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Question
Compare the stability of +2 oxidation state for the elements of the first transition series.
Solution 1
| Sc | +3 | ||||||
| Ti | +1 | +2 | +3 | +4 | |||
| V | +1 | +2 | +3 | +4 | +5 | ||
| Cr | +1 | +2 | +3 | +4 | +5 | +6 | |
| Mn | +1 | +2 | +3 | +4 | +5 | +6 | +7 |
| Fe | +1 | +2 | +3 | +4 | +5 | +6 | |
| Co | +1 | +2 | +3 | +4 | +5 | ||
| Ni | +1 | +2 | +3 | +4 | |||
| Cu | +1 | +2 | +3 | ||||
| Zn | +2 |
From the above table, it is evident that the maximum number of oxidation states is shown by Mn, varying from +2 to +7. The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons. The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.
Solution 2
Firstly, in the first half of the transition series, the sum of the first and second ionization enthalpies increases with increasing atomic number. Hence, the standard reducing potential (EΘ) is low and negative. Hence, the tendency to form M2+ ions decreases. Hence, the +2 oxidation state is more stable in the first half. The greater stability of the +2 oxidation state is due to half-filled d-subshells (d5) in Mn2+, fully filled d-subshells (d10) in Zn2+ and high negative hydration enthalpy in nickel.
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