Determine the number of unpaired electrons and the spin‑only magnetic moment (in Bohr magnetons) for the octahedral complex $[M n (H_{2} O)_{6}]^{2 +}$ (Mn $^{2 +}$ , $d^{5}$ ).

medium 2 marks Class 12 · Chemistry · d- and f-Block Elements

Answer

5 unpaired electrons; spin‑only magnetic moment

5.92; m u_{B}

The correct answer is 5 unpaired electrons; spin‑only magnetic moment $5.92; m u_{B}$ — Substitute $n = 5$ : $μ_{so} = 5 \times 7 = 35 = 5.92; μ_{B}$ .

Step-by-step solution

Mn $^{2 +}$ has electron configuration $d^{5}$ . In an octahedral ligand field with weak field ( $H_{2} O$ ), it is high‑spin: $t_{2 g}^{3} e_{g}^{2}$ .
Each of the 5 $d$ electrons occupies a different orbital, so there are 5 unpaired electrons.
The spin‑only magnetic moment is calculated as $m u_{so} = n (n + 2)$ where $n$ is the number of unpaired electrons.
Substitute $n = 5$ : $μ_{so} = 5 \times 7 = 35 = 5.92; μ_{B}$ .

Common mistake: Using formula for total angular momentum instead of spin‑only

Concept tested: Magnetic properties of high‑spin complexes

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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Determine the number of unpaired electrons and the spin‑only magnetic moment (in Bohr magnetons) for the octahedral complex [Mn(H2​O)6​]2+ (Mn2+, d5).

Step-by-step solution

Determine the number of unpaired electrons and the spin‑only magnetic moment (in Bohr magnetons) for the octahedral complex $[M n (H_{2} O)_{6}]^{2 +}$ (Mn $^{2 +}$ , $d^{5}$ ).