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Magnetic Moment

Mediumchemistry

A tetrahedral nickel complex has two unpaired electrons; using the spin-only formula, what is its approximate magnetic moment in Bohr magnetons?

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About This Question

Subject
chemistry
Chapter
coordination compounds
Topic
magnetic moment
Difficulty
Medium
Year
2025
Tags
spin only formulamagnetic momentunpaired electronstetrahedral nickelparamagnetism

Solution

Correct Answer:

2.83 BM

The magnetic behaviour of many transition-metal complexes is estimated from the spin-only formula, μ = √(n(n+1)) Bohr magnetons, where n is the number of unpaired electrons. This relation ignores orbital contributions, which is a good approximation for first-row transition metals. For a tetrahedral nickel(II) species the d^8 ion typically keeps two unpaired electrons, so n = 2. Substituting gives μ = √(2 × 3) = √6 ≈ 2.83 BM. The value 1.73 BM corresponds to a single unpaired electron, 3.87 BM corresponds to three unpaired electrons, and 0 BM would indicate a diamagnetic species with all electrons paired, none of which match n = 2. Tetrahedral nickel(II) complexes are commonly paramagnetic, in line with NCERT's treatment of crystal field effects on geometry and spin. Plausibility check: the computed 2.83 BM lies between the one-electron and three-electron values, exactly where a two-electron paramagnet should fall, confirming the result is reasonable.

This medium difficulty chemistry question is from the chapter coordination compounds, covering the topic of magnetic moment. It appeared in the 2025 exam.

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