alleryThe spin-only magnetic moment μ in BM for n unpaired electrons is given by:
A√(n(n+2)) BM
Bn + 1 BM
Cn² BM
Dn × 1.73 BM
Answer & Solution
Correct answer: A. √(n(n+2)) BM
**μ_spin-only = √(n(n+2)) BM**.
**Examples:** n=1 → 1.73
n=2 → 2.84; n=3 → 3.87
n=4 → 4.90; n=5 → 5.92 BM.
Used to estimate the number of unpaired electrons in a transition-metal ion from experimental magnetic moment.
Keep practising d- and f-Block Elements
Browse all d- and f-Block Elements questions for AP Intermediate 2nd Year, or explore more Chemistry topics.
Related questions
Why is Sc³⁺ colorless?Eu²⁺ is more stable than other Ln²⁺ because:Number of unpaired electrons in [Mn(CN)₆]³⁻ (Mn(III), d⁴, strong field CN⁻):Magnetic moment of [Fe(H₂O)₆]³⁺ (Fe³⁺ is d⁵, weak field H₂O → high spin):Coordination number of Ag in [Ag(CN)₂]⁻:K₂Cr₂O₇ (potassium dichromate) is a strong:Magnetic moment of complex with 5 unpaired d-electrons (high spin):Which transition metal forms unstable highest oxidation state +7?