Question

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

Answer 1

In copper sulfate, we got `Cu(+II)`.

Explanation:

Copper sulfate is an ionic salt composed of `Cu^(2+)` and `SO_4^(2-)` ions. The oxidation number of copper is the charge on its ion, i.e. `Cu(+II)`.

For sulfate anion, `SO_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_"oxidation no."+4xxO_"oxidation number"=-2`

Sonce the oxidation number of oxygen in its compounds is usually `-II`, and it is here, `S_"oxidation no."=+VI`.

For `FeHCl` (which would be very to isolate) we gots `Fe(+II)`; hydrogen binds here as a hydride, i.e. `H^-`.