# What are the oxidation states of the metal in CuSO_4, and FeHCl?

Aug 12, 2017

In copper sulfate, we got $C u \left(+ I I\right)$.

#### Explanation:

Copper sulfate is an ionic salt composed of $C {u}^{2 +}$ and $S {O}_{4}^{2 -}$ ions. The oxidation number of copper is the charge on its ion, i.e. $C u \left(+ I I\right)$.

For sulfate anion, $S {O}_{4}^{2 -}$, the SUM of the individual oxidation numbers of each element is equal to the charge of the ion, i.e. $- 2$.

And thus ${S}_{\text{oxidation no."+4xxO_"oxidation number}} = - 2$

Sonce the oxidation number of oxygen in its compounds is usually $- I I$, and it is here, ${S}_{\text{oxidation no.}} = + V I$.

For $F e H C l$ (which would be very to isolate) we gots $F e \left(+ I I\right)$; hydrogen binds here as a hydride, i.e. ${H}^{-}$.