CUET · CHEMISTRY · PYQ PAPER 2025
Match List-l with List-ll
| List-l (Electronic configuration) | List-II (lon) |
| (A) \([ Ar ] 3 d^{10}\) | (I) \(Ce ^{4+}\) |
| (B) \(\left[ X _{ e }\right]\) | (II) \(Cu ^{+}\) |
| (C) \([ Rn ]\) | (III) \(Th ^{4+}\) |
| (D) \([ Xe ] 4 f^{11} 5 d^1\) | (IV) \(Lu ^{2+}\) |
- A (А)-(II), (B)-(I), (C)-(III), (D)-(IV)
- B (А)-(І), (В)-(III), (С)-(II), (D)-(IV)
- C (А)-(I), (В)-(I), (C)-(IV), (D)-(III)
- D (А)-(III), (В)-(IV), (C)-(I), (D)-(II)
Answer & Solution
Correct Answer
(A) (А)-(II), (B)-(I), (C)-(III), (D)-(IV)
Step-by-step Solution
Detailed explanation
A: \( [Ar] 3d^{10} \) (28 electrons). \( Cu^+ \) (Z=29) has \( 29-1=28 \) electrons. So, (A)-(II). B: \( [Xe] \) (54 electrons). \( Ce^{4+} \) (Z=58) has \( 58-4=54 \) electrons. So, (B)-(I). C: \( [Rn] \) (86 electrons). \( Th^{4+} \) (Z=90) has \( 90-4=86 \) electrons. So,…
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