Explain why the substitution of the chloride ligand in $[C o (N H_{3})_{5} C l]^{2 +}$ by water occurs very slowly. In your answer, discuss the electronic configuration, the type of substitution mechanism, write the overall balanced reaction, and give the rate law for this reaction.

hard 5 marks Class 12 · Chemistry · Coordination Compounds

Answer

The substitution follows an associative mechanism; rate = k[Co(NH₃)₅Cl]²⁺; reaction is slow due to high CFSE and strong Co–Cl bond.

Step-by-step solution

State the complex: $[C o (N H_{3})_{5} C l]^{2 +}$ is a $C o^{I I I}$ octahedral complex.
Co $^{I I I}$ has a $d^{6}$ low‑spin configuration ( $t_{2 g}^{6} e_{g}^{0}$ ) giving a large crystal‑field stabilization energy, which makes the complex kinetically inert.
The substitution of Cl⁻ by H₂O requires breaking the strong Co–Cl bond and forming a Co–OH₂ bond.
Because Co(III) can transiently expand its coordination number, the reaction proceeds by an associative (A) mechanism: water attacks the complex forming a seven‑coordinate intermediate $[C o (N H_{3})_{5} C l (H_{2} O)]^{3 +}$ .
The overall balanced equation is: $[C o (N H_{3})_{5} C l]^{2 +} + H_{2} O \to [C o (N H_{3})_{5} (H_{2} O)]^{3 +} + C l^{-}$ .
For an associative first‑order reaction, the rate depends only on the concentration of the reactant complex: rate $= k [C o (N H_{3})_{5} C l]^{2 +}$ .
The reaction is slow because (i) the high CFSE stabilises the $t_{2 g}^{6}$ ground state, (ii) the Co–Cl bond is strong, and (iii) the activation energy for forming the seven‑coordinate transition state is large.
Thus kinetic inertness of $C o^{I I I}$ complexes explains the observed slow substitution.

Common mistake: Assuming a dissociative mechanism and writing a second‑order rate law

Concept tested: Kinetics and mechanism of substitution in octahedral complexes