A divalent ion of 'V' (Atomic number- 23) in aqueous solution has magnetic moment of:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{15}BM\)
The spin-only magnetic moment is μ = √[n(n + 2)] BM, where n is the number of unpaired electrons. Vanadium (Z = 23) is [Ar] 3d3 4s2; the two 4s electrons are removed first, so \( \ce{V^{2+}} \) is [Ar] 3d3. By Hund’s rule the three electrons occupy three separate 3d orbitals:
3d: [↑] [↑] [↑] [ ] [ ], n = 3
\( \displaystyle \mu = \sqrt{3(3+2)} = \sqrt{15}\ \text{BM} \approx 3.87\ \text{BM} \)
NCERT (2026-27 edition):
Magnetic Properties (Table 4.7: Calculated and Observed Magnetic Moments (BM))
| Ion | Configuration | Unpaired electron(s) | Calculated | Observed |
|---|---|---|---|---|
| Sc3+ | 3d0 | 0 | 0 | 0 |
| Ti3+ | 3d1 | 1 | 1.73 | 1.75 |
| Ti2+ | 3d2 | 2 | 2.84 | 2.76 |
| V2+ | 3d3 | 3 | 3.87 | 3.86 |
| Cr2+ | 3d4 | 4 | 4.90 | 4.80 |
| Mn2+ | 3d5 | 5 | 5.92 | 5.96 |
| Fe2+ | 3d6 | 4 | 4.90 | 5.3 – 5.5 |
| Co2+ | 3d7 | 3 | 3.87 | 4.4 – 5.2 |
| Ni2+ | 3d8 | 2 | 2.84 | 2.9 – 3.4 |
| Cu2+ | 3d9 | 1 | 1.73 | 1.8 – 2.2 |
| Zn2+ | 3d10 | 0 | 0 |