Question #da3f8

Question

1069 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-13T19:06:29

Here's the reason for that.

The electronic configuration of Iron #(Fe)# in noble electronic configuration is #rarr#

#Fe_26rarr[Ar]_(18)3d^(6)4s^2#.

Now, to reach #Fe^(2+)# state, the atom need to loose #2# electrons from the valance #4s# subshell.

Hence,

#Fe_(26)^(2+)rarr[Ar]_(18)3d^6#.

Now, again to reach the #Fe^(3+)# state, the ferrous ion would loose one more electron from the #3d# subshell, which is diffuse in nature.

Hence,

#Fe^(3+)rarr[Ar]_(18)3d^5#.

Now, the #3d# subshell of the ferric ion is half filled having the state #3d^5#.

For info, the fully filled & the half filled #d# orbitals are highly stable.

Hence, the configuration of ferric ion is highly stable.

Hence, to reach to the highly stable ferric ion, the ferrous ion would loose electron from it's #3d# subshell.

Hope it Helps:)