The Lanthanide Ion Having Four Unpaired Electrons

Question

The lanthanide ion having four unpaired electrons is:

1. Tb³⁺
2. Ho³⁺
3. Nd³⁺
4. Ce³⁺

Given: Atomic numbers
Ce = 58, Nd = 60, Tb = 65, Ho = 67

---

Step 1: Write the electronic configuration of Ln³⁺ ions

In lanthanides, the 6s electrons and one 5d electron (if present) are removed first to form the +3 ion.

Ce³⁺ (Z = 58)

Ce = [Xe] 4f¹5d¹6s²

Ce³⁺ = [Xe] 4f¹

Number of unpaired electrons = 1

---

Nd³⁺ (Z = 60)

Nd = [Xe] 4f⁴6s²

Nd³⁺ = [Xe] 4f³

Number of unpaired electrons = 3

---

Tb³⁺ (Z = 65)

Tb = [Xe] 4f⁹6s²

Tb³⁺ = [Xe] 4f⁸

For 4f⁸ configuration:

• Seven f-orbitals are first singly occupied.
• The eighth electron pairs up in one orbital.

Therefore, unpaired electrons = 6

---

Ho³⁺ (Z = 67)

Ho = [Xe] 4f¹¹6s²

Ho³⁺ = [Xe] 4f¹⁰

For 4f¹⁰ configuration:

• Seven electrons occupy seven orbitals singly.
• Three additional electrons pair up in three orbitals.

Therefore, unpaired electrons = 4

---

Step 2: Compare the number of unpaired electrons

Ion	Configuration	Unpaired Electrons
Ce³⁺	4f¹	1
Nd³⁺	4f³	3
Tb³⁺	4f⁸	6
Ho³⁺	4f¹⁰	4

---

Answer

Ho³⁺ has four unpaired electrons.

Correct Option: (2) Ho³⁺

Quick Trick: For an fⁿ configuration,

• If n ≤ 7, unpaired electrons = n
• If n > 7, unpaired electrons = 14 − n

For Ho³⁺: 4f¹⁰ → 14 − 10 = 4 unpaired electrons. ✔️