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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 1
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).
Solution 2
Losing two electrons from s-orbital leads to an increase in effective nuclear charge as the atomic number increases. The ion size decreases, which leads to more stability. The stability is reduced in the beginning because there are few electrons to lose or exchange.
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