Question

Among the elements of lanthanoid series (58 Ce_71 Lu), the total number of elements which have odd number of electrons in f-orbitals in their ground state configuration

Answer 1

8

Answer 2

The electron configurations (f‐electron counts) for the lanthanoid series (from Ce to Lu) are as follows (noting that the d‐electron does not count toward f electrons):

$$\begin{array}{lcl} \text{Ce (Z = 58)} & : & 4f^1\,5d^1\,6s^2 \quad \text{(1, odd)}\\[6pt] \text{Pr (Z = 59)} & : & [\text{Xe}]\,4f^3\,6s^2 \quad \text{(3, odd)}\\[6pt] \text{Nd (Z = 60)} & : & [\text{Xe}]\,4f^4\,6s^2 \quad \text{(4, even)}\\[6pt] \text{Pm (Z = 61)} & : & [\text{Xe}]\,4f^5\,6s^2 \quad \text{(5, odd)}\\[6pt] \text{Sm (Z = 62)} & : & [\text{Xe}]\,4f^6\,6s^2 \quad \text{(6, even)}\\[6pt] \text{Eu (Z = 63)} & : & [\text{Xe}]\,4f^7\,6s^2 \quad \text{(7, odd)}\\[6pt] \text{Gd (Z = 64)} & : & [\text{Xe}]\,4f^7\,5d^1\,6s^2 \quad \text{(7, odd)}\\[6pt] \text{Tb (Z = 65)} & : & [\text{Xe}]\,4f^9\,6s^2 \quad \text{(9, odd)}\\[6pt] \text{Dy (Z = 66)} & : & [\text{Xe}]\,4f^{10}\,6s^2 \quad \text{(10, even)}\\[6pt] \text{Ho (Z = 67)} & : & [\text{Xe}]\,4f^{11}\,6s^2 \quad \text{(11, odd)}\\[6pt] \text{Er (Z = 68)} & : & [\text{Xe}]\,4f^{12}\,6s^2 \quad \text{(12, even)}\\[6pt] \text{Tm (Z = 69)} & : & [\text{Xe}]\,4f^{13}\,6s^2 \quad \text{(13, odd)}\\[6pt] \text{Yb (Z = 70)} & : & [\text{Xe}]\,4f^{14}\,6s^2 \quad \text{(14, even)}\\[6pt] \text{Lu (Z = 71)} & : & [\text{Xe}]\,4f^{14}\,5d^1\,6s^2 \quad \text{(14, even)} \end{array}$$

Counting the elements with an odd number of f electrons:

• Ce (1)
• Pr (3)
• Pm (5)
• Eu (7)
• Gd (7)
• Tb (9)
• Ho (11)
• Tm (13)

There are 8 elements with an odd number of electrons in the f–orbitals.