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):
Step 1: Ti\(^{3+}\) has just one 3d electron (d\(^1\)). Any colour of a transition-metal complex needs a partly filled d-subshell so that electrons can jump between split d-levels.
Step 2: Six water molecules produce an octahedral ligand field. This raises the two orbitals pointing at the ligands (\(e_g\)) and lowers the three pointing between them (\(t_{2g}\)), the gap being \(\Delta_o\). The lone electron rests in \(t_{2g}\).
Step 3: White light passing through the solution loses the yellow-green band whose energy matches \(\Delta_o\); that photon lifts the electron \(t_{2g}\rightarrow e_g\). What comes out to our eyes is the leftover (complementary) colour, which is violet. Remove the d electron (as in colourless Ti\(^{4+}\)) and no such absorption is possible.

Part (ii):
Step 1: Both centres are d\(^8\) Ni\(^{2+}\). The deciding factor is the ligand field strength in the spectrochemical series: Cl\(^-\) is weak, CN\(^-\) is strong.
Step 2: With weak Cl\(^-\), no pairing occurs; Ni\(^{2+}\) hybridises \(sp^3\) and adopts a tetrahedral geometry keeping 2 unpaired electrons. Unpaired electrons mean it is attracted by a magnetic field, i.e. paramagnetic (magnetic moment \(\approx 2.83\) BM).
Step 3: With strong CN\(^-\), the two unpaired electrons are squeezed into one orbital (paired). A vacant inner 3d orbital allows \(dsp^2\) hybridisation and square-planar geometry. Zero unpaired electrons make it diamagnetic (magnetic moment \(= 0\)).