Why does chromium form a #"Cr"^(2+)# ion?

1 Answer
Truong-Son N.
Sep 9, 2017

Because it can? It can also form #"Cr"^(3+)# and #"Cr"^(6+)# ions quite often, and in fact, more often. I would say the prevalent cation depends on the environment.

It is usually easier to only lose #2# electrons if there are few strong oxidizers nearby, like #"F"_2# or #"O"_2#. In isolation, the #+2# cation is most stable because we have put in the least ionization energy, increasing its energy the least.

However, since oxidizing environments are generally rather common (we have plenty of oxygen in the air), I would say that that is why the #+3# and #+6# oxidation states are stabilized and therefore more common in reality, while the #+2# could occur in more reducing environments and is more stable in isolation.

Many transition metals take on variable oxidation states depending on the context... Their #(n-1)d# orbitals are close in energy to their #ns# orbitals.

Examples for chromium are:

  • #"CrBr"_2#, #"CrO"#, etc. #" "" "" "" "" "#(#"Cr"^(+2)#, a #3d^4# configuration)
  • #"Cr"("NO"_3)_3#, #"Cr""PO"_4#, etc. #" "" "" "#(#"Cr"^(+3)#, a #3d^3# configuration)
  • #"CrO"_3#, #("NH"_4)_2"Cr"_2"O"_7#, etc. #" "" "#(#"Cr"^(+6)#, a noble gas configuration)

In fact, the #+3# and #+6# oxidation states have been observed more often than the #+2# for #"Cr"#. But the higher oxidation states, if you notice, occur in highly oxidizing environments.

Related questions
See all questions in Net Charge
Impact of this question
7020 views around the world
You can reuse this answer
Creative Commons License