Which element has a ground-state electron configuration that ends in $n^1{f}^{14}{n}^2{d}^{10}{n}^3{p}^6$ (i.e., a filled $f$ subshell)?

• A. Cerium (Ce)
• B. Lutetium (Lu)
• C. Gadolinium (Gd)
• D. Ytterbium (Yb)

Answer: (D)

Verified step by step guidance

1. Understand that the problem is asking for an element whose ground-state electron configuration ends with a completely filled f subshell, specifically \(n^1f^{14} n^2d^{10} n^3p^6\), indicating a filled f subshell with 14 electrons.
2. Recall that the f subshell can hold a maximum of 14 electrons, and elements with a filled f subshell are typically the lanthanides or actinides, where the 4f or 5f orbitals are being filled.
3. Identify that the notation \(n^1f^{14}\) refers to the f subshell being fully occupied, and the subsequent \(n^2d^{10} n^3p^6\) indicates that the d and p subshells in higher principal energy levels are also filled, suggesting a stable, noble gas-like configuration after the f block.
4. Review the electron configurations of the given elements (Cerium, Lutetium, Gadolinium, and Ytterbium) to determine which one has a fully filled f subshell (14 electrons). For example, Ytterbium (Yb) has the electron configuration ending in \$4f^{14} 6s^2$, indicating a filled 4f subshell.
5. Conclude that the element with the ground-state electron configuration ending in a filled f subshell (\(f^{14}\)) is the one whose electron configuration matches this pattern, which is Ytterbium (Yb).