$F{e^{ + 3}}$ 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

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
$1{S^2}2{S^2}2{P^6}3{S^2}3{P^6}3{D^5}$.
And the electronic configuration of $F{e^{3 + }}$ is
$1{S^2}2{S^2}2{P^6}3{S^2}3{P^6}3{D^5}$ .
Now this can we understand as follows :
$F{e^{3 + }}$ has $3{d^5}$half field configuration whereas $F{e^{2 + }}$ has $3{d^6}$ configuration . Due to half filled , $3{d^5}$ 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 .