The electronic configurations of bivalent europium and trivalent cerium are
(atomic number : Xe = 54, Ce = 58, Eu = 63)

Solution

Chemistry ma’am · 3 steps

Write the neutral atom first, then take electrons from 6s, then 5d, and only then from 4f.

1. Removal order. When a lanthanide forms an ion, the outermost electrons leave first. The two 6s electrons go first, then any 5d electron, and only after that a 4f electron.

   Species	Electrons	Neutral atom	Electrons removed	Ion
   Eu²⁺	63 - 2 = 61	[Xe] 4f⁷ 6s²	both 6s	[Xe] 4f⁷
   Ce³⁺	58 - 3 = 55	[Xe] 4f¹ 5d¹ 6s²	both 6s and the 5d	[Xe] 4f¹

2. Quick check. Xenon accounts for 54 electrons. Eu²⁺ has 61, so 7 are left over and all sit in 4f. Ce³⁺ has 55, so exactly 1 is left over, in 4f.
3. So. Bivalent europium is [Xe] 4f⁷ and trivalent cerium is [Xe] 4f¹, which is option (d).

The idea to keep. For any lanthanide ion, subtract 54 for xenon from the ion's electron count. Whatever is left sits in 4f, because the 6s and 5d electrons always leave first.

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