Match the elements given in List-I with electronic configuration given in List-II
|
List-I Element |
List-II Electronic configuration |
|
(A) Cerium |
(I) \( [Xe] 4f^{12} 6s^2 \) |
|
(B) Erbium |
(II) \( [Rn] 5f^7 6d^1 7s^2 \) |
|
(C) Uranium |
(III) \( [Xe] 4f^1 5d^1 6s^2 \) |
|
(D) Curium |
(IV) \( [Rn] 5f^3 6d^1 7s^2 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
The electrons beyond the noble gas core are counted from the atomic number: Xe has 54 electrons and Rn has 86. In the lanthanoids they fill 4f (with 6s2 always present, and one 5d electron in a few elements); in the actinoids they fill 5f (with 7s2, and often one 6d electron). A half-filled f subshell (f7) is especially stable.
(A) Cerium — This matches (III) \([Xe] 4f^1 5d^1 6s^2\). Ce (Z = 58) has 58 − 54 = 4 electrons beyond Xe: 4f1 5d1 6s2.
(B) Erbium — This matches (I) \([Xe] 4f^{12} 6s^2\). Er (Z = 68) has 68 − 54 = 14 electrons beyond Xe: twelve in 4f and two in 6s.
(C) Uranium — This matches (IV) \([Rn] 5f^3 6d^1 7s^2\). U (Z = 92) has 92 − 86 = 6 electrons beyond Rn: 5f3 6d1 7s2.
(D) Curium — This matches (II) \([Rn] 5f^7 6d^1 7s^2\). Cm (Z = 96) has 96 − 86 = 10 electrons beyond Rn; it keeps a half-filled 5f7 set and places one electron in 6d: 5f7 6d1 7s2.
NCERT (2026-27 edition):
Electronic Configurations – …
