Answer: mu = sqrt(n(n+2)) Bohr magnetons, where n = number of unpaired electrons.
- A mu = sqrt(n(n+2)) Bohr magnetons, where n = number of unpaired electrons
- B mu = n Bohr magnetons, scaling linearly with unpaired electron count
- C mu = n<sup>2</sup> Bohr magnetons, scaling with the square of electron count
- D mu = 2n Bohr magnetons, simply doubling the unpaired electron count
Correct answer: A. mu = sqrt(n(n+2)) Bohr magnetons, where n = number of unpaired electrons
Explanation: The spin-only magnetic moment: mu = sqrt(n(n+2)) BM. For n=1: mu=1.73 BM; n=2: 2.83 BM; n=3: 3.87 BM; n=4: 4.90 BM; n=5: 5.92 BM.
Aqueous solutions of transition metal ions show characteristic colours caused by electrons absorbing visible light to jump between split d-orbitals; the exact colour depends on the metal, its oxidation state, and surrounding ligands.
Concept context
Transition metals (d-block) and lanthanides/actinides (f-block). Known for coloured compounds, variable oxidation states, complex formation, and catalytic properties. Frequently tested in JEE and NEET.