Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
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 | |
| +2 | +2 | +2 | +2 | ||
| +3 | +3 | +3 | +3 | +3 | |
| Oxidation state | +4 | +4 | +4 | +4 | |
| +5 | +5 | +6 | |||
| +6 | +7 |
It can be easily observed that except Sc, all others metals display +2 oxidation state. 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 4 selectrons 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 half-filled d shell, which is highly stable).
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Most stable element of lanthanide 2+
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