Find out the oxidation number of S in (i) `SO_(2)Cl_(2)` and (ii) `Na_(2)S_(2)O_(3)`

Explanation

(i) `overset(x)(S)overset(-2)(O_(2))overset(-1)(Cl_(2))" "therefore" "x+2(-2)+2(-1)=0 or x= +6`
(ii) `Na^(+).^(-)O-underset(O)underset(||)overset(S)overset(uarr)(S)-O^(-)Na^(+)`
Since there is a coordinate bond between the two S atoms, therefore, acceptor S atom has an O.N. of -2. The O.N. of the other S atom an be calculated as follows :
`underset(("for Na"^(+)))(2 xx (+1)) underset(("for O atoms"))(+3 xx (-2)) + x underset(("for coordinate S"))(+1 xx (-2)) or +2 - 6 + x - 2 = 0 or x = +6 `
Thus, the two S atoms in `Na_(2)S_(2)O_(3)` have oxidation numbers of -2 and +6.