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

1 Answer
Aug 12, 2017

Answer:

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^-#.