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)
|
List-I Element |
List-II Electronic configuration |
|
(A) Cerium |
(III) $[Xe] 4f^1 5d^1 6s^2$ |
|
(B) Erbium |
(I) $[Xe] 4f^{12} 6s^2$ |
|
(C) Uranium |
(IV) $[Rn] 5f^3 6d^1 7s^2$ |
|
(D) Curium |
(II) $[Rn] 5f^7 6d^1 7s^2$ |
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(A) Cerium (Ce, Z=58): As the first element of the lanthanide series after Lanthanum, it retains one electron in the $5d$ orbital.
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Configuration: $[Xe] 4f^1 5d^1 6s^2$ → (III)
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(B) Erbium (Er, Z=68): This is a later lanthanide. By this point, the $4f$ orbital is being filled, and there are no electrons in the $5d$ orbital for the ground state.
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Configuration: $[Xe] 4f^{12} 6s^2$ → (I)
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(C) Uranium (U, Z=92): An actinide. It has three electrons in the $5f$ orbital and one in the $6d$ orbital.
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Configuration: $[Rn] 5f^3 6d^1 7s^2$ → (IV)
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(D) Curium (Cm, Z=96): This is a key exception in the actinide series. To achieve a stable half-filled $5f^7$ configuration, one electron resides in the $6d$ orbital.
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Configuration: $[Rn] 5f^7 6d^1 7s^2$ → (II)
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