Ferrites and ionic moments — magnetic moment of a ferrous (Fe2+) ion in ferrites What is the magnetic moment, expressed in Bohr magnetons (μB), typically attributed to a ferrous ion (Fe2+) within ferrite materials under the spin-only approximation?
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Azero
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B2
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C4
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D6
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E5
Answer
Correct Answer: 4
Explanation
Introduction / Context:Ferrites are mixed iron oxides where magnetic moments arise from Fe2+ and Fe3+ ions located on tetrahedral and octahedral sites. The effective ionic moments used in basic ferrite magnetism often follow spin-only values rounded to integral μB to simplify net moment calculations.
Given Data / Assumptions:
- High-spin Fe2+ (3d^6) in octahedral coordination typical for many ferrites.
- Spin-only magnetic moment approximation.
- Orbital contribution largely quenched in crystalline fields.
Concept / Approach:The spin-only moment (in μB) is μ ≈ √(n(n+2)), where n is the number of unpaired electrons. For Fe2+ (high spin d^6), n = 4, giving μ ≈ √(24) ≈ 4.9 μB. In introductory ferrite analyses and many tabulations, Fe2+ is often assigned an idealized value of approximately 4 μB (while Fe3+ is taken as about 5 μB). The question aligns with this conventional rounding used for magnetic moment bookkeeping in ferrites.
Step-by-Step Solution:Identify Fe2+ high-spin state with 4 unpaired electrons.Compute spin-only moment as √(n(n+2)) ≈ 4.9 μB.Use ferrite convention that approximates Fe2+ as ~4 μB for quick lattice moment sums; select 4 μB from the choices.
Verification / Alternative check:More precise values depend on site symmetry and covalency; however, typical ferrite problems adopt 4 μB (Fe2+) and 5 μB (Fe3+) for sublattice calculations.
Why Other Options Are Wrong:0 μB contradicts unpaired spins; 2 μB and 6 μB do not match the spin-only estimate; 5 μB corresponds to Fe3+ rather than Fe2+ in these simplified models.
Common Pitfalls:
- Confusing Fe2+ with Fe3+ tabulated moments.
- Forgetting that orbital quenching justifies spin-only approximations in many oxides.
Final Answer:4