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Concept 02 / 06The d and f Block Elements

Oxidation States and Standard Electrode Potentials of Transition Metals

Variable oxidation states of 3d metals and their trends, highest oxidation states in oxides and fluorides, stability of oxidation states, standard electrode potentials E(M2+/M) and E(M3+/M2+) with their irregularities, stability of higher and lower oxidation states, and disproportionation of Cu+ and Mn(VI).

Figure 1
Variable oxidation states. Because the nsns and (n−1)d(n-1)d electrons have similar energies, transition metals can lose different numbers of electrons (see figure). Successive oxidation states usually differ by one (Fe2+\mathrm{Fe^{2+}}, Fe3+\mathrm{Fe^{3+}}), unlike p-block elements where they often differ by two. The number of oxidation states increases up to the middle of the series: Sc shows only +3+3; Mn shows +2+2 to +7+7 (the maximum, using all 4s4s and 3d3d electrons). After Mn, pairing of d electrons reduces the number of available states. At the end, Cu shows +1+1 and +2+2, and Zn only +2+2. The highest oxidation states occur in oxides and fluorides (and oxoanions): Mn2O7\mathrm{Mn_{2}O_{7}}, MnO4−\mathrm{MnO_{4}^{-}}, CrO3\mathrm{CrO_{3}}, Cr2O72−\mathrm{Cr_{2}O_{7}^{2-}}, V2O5\mathrm{V_{2}O_{5}}, VF5\mathrm{VF_{5}}, CrF6\mathrm{CrF_{6}}. Oxygen stabilises high states better than fluorine because it can form multiple bonds with the metal (so Mn reaches +7+7 in Mn2O7\mathrm{Mn_{2}O_{7}} but only +4+4 in MnF4\mathrm{MnF_{4}}). Low oxidation states (0, +1+1) are found in complexes with π\pi-acceptor ligands, such as Ni(CO)4\mathrm{Ni(CO)_{4}} and Fe(CO)5\mathrm{Fe(CO)_{5}}. Down a group, higher oxidation states become more stable (opposite to the p-block): Cr6+\mathrm{Cr^{6+}} in dichromate is a strong oxidant, while Mo6+\mathrm{Mo^{6+}} and W6+\mathrm{W^{6+}} are stable. Standard electrode potentials E∘(M2+/M)E^{\circ}(\mathrm{M^{2+}/M}) (V): Ti −1.63-1.63, V −1.18-1.18, Cr −0.90-0.90, Mn −1.18-1.18, Fe −0.44-0.44, Co −0.28-0.28, Ni −0.25-0.25, Cu +0.34+0.34, Zn −0.76-0.76. They become less negative across the series (increasing sum of first and second ionisation enthalpies). Irregularities: Mn, Ni and Zn are more negative than expected, due to the stability of half-filled d5d^{5} (Mn2+\mathrm{Mn^{2+}}), the very high hydration enthalpy of Ni2+\mathrm{Ni^{2+}}, and filled d10d^{10} (Zn2+\mathrm{Zn^{2+}}). Copper is the only 3d metal with a positive value, because its high enthalpy of atomisation and ionisation are not compensated by the hydration enthalpy of Cu2+\mathrm{Cu^{2+}}; so Cu does not liberate H2\mathrm{H_{2}} from dilute acids. E∘(M3+/M2+)E^{\circ}(\mathrm{M^{3+}/M^{2+}}) (V): Ti −0.37-0.37, V −0.26-0.26, Cr −0.41-0.41, Mn +1.57+1.57, Fe +0.77+0.77, Co +1.97+1.97. The low value for Sc (only +3+3) and for Cr reflects the stability of Sc3+\mathrm{Sc^{3+}} (d0d^{0}) and Cr3+\mathrm{Cr^{3+}} (t2g3t_{2g}^{3}); so Cr2+\mathrm{Cr^{2+}} is a