Match List I with List II
| List I | List II |
| A. \(Cr^{3+}\) | I. \(d^3\) configuration |
| B. \(Mn^{5+}\) | II. \(d^{10}\) configuration |
| C. \(Cu^{+}\) | III. Paramagnetic |
| D. \(Ti^{4+}\) | IV. Diamagnetic |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-I, B-III, C-II, D-IV
The d-electron count of an ion follows from its atomic number: the 4s electrons are removed first, then 3d. An ion with unpaired electrons is paramagnetic; one with all electrons paired (d0 or d10) is diamagnetic.
(A) Cr3+ — This matches (I) d3 configuration. Cr (Z = 24) is [Ar]3d54s1; removing one 4s and two 3d electrons gives 3d3.
(B) Mn5+ — This matches (III) Paramagnetic. Mn (Z = 25) is [Ar]3d54s2; removing two 4s and three 3d electrons gives 3d2, with 2 unpaired electrons.
(C) Cu+ — This matches (II) d10 configuration. Cu (Z = 29) is [Ar]3d104s1; removing the 4s electron gives 3d10.
(D) Ti4+ — This matches (IV) Diamagnetic. Ti (Z = 22) is [Ar]3d24s2; removing all four outer electrons gives 3d0, with no unpaired electron.
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 02-06 S3 31-05 S1 | – |