Kurnakov's Test
The Kurnakov test is also known as Kurnakov's reaction and is an interesting application of the trans effect is used to distinguish between cis-and trans isomers of the complexes of the type [Pt(A)2 X2 ] (where A = ammine and X = halide). This test was devised by Russian Chemist Nikolai Semenovich Kurnakov in 1893. The test procedure is as follows:
cis-[Pt(NH3)2(Cl2] when added with thiourea [(NH2)CS] (tu) gives the final product [Pt(tu)4]2+ by replacement of all the former ligands of the parent complex.
In step (i), one of the Cl trans to NH3 is replaced by tu because of the inherent lability of the M–Cl bond, whilst in step (ii), NH3 trans to tu is replaced by another tu molecule because the trans effect of tu > NH3. In step (iii), the product shown is formed because of the inherent lability of the M–Cl bond. In step (iv), NH3 trans to tu is replaced by another tu molecule because the trans effect of tu is greater than NH3.
Contrary to the above, in the trans isomer of [Pt(NH3)2Cl2], replacement stops after the two chloride ions have been replaced. The explanation behind this is as follows:
In step (i), any of the two Cl atoms, trans to each other is replaced by tu molecule. In step (ii), Cl trans to tu is replaced by another tu molecule because trans effect of tu > Cl−. Since, the trans ammonia molecules do not labilize each other, no replacement is further possible by any number of tu molecules.
Source: Inorganic Chemistry by R.C. Maurya
