Variable Oxidation States in Transition Elements
Electron Configuration
- The full electronic configuration of the first-row transition metals is shown in the table below
- Following the Aufbau Principle electrons occupy the lowest energy subshells first
- The 4s overlaps with the 3d subshell so the 4s is filled first
- Remember: You can abbreviate the first five subshells, 1s-3p, to [Ar] representing the configuration of argon (known as the argon core)
Table showing the electronic configuration of the first d-series transition elements
Transition metal | Electronic configuration | Noble gas core electronic configuration |
Ti | 1s2 2s2 2p6 3s2 3p6 3d2 4s2 | [Ar] 3d2 4s2 |
V | 1s2 2s2 2p6 3s2 3p6 3d3 4s2 | [Ar] 3d3 4s2 |
Cr | 1s2 2s2 2p6 3s2 3p6 3d5 4s1 | [Ar] 3d5 4s1 |
Mn | 1s2 2s2 2p6 3s2 3p6 3d5 4s2 | [Ar] 3d5 4s2 |
Fe | 1s2 2s2 2p6 3s2 3p6 3d6 4s2 | [Ar] 3d6 4s2 |
Co | 1s2 2s2 2p6 3s2 3p6 3d7 4s2 | [Ar] 3d7 4s2 |
Ni | 1s2 2s2 2p6 3s2 3p6 3d8 4s2 | [Ar] 3d8 4s2 |
Cu | 1s2 2s2 2p6 3s2 3p6 3d10 4s1 | [Ar] 3d10 4s1 |
- There are two exceptions to the Aufbau Principle in the first row of the d-block:
- Chromium
- Copper
- In both cases, an electron is promoted from 4s to 3d to achieve a half-full and full d-subshell, respectively
- Chromium and copper have the following electron configurations, which are different to what you may expect:
- Cr is [Ar] 3d5 4s1 not [Ar] 3d4 4s2
- Cu is [Ar] 3d10 4s1 not [Ar] 3d9 4s2
- This is because the [Ar] 3d5 4s1 and [Ar] 3d10 4s1 configurations are energetically more stable and are preferred configurations
- When forming cations, remove the 4s electrons first
Worked example
Deducing the electronic configuration of transition element ions
State the full electronic configuration of:
- Cu
- Mn(III) ions
- V4+
Answer 1 - Cu:
- Cu atomic number = 29
- 1s22s22p63s23p63d104s1 OR 1s22s22p63s23p64s13d10
- Remember: Copper atoms prefer a complete d subshell
Answer 2 - Mn(III):
- Step 1: Write out the electron configuration of the atom first:
- Mn atomic number = 25
- 1s22s22p63s23p64s23d5
- Step 2: Subtract the appropriate number of electrons starting from the 4s subshell
- Mn(III) = 22 electrons
- 1s22s22p63s23p63d4
Answer 3 - V4+:
- Step 1: Write out the electron configuration of the atom first:
- V atomic number = 23
- 1s2 2s2 2p6 3s2 3p6 3d3 4s2
- Step 2: Subtract the appropriate number of electrons starting from the 4s subshell
- V4+ = 19 electrons
- 1s22s22p63s23p63d1
- When transition elements forms ions they lose electrons from the 4s subshell first
- This is because when the orbitals are occupied, the repulsion between electrons pushes the 4s into a higher energy state so that it now becomes slightly higher in energy than the 3d subshell
- The 4s is now the outer shell and loses electrons first
- The loss of the 4s electrons means that +2 is a common oxidation state in transition metals
- The reason why the transition metals have variable oxidation states all comes down to energy
Table showing the common oxidation states of transition elements
Element | Most common oxidation states |
Ti | +3, +4 |
V | +3, +5 |
Cr | +3, +6 |
Mn | +2, +4, +7 |
Fe | +2, +3 |
Co | +2, +3 |
Ni | +2 |
Cu | +1, +2 |
Explaining variable oxidation states using successive ionisation energies
- Using titanium and vanadium as examples, the graph below shows that the first few ionisation energies are relatively small and relatively close together
- This means that the energy difference associated with removing a small number of electrons enables transition metals to vary their oxidation state with ease
Graph of titanium and vanadium ionisation energies
Ionisation energies increase for the removal of successive electrons in titanium and vanadium
- The +2 and +3 oxidation states are shown by all the transition elements
- The +3 state is more stable in the elements up to chromium
- The +2 state is more stable in the later elements
- Transition metal ions with oxidation state +3 and above tend to be polarising and have a degree of covalent character in the bonds they form
- The ions have a high charge density and pull electrons towards themselves
- The maximum oxidation state possible corresponds to the total number of electrons in the 4s and 3d which reaches a maximum at manganese
- An example you may be familiar with is the manganate(VII) ion, MnO4– which is a powerful oxidising agent
Examiner Tip
- You may sometimes see electronic configurations with:
- 3d electrons written before 4s
- 4s electrons written before 3d
- Both ways are acceptable although putting the 3d electrons first is more conventional, even though 4s fills before 3d