JEE Main202228 Jun 2022Morning ShiftChemistryd and f Block ElementsActual
Which one of the lanthanoids given below is the most stable in divalent form?
Options
- AYb (Atomic Number 70 )
- BSm (Atomic Number 62 )
- CEu (Atomic Number 63 )
- DCe (Atomic Number 58 )
Correct answer
C. Eu (Atomic Number 63 )
Step-by-step solution
Ce + 2 : xe 4 f 2 Eu + 2 : xe 4 f 7 Sm + 2 : xe 4 f 6 Yb + 2 : xe 4 f 14 As Yb + 2   and   Eu + 2 has completely filled and half filled 4 f orbital, so they would be the most stable. But among the these two Eu + 2 is more stable based on reduction potential value. E M 3 + / M 2 + ° ⇒ Eu Yb - 0 . 35 - 1 . 05 Hence, due to more reduction potential in Eu as compared to Yb , it can concluded that Eu 2 + is more stable than Yb 2 + .