High Spin And Low Spin

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When the same metal ion binds a weak-field ligand instead of a strong-field ligand in an octahedral complex, how does the number of unpaired electrons usually change?

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Whether an octahedral complex is high spin or low spin depends on the competition between the crystal field splitting energy and the electron pairing energy. A weak-field ligand produces a small splitting, so it is energetically cheaper for electrons to occupy the higher e_g orbitals singly than to pair up in the lower t_{2g} set. This gives the high-spin configuration with the maximum number of unpaired electrons. A strong-field ligand produces a large splitting that exceeds the pairing energy, so electrons pair in t_{2g} first, giving the low-spin arrangement with fewer unpaired electrons. Therefore switching from a strong-field to a weak-field ligand generally increases the number of unpaired electrons. The option that the count never changes ignores the field-strength dependence. The option that weak fields force pairing reverses the correct logic. Reaching zero unpaired electrons is a low-spin, not weak-field, outcome. This balance is the core NCERT criterion for spin state. Plausibility check: weak-field [CoF6]^{3-} is paramagnetic while strong-field [Co(NH3)6]^{3+} is diamagnetic, exactly matching the predicted increase in unpaired electrons.

High Spin And Low Spin

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