Match List-I with List-II
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List-I (Electronic configuration) |
List-II (Ion) |
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(A) \( [Ar] 3d^{10} \) |
(I) \( Ce^{4+} \) |
|
(B) \( [Xe] \) |
(II) \( Cu^+ \) |
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(C) \( [Rn] \) |
(III) \( Th^{4+} \) |
|
(D) \( [Xe] 4f^{14} 5d^1 \) |
(IV) \( Lu^{2+} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
To find the configuration of an ion, write the configuration of the neutral atom from its atomic number, then remove the electrons lost: the outermost s electrons first, then the d electrons, then the f electrons.
(A) \([Ar] 3d^{10}\) — This matches (II) \(Cu^+\). Cu (Z = 29) is \( \mathrm{[Ar]\,3d^{10}\,4s^{1}} \); losing the 4s electron gives \( \ce{Cu^+} \): \( \mathrm{[Ar]\,3d^{10}} \).
(B) \([Xe]\) — This matches (I) \(Ce^{4+}\). Ce (Z = 58) is \( \mathrm{[Xe]\,4f^{1}\,5d^{1}\,6s^{2}} \); losing the two 6s, the 5d and the 4f electron gives \( \ce{Ce^4+} \): \( \mathrm{[Xe]} \), the noble gas configuration that makes +4 a stable state of cerium.
(C) \([Rn]\) — This matches (III) \(Th^{4+}\). Th (Z = 90) is \( \mathrm{[Rn]\,6d^{2}\,7s^{2}} \); losing the two 7s and the two 6d electrons gives \( \ce{Th^4+} \): \( \mathrm{[Rn]} \).
(D) \([Xe] 4f^{14} 5d^1\) — This matches (IV) \(Lu^{2+}\). Lu (Z = 71) is \( \mathrm{[Xe]\,4f^{14}\,5d^{1}\,6s^{2}} \); losing the two 6s electrons gives \( \ce{Lu^2+} \): \( \mathrm{[Xe]\,4f^{14}\,5d^{1}} \).
Shortcut: (A)-(II) appears only in Option (1).