Elements with Same Number of 4f Electrons

To determine which pairs of elements have the same number of electrons in their respective $4\text{f}$ orbitals, let us write out the ground-state electronic configurations for each given element:

1. Europium ($\text{Eu}, Z = 63$): $\text{Eu}: [\text{Xe}]\, 4\text{f}^7\, 5\text{d}^0\, 6\text{s}^2 \implies \text{Number of } 4\text{f} \text{ electrons} = 7$

2. Gadolinium ($\text{Gd}, Z = 64$): $\text{Gd}: [\text{Xe}]\, 4\text{f}^7\, 5\text{d}^1\, 6\text{s}^2 \implies \text{Number of } 4\text{f} \text{ electrons} = 7$

3. Dysprosium ($\text{Dy}, Z = 66$): $\text{Dy}: [\text{Xe}]\, 4\text{f}^{10}\, 5\text{d}^0\, 6\text{s}^2 \implies \text{Number of } 4\text{f} \text{ electrons} = 10$

4. Holmium ($\text{Ho}, Z = 67$): $\text{Ho}: [\text{Xe}]\, 4\text{f}^{11}\, 5\text{d}^0\, 6\text{s}^2 \implies \text{Number of } 4\text{f} \text{ electrons} = 11$

5. Thulium ($\text{Tm}, Z = 69$): $\text{Tm}: [\text{Xe}]\, 4\text{f}^{13}\, 5\text{d}^0\, 6\text{s}^2 \implies \text{Number of } 4\text{f} \text{ electrons} = 13$

6. Ytterbium ($\text{Yb}, Z = 70$): $\text{Yb}: [\text{Xe}]\, 4\text{f}^{14}\, 5\text{d}^0\, 6\text{s}^2 \implies \text{Number of } 4\text{f} \text{ electrons} = 14$

7. Lutetium ($\text{Lu}, Z = 71$): $\text{Lu}: [\text{Xe}]\, 4\text{f}^{14}\, 5\text{d}^1\, 6\text{s}^2 \implies \text{Number of } 4\text{f} \text{ electrons} = 14$

8. Hafnium ($\text{Hf}, Z = 72$): $\text{Hf}: [\text{Xe}]\, 4\text{f}^{14}\, 5\text{d}^2\, 6\text{s}^2 \implies \text{Number of } 4\text{f} \text{ electrons} = 14$

Now, let's analyze the given pairs:

• Pair A (Eu and Gd): Both have 7 $4\text{f}$ electrons. (Correct)
• Pair B (Dy and Ho): Dy has 10 and Ho has 11 $4\text{f}$ electrons. (Incorrect)
• Pair C (Yb and Hf): Both have 14 $4\text{f}$ electrons. (Correct)
• Pair D (Lu and Tm): Lu has 14 and Tm has 13 $4\text{f}$ electrons. (Incorrect)

Hence, pairs A and C have the same number of electrons in their $4\text{f}$ orbitals.

Thus, the correct answer is A and C Only.