- 2067 Chaitra · 5 marks
VBT: magnetism of [Fe(CN)₆]⁴⁻/³⁻ and [Ni(CO)₄]
Short answer
[Fe(CN)₆]⁴⁻ has Fe²⁺ (d⁶), CN⁻ causes pairing → diamagnetic. [Fe(CN)₆]³⁻ has Fe³⁺ (d⁵) with one unpaired electron → paramagnetic. [Ni(CO)₄] has Ni(0) 3d¹⁰, sp³ hybridization giving tetrahedral shape, all electrons paired → diamagnetic.
How does the valence bond theory account for the following facts? a) [Fe(CN)₆]⁴⁻ ion is diamagnetic but [Fe(CN)₆]³⁻ is paramagnetic b) [Ni(CO)₄]⁰ is diamagnetic and tetrahedral.
a) [Fe(CN)₆]⁴⁻ diamagnetic but [Fe(CN)₆]³⁻ paramagnetic
- [Fe(CN)₆]⁴⁻ contains Fe²⁺ (3d⁶). CN⁻ is strong-field, so all six electrons pair in three 3d orbitals; d²sp³ hybridisation. No unpaired electron: diamagnetic.
- [Fe(CN)₆]³⁻ contains Fe³⁺ (3d⁵). The electrons pair up as far as possible in the three lower orbitals, but five electrons cannot all pair. One unpaired electron remains: paramagnetic (1.73 BM), also d²sp³.
Fe²⁺: [↑↓][↑↓][↑↓][ ][ ] Fe³⁺: [↑↓][↑↓][↑ ][ ][ ]
b) [Ni(CO)₄]⁰ diamagnetic and tetrahedral
Ni(0) is 3d⁸ 4s². The strong-field ligand CO causes the 4s electrons to shift into 3d, giving 3d¹⁰ (all paired). The empty 4s and three 4p orbitals form sp³ hybrids, giving a tetrahedral shape. No unpaired electron, hence diamagnetic.
3d: [↑↓][↑↓][↑↓][↑↓][↑↓] sp³ (4s + 3×4p)
✓ Answer checked by Claude Opus 5.5 (high effort)
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