strong reducing agent. The high value for Mn shows that Mn2+\mathrm{Mn^{2+}} (d5d^{5}) is very stable and Mn3+\mathrm{Mn^{3+}} a strong oxidant. The comparatively low value for Fe reflects the stability of Fe3+\mathrm{Fe^{3+}} (d5d^{5}). Co3+\mathrm{Co^{3+}} is a very strong oxidant in water (though stabilised in complexes). Disproportionation. Cu+\mathrm{Cu^{+}} is unstable in water: 2Cu+(aq)→Cu2+(aq)+Cu(s)\mathrm{2Cu^{+}(aq) \rightarrow Cu^{2+}(aq) + Cu(s)}, because the much larger hydration enthalpy of Cu2+\mathrm{Cu^{2+}} more than compensates for the second ionisation enthalpy. Similarly, manganate(VI) disproportionates in acid: 3MnO42−+4H+→2MnO4−+MnO2+2H2O\mathrm{3MnO_{4}^{2-} + 4H^{+} \rightarrow 2MnO_{4}^{-} + MnO_{2} + 2H_{2}O}. Key Takeaways: - Oxidation states differ by one; maximum +7+7 at Mn; fewer after Mn. - Highest states in oxides and fluorides; O stabilises higher states than F. - Cu is the only 3d metal with positive E∘(M2+/M)E^{\circ}(\mathrm{M^{2+}/M}). - Cr2+\mathrm{Cr^{2+}} reductant, Mn3+\mathrm{Mn^{3+}} and Co3+\mathrm{Co^{3+}} oxidants (from E∘(M3+/M2+)E^{\circ}(\mathrm{M^{3+}/M^{2+}})). - Cu+\mathrm{Cu^{+}} disproportionates in water. - Pitfall: in the d-block, heavier elements favour higher oxidation states, the opposite of the inert pair trend in the p-block. Brain-Teaser 1: Why is Cr2+\mathrm{Cr^{2+}} reducing and Mn3+\mathrm{Mn^{3+}} oxidising, though both have a d4d^{4} configuration? Solution: Cr2+\mathrm{Cr^{2+}} (d4d^{4}) is oxidised to Cr3+\mathrm{Cr^{3+}} (d3d^{3}, half-filled t2gt_{2g}, very stable), so it gives up electrons readily. Mn3+\mathrm{Mn^{3+}} (d4d^{4}) is reduced to Mn2+\mathrm{Mn^{2+}} (d5d^{5}, half-filled d, extra stable), so it accepts electrons readily. E∘E^{\circ}: Cr3+/Cr2+=−0.41 V\mathrm{Cr^{3+}/Cr^{2+}} = -0.41\,\mathrm{V}, Mn3+/Mn2+=+1.57 V\mathrm{Mn^{3+}/Mn^{2+}} = +1.57\,\mathrm{V}. Brain-Teaser 2: Why does manganese show its highest oxidation state in an oxide (Mn2O7\mathrm{Mn_{2}O_{7}}) rather than in a fluoride? Solution: Oxygen can form multiple (π\pi) bonds with the metal, so fewer O atoms are needed to reach a high oxidation state: in Mn2O7\mathrm{Mn_{2}O_{7}}, each Mn is tetrahedrally surrounded by oxygens with Mn=O bonds. A fluoride would need seven F atoms around one small Mn, which is sterically impossible; the highest fluoride is MnF4\mathrm{MnF_{4}}. Brain-Teaser 3: Using E∘(Cu2+/Cu+)=0.15 VE^{\circ}(\mathrm{Cu^{2+}/Cu^{+}}) = 0.15\,\mathrm{V} and E∘(Cu+/Cu)=0.52 VE^{\circ}(\mathrm{Cu^{+}/Cu}) = 0.52\,\mathrm{V}, show that Cu+\mathrm{Cu^{+}} disproportionates in water. Solution: For 2Cu+→Cu2++Cu\mathrm{2Cu^{+} \rightarrow Cu^{2+} + Cu}: reduction half Cu+→Cu\mathrm{Cu^{+} \rightarrow Cu} (0.520.52), oxidation half Cu+→Cu2+\mathrm{Cu^{+} \rightarrow Cu^{2+}} (reverse of 0.150.15). Ecell∘=0.52−0.15=+0.37 VE^{\circ}_{\text{cell}} = 0.52 - 0.15 = +0.37\,\mathrm{V}, positive, so the disproportionation is spontaneous.

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