AP EAPCET 21 May 2025 Shift 2 Paper

Step 1. Recall that 4f7 means half-filled 4 f shell (7 electrons).
Step 2. Check each ion
(A) Pr3+
Praseodymium (Pr) = atomic number 59
Neutral configuration: [Xe]4f36s2
When it loses 3 electrons ⟶ remove 6s2 and one 4f :
Pr3+:[Xe]4f2
Not 4f7.
(B) Lu3+
Lutetium (Lu) = atomic number 71
Neutral configuration: [Xe]4f145d16s2
Removing three electrons ⟶6s2 and 5d1 :
Lu3+:[Xe]4f14
Not 4f7.
(C) Eu2+
Europium (Eu)= atomic number 63
Neutral configuration: [Xe]4f76s2
Losing two electrons ⟶ remove both 6s electrons:
Eu2+:[Xe]4f7
、 Matches 4f7.
(D) Ce4+
Cerium (Ce) = atomic number 58
Neutral: [Xe]4f15d16s2
Losing 4 electrons ⟶4f15d16s2⟶ remove all ⟶
Ce4+:[Xe]
Not‌4‌f7.
Correct Answer: Option C - Eu ‌2+