What is the oxidation number of chromium in #[Cr(NH_3)_4Cl_2]Cl#?

1 Answer
Jul 4, 2016

Answer:

This is clearly #Cr(III+)", chromic ion"#. That is the metal has a #+III# oxidation state.

Explanation:

So how do we know this?

First let's remove the chloride counterion:

#[Cr(NH_3)_4Cl_2]^+#

Chromium is still 6-coordinate, but because we conserve charge when we remove #Cl^-# we are left with a cationic species.

And from #[Cr(NH_3)_4Cl_2]^+# we can remove 4 neutral ammine ligands:

#[CrCl_2]^+ +4NH_3#

And then we remove #2xxCl^-#, i.e. 2 negatively charged ligands:

#Cr^(3+) + 2xxCl^-#.

And thus we have #Cr(III)#.

Now obviously I have made a meal of this as a method of emphasis. All I have really done is conserve charge througout.

Can you tell me the oxidation number of the metal in #[Mn(CO)_5Br]#, #Pt(NH_3)_2Cl_2#, and #KMnO_4#?