Calculate the oxidation number of
(i) `S` in `H_(2)S`,

(ii) `C` in `CO_(2)`
(iii)`C` in `CH_(2)CI_(2)`,
(iv) `N` in `(NH_(4))SO_(4)`
(v)`P` in `Na_(3)PO_(4)`

Explanation

(i) `S in H_(2)S` Let the oxidation number of S in `H_(2)S` be x writing the oxidation number of each atom above its symbol
`overset(+1)H_(2) overset(x)S`
sum of oxidation number of various atoms in `H_(2)S=2(+1)+x=2+x`
But the sum fo the oxidation numbers of various atoms in `H_(2)S` (neutral) is zero (Rule 7)
`therefore` 2+x=0 or x=-2
Thus the oxidation numbr of S in `H_(2)S is -2`
(ii) `C in CO_(2)` Let the oxidaion number of C in `CO_(2)` be x writing the oxidaion number o feach atom above its symbol
`therefore` x-4=0 or x=+4
Thus the oxidation number of C in `CO_(2)` is +4
(iii)`C in CH_(2)CI_(2)` Let the oxidaition number of C in `CH_(2)CI_(2)` be x writing the oxidatin number of each atom above its sysmbol `overset(x)C overset(+1)H_(2) overset(-1)CI_(2)`
`therefore` sum of hte oxidation numbers of various atoms in `CH_(2)CI_(2)` (neutral ) is zero (Rule 7)
`therefore` x=0
Thus the oxidation number of C in `CH_(2)CI_(2)` is zero
(iv) N in `(NH_(4))SO_(4)` Let the oxdation number of nitrogen in `(NH_(4))_(2)` be x writing the oxdation number o fhydrogen above its sysmbol and that of `SO_(4)^(2-)` ion above its fromula
`therefore` Sum of oxidation numbers of all the atoms in `(NH_(4))SO_(4)=2x+2(+1xx4)+(-2)=2x+6`
But the sum of oxidation numbers of all the atoms in `(NH_(4))_(2)SO_(4)` (neutral ) is zero (Rule 7)
`therefore 2x+6=0 or x =-3`
Thus the oxdation number of nitrogen in `(NH_(4)^(2))` is -3
(v)`P in Na_(3)PO_(4)` Let the oxidation number of P in `Na_(3)PO_(4)` be x writing hte oxidation number of each atom above its symbol `overset(+1)Na_(3) overset(x)P overset(-2)O_(4)`
Sum of the oxidation numbers of various of various atoms in `Na_(3)PO_(4)=3(+1)+x4(-2)=x-5`