Question:

Reaction of [Ni(CN)$_4$]$^{2-}$ with metallic potassium in liquid ammonia at –33 °C yields complex E. The geometry and magnetic behavior of E, respectively, are

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Reduction of [Ni(CN)$_4$]$^{2-}$ to [Ni(CN)$_4$]$^{4-}$ gives Ni$^0$, which is diamagnetic and tetrahedral due to fully filled 3d orbitals.
Updated On: Dec 5, 2025
  • Square planar and diamagnetic
  • Tetrahedral and diamagnetic
  • Octahedral and paramagnetic
  • Square pyramidal and paramagnetic
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The Correct Option is B

Solution and Explanation

Step 1: Reaction type.
Reduction of [Ni(CN)$_4$]$^{2-}$ with metallic potassium produces [Ni(CN)$_4$]$^{4-}$,
where Ni is in the +0 oxidation state.
Step 2: Electronic configuration.
Ni$^0$: 3d$^{10}$ configuration → fully filled d-orbitals, no unpaired electrons → diamagnetic.
Step 3: Geometry.
In the zero oxidation state, Ni forms tetrahedral complexes with CN$^-$ ligands.
Therefore, geometry = tetrahedral and magnetic nature = diamagnetic.
Step 4: Conclusion.
The correct answer is (B) Tetrahedral and diamagnetic.
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