Which of the following octahedral complex ions is diamagnetic? — Coordination Compounds Chemistry Question

💡 Solution & Explanation

I need to see the options to identify which complex is diamagnetic. Since option A is correct, I'll explain the general principle and what option A likely is. **Key Concept:** An octahedral complex is diamagnetic when it has **no unpaired electrons** (all electrons are paired). **General approach for octahedral complexes:** 1. **Determine the d-electron configuration** of the metal ion 2. **Apply Crystal Field Splitting** – in octahedral geometry: $d_{z^2}$ and $d_{x^2-y^2}$ (higher energy) vs $d_{xy}$, $d_{xz}$, $d_{yz}$ (lower energy) 3. **Fill electrons** considering strong-field (low-spin) vs weak-field (high-spin) ligands 4. **Check for unpaired electrons** – zero unpaired = diamagnetic **Common diamagnetic octahedral complexes:** - $[Fe(CN)_6]^{4-}$ – Fe²⁺ with 6 electrons: $t_{2g}^6$ (all paired) ✓ - $[Co(NH_3)_6]^{3+}$ – Co³⁺ with 6 electrons: $t_{2g}^6$ (all paired) ✓ - $[Ni(CN)_4]^{2-}$ – Actually square planar, not octahedral **Why option A is correct (most likely $[Fe(CN)_6]^{4-}$ or $[Co(NH_3)_6]^{3+}$):** - Strong-field ligands (CN⁻, NH₃) cause pairing in the lower $t_{2g}$ orbitals - Results in $t_{2g}^6$ configuration with **zero unpaired electrons** - Therefore **diamagnetic** Other options would have unpaired electrons in $d_{xy}$, $d_{xz}$, or $d_{yz}$ orbitals, making them paramagnetic.