Question

(i) Discuss the violet colour of the complex \( [\text{Ti}(\text{H}_2\text{O})_6]^{3+} \) on the basis of crystal field theory.
(ii) Explain with reason that \( [\text{NiCl}_4]^{2-} \) is paramagnetic while \( [\text{Ni}(\text{CN})_4]^{2-} \) is diamagnetic.

Hint:

Ti\(^{3+}\) is d\(^1\); the single electron makes a d-d transition (\(t_{2g}\rightarrow e_g\)) absorbing yellow-green light so violet is seen. For Ni: Cl\(^-\) weak field (tetrahedral, unpaired) vs CN\(^-\) strong field (square planar, paired).

Answer 1

Part (i): Violet colour of \( [\text{Ti}(\text{H}_2\text{O})_6]^{3+} \)
Step 1 (Oxidation state and configuration): In this complex titanium is in the +3 state. Ti (Z = 22) is [Ar]3d\(^2\)4s\(^2\); removing 3 electrons gives Ti\(^{3+}\) = 3d\(^1\), i.e. one d-electron.
Step 2 (Crystal field splitting): The six H\(_2\)O ligands form an octahedral field which splits the five d-orbitals into a lower set \( t_{2g} \) (three orbitals) and a higher set \( e_g \) (two orbitals), separated by the crystal field splitting energy \( \Delta_o \). In the ground state the single electron sits in a \( t_{2g} \) orbital.
Step 3 (d-d transition): When visible (white) light passes through the solution, this electron absorbs a photon of energy equal to \( \Delta_o \) and is promoted from \( t_{2g} \) to \( e_g \) (a d-d transition). The energy absorbed corresponds to yellow-green light (around 500 nm, \( \approx 20300\ \text{cm}^{-1} \)).
Step 4 (Complementary colour): Since yellow-green light is absorbed, the transmitted light is its complementary colour, violet (purple). Hence the solution looks violet.

Part (ii): Magnetic behaviour of Ni complexes
Step 1 (Oxidation state): In both complexes nickel is Ni\(^{2+}\), which is 3d\(^8\) (two unpaired electrons in the free ion).
Step 2 (\( [\text{NiCl}_4]^{2-} \)): Cl\(^-\) is a weak field ligand, so it cannot pair up the d-electrons. The ion uses \( sp^3 \) hybridisation giving a tetrahedral shape; the two unpaired electrons remain, so the complex is paramagnetic.
Step 3 (\( [\text{Ni}(\text{CN})_4]^{2-} \)): CN\(^-\) is a strong field ligand, so it forces the two unpaired 3d electrons to pair up, freeing one 3d orbital. The ion then uses \( dsp^2 \) hybridisation giving a square planar shape with no unpaired electrons, so the complex is diamagnetic.