Calculate the oxidation number of all the atoms in the following compounds and ions:
a. `PbSO_(4)`, b. `CrO_(4)^(2-)`, c. `Sb_(2)O_(5)`, d. `NH_(4)_(2)SO_(4)`

Explanation

a. In `PbSO_(4)` or `Pb^(2+)(SO_(4))^(2-)`
Oxidation number of `Pb= +2`
Oxidation number of each `O` atom `=-2`
Let oxidation number of `S=x`
`:. +2+x+4(-2)=0implies2+x-8=0impliesx= +6`
Hence, oxidation number of `S` in `PbSO_(4)= +6`
b. In `CrO_(4)^(2-)`
Oxidation number of each `O` atom `= -2`
Let oxidation number of `Cr=x`
`x+4(-2)= -2 impliesx-8= -2 impliesx=6`
Hence, oxidation number of `Cr` in `CrO_(4)^(2-)= +6`.
c. In `Sb_(2)O_(5)`
Oxidation number of each `O` atom `= -2`
Let oxidation number of `Sb=x`
`2x+5(-2)=0implies2x-10=0impliesx= +5`
Hence, oxidation number of `Sb` in `Sb_(2)O_(5)= +5`
d. In `(NH)_(4))_(2)SO_(4)` or `(NH_(4)^(o+))_(2)SO_(4)^(2-)`
Let oxidation number of `N` in `NH_(4)^(o+)=x`
Oxidation number of each `H` atom `= +1`
`x+4(+1)= +1` (taking `NH_(4)^(o+)`)
`x= -3`
Hence, oxidation number of `N` in `(NH_(4))_(2)SO_(4)= -3`
and oxidation number of each `O` atom `= -2`
Let oxidation number of `S=x`
`:. x=4(-2)+2`( taking `SO_(4)^(2-)`)
`x-8= -2 impliesor x=6`
Hence, oxidation number of `S` in `(NH_(4))_(2)SO_(4)=6`.