Nd²⁺ = __________

Solution

Chemistry ma’am · 3 steps

Write neutral Nd first, then take both electrons from 6s, never from 4f.

1. Neutral neodymium. Nd has Z = 60, which is 6 more than xenon. Its configuration is [Xe] 4f⁴ 6s², so it has 4 electrons in 4f and 2 in 6s.
2. Remove outermost first. When a lanthanoid forms a cation, it loses the 6s electrons first. The 6s shell has the highest principal quantum number, so those electrons are the outermost and the easiest to remove.

   Nd [Xe] 4f⁴ 6s² → Nd²⁺ [Xe] 4f⁴ + 2e⁻

3. So. Nd²⁺ is [Xe] 4f⁴. The 4f count stays at 4, which is option (c).

The idea to keep. For lanthanoid ions, empty the 6s shell first. Ln²⁺ keeps all the 4f electrons of the atom, and Ln³⁺ loses one more, from 4f.

Answer from the NTA final answer key. Solution by PrepWiser.