CBSE · Class 12 · Chemistry · Coordination CompoundsWhat is the spin-only magnetic moment of Mn(H2O)62+ in Bohr Magnetons (BM)?
Step-by-Step Solution
In [Mn(H2O)6]2+, Manganese is in +2 oxidation state (Mn2+, electronic configuration 3d5). Water is a weak field ligand, so the electrons do not pair up. Thus, there are 5 unpaired electrons (n = 5). Using the spin-only magnetic moment formula μ = √(n(n+2)) BM, we get μ = √(5(5+2)) = √35 ≈ 5.92 BM.
Detailed Options Breakdown
Option : 1.73 BM
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Coordination Compounds.
Option 1: 3.87 BM
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Coordination Compounds.
Option 2: 4.90 BM
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Coordination Compounds.
Option 3: 5.92 BM (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
💡 Study Guide: This question tests core syllabus concepts from Coordination Compounds. For formulas, key summaries, and mock exam reference guides, read the full Coordination Compounds Revision Notes.