configuration $ 2,8,14,2 $ not $ 2,8,8,8 $ .
Let us understand this in simple term:
We know that only a certain number of electrons can revolve in certain orbits . For example in K shells almost $ 2 $ electrons can be there. Through some experiments it is found that in the valence shell almost $ 8 $ electrons can be there.
So if Iron $ \left( {Fe} \right) $ electronic configuration would be like $ 2,8,16 $ , then we can see that the valence shell would not be a stable one.
So the electronic configuration goes like
$ 2,8,14,2 $ .
Here we can see that the atoms is table since it does not exceed the number
$ 8 $ .
Hence this is the reason for the Iron configuration to be like $ 2,8,14,2 $ .
Note:
We should know that iron having $ + 3 $ oxidation is called a ferric ion and the one having $ + 2 $ oxidation rate is known as a ferrous ion. We should note that removal of ions will always take place from the last filled orbitals regardless of whether it is half filled, partially filled and completely filled.
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