Question

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

Answer 1

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`?