The first ammonia is a weak field ligand and will not cause zinc rearrangement. Of course, the zinc ion is 3d 10 and cannot be rearranged. D orbital is full, and hybridization can only be sp3! Tetrahedron! The second and third D orbits of silver and mercury are also full! So hybridization does not involve d orbitals!

(1) The bonding electron pairs (the coordination bond is provided by the ligand unilaterally, and the common valence bond atom provides one each) are as far away as possible, and the Coulomb repulsion is the smallest and the repulsive potential energy is the smallest.

(2) When chemical bonds are formed, the orbital overlap is the largest. The latter factor is not directly related to the landlord's problem. What is hybridization? Is to combine different atomic orbits in a certain proportion. For example, sp3 hybrid orbitals consist of 1/4 S orbitals and 3/4 P orbitals. One S orbit and three P orbits * * * can be combined into four sp3 orbits (details will not be known until the junior year of chemistry), and the distance between these four orbits should be as far as possible (tetrahedron configuration). Using such hybrid orbitals to bond with lone pair orbitals of ligands can not only maximize orbital overlap, but also make the four bonding electron pairs as far away as possible, thus achieving the lowest energy.