Match List I with List II
|
List I (Ions) |
List II Magnetic Moment (BM) |
| A. \(V^{2+}\) | I. 1.73 |
| B. \(Ni^{2+}\) | II. 3.87 |
| C. \(Cu^{2+}\) | III. 4.90 |
| D. \(Cr^{2+}\) | IV. 2.84 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-IV, C-I, D-III
For the ions of the first transition series the magnetic moment comes from the unpaired electrons alone and is given by the spin-only formula
\( \mu = \sqrt{n(n+2)}\ \text{BM} \)
n is the number of unpaired electrons. Electrons are removed from 4s before 3d when the ion is formed, and the 3d electrons fill the five 3d orbitals by Hund’s rule.
A. V2+ — This matches II. 3.87. V (Z = 23) is [Ar] 3d3 4s2; removing the two 4s electrons gives \( \ce{V^2+} \) = [Ar] 3d3: [↑] [↑] [↑] [ ] [ ], n = 3, μ = √15 = 3.87 BM.
B. Ni2+ — This matches IV. 2.84. Ni (Z = 28) is [Ar] 3d8 4s2; removing the two 4s electrons gives \( \ce{Ni^2+} \) = [Ar] 3d8: [↑↓] [↑↓] [↑↓] [↑] [↑], n = 2, μ = √8 = 2.84 BM.
C. Cu2+ — This matches I. 1.73. Cu (Z = 29) is [Ar] 3d10 4s1; removing the 4s electron and one 3d electron gives \( \ce{Cu^2+} \) = [Ar] 3d9: [↑↓] [↑↓] [↑↓] [↑↓] [↑], n = 1, μ = √3 = 1.73 BM.
D. Cr2+ — This matches III. 4.90. Cr (Z = 24) is [Ar] 3d5 4s1; removing the 4s electron and one 3d electron gives \( \ce{Cr^2+} \) = [Ar] 3d4: [↑] [↑] [↑] [↑] [ ], n = 4, μ = √24 = 4.90 BM.
NCERT (2026-27 edition):
Magnetic Properties – …
