SELECT * FROM question_mgmt as q WHERE id=1532 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=63 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

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?

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).


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