Question

Answer the following:
(i) What happens to the colour of the complex \([ \text{Ti(H}_2\text{O)}_6 ]^{3+}\) when heated gradually?
(ii) Write the electronic configuration for a \(d^5\) ion if \(\Delta_o < P\).
(iii) Write the hybridization and magnetic behaviour of the complex \([ \text{Ni(CO)}_4 ]\).
(Atomic number: Ni = 28)

Answer 1

Coordination Chemistry – Question and Answer

(i) What happens to the colour of the complex \([ \text{Ti(H}_2\text{O)}_6 ]^{3+}\) when heated gradually?

Answer: The colour of the complex \([ \text{Ti(H}_2\text{O)}_6 ]^{3+}\) deepens when it is heated gradually. This occurs because thermal energy promotes electronic transitions between the metal ion and the ligands. As temperature increases, the ligand field interaction may change slightly, shifting the absorption spectrum and resulting in a more intensely coloured complex.

(ii) Write the electronic configuration for a \(d^5\) ion if \(\Delta_o < P\).

Answer: When \(\Delta_o < P\), the complex forms a high-spin configuration. Thus, the electronic configuration is:

\[ t_{2g}^3 e_g^2 \]

This means three electrons occupy the lower-energy \(t_{2g}\) orbitals and two occupy the higher-energy \(e_g\) orbitals, following Hund’s rule of maximum multiplicity.

(iii) Write the hybridization and magnetic behaviour of the complex \([ \text{Ni(CO)}_4 ]\).
(Atomic number: Ni = 28)

Answer: In \([ \text{Ni(CO)}_4 ]\), nickel is in the zero oxidation state. The electron configuration of Ni is:

\[ [\text{Ar}]\, 3d^8 4s^2 \]

CO is a strong field ligand, so it causes pairing of the \(3d\) electrons. All electrons pair up, and the \(4s\) and \(4p\) orbitals hybridize to form four equivalent \(sp^3\) orbitals, giving:

• Hybridization: \(sp^3\)
• Geometry: Tetrahedral
• Magnetic behaviour: Diamagnetic (no unpaired electrons)