- 2072 Magh · 3 marks
VBT for [Ni(CN)₄]²⁻ and its magnetism
Short answer
Ni2+ (3d8) with strong CN- forces electron pairing, vacating one 3d orbital. dsp2 hybridization yields square planar geometry. No unpaired electrons, so the complex is diamagnetic.
Explain the formation of [Ni(CN)₄]²⁻ on the basis of VBT of coordination compounds and predict its magnetic property.
Ni (Z = 28); Ni²⁺ is 3d⁸. CN⁻ is a strong-field ligand.
Formation: the two unpaired 3d electrons are forced to pair, which vacates one 3d orbital. One 3d, one 4s and two 4p orbitals hybridise to form four dsp² orbitals, which accept four CN⁻ lone pairs.
Ni²⁺ 3d: [↑↓][↑↓][↑↓][↑↓][ ] 4s [ ] 4p [ ][ ][ ]
\______ dsp² ______/
- Geometry: square planar.
- Magnetism: no unpaired electron, so the complex is diamagnetic.
✓ Answer checked by Claude Opus 5.5 (high effort)
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