Calculating oxidation states in compounds

The algebraic sum of the oxidation numbers equals the charge on the species (zero for a neutral compound); oxygen is −2, alkali metals +1.

Worked example 1. Find the oxidation number of the transition metal in (a) K₂Cr₂O₇, (b) KMnO₄, (c) K₂MnO₄, (d) V₂O₅, (e) Fe₃O₄. (a) 2(+1) + 2x + 7(−2) = 0 → 2x = 12 → Cr = +6. (b) +1 + x + 4(−2) = 0 → Mn = +7. (c) 2(+1) + x + 4(−2) = 0 → Mn = +6. (d) 2x + 5(−2) = 0 → V = +5. (e) 3x + 4(−2) = 0 → x = +8/3 on average; the formula is the mixed oxide FeO·Fe₂O₃, containing one Fe(II) and two Fe(III) (+2 + 2×3 = +8 for three atoms). In (a), (b) and (d) the metal has lost all its 4s and 3d valence electrons (it is d⁰), so those compounds are diamagnetic and do not themselves absorb visible light through d–d transitions; they are often strongly oxidizing since the metal is in its highest state.