give the oxidation number of each kind of of atom or ion 1. Sulfate 2.sn 3.S^2- 4.Fe^3+ 5. Sn^4+ Solution For a neutral ion, it\'s formal charge, and therefore, its ON = 0 For a monoatomic ion, the formal charge and therefore, its ON = it\'s ionic charge For a polyatomic ion, ON needs to be calculated using algebraic method( equating the formal charge of various ions equal to its ionic charge) 1) Sulfate: SO 4 2 - : Let FC of S = X FC of O = -2 Hence, calculating : X -2*4 = -2 => X = 6 2) Sn: As it\'s neutral, it\'s ON = 0 3) S 2- : As it\'s monoatomic, ON = its FC = -2 4)Fe +3 : As it\'s monoatomic, it\'s ON = its FC = +3 5) Sn +4 : As it\'s monoatomic, it\'s ON = its FC = +4 Please leave a positive rating in case I was helpful. If not, leave your doubt in the comments. I\'ll get back to you ASAP :) .