Question

$F{e}^{+3}$ is more stable than $F{e}^{+2}$, the reason is/are :
This question has multiple correct options
(A). 1st and 2nd I.P. difference is less then 11.0 $eV$
(B). core of $F{e}^{+3}$ is more stable
(C). 2nd and 3rd I.P. difference is less then 11.0 $eV$
(D). I.P. of $F{e}^{+3}$ is high

Answer 1

Hint: $F{e}^{2+}$ and $F{e}^{3+}$ have some differences like in the number of electrons , which in turn result in different properties . And the stability of atoms depends on their outermost shell . If the outer shell is filled , then the atom is stable . And if the atoms with unfilled outer shells are called unstable .

Complete step by step answer:
As we learnt previously about the electronic configurations .
So here the electronic configuration of $F{e}^{2+}$ is
$1S^{2}2S^{2}2P^{6}3S^{2}3P^{6}3D^5$.
And the electronic configuration of $F{e}^{3+}$ is
$1S^{2}2S^{2}2P^{6}3S^{2}3P^{6}3D^5$ .
Now this can we understand as follows :
$F{e}^{3+}$ has $3d^5$half field configuration whereas $F{e}^{2+}$ has $3d^6$ configuration . Due to half filled , $3d^5$ has a more stable configuration . Due to the presence of half filled $d$ orbital . Thus , $F{e}^{3+}$ is more stable than $F{e}^{2+}$ .
Moreover , $F{e}^{3+}$ is more reactive since it has higher valency state and $F{e}^{2+}$ is easy to oxidize to $F{e}^{3+}$ because of removing the electron which results in a half filled s subshell . Therefore , $F{e}^{2+}$ is easily oxidized to $F{e}^{3+}$ .
Hence the correct option is B and C.


Note:
$F{e}^{2+}$ is oxidized to $F{e}^{3+}$ as dimethyl sulphoxide (DMSO) is normally used in organic oxidation reaction. So , it can oxidize ferrous salt to ferric .