The equilibrium constant values for the reaction of Ni+2 in aqueous solution with an amine to form the corresponding ammonia, en, and dien complexes increase in the order dien>en>ammonia. Why is this the case?

Relative crystal field splitting of the ligands

Explanation:

The formation constant of a complex is the equilibrium constant that shows the stability of the complex. This equilibrium constant is found to depend on the identity of the central metal atom/ion and the identity of the ligands.

Some ligands are called strong field ligands while others are known as weak field ligands. Other ligands lie somewhere in between the two extremes in the spectrochemical series. The spectrochemical series arranges ligands in order of their relative magnitude of crystal field splitting. The greater the magnitude of crystal field splitting, the greater the formation constant or equilibrium constant of the complex.

Strong field ligands are able to participate in ligand to metal charge transfer or metal to ligand charge transfer as the case may be. Hence strong field ligands possess pi-orbitals which interact with metal pi orbitals.

The order of decreasing formation constant in the question reflects the relative positions of the ligands in the spectrochemical series. Ammonia is lower than ethylene diammine in the series which is also lower than a diene. A diene has pi orbitals capable of interaction with metal pi orbitals hence the highest crystal field splitting and equilibrium constant.

the answers would be

the first one and the second one