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Question
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
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Solution
The oxidation states displayed by the first half of the first row of transition metals are given in the table below.
| Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn |
| +2 | +2 | +2 | +2 | +2 | +2 | +2 | +1 | +2 | |
| +3 | +3 | +3 | +3 | +3 | +3 | +3 | +3 | +2 | |
| +4 | +4 | +4 | +4 | +4 | +4 | +4 | |||
| +5 | +5 | +5 | |||||||
| +6 | +6 | +6 | |||||||
| +7 |
It can be easily observed that except Sc, all other metals display +2 oxidation states. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.
| Sc (+2) = d1 |
| Ti (+2) = d2 |
| V (+2) = d3 |
| Cr (+2) = d4 |
| Mn (+2) = d5 |
+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in (+2) state also increases from Ti (+2) to Mn (+2), the stability of +2 state increases (as d-orbital is becoming more and more half-filled). Mn (+2) has d5 electrons (that is, a half-filled d shell, which is highly stable).
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