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ChemistryEdu Logo d-block Elements | Oxidation States#

Oxidation states of d-block elements#

  • +2 is the most stable oxidation state among d-block elements due to high value of third ionization enthalpy.
  • The elements which show large number of oxidation states occur in or near the middle of the series. For example, manganese exhibits all oxidation states from +2 to +7.
  • Transition elements show oxidation states which differ by unity due to incompletely filled d-orbitals.
  • The elements at the extreme ends exhibit lesser number of oxidation states due to: (i) too few electrons to lose or share and (ii) too many d-electrons (fewer orbitals available for sharing of electrons).
  • Heavier elements favour higher oxidation states.
  • Low oxidation states are found when a complex has ligands capable of π-acceptor character in addition to σ- bonding. For example, in Ni(CO)4 and Fe(CO5), oxidation states of Ni and Fe = 0.

  • The highest oxidation numbers are achieved in case of TiX4 (tetrahalides), VF5 and CrF6.

  • The +7 oxidation state for Mn is not represented in simple halides but MnO3F is known, and beyond Mn no metal has a trihalide except FeX3 and CoF3.

  • Fluorine has the ability to stabilize higher oxidation state due to: (i) high lattice energy (in case of CoF3) and (ii) higher bond enthalpy terms for higher covalent compounds (VF5 and CrF6).

  • Fluorides are unstable in low oxidation states. Example: VX2, X = Cl, Br or I

  • Cu(II) halides are known except iodide. In this case, Cu2+ oxidizes I- to I2.

\[2Cu^{2+} + 4I^- → Cu_2I_{2(s)} + I_2\]
  • Many copper(I) compounds are unstable in aqueous solution and undergo disproportionation.
\[2Cu^+ → Cu^{2+} + Cu\]

  • Cu2+ is more stable than Cu+ due to more negative value of hydration enthalpy of Cu2+ which more than compensates for the second ionization enthalpy of copper.

  • The ability of oxygen to stabilize higher oxidation states is more than fluorine. It is because oxygen has the ability to form multiple bonds to metals.

  • The E0 value of M2+/M becomes less negative across the series due to increase in the sum of first and second ionization enthalpies.

  • The value of E0(M2+/M) for Mn, Ni and Zn are more negative than expected. For Mn and Zn, it is due to stability of half filled and fully filled d-orbitals respectively. For Ni, it is due to the highest negative hydration enthalpy.

  • The value of E0(M2+/M) for copper is positive because hydration energy and lattice energy of Cu2+ is more than that of Cu.

Questions#

Name a transition element which does not exhibit variable oxidation states.

Scandium (Z = 21) does not exhibit variable oxidation states.

Note: Zn also does not exhibit variable oxidation state but it is not a transition element.

Which of the 3d series of the transition metal exhibits the largest number of oxidation states and why?

Mn exhibits the largest number of oxidation states because it has total 7 electrons in s- and d-orbitals which can take part in bonding.

Cr(VI) in the form of dichromate in acidic medium is a strong oxidizing agent whereas MoO3 and WO3 are not. Explain.

It is because Mo(VI) and W(VI) are found to be more stable than Cr(VI). [Heavier elements favour higher oxidation state.]

The E0(M2+/M) value of copper is positive (+0.34 V). What is possible reason for this?

The E0(M2+/M) value of copper is positive (+0.34 V) because hydration energy and lattice energy of Cu2+ is more than that of Cu.

Only oxidizing acids like HNO3 and hot conc. H2SO4 react with Cu, the acids being reduced. Explain.

Because of positive E0 value of Cu, it cannot liberate H2 from acids. So, only oxidizing acids like HNO3 and hoyt conc. H2SO4 react with Cu, the acids being reduced.

Why E0 values for Mn, Ni and Zn are more negative than expected?

The value of E0(M2+/M) for Mn, Ni and Zn are more negative than expected. For Mn and Zn, it is due to stability of half filled and fully filled d-orbitals respectively. For Ni, it is due to highest negative hydration enthalpy.

Why is Cr2+ reducing and Mn3+ oxidizing when both have d4 configuration?

Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half filled t2g level. On the other hand, the change from Mn2+ to Mn3+ results in the half filled d5 configuration which has extra stability.

How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?

It is due to different stability of their electronic configuration.

How would you account for the increasing oxidizing power in the series VO2+ < Cr2O72- < MnO4-?

This is due to the increasing stability of the lower species to which they are reduced.

The highest Mn fluoride known is MnF4 whereas the highest oxide is Mn2O7. Explain.

It is due to the fact that the ability of oxygen to stabilize the high oxidation states exceeds that of fluorine.

Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

It is because O2 and F2 are strong oxidizing agents due to high electronegativity.

Which is a stronger reducing agent Cr2+ or Fe2+ and why?

Fe2+ is strong reducing agent because it will get oxidized to Fe3+ which has half filled d-orbitals which is more stable.

For the first row of transition metals, the E0 values are:

ElementVCrMnFeCoNiCu
E0(M2+/M)-1.18-0.91-1.18-0.44-0.28-0.25+0.34

Explain the irregularity in above values.

The E0(M2+/M) values are not regular which can be explained from the irregular variation of ionization enthalpies and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

Why is the E0 values for the Mn3+/Mn2+ couple much more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+?

It is due to much larger third ionization energy of Mn (where the required change is d5 to d4). This also explains why the +3 oxidation state of Mn is of little importance.

What is meant by disproportionation of an oxidation state? Give an example.

When a particular oxidation state becomes less stable relative to other oxidation states: one lower and one higher, it is said to undergo disproportionation.

Example: Manganese (VI) becomes unstable relative to Manganese (VII) and Manganese (IV) in acidic solution.

\[3MnO_4^{2-} → 2MnO_4^- + MnO_2 + 2H_2O\]

Explain why Cu+ ion is not stable in aqueous solutions?

It is because Cu2+ ion is more stable as hydration energy overcomes second ionization energy.