Cue Column | Note Column

Free element rule? | Any element in its free (uncombined) state has oxidation number = 0. e.g., Fe(s), $O_{2}$(g), $Cl_{2}$(g), Na(s) → all = 0.

F always? | Fluorine = −1 in ALL compounds (no exception; most electronegative element).

O exceptions? | Usually −2. Exceptions: Peroxides (H_{2}$$O_{2}, Na_{2}$$O_{2}) = −1; $OF_{2}$ = +2; Superoxides (K$O_{2}$) = −1/2.

H exceptions? | Usually +1. Exception: Metal hydrides (NaH, Ca$H_{2}$, LiH) = −1.

Sum rule? | For neutral molecule: sum = 0. For ion: sum = charge on ion.

Common examples? | Cr in $K_{2}Cr_{2}O_{7}$: +6 (2 + 2x − 14 = 0 → x = +6); Mn in $KMnO_{4}$: +7 (1 + x − 8 = 0 → x = +7); S in $SO_{4}^{2-}$: +6; N in $NO_{3}^{-}$: +5; S in H_{2}$$SO_{3}: +4.

Highest oxidation state? | Usually = group number in periodic table. Cl can be +7 (Cl_{2}$$O_{7}), S can be +6 (H_{2}$$SO_{4}), Cr can be +6, Mn can be +7.

Identifying redox? | If any element changes oxidation number → redox reaction. Oxidized species: ox. no. increases (loses $e^{-}$). Reduced species: ox. no. decreases (gains $e^{-}$).

Summary

Key exceptions to memorize: F = always −1; O = −1 in peroxides, +2 in $OF_{2}$; H = −1 in metal hydrides. For any compound, assign known elements first, then solve algebraically for unknown. Oxidation = increase in oxidation number = loss of electrons.