- 2073 Bhadra · 5 marks
VBT: [Ni(CN)₄]²⁻ and [Ni(CO)₄]
Short answer
[Ni(CN)4]2- is square planar (dsp2) and [Ni(CO)4] is tetrahedral (sp3). Both have all electrons paired and are diamagnetic, so magnetic behavior is the same but structures differ due to hybridization.
Using VBT, explain as to show the two complexes [Ni(CN)₄]²⁻ and [Ni(CO)₄] have different structure but do not differ in their magnetic behavior.
[Ni(CN)₄]²⁻
Ni²⁺ is 3d⁸. CN⁻ is a strong-field ligand, so the two unpaired 3d electrons pair up and one 3d orbital becomes empty. One 3d, one 4s and two 4p orbitals mix to give dsp² hybrid orbitals.
3d: [↑↓][↑↓][↑↓][↑↓][ ] 4s [ ] 4p [ ][ ] → dsp²
Geometry square planar; no unpaired electrons, diamagnetic.
[Ni(CO)₄]
Ni is in the 0 oxidation state: 3d⁸ 4s². CO is a strong-field ligand, so the 4s electrons shift into 3d, giving 3d¹⁰ (all paired). One 4s and three 4p orbitals are vacant and form sp³ hybrid orbitals.
3d: [↑↓][↑↓][↑↓][↑↓][↑↓] 4s [ ] 4p [ ][ ][ ] → sp³
Geometry tetrahedral; no unpaired electrons, diamagnetic.
Both complexes have no unpaired electrons and so show the same (diamagnetic) behaviour, but the different hybridisation (dsp² vs sp³) gives different shapes.
✓ Answer checked by Claude Opus 5.5 (high effort)
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