Question

Which element among the following exhibits electronic configuration as [Xe]4f⁰ in + 4 oxidation state?

पर्याय

• Dysprosium (Z = 66)

• Praseodymium (Z = 59)

• Neodymium (Z = 60)

• Cerium (Z = 58)

Answer 1

Cerium (Z = 58)

Explanation:

1. Dysprosium (66) = [Xe] 4f¹⁰ 6s²
2. Praseodymium (59) = [Xe] 4f³ 6s²
3. Neodymium (60) = [Xe] 4f⁴ 6s²
4. Cerium (58) = [Xe] 4f¹ 5d¹ 6s²

Cerium exhibits +4 oxidation state with configuration 4f⁰.