Question

Why are low spin tetrahedral complexes rarely formed ?

Hint:

Δ_t is too small to pair electrons, so tetrahedral complexes stay high-spin.

Answer 1

Both complexes are $\mathrm{[NiX_4]^{2-}}$ with Ni in the $+2$ state, so Ni has a $\mathrm{3d^8}$ arrangement. What changes everything is the ligand: $\mathrm{Cl^-}$ is a weak-field ligand (it does not push the d-electrons to pair up), while $\mathrm{CN^-}$ is a strong-field ligand (it forces the d-electrons to pair up). Keeping this one idea in mind answers all three parts.

(i) answer: In $\mathrm{[NiCl_4]^{2-}}$ the weak-field $\mathrm{Cl^-}$ leaves the 3d electrons spread out as they are, so Ni uses its outer orbitals (one 4s and three 4p) to bond, giving $\mathrm{sp^3}$ hybridisation. In $\mathrm{[Ni(CN)_4]^{2-}}$ the strong-field $\mathrm{CN^-}$ pairs up the 3d electrons, freeing one inner 3d orbital, so Ni uses one 3d, one 4s and two 4p, giving $\mathrm{dsp^2}$ hybridisation.

(ii) answer: A complex that uses an inner (3d) orbital is called an inner-orbital complex, and one that uses only outer (4s, 4p) orbitals is an outer-orbital complex. So $\mathrm{[Ni(CN)_4]^{2-}}$ ($\mathrm{dsp^2}$, uses a 3d orbital) is the inner-orbital complex, and $\mathrm{[NiCl_4]^{2-}}$ ($\mathrm{sp^3}$, uses only 4s and 4p) is the outer-orbital complex.

(iii) answer: Magnetic behaviour depends on unpaired electrons. In $\mathrm{[NiCl_4]^{2-}}$ the $\mathrm{3d^8}$ electrons stay as they are, leaving 2 unpaired electrons, so it is paramagnetic (attracted by a magnet). In $\mathrm{[Ni(CN)_4]^{2-}}$ the strong $\mathrm{CN^-}$ forces all the electrons to pair, leaving 0 unpaired, so it is diamagnetic (not attracted).