Which of the following metal ions cannot form both high spin and low spin octahedral complexes?

Question

BlackTom AI · Accepted Answer

A metal ion must have d-electrons that can be distributed differently between t2g and eg levels (competing crystal-field splitting Δoct vs pairing energy P) for high-spin/low-spin behaviour to exist.

1) Relevant knowledge points
- Octahedral crystal-field splits d orbitals into lower-energy t2g (3) and higher-energy eg (2).
- High- vs low-spin depends on whether electrons pair in t2g (large Δoct favors pairing → low-spin) or occupy eg to remain unpaired (small Δoct favors high-spin).
- Only ions with appropriate d-electron counts (commonly d4–d7 for first-row transition metals) can show both possibilities. d0 has no d electrons; d9 has effectively a single unpaired electron regardless of Δoct.

2) Step-by-step reasoning (electron counts)
- Ti4+: atomic Ti is [Ar]3d2 4s2; Ti4+ removes all 4 valence electrons → d0.
  - d-electron count = 0. No electrons to occupy t2g or eg; no pairing choices exist.
- Fe2+: Fe is [Ar]3d6 4s2 → Fe2+ is d6.
  - d6 can be high-spin (weak field) t2g^4 eg^2 (4 unpaired) or low-spin (strong field) t2g^6 eg^0 (0 unpaired).
- Co2+: Co is [Ar]3d7 4s2 → Co2+ is d7.
  - d7 can be high-spin t2g^5 eg^2 (3 unpaired) or low-spin t2g^6 eg^1 (1 unpaired) depending on ligand strength.
- Cu2+: Cu is [Ar]3d10 4s1 → Cu2+ is d9.
  - d9 final arrangement in octahedral field is t2g^6 eg^3, which leaves one unpaired electron (S = 1/2). There is no alternative pairing pattern that changes the net number of unpaired electrons, so Cu2+ does not exhibit a high/low spin crossover.

3) Intermediate steps shown above: electron configurations and the resulting t2g/eg occupations for each ion.

4) Why the correct answer is correct
Ti4+ (d0) has no d-electrons to redistribute between t2g and eg, so it cannot exhibit either a high-spin or a low-spin octahedral electronic configuration. It is always diamagnetic with no d-electron-based spin states.

5) Why other options are incorrect
- Co2+ (d7) and Fe2+ (d6) are classic examples that can be either high-spin or low-spin depending on ligand field strength.
- Cu2+ (d9) does have d-electrons but its electron count yields a single unpaired electron in any octahedral splitting pattern; it therefore does not show a spin-crossover, but unlike Ti4+ it is not d0 (it still has a d-electron-derived magnetic moment).

6) Conclusion
Therefore Ti4+ cannot form both high-spin and low-spin octahedral complexes; the final answer is (a) Ti4+. This matches the provided answer.

Which of the following metal ions cannot form both high spin and low spin octahedral complexes?单项选择题

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