Question 3: Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
Answer:
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).
Comments
Most stable element of lanthanide 2+