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

Jul 4, 2016

This is clearly $C r \left(I I I +\right) \text{, chromic ion}$. That is the metal has a $+ I I I$ oxidation state.

#### Explanation:

So how do we know this?

First let's remove the chloride counterion:

${\left[C r {\left(N {H}_{3}\right)}_{4} C {l}_{2}\right]}^{+}$

Chromium is still 6-coordinate, but because we conserve charge when we remove $C {l}^{-}$ we are left with a cationic species.

And from ${\left[C r {\left(N {H}_{3}\right)}_{4} C {l}_{2}\right]}^{+}$ we can remove 4 neutral ammine ligands:

${\left[C r C {l}_{2}\right]}^{+} + 4 N {H}_{3}$

And then we remove $2 \times C {l}^{-}$, i.e. 2 negatively charged ligands:

$C {r}^{3 +} + 2 \times C {l}^{-}$.

And thus we have $C r \left(I I I\right)$.

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 $\left[M n {\left(C O\right)}_{5} B r\right]$, $P t {\left(N {H}_{3}\right)}_{2} C {l}_{2}$, and $K M n {O}_{4}$?