Question

what is oxidation state of Fe in FeSO4​

Answer 1

The oxidation number of SO4 group is -2,

Since oxidation number of oxygen is -2 and the oxidation number of sulphur is +6(atomic number of sulphur is 16 so electronic configuration is 1s2 2s2 2p6 3s2 3p6 , in outermost shell there are 6 valence electrons).

Now to form stable compound the net charge on FeSO4 must equal to zero.

So ,

1*(o.s. of Fe)+1*(o.s. of S)+4*(o.s. of O)=0

1*(o.s. of Fe)+1(+6)+4(-2)=0

(O.s. of Fe) +6–8=0

O.s. of Fe =8–6

O.s.of Fe=+2.

Hence oxidation state /oxidation number of Fe in FeSO4 is +2.

Similar example is MgSO4.

Answer 2

Given: Given compound is $FeSO_4$.

To find: We have to find the oxidation state of $Fe$ in $FeSO_4$

Solution:

• The chemical name of $FeSO_4$ is ferrous sulphate.
• Oxidation state of a atom defines the charge on the atom.

Here $FeSO_4$ is a neutral compound so its overall charge is $0$.

Now $S$ has a oxidation state $+6$ and $O$ has a oxidation state $-2$.

Now let oxidation state of $Fe$ is $x$.

We can write-

$+x-8+6=0$

$x=+8-6$

$+2$.

So, the oxidation state of $Fe$ is $+2$.