Find the oxidation number of
(a) `S "in" SO_(4)^(2-)`ion (b) `S "in"HSO_(3)^(-)`ion
(c ). Pt in `(PtCl_(6)^(2-))`ion (d). Mn in `(MnO_(4))^(-)` ion

Explanation

Let the oxidation number of S be x.
We know that Ox. No. of O=-2
So, Ox. No. S+4(Ox. No.O)=-2
or `{:(,x,+,4(-2),=-2),(,x,,-8,=-2):}`
x=+8-2=+6
The oxidation number of S in `SO_(4)^(2-) "ion is" +6`
(b). Let the oxidation number of S be x in `HSO_(3)^(-)` ion.
We know that, Ox. no. of H=+1
Ox. no. of O=-2
So, Ox. no. H+Ox. no. S+3(Ox. no. O)=-1
`{:(,+1,+,x,+,3(-2),=-1),(,+1,+,x,-,6,=-1):}`
or x-5=-1
x=+5-1=+4
The oxidation number of S in `HSO_(3)^(-)` ion is +4.
(c ). THe oxidation number of Pt be x.
We know that Ox. no. of Cl=-1
Ox. no. Pt+6(Ox.no. Cl)=-2
`{:(,x,6(-1),=-2),(,x-,6,=-2):}`
or x=+6-2=4
The oxidation number of Pt in `[Pt(Cl)_(6)]^(2-)` ion is +4
(d). Let the oxidation number of Mn be x,
We know that, Ox. no. of O=-2
So, Ox. no. Mn+4(Ox. no. O) io=-1
`{:(,x,4(-2),=-1),(,x-,8,=-1):}`
or x=+8-1=7
The oxidation number of Mn in `[MnO_(4)]^(-)` ion is +7