Skip to main content

Unpaired electrons and paramagnetism; [NiCl₄]²⁻, [Co(NH₃)₆]³⁺

Short answer

Unpaired electrons produce a net magnetic moment because opposite spins do not cancel. [NiCl₄]²⁻ is tetrahedral (sp³), 2 unpaired electrons, paramagnetic. [Co(NH₃)₆]³⁺ is octahedral (d²sp³), no unpaired electrons, diamagnetic.

Why does the presence of unpaired electron make a substance paramagnetic in nature? Show geometry and magnetic properties of [NiCl₄]²⁻ and [Co(NH₃)₆]³⁺.

Unpaired electrons and paramagnetism

An electron has spin and so behaves like a tiny magnet. In a pair, the two opposite spins cancel each other's magnetic effect. An unpaired electron has no partner to cancel it, so the atom has a net magnetic moment and is attracted into an external magnetic field. Such a substance is paramagnetic. Magnetic moment μ = √(n(n+2)) BM, where n = number of unpaired electrons.

[NiCl₄]²⁻

Ni²⁺ is 3d⁸; Cl⁻ is a weak-field ligand, so no pairing. sp³ hybridisation.

3d: [↑↓][↑↓][↑↓][↑ ][↑ ]  4s 4p → sp³

Geometry: tetrahedral. Two unpaired electrons, so paramagnetic (μ = 2.83 BM).

[Co(NH₃)₆]³⁺

Co³⁺ is 3d⁶; NH₃ is strong-field and forces pairing, so d²sp³.

3d: [↑↓][↑↓][↑↓][  ][  ]  4s 4p → d²sp³

Geometry: octahedral. No unpaired electron, so diamagnetic.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Engineering Chemistry chapter 6 questions →