(a) Using Valence Bond theory prove that $[Ni(CN)_4]^{2-}$ is diamagnetic, whereas $[Ni(NH_3)_4]^{2+}$ is paramagnetic.

(b) Explain Radiocarbon dating.

Question

(a) Using Valence Bond theory prove that $[Ni(CN)_4]^{2-}$ is diamagnetic, whereas $[Ni(NH_3)_4]^{2+}$ is paramagnetic.

(b) Explain Radiocarbon dating.

Accepted Answer

> [!TLDR]
> (a) Valence Bond theory explains the magnetic behaviour of the two nickel(II) complexes through their hybridisation: $CN^-$ (strong field) pairs up Ni$^{2+}$'s d-electrons giving a diamagnetic $dsp^2$ square-planar $[Ni(CN)_4]^{2-}$, while $NH_3$ (weaker field) leaves them unpaired, giving a paramagnetic $sp^3$ tetrahedral $[Ni(NH_3)_4]^{2+}$. (b) Radiocarbon dating exploits the fixed decay rate of atmospherically-generated $^{14}C$ to estimate the age of once-living material.
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> **(a) VB theory: $[Ni(CN)_4]^{2-}$ vs $[Ni(NH_3)_4]^{2+}$**
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> Nickel, $Z=28$: $[Ar]\,3d^8\,4s^2$. On forming $Ni^{2+}$, the two $4s$ electrons are lost first, giving the configuration $3d^8$. Distributed over the five 3d orbitals by Hund's rule, the free-ion ground state has 3 orbitals doubly occupied and 2 orbitals singly occupied — i.e. **2 unpaired electrons** ($\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ \uparrow$).
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> **$[Ni(CN)_4]^{2-}$:** $CN^-$ is a strong-field ligand. Its strong ligand field forces the two unpaired 3d electrons of $Ni^{2+}$ to pair up within the existing d-orbitals, which empties one d orbital (the $d_{x^2-y^2}$). This vacated d-orbital, together with one $4s$ and two $4p$ orbitals, undergoes $dsp^2$ hybridisation, giving four equivalent hybrid orbitals directed towards the corners of a **square plane**, each accepting a lone pair from a $CN^-$ ligand. Because all 8 d-electrons are now paired, $[Ni(CN)_4]^{2-}$ has **zero unpaired electrons** and is **diamagnetic**.
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> $$Ni^{2+}(3d^8,\ 	ext{paired by }CN^-) 	o dsp^2 \ 	ext{hybridisation} 	o 	ext{square planar, diamagnetic}$$
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> **$[Ni(NH_3)_4]^{2+}$:** $NH_3$ is a comparatively weak/moderate-field ligand and does not supply enough crystal-field stabilisation to force pairing of the Ni$^{2+}$ 3d electrons. The two unpaired electrons remain in the 3d orbitals, so the metal must use its outer, unoccupied $4s$ and three $4p$ orbitals for bonding — $sp^3$ hybridisation —

Q.(a) Using Valence Bond theory prove that $[Ni(CN)_4]^{2-}$ is diamagnetic, whereas $[Ni(NH_3)_4]^{2+}$ is paramagnetic.

Concept understanding — Valence Bond Theory of Coordination Compounds

Q.(a) Using Valence Bond theory prove that [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2− is diamagnetic, whereas [Ni(NH3)4]2+[Ni(NH_3)_4]^{2+}[Ni(NH3​)4​]2+ is paramagnetic.

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Q.(a) Using Valence Bond theory prove that $[Ni(CN)_4]^{2-}$ is diamagnetic, whereas $[Ni(NH_3)_4]^{2+}$ is paramagnetic.

